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Part 1. Background Material
In this portion of the text, most of the topics that are appropriate to an
undergraduate reader are covered. Many of these subjects are subsequently discussed
again in Chapter 5, where a broad perspective of what theoretical chemistry is about is
offered. They are treated again in greater detail in Chapters 6-8 where the three main
disciplines of theory are covered in depth appropriate to a graduate-student reader.
Chapter 1. The Basics of Quantum Mechanics
Why Quantum Mechanics is Necessary for Describing Molecular Properties.
We know that all molecules are made of atoms which, in turn, contain nuclei and
electrons. As I discuss in this introductory section, the equations that govern the motions
of electrons and of nuclei are not the familiar Newton equations
F = m a
but a new set of equations called Schr?dinger equations. When scientists first studied the
behavior of electrons and nuclei, they tried to interpret their experimental findings in
terms of classical Newtonian motions, but such attempts eventually failed. They found
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that such small light particles behaved in a way that simply is not consistent with the
Newton equations. Let me now illustrate some of the experimental data that gave rise to
these paradoxes and show you how the scientists of those early times then used these data
to suggest new equations that these particles might obey. I want to stress that the
Schr?dinger equation was not derived but postulated by these scientists. In fact, to date,
no one has been able to derive the Schr?dinger equation.
From the pioneering work of Bragg on diffraction of x-rays from planes of atoms
or ions in crystals, it was known that peaks in the intensity of diffracted x-rays having
wavelength λ would occur at scattering angles θ determined by the famous Bragg
equation:
n λ = 2 d sinθ,
where d is the spacing between neighboring planes of atoms or ions. These quantities are
illustrated in Fig. 1.1 shown below. There are may such diffraction peaks, each labeled by
a different value of the integer n (n = 1, 2, 3, …). The Bragg formula can be derived by
considering when two photons, one scattering from the second plane in the figure and the
second scattering from the third plane, will undergo constructive interference. This
condition is met when the “extra path length” covered by the second photon (i.e., the
length from points A to B to C) is an integer multiple of the wavelength of the photons.
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Figure 1.1. Scattering of two beams at angle θ from two planes in a crystal spaced by d.
The importance of these x-ray scattering experiments to electrons and nuclei
appears in the experiments of Davisson and Germer in 1927 who scattered electrons of
(reasonably) fixed kinetic energy E from metallic crystals. These workers found that plots
of the number of scattered electrons as a function of scattering angle θ displayed “peaks”
at angles θ that obeyed a Bragg-like equation. The startling thing about this observation is
that electrons are particles, yet the Bragg equation is based on the properties of waves.
An important observation derived from the Davisson-Germer experiments was that the
scattering angles θ observed for electrons of kinetic energy E could be fit to the Bragg n
λ = 2d sinθ equation if a wavelength were ascribed to these electrons that was defined by
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λ = h/(2me E)1/2,
where me is the mass of the electron and h is the constant introduced by Max Planck and
Albert Einstein in the early 1900s to relate a photon’s energy E to its frequency ν via E =
hν. These amazing findings were among the earliest to suggest that electrons, which had
always been viewed as particles, might have some properties usually ascribed to waves.
That is, as de Broglie has suggested in 1925, an electron seems to have a wavelength
inversely related to its momentum, and to display wave-type diffraction. I should mention
that analogous diffraction was also observed when other small light particles (e.g.,
protons, neutrons, nuclei, and small atomic ions) were scattered from crystal planes. In all
such cases, Bragg-like diffraction is observed and the Bragg equation is found to govern
the scattering angles if one assigns a wavelength to the scattering particle according to
λ = h/(2 m E)1/2
where m is the mass of the scattered particle and h is Planck’s constant (6.62 x10-27 erg
sec).
The observation that electrons and other small light particles display wave like
behavior was important because these particles are what all atoms and molecules are
made of. So, if we want to fully understand the motions and behavior of molecules, we
must be sure that we can adequately describe such properties for their constituents.
Because the classical Newtonian equations do not contain factors that suggest wave
properties for electrons or nuclei moving freely in space, the above behaviors presented
significant challenges.
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Another problem that arose in early studies of atoms and molecules resulted from
the study of the photons emitted from atoms and ions that had been heated or otherwise
excited (e.g., by electric discharge). It was found that each kind of atom (i.e., H or C or
O) emitted photons whose frequencies ν were of very characteristic values. An example
of such emission spectra is shown in Fig. 1.2 for hydrogen atoms.
Figure 1.2. Emission spectrum of atomic hydrogen with some lines repeated below to
illustrate the series to which they belong.
In the top panel, we see all of the lines emitted with their wave lengths indicated in nano-
meters. The other panels show how these lines have been analyzed (by scientists whose
names are associated) into patterns that relate to the specific energy levels between which
transitions occur to emit the corresponding photons.
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In the early attempts to rationalize such spectra in terms of electronic motions,
one described an electron as moving about the atomic nuclei in circular orbits such as
shown in Fig. 1. 3.
Figure 1. 3. Characterization of small and large stable orbits for an electron moving
around a nucleus.
A circular orbit was thought to be stable when the outward centrifugal force characterized
by radius r and speed v (me v2/r) on the electron perfectly counterbalanced the inward
attractive Coulomb force (Ze2/r2) exerted by the nucleus of charge Z:
me v2/r = Ze2/r2
r2
Two circular orbits of radii r1 and r2.
r1
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This equation, in turn, allows one to relate the kinetic energy 1/2 me v2 to the Coulombic
energy Ze2/r, and thus to express the total energy E of an orbit in terms of the radius of
the orbit:
E = 1/2 me v2 – Ze2/r = -1/2 Ze2/r.
The energy characterizing an orbit or radius r, relative to the E = 0 reference of
energy at r → ∞, becomes more and more negative (i.e., lower and lower) as r becomes
smaller. This relationship between outward and inward forces allows one to conclude that
the electron should move faster as it moves closer to the nucleus since v2 = Ze2/(r me).
However, nowhere in this model is a concept that relates to the experimental fact that
each atom emits only certain kinds of photons. It was believed that photon emission
occurred when an electron moving in a larger circular orbit lost energy and moved to a
smaller circular orbit. However, the Newtonian dynamics that produced the above
equation would allow orbits of any radius, and hence any energy, to be followed. Thus, it
would appear that the electron should be able to emit photons of any energy as it moved
from orbit to orbit.
The breakthrough that allowed scientists such as Niels Bohr to apply the circular-
orbit model to the observed spectral data involved first introducing the idea that the
electron has a wavelength and that this wavelength λ is related to its momentum by the
de Broglie equation λ = h/p. The key step in the Bohr model was to also specify that the
radius of the circular orbit be such that the circumference of the circle 2pi r equal an
integer (n) multiple of the wavelength λ. Only in this way will the electron’s wave
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experience constructive interference as the electron orbits the nucleus. Thus, the Bohr
relationship that is analogous to the Bragg equation that determines at what angles
constructive interference can occur is
2 pi r = n λ.
Both this equation and the analogous Bragg equation are illustrations of what we call
boundary conditions; they are extra conditions placed on the wavelength to produce some
desired character in the resultant wave (in these cases, constructive interference). Of
course, there remains the question of why one must impose these extra conditions when
the Newton dynamics do not require them. The resolution of this paradox is one of the
things that quantum mechanics does.
Returning to the above analysis and using λ = h/p = h/(mv), 2pi r = nλ, as well as
the force-balance equation me v2/r = Ze2/r2, one can then solve for the radii that stable
Bohr orbits obey:
r = (nh/2pi) 1/(me Z e2)
and, in turn for the velocities of electrons in these orbits
v = Z e2/(nh/2pi).
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These two results then allow one to express the sum of the kinetic (1/2 me v2) and
Coulomb potential (-Ze2/r) energies as
E = -1/2 me Z2 e4/(nh/2pi)2.
Just as in the Bragg diffraction result, which specified at what angles special high
intensities occurred in the scattering, there are many stable Bohr orbits, each labeled by a
value of the integer n. Those with small n have small radii, high velocities and more
negative total energies (n.b., the reference zero of energy corresponds to the electron at r
= ∞ , and with v = 0). So, it is the result that only certain orbits are “allowed” that causes
only certain energies to occur and thus only certain energies to be observed in the emitted
photons.
It turned out that the Bohr formula for the energy levels (labeled by n) of an
electron moving about a nucleus could be used to explain the discrete line emission
spectra of all one-electron atoms and ions (i.e., H, He+, Li+2, etc.) to very high precision.
In such an interpretation of the experimental data, one claims that a photon of energy
hν = R (1/nf2 – 1/ni2)
is emitted when the atom or ion undergoes a transition from an orbit having quantum
number ni to a lower-energy orbit having nf. Here the symbol R is used to denote the
following collection of factors:
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R = 1/2 me Z2 e4/(h/2pi)2.
The Bohr formula for energy levels did not agree as well with the observed pattern of
emission spectra for species containing more than a single electron. However, it does
give a reasonable fit, for example, to the Na atom spectra if one examines only transitions
involving only the single valence electron. The primary reason for the breakdown of the
Bohr formula is the neglect of electron-electron Coulomb repulsions in its derivation.
Nevertheless, the success of this model made it clear that discrete emission spectra could
only be explained by introducing the concept that not all orbits were “allowed”. Only
special orbits that obeyed a constructive-interference condition were really accessible to
the electron’s motions. This idea that not all energies were allowed, but only certain
“quantized” energies could occur was essential to achieving even a qualitative sense of
agreement with the experimental fact that emission spectra were discrete.
In summary, two experimental observations on the behavior of electrons that were
crucial to the abandonment of Newtonian dynamics were the observations of electron
diffraction and of discrete emission spectra. Both of these findings seem to suggest that
electrons have some wave characteristics and that these waves have only certain allowed
(i.e., quantized) wavelengths.
So, now we have some idea about why Newton’s equations fail to account for the
dynamical motions of light and small particles such as electrons and nuclei. We see that
extra conditions (e.g., the Bragg condition or constraints on the de Broglie wavelength)
could be imposed to achieve some degree of agreement with experimental observation.
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However, we still are left wondering what the equations are that can be applied to
properly describe such motions and why the extra conditions are needed. It turns out that
a new kind of equation based on combining wave and particle properties needed to be
developed to address such issues. These are the so-called Schr?dinger equations to which
we now turn our attention.
As I said earlier, no one has yet shown that the Schr?diger equation follows
deductively from some more fundamental theory. That is, scientists did not derive this
equation; they postulated it. Some idea of how the scientists of that era “dreamed up” the
Schr?dinger equation can be had by examining the time and spatial dependence that
characterizes so-called travelling waves. It should be noted that the people who worked
on these problems knew a great deal about waves (e.g., sound waves and water waves)
and the equations they obeyed. Moreover, they knew that waves could sometimes display
the characteristic of quantized wavelengths or frequencies (e.g., fundamentals and
overtones in sound waves). They knew, for example, that waves in one dimension that
are constrained at two points (e.g., a violin string held fixed at two ends) undergo
oscillatory motion in space and time with characteristic frequencies and wavelengths. For
example, the motion of the violin string just mentioned can be described as having an
amplitude A(x,t) at a position x along its length at time t given by
A(x,t) = A(x,o) cos(2pi ν t),
where ν is its oscillation frequency. The amplitude’s spatial dependence also has a
sinusoidal dependence given by
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A(x,0) = A sin (2pi x/λ)
where λ is the crest-to-crest length of the wave. Two examples of such waves in one
dimension are shown in Fig. 1. 4.
Figure 1.4. Fundamental and first overtone notes of a violin string.
In these cases, the string is fixed at x = 0 and at x = L, so the wavelengths belonging to
the two waves shown are λ = 2L and λ = L. If the violin string were not clamped at x = L,
the waves could have any value of λ. However, because the string is attached at x = L,
the allowed wavelengths are quantized to obey
A(x,0)
x
sin(1pix/L)
sin(2pix/L)
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λ = L/n,
where n = 1, 2, 3, 4, ... .The equation that such waves obey, called the wave equation,
reads:
d2Α(x,t)/dt2 = c2 d2Α/dx2
where c is the speed at which the wave travels. This speed depends on the composition of
the material from which the violin string is made. Using the earlier expressions for the x-
and t- dependences of the wave A(x,t), we find that the wave’s frequency and wavelength
are related by the so-called dispersion equation:
ν2 = (c/λ)2,
or
c = λ ν.
This relationship implies, for example, that an instrument string made of a very stiff
material (large c) will produce a higher frequency tone for a given wavelength (i.e., a
given value of n) than will a string made of a softer material (smaller c).
For waves moving on the surface of, for example, a rectangular two-dimensional
surface of lengths Lx and Ly, one finds
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A(x,y,t) = sin(nx pix/Lx) sin(ny piy/Ly) cos(2pi νt).
Hence, the waves are quantized in two dimensions because their wavelengths must be
constrained to cause A(x,y,t) to vanish at x = 0 and x = Lx as well as at y = 0 and y = Ly
for all times t. Let us now return to the issue of waves that describe electrons moving.
The pioneers of quantum mechanics examined functional forms similar to those
shown above. For example, forms such as A = exp[±2pii(ν t – x/λ)] were considered
because they correspond to periodic waves that evolve in x and t under no external x- or
t- dependent forces. Noticing that
d2A/dx2 = - (2pi/λ)2 A
and using the de Broglie hypothesis λ = h/p in the above equation, one finds
d2A/dx2 = - p2 (2pi/h)2 A.
If A is supposed to relate to the motion of a particle of momentum p under no external
forces (since the waveform corresponds to this case), p2 can be related to the energy E of
the particle by E = p2/2m. So, the equation for A can be rewritten as:
d2A/dx2 = - 2m E (2pi/h)2 A,
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or, alternatively,
- (h/2pi)2 d2A/dx2 = E A.
Returning to the time-dependence of A(x,t) and using ν = E/h, one can also show that
i (h/2pi) dA/dt = E A,
which, using the first result, suggests that
i (h/2pi) dA/dt = - (h/2pi)2 d2A/dx2.
This is a primitive form of the Schr?dinger equation that we will address in much more
detail below. Briefly, what is important to keep in mind that the use of the de Broglie and
Planck/Einstein connections (λ = h/p and E = h ν), both of which involve the constant h,
produces suggestive connections between
i (h/2pi) d/dt and E
and between
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p2 and – (h/2pi)2 d2/dx2
or, alternatively, between
p and – i (h/2pi) d/dx.
These connections between physical properties (energy E and momentum p) and
differential operators are some of the unusual features of quantum mechanics.
The above discussion about waves and quantized wavelengths as well as the
observations about the wave equation and differential operators are not meant to provide
or even suggest a derivation of the Schr?dinger equation. Again the scientists who
invented quantum mechanics did not derive its working equations. Instead, the equations
and rules of quantum mechanics have been postulated and designed to be consistent with
laboratory observations. My students often find this to be disconcerting because they are
hoping and searching for an uderlying fundamental basis from which the basic laws of
quantum mechanics follows logically. I try to remind them that this is not how theory
works. Instead, one uses experimental observation to postulate a rule or equation or
theory, and one then tests the theory by making predictions that can be tested by further
experiments. If the theory fails, it must be “refined”, and this process continues until one
has a better and better theory. In this sense, quantum mechanics, with all of its unusual
mathematical constructs and rules, should be viewed as arising from the imaginations of
scientists who tried to invent a theory that was consistent with experimental data and
which could be used to predict things that could then be tested in the laboratory. Thus far,
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this theory has proven to be reliable, but, of course, we are always searching for a “new
and improved” theory that describes how small light particles move.
If it helps you to be more accepting of quantum theory, I should point out that the
quantum description of particles will reduce to the classical Newton description under
certain circumstances. In particular, when treating heavy particles (e.g., macroscopic
masses and even heavier atoms), it is often possible to use Newton dynamics. Briefly, we
will discuss in more detail how the quantum and classical dynamics sometimes coincide
(in which case one is free to use the simpler Newton dynamics). So, let us now move on
to look at this strange Schr?dinger equation that we have been digressing about for so
long.
I. The Schr?dinger Equation and Its Components
It has been well established that electrons moving in atoms and molecules do not
obey the classical Newton equations of motion. People long ago tried to treat electronic
motion classically, and found that features observed clearly in experimental
measurements simply were not consistent with such a treatment. Attempts were made to
supplement the classical equations with conditions that could be used to rationalize such
observations. For example, early workers required that the angular momentum L = r x p
be allowed to assume only integer mulitples of h/2pi (which is often abbreviated as h),
which can be shown to be equivalent to the Borh postulate n λ = 2 pir. However, until
scientists realized that a new set of laws, those of quantum mechanics, applied to light
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microscopic particles, a wide gulf existed between laboratory observations of molecule-
level phenomena and the equations used to describe such behavior.
Quantum mechanics is cast in a language that is not familiar to most students of
chemistry who are examining the subject for the first time. Its mathematical content and
how it relates to experimental measurements both require a great deal of effort to master.
With these thoughts in mind, I have organized this material in a manner that first provides
a brief introduction to the two primary constructs of quantum mechanics- operators and
wave functions that obey a Schr?dinger equation. Next, I demonstrate the application of
these constructs to several chemically relevant model problems. By learning the solutions
of the Schr?dinger equation for a few model systems, the student can better appreciate
the treatment of the fundamental postulates of quantum mechanics as well as their
relation to experimental measurement for which the wave functions of the known model
problems offer important interpretations.
A. Operators
Each physically measurable quantity has a corresponding operator. The
eigenvalues of the operator tell the only values of the corresponding physical property
that can be observed.
Any experimentally measurable physical quantity F (e.g., energy, dipole moment,
orbital angular momentum, spin angular momentum, linear momentum, kinetic energy)
has a classical mechanical expression in terms of the Cartesian positions {qi} and
momenta {pi} of the particles that comprise the system of interest. Each such classical
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expression is assigned a corresponding quantum mechanical operator F formed by
replacing the {pi} in the classical form by the differential operator -ih?/?qj and leaving
the coordinates qj that appear in F untouched. For example, the classical kinetic energy of
N particles (with masses ml) moving in a potential field containing both quadratic and
linear coordinate-dependence can be written as
F=Σl=1,N (pl2/2ml + 1/2 k(ql-ql0)2 + L(ql-ql0)).
The quantum mechanical operator associated with this F is
F=Σl=1,N (- h2/2ml ?2/?ql2 + 1/2 k(ql-ql0)2 + L(ql-ql0)).
Such an operator would occur when, for example, one describes the sum of the kinetic
energies of a collection of particles (the Σl=1,N (pl2/2ml ) term), plus the sum of "Hookes'
Law" parabolic potentials (the 1/2 Σl=1,N k(ql-ql0)2), and (the last term in F) the
interactions of the particles with an externally applied field whose potential energy varies
linearly as the particles move away from their equilibrium positions {ql0}.
Let us try more examples. The sum of the z-components of angular momenta
(recall that vector angular momentum L is defined as L = r x p) of a collection of N
particles has the following classical expression
F=Σj=1,N (xjpyj - yjpxj),
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and the corresponding operator is
F=-ih Σj=1,N (xj?/?yj - yj?/?xj).
If one transforms these Cartesian coordinates and derivatives into polar coordinates, the
above expression reduces to
F = -i h Σj=1,N ?/?φj.
The x-component of the dipole moment for a collection of N particles has a classical
form of
F=Σj=1,N Zjexj,
for which the quantum operator is
F=Σj=1,N Zjexj ,
where Zje is the charge on the jth particle. Notice that in this case, classical and quantum
forms are identical because F contains no momentum operators.
The mapping from F to F is straightforward only in terms of Cartesian
coordinates. To map a classical function F, given in terms of curvilinear coordinates
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(even if they are orthogonal), into its quantum operator is not at all straightforward. The
mapping can always be done in terms of Cartesian coordinates after which a
transformation of the resulting coordinates and differential operators to a curvilinear
system can be performed.
The relationship of these quantum mechanical operators to experimental
measurement lies in the eigenvalues of the quantum operators. Each such operator has a
corresponding eigenvalue equation
F χj = αj χj
in which the χj are called eigenfunctions and the (scalar numbers) αj are called
eigenvalues. All such eigenvalue equations are posed in terms of a given operator (F in
this case) and those functions {χj} that F acts on to produce the function back again but
multiplied by a constant (the eigenvalue). Because the operator F usually contains
differential operators (coming from the momentum), these equations are differential
equations. Their solutions χj depend on the coordinates that F contains as differential
operators. An example will help clarify these points. The differential operator d/dy acts
on what functions (of y) to generate the same function back again but multiplied by a
constant? The answer is functions of the form exp(ay) since
d (exp(ay))/dy = a exp(ay).
So, we say that exp(ay) is an eigenfunction of d/dy and a is the corresponding eigenvalue.
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As I will discuss in more detail shortly, the eigenvalues of the operator F tell us
the only values of the physical property corresponding to the operator F that can be
observed in a laboratory measurement. Some F operators that we encounter possess
eigenvalues that are discrete or quantized. For such properties, laboratory measurement
will result in only those discrete values. Other F operators have eigenvalues that can take
on a continuous range of values; for these properties, laboratory measurement can give
any value in this continuous range.
B. Wave functions
The eigenfunctions of a quantum mechanical operator depend on the coordinates
upon which the operator acts. The particular operator that corresponds to the total
energy of the system is called the Hamiltonian operator. The eigenfunctions of this
particular operator are called wave functions
A special case of an operator corresponding to a physically measurable quantity is
the Hamiltonian operator H that relates to the total energy of the system. The energy
eigenstates of the system Ψ are functions of the coordinates {qj} that H depends on and
of time t. The function |Ψ(qj,t)|2 = Ψ*Ψ gives the probability density for observing the
coordinates at the values qj at time t. For a many-particle system such as the H2O
molecule, the wave function depends on many coordinates. For H2O, it depends on the x,
y, and z (or r,θ, and φ) coordinates of the ten electrons and the x, y, and z (or r,θ, and φ)
coordinates of the oxygen nucleus and of the two protons; a total of thirty-nine
coordinates appear in Ψ.
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In classical mechanics, the coordinates qj and their corresponding momenta pj are
functions of time. The state of the system is then described by specifying qj(t) and pj(t).
In quantum mechanics, the concept that qj is known as a function of time is replaced by
the concept of the probability density for finding qj at a particular value at a particular
time |Ψ(qj,t)|2. Knowledge of the corresponding momenta as functions of time is also
relinquished in quantum mechanics; again, only knowledge of the probability density for
finding pj with any particular value at a particular time t remains.
The Hamiltonian eigenstates are especially important in chemistry because many
of the tools that chemists use to study molecules probe the energy states of the molecule.
For example, most spectroscopic methods are designed to determine which energy state a
molecule is in. However, there are other experimental measurements that measure other
properties (e.g., the z-component of angular momentum or the total angular momentum).
As stated earlier, if the state of some molecular system is characterized by a wave
function Ψ that happens to be an eigenfunction of a quantum mechanical operator F, one
can immediately say something about what the outcome will be if the physical property F
corresponding to the operator F is measured. In particular, since
F χj = λj χj,
where λj is one of the eigenvalues of F, we know that the value λj will be observed if the
property F is measured while the molecule is described by the wave function Ψ = χj. In
fact, once a measurement of a physical quantity F has been carried out and a particular
eigenvalue λj has been observed, the system's wave function Ψ becomes the
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eigenfunction χj that corresponds to that eigenvalue. That is, the act of making the
measurement causes the system's wave function to become the eigenfunction of the
property that was measured.
What happens if some other property G, whose quantum mechanical operator is G
is measured in such a case? We know from what was said earlier that some eigenvalue μk
of the operator G will be observed in the measurement. But, will the molecule's wave
function remain, after G is measured, the eigenfunction of F, or will the measurement of
G cause Ψ to be altered in a way that makes the molecule's state no longer an
eigenfunction of F? It turns out that if the two operators F and G obey the condition
F G = G F,
then, when the property G is measured, the wave function Ψ = χj will remain unchanged.
This property that the order of application of the two operators does not matter is called
commutation; that is, we say the two operators commute if they obey this property. Let us
see how this property leads to the conclusion about Ψ remaining unchanged if the two
operators commute. In particular, we apply the G operator to the above eigenvalue
equation:
G F χj = G λj χj.
Next, we use the commutation to re-write the left-hand side of this equation, and use the
fact that λj is a scalar number to thus obtain:
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F G χj = λj G χj.
So, now we see that (Gχj) itself is an eigenfunction of F having eigenvalue λj. So, unless
there are more than one eigenfunction of F corresponding to the eigenvalue λj (i.e., unless
this eigenvalue is degenerate), Gχj must itself be proportional to χj. We write this
proportionality conclusion as
G χj = μj χj,
which means that χj is also an eigenfunction of G. This, in turn, means that measuring the
property G while the system is described by the wave function Ψ = χj does not change the
wave function; it remains χj.
So, when the operators corresponding to two physical properties commute, once
one measures one of the properties (and thus causes the system to be an eigenfunction of
that operator), subsequent measurement of the second operator will (if the eigenvalue of
the first operator is not degenerate) produce a unique eigenvalue of the second operator
and will not change the system wave function.
If the two operators do not commute, one simply can not reach the above
conclusions. In such cases, measurement of the property corresponding to the first
operator will lead to one of the eigenvalues of that operator and cause the system wave
function to become the corresponding eigenfuction. However, subsequent measurement
of the second operator will produce an eigenvalue of that operator, but the system wave
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function will be changed to become an eigenfuntion of the second operator and thus no
longer the eigenfunction of the first.
C. The Schr?dinger Equation
This equation is an eigenvalue equation for the energy or Hamiltonian operator;
its eigenvalues provide the only allowed energy levels of the system
1. The Time-Dependent Equation
If the Hamiltonian operator contains the time variable explicitly, one must solve
the time-dependent Schr?dinger equation
Before moving deeper into understanding what quantum mechanics 'means', it is
useful to learn how the wave functions Ψ are found by applying the basic equation of
quantum mechanics, the Schr?dinger equation, to a few exactly soluble model problems.
Knowing the solutions to these 'easy' yet chemically very relevant models will then
facilitate learning more of the details about the structure of quantum mechanics.
The Schr?dinger equation is a differential equation depending on time and on all
of the spatial coordinates necessary to describe the system at hand (thirty-nine for the
H2O example cited above). It is usually written
H Ψ = i h ?Ψ/?t
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where Ψ(qj,t) is the unknown wave function and H is the operator corresponding to the
total energy of the system. This operator is called the Hamiltonian and is formed, as
stated above, by first writing down the classical mechanical expression for the total
energy (kinetic plus potential) in Cartesian coordinates and momenta and then replacing
all classical momenta pj by their quantum mechanical operators pj = - ih?/?qj .
For the H2O example used above, the classical mechanical energy of all thirteen
particles is
E = Σi { pi2/2me + 1/2 Σj e2/ri,j - Σa Zae2/ri,a }
+ Σa {pa2/2ma + 1/2 Σb ZaZbe2/ra,b },
where the indices i and j are used to label the ten electrons whose thirty Cartesian
coordinates are {qi} and a and b label the three nuclei whose charges are denoted {Za},
and whose nine Cartesian coordinates are {qa}. The electron and nuclear masses are
denoted me and {ma}, respectively. The corresponding Hamiltonian operator is
H = Σi { - (h2/2me) ?2/?qi2 + 1/2 Σj e2/ri,j - Σa Zae2/ri,a }
+ Σa { - (h2/2ma) ?2/?qa2+ 1/2 Σb ZaZbe2/ra,b }.
Notice that H is a second order differential operator in the space of the thirty-nine
28
Cartesian coordinates that describe the positions of the ten electrons and three nuclei. It is
a second order operator because the momenta appear in the kinetic energy as pj2 and pa2,
and the quantum mechanical operator for each momentum p = -ih ?/?q is of first order.
The Schr?dinger equation for the H2O example at hand then reads
Σi { - (h2/2me) ?2/?qi2 + 1/2 Σj e2/ri,j - Σa Zae2/ri,a } Ψ
+ Σa { - (h2/2ma) ?2/?qa2+ 1/2 Σb ZaZbe2/ra,b } Ψ = i h ?Ψ/?t.
The Hamiltonian in this case contains t nowhere. An example of a case where H does
contain t occurs when the an oscillating electric field E cos(ωt) along the x-axis interacts
with the electrons and nuclei and a term
Σa Zze Xa E cos(ωt) - Σj e xj E cos(ωt)
is added to the Hamiltonian. Here, Xa and xj denote the x coordinates of the ath nucleus
and the jth electron, respectively.
2. The Time-Independent Equation
If the Hamiltonian operator does not contain the time variable explicitly, one can
solve the time-independent Schr?dinger equation
In cases where the classical energy, and hence the quantum Hamiltonian, do not
29
contain terms that are explicitly time dependent (e.g., interactions with time varying
external electric or magnetic fields would add to the above classical energy expression
time dependent terms), the separations of variables techniques can be used to reduce the
Schr?dinger equation to a time-independent equation.
In such cases, H is not explicitly time dependent, so one can assume that Ψ(qj,t) is
of the form (n.b., this step is an example of the use of the separations of variables method
to solve a differential equation)
Ψ(qj,t) = Ψ(qj) F(t).
Substituting this 'ansatz' into the time-dependent Schr?dinger equation gives
Ψ(qj) i h ?F/?t = F(t) H Ψ(qj) .
Dividing by Ψ(qj) F(t) then gives
F-1 (i h ?F/?t) = Ψ-1 (H Ψ(qj) ).
Since F(t) is only a function of time t, and Ψ(qj) is only a function of the spatial
coordinates {qj}, and because the left hand and right hand sides must be equal for all
values of t and of {qj}, both the left and right hand sides must equal a constant. If this
constant is called E, the two equations that are embodied in this separated Schr?dinger
equation read as follows:
30
H Ψ(qj) = E Ψ(qj),
ih dF(t)/dt = E F(t).
The first of these equations is called the time-independent Schr?dinger equation; it is a
so-called eigenvalue equation in which one is asked to find functions that yield a constant
multiple of themselves when acted on by the Hamiltonian operator. Such functions are
called eigenfunctions of H and the corresponding constants are called eigenvalues of H.
For example, if H were of the form (- h2/2M) ?2/?φ2 = H , then functions of the form
exp(i mφ) would be eigenfunctions because
{ - h2/2M ?2/?φ2} exp(i mφ) = { m2 h2 /2M } exp(i mφ).
In this case, m2 h2 /2M is the eigenvalue. In this example, the Hamiltonian contains the
square of an angular momentum operator (recall earlier that we showed the z-component
of angular momentum is to equal – i h d/dφ).
When the Schr?dinger equation can be separated to generate a time-independent
equation describing the spatial coordinate dependence of the wave function, the
eigenvalue E must be returned to the equation determining F(t) to find the time dependent
part of the wave function. By solving
ih dF(t)/dt = E F(t)
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once E is known, one obtains
F(t) = exp( -i Et/ h),
and the full wave function can be written as
Ψ(qj,t) = Ψ(qj) exp (-i Et/ h).
For the above example, the time dependence is expressed by
F(t) = exp ( -i t { m2 h2 /2M }/ h).
In summary, whenever the Hamiltonian does not depend on time explicitly, one
can solve the time-independent Schr?dinger equation first and then obtain the time
dependence as exp(-i Et/ h) once the energy E is known. In the case of molecular
structure theory, it is a quite daunting task even to approximately solve the full
Schr?dinger equation because it is a partial differential equation depending on all of the
coordinates of the electrons and nuclei in the molecule. For this reason, there are various
approximations that one usually implements when attempting to study molecular
structure using quantum mechanics.
3. The Born-Oppenheimer Approximation
32
One of the most important approximations relating to applying quantum
mechanics to molecules is known as the Born-Oppenheimer (BO) approximation.
The basic idea behind this approximation involves realizing that in the full electrons-plus-
nuclei Hamiltonian operator introduced above
H = Σi { - (h2/2me) ?2/?qi2 + 1/2 Σj e2/ri,j - Σa Zae2/ri,a }
+ Σa { - (h2/2ma) ?2/?qa2+ 1/2 Σb ZaZbe2/ra,b }
the time scales with which the electrons and nuclei move are generally quite different. In
particular, the heavy nuclei (i.e., even a H nucleus weighs nearly 2000 times what an
electron weighs) move (i.e., vibrate and rotate) more slowly than do the lighter electrons.
Thus, we expect the electrons to be able to “adjust” their motions to the much more
slowly moving nuclei. This observation motivates us to solve the Schr?dinger equation
for the movement of the electrons in the presence of fixed nuclei as a way to represent the
fully-adjusted state of the electrons at any fixed positions of the nuclei.
The electronic Hamiltonian that pertains to the motions of the electrons in the
presence of so-called clamped nuclei
H = Σi { - (h2/2me) ?2/?qi2 + 1/2 Σj e2/ri,j - Σa Zae2/ri,a }
produces as its eigenvalues through the equation
33
H ψJ(qj|qa) = EJ(qa) ψJ(qj|qa)
energies EK(qa) that depend on where the nuclei are located (i.e., the {qa} coordinates). As
its eigenfunctions, one obtains what are called electronic wave functions {ψK(qi|qa)}
which also depend on where the nuclei are located. The energies EK(qa) are what we
usually call potential energy surfaces. An example of such a surface is shown in Fig. 1.5.
Figure 1. 5. Two dimensional potential energy surface showing local minima, transition
states and paths connecting them.
This surface depends on two geometrical coordinates {qa} and is a plot of one particular
eigenvalue EJ(qa) vs. these two coordinates.
34
Although this plot has more information on it than we shall discuss now, a few
features are worth noting. There appear to be three minima (i.e., points where the
derivative of EJ with respect to both coordinates vanish and where the surface has
positive curvature). These points correspond, as we will see toward the end of this
introductory material, to geometries of stable molecular structures. The surface also
displays two first-order saddle points (labeled transition structures A and B) that connect
the three minima. These points have zero first derivative of EJ with respect to both
coordinates but have one direction of negative curvature. As we will show later, these
points describe transition states that play crucial roles in the kinetics of transitions among
the three stable geometries.
Keep in mind that Fig. 1. 5 shows just one of the EJ surfaces; each molecule has a
ground-state surface (i.e., the one that is lowest in energy) as well as an infinite number
of excited-state surfaces. Let’s now return to our discussion of the BO model and ask
what one does once one has such an energy surface in hand.
The motion of the nuclei are subsequently, within the BO model, assumed to obey
a Schr?dinger equation in which Σa { - (h2/2ma) ?2/?qa2+ 1/2 Σb ZaZbe2/ra,b } + EK(qa)
defines a rotation-vibration Hamiltonian for the particular energy state EK of interest. The
rotational and vibrational energies and wave functions belonging to each electronic state
(i.e., for each value of the index K in EK(qa)) are then found by solving a Schr?dinger
equation with such a Hamiltonian.
This BO model forms the basis of much of how chemists view molecular
structure and molecular spectroscopy. For example as applied to formaldehyde H2C=O,
we speak of the singlet ground electronic state (with all electrons spin paired and
35
occupying the lowest energy orbitals) and its vibrational states as well as the n→ pi* and
pi → pi* electronic states and their vibrational levels. Although much more will be said
about these concepts later in this text, the student should be aware of the concepts of
electronic energy surfaces (i.e., the {EK(qa)}) and the vibration-rotation states that belong
to each such surface.
Having been introduced to the concepts of operators, wave functions, the
Hamiltonian and its Schr?dinger equation, it is important to now consider several
examples of the applications of these concepts. The examples treated below were chosen
to provide the reader with valuable experience in solving the Schr?dinger equation; they
were also chosen because they form the most elementary chemical models of electronic
motions in conjugated molecules and in atoms, rotations of linear molecules, and
vibrations of chemical bonds.
II. Your First Application of Quantum Mechanics- Motion of a Particle in One
Dimension.
This is a very important problem whose solutions chemists use to model a wide
variety of phenomena.
Let’s begin by examining the motion of a single particle of mass m in one direction
which we will call x while under the influence of a potential denoted V(x). The classical
expression for the total energy of such a system is E = p2/2m + V(x), where p is the
momentum of the particle along the x-axis. To focus on specific examples, consider how
36
this particle would move if V(x) were of the forms shown in Fig. 1. 6, where the total
energy E is denoted by the position of the horizontal line.
Figure 1. 6. Three characteristic potentials showing left and right classical turning points
at energies denoted by the horizontal lines.
A. The Classical Probability Density
I would like you to imagine what the probability density would be for this particle
moving with total energy E and with V(x) varying as the above three plots illustrate. To
conceptualize the probability density, imagine the particle to have a blinking lamp
attached to it and think of this lamp blinking say 100 times for each time it takes for the
particle to complete a full transit from its left turning point, to its right turning point and
back to the former. The turning points xL and xR are the positions at which the particle, if
xL xR
37
it were moving under Newton’s laws, would reverse direction (as the momentum changes
sign) and turn around. These positions can be found by asking where the momentum goes
to zero:
0 = p = (2m(E-V(x))1/2.
These are the positions where all of the energy appears as potential energy E = V(x) and
correspond in the above figures to the points where the dark horizontal lines touch the
V(x) plots as shown in the central plot.
The probability density at any value of x represents the fraction of time the
particle spends at this value of x (i.e., within x and x+dx). Think of forming this density
by allowing the blinking lamp attached to the particle to shed light on a photographic
plate that is exposed to this light for many oscillations of the particle between xL and xR.
Alternatively, one can express this probability amplitude P(x) by dividing the spatial
distance dx by the velocity of the particle at the point x:
P(x) = (2m(E-V(x))-1/2 m dx.
Because E is constant throughout the particle’s motion, P(x) will be small at x values
where the particle is moving quickly (i.e., where V is low) and will be high where the
particle moves slowly (where V is high). So, the photographic plate will show a bright
region where V is high (because the particle moves slowly in such regions) and less
brightness where V is low.
38
The bottom line is that the probability densities anticipated by analyzing the
classical Newtonian dynamics of this one particle would appear as the histogram plots
shown in Fig. 1.7 illustrate.
Figure 1. 7 Classical probability plots for the three potentials shown
Where the particle has high kinetic energy (and thus lower V(x)), it spends less time and
P(x) is small. Where the particle moves slowly, it spends more time and P(x) is larger.
For the plot on the right, V(x) is constant within the “box”, so the speed is constant,
hence P(x) is constant for all x values within this one-dimensional box. I ask that you
keep these plots in mind because they are very different from what one finds when one
solves the Schr?dinger equation for this same problem. Also please keep in mind that
these plots represent what one expects if the particle were moving according to classical
Newtonian dynamics (which we know it is not!).
xL xR
39
B. The Quantum Treatment
To solve for the quantum mechanical wave functions and energies of this same
problem, we first write the Hamiltonian operator as discussed above by replacing p by
-i h d/dx :
H = - h2/2m d2/dx2 + V(x).
We then try to find solutions ψ(x) to Hψ = Eψ that obey certain conditions. These
conditions are related to the fact that |ψ (x)|2 is supposed to be the probability density for
finding the particle between x and x+dx. To keep things as simple as possible, let’s focus
on the “box” potential V shown in the right side of Fig. B. 7. This potential, expressed as
a function of x is: V(x) = ∞ for x< 0 and for x> L; V(x) = 0 for x between 0 and L.
The fact that V is infinite for x< 0 and for x> L, and that the total energy E must
be finite, says that ψ must vanish in these two regions (ψ = 0 for x< 0 and for x> L). This
condition means that the particle can not access these regions where the potential is
infinite. The second condition that we make use of is that ψ (x) must be continuous; this
means that the probability of the particle being at x can not be discontinuously related to
the probability of it being at a nearby point.
C. The Energies and Wave functions
The second-order differential equation
40
- h 2/2m d2ψ/dx2 + V(x) ψ = E ψ
has two solutions (because it is a second order equation) in the region between x= 0 and
x= L:
ψ = sin(kx) and ψ = cos(kx), where k is defined as k=(2mE/h 2)1/2.
Hence, the most general solution is some combination of these two:
ψ = A sin(kx) + B cos(kx).
The fact that ψ must vanish at x= 0 (n.b., ψ vanishes for x< 0 and is continuous, so it
must vanish at the point x= 0) means that the weighting amplitude of the cos(kx) term
must vanish because cos(kx) = 1 at x = 0. That is,
B = 0.
The amplitude of the sin(kx) term is not affected by the condition that ψ vanish at x= 0,
since sin(kx) itself vanishes at x= 0. So, now we know that ψ is really of the form:
ψ (x) = A sin(kx).
41
The condition that ψ also vanish at x= L has two possible implications. Either A = 0 or k
must be such that sin(kL) = 0. The option A = 0 would lead to an answer ψ that vanishes
at all values of x and thus a probability that vanishes everywhere. This is unacceptable
because it would imply that the particle is never observed anywhere.
The other possibility is that sin(kL) = 0. Let’s explore this answer because it
offers the first example of energy quantization that you have probably encountered. As
you know, the sin function vanishes at integral multiples of pi. Hence kL must be some
multiple of pi; let’s call the integer n and write L k = n pi (using the definition of k) in the
form:
L (2mE/h2)1/2 = n pi.
Solving this equation for the energy E, we obtain:
E = n2 pi2 h2/(2mL2)
This result says that the only energy values that are capable of giving a wave function ψ
(x) that will obey the above conditions are these specific E values. In other words, not all
energy values are “allowed” in the sense that they can produce ψ functions that are
continuous and vanish in regions where V(x) is infinite. If one uses an energy E that is
not one of the allowed values and substitutes this E into sin(kx), the resultant function
will not vanish at x = L. I hope the solution to this problem reminds you of the violin
42
string that we discussed earlier. Recall that the violin string being tied down at x = 0 and
at x = L gave rise to quantization of the the wavelength just as the conditions that ψ be
continuous at x = 0 and x = L gave energy quantization.
Substituting k = npi/L into ψ = A sin(kx) gives
ψ (x) = A sin(npix/L).
The value of A can be found by remembering that |Ψ|2 is supposed to represent the
probability density for finding the particle at x. Such probability densities are supposed to
be normalized, meaning that their integral over all x values should amount to unity. So,
we can find A by requiring that
1 = ∫ |ψ(x)|2 dx = |A|2 ∫ sin2(npix/L) dx
where the integral ranges from x = o to x = L. Looking up the integral of sin2(ax) and
solving the above equation for the so-called normalization constant A gives
A = (2/L)1/2 and so
ψ(x) = (2/L)1/2 sin(npix/L).
The values that n can take on are n = 1, 2, 3, ….; the choice n = 0 is unacceptable because
it would produce a wave function ψ(x) that vanishes at all x.
43
The full x- and t- dependent wave functions are then given as
Ψ(x,t) = (2/L)1/2 sin(npix/L) exp[-it n2 pi2 h2/(2mL2)/ h].
Notice that the spatial probability density |Ψ(x,t)|2 is not dependent on time and is equal
to |ψ(x)|2 because the complex exponential disappears when Ψ*Ψ is formed. This means
that the probability of finding the particle at various values of x is time-independent.
Another thing I want you to notice is that, unlike the classical dynamics case, not
all energy values E are allowed. In the Newtonian dynamics situation, E could be
specified and the particle’s momentum at any x value was then determined to within a
sign. In contrast, in quantum mechanics, one must determine, by solving the Schr?dinger
equation, what the allowed values of E are. These E values are quantized, meaning that
they occur only for discrete values E = n2 pi2h2/(2mL2) determined by a quantum number
n, by the mass of the particle m, and by characteristics of the potential (L in this case).
D. The Probability Densities
Let’s now look at some of the wave functions Ψ(x) and compare the probability
densities |Ψ(x)|2 that they represent to the classical probability densities discussed earlier.
The n = 1 and n = 2 wave functions are shown in the top of Fig. 1.8. The corresponding
probability densities are shown below the wave functions in two formats
(as x-y plots and shaded plots that could relate to the flashing light way of monitoring the
particle’s location that we discussed earlier).
44
Figure 1. 8. The two lowest wave functions and probability densities
A more complete set of wave functions (for n ranging from 1 to 7) are shown in Fig. 1. 9.
45
Figure 1. 9. Seven lowest wave functions and energies
Notice that as the quantum number n increases, the energy E also increases
(quadratically with n in this case) and the number of nodes in Ψ also increases. Also
notice that the probability densities are very different from what we encountered earlier
46
for the classical case. For example, look at the n = 1 and n = 2 densities and compare
them to the classical density illustrated in Fig. 1.10.
Figure 1. 10. Classical probability density for potential shown
The classical density is easy to understand because we are familiar with classical
dynamics. In this case, we say that P(x) is constant within the box because the fact that
V(x) is constant causes the kinetic energy and hence the speed of the particle to remain
constant. In contrast, the n = 1 quantum wave function’s P(x) plot is peaked in the middle
of the box and falls to zero at the walls. The n = 2 density P(x) has two peaks (one to the
left of the box midpoint, and one to the right), a node at the box midpoint, and falls to
zero at the walls. One thing that students often ask me is “how does the particle get from
being in the left peak to being in the right peak if it has zero chance of ever being at the
midpoint where the node is?” The difficulty with this question is that it is posed in a
terminology that asks for a classical dynamics answer. That is, by asking “how does the
particle get...” one is demanding an answer that involves describing its motion (i.e, it
47
moves from here at time t1 to there at time t2). Unfortunately, quantum mechanics does
not deal with issues such as a particle’s trajectory (i.e., where it is at various times) but
only with its probabilty of being somewhere (i.e., |Ψ|2). The next section will treat such
paradoxical issues even further.
E. Classical and Quantum Probability Densities
As just noted, it is tempting for most beginning students of quantum mechanics to
attempt to interpret the quantum behavior of a particle in classical terms. However, this
adventure is full of danger and bound to fail because small light particles simply do not
move according to Newton’s laws. To illustrate, let’s try to “understand” what kind of
(classical) motion would be consistent with the n = 1 or n = 2 quantum P(x) plots shown
in Fig. B. 8. However, as I hope you anticipate, this attempt at gaining classical
understanding of a quantum result will not “work” in that it will lead to nonsensical
results. My point in leading you to attempt such a classical understanding is to teach you
that classical and quantum results are simply different and that you must resist the urge to
impose a classical understanding on quantum results.
For the n = 1 case, we note that P(x) is highest at the box midpoint and vanishes
at x = 0 and x = L. In a classical mechanics world, this would mean that the particle
moves slowly near x = L/2 and more quickly near x = 0 and x = L. Because the particle’s
total energy E must remain constant as it moves, in regions where it moves slowly, the
potential it experiences must be high, and where it moves quickly, V must be small. This
analysis (n.b., based on classical concepts) would lead us to conclude that the n =1 P(x)
48
arises from the particle moving in a potential that is high near x = L/2 and low as x
approaches 0 or L.
A similar analysis of the n = 2 P(x) plot would lead us to conclude that the
particle for which this is the correct P(x) must experience a potential that is high midway
between x = 0 and x = L/2, high midway between x = L/2 and x = L,. and very low near x
= L/2 and near x = 0 and x = L. These conclusions are “crazy” because we know that the
potential V(x) for which we solved the Schr?dinger equation to generate both of the wave
functions (and both probability densities) is constant between x = 0 and x = L. That is, we
know the same V(x) applies to the particle moving in the n = 1 and n = 2 states, whereas
the classical motion analysis offered above suggests that V(x) is different for these two
cases.
What is wrong with our attempt to understand the quantum P(x) plots? The
mistake we made was in attempting to apply the equations and concepts of classical
dynamics to a P(x) plot that did not arise from classical motion. Simply put, one can not
ask how the particle is moving (i.e., what is its speed at various positions) when the
particle is undergoing quantum dynamics. Most students, when first experiencing
quantum wave functions and quantum probabilities, try to think of the particle moving in
a classical way that is consistent with the quantum P(x). This attempt to retain a degree of
classical understanding of the particle’s movement is always met with frustration, as I
illustrated with the above example and will illustrate later in other cases.
Continuing with this first example of how one solves the Schr?dinger equation
and how one thinks of the quantized E values and wave functions Ψ, let me offer a little
more optimistic note than offered in the preceding discussion. If we examine the Ψ(x)
49
plot shown in Fig. B.9 for n = 7, and think of the corresponding P(x) = |Ψ(x)|2, we note
that the P(x) plot would look something like that shown in Fig. 1. 11.
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sin(7 pi x/L)^2
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Figure 1. 11. Quantum probability density for n = 7 showing seven peaks and six nodes
It would have seven maxima separated by six nodes. If we were to plot |Ψ(x)|2 for a very
large n value such as n = 55, we would find a P(x) plot having 55 maxima separated by
54 nodes, with the maxima separated approximately by distances of (1/55L). Such a plot,
when viewed in a “coarse grained” sense (i.e., focusing with somewhat blurred vision on
the positions and heights of the maxima) looks very much like the classical P(x) plot in
which P(x) is constant for all x. In fact, it is a general result of quantum mechanics that
the quantum P(x) distributions for large quantum numbers take on the form of the
50
classical P(x) for the same potential V that was used to solve the Schr?dinger equation. It
is also true that classical and quantum results agree when one is dealing with heavy
particles. For example, a given particle-in-a-box energy En = n2 h2/(2mL2) would be
achieved for a heavier particle at higher n-values than for a lighter particle. Hence,
heavier particles, moving with a given energy E, have higher n and thus more classical
probability distributions.
We will encounter this so-called quantum-classical correspondence principal
again when we examine other model problems. It is an important property of solutions to
the Schr?dinger equation because it is what allows us to bridge the “gap” between using
the Schr?dinger equation to treat small light particles and the Newton equations for
macroscopic (big, heavy) systems.
Another thing I would like you to be aware of concerning the solutions ψ and E to
this Schr?dinger equation is that each pair of wave functions ψn and ψn’ belonging to
different quantum numbers n and n’ (and to different energies) display a property termed
orthonormality. This property means that not only are ψn and ψn’ each normalized
1= ∫ |ψn|2 dx = ∫ |ψn’|2 dx,
but they are also orthogonal to each other
0 = ∫ (ψn)* ψn’ dx
51
where the complex conjugate * of the first function appears only when the ψ solutions
contain imaginary components (you have only seen one such case thus far- the exp(imφ)
eigenfunctions of the z-component of angular momentm). It is common to write the
integrals displaying the normalization and orthogonality conditions in the following so-
called Dirac notation
1 = <ψn | ψn> 0 = <ψn | ψn’>,
where the | > and < | symbols represent ψ and ψ*, respectively, and putting the two
together in the < | > construct implies the integration over the variable that ψ depends
upon.
The orthogonality conditon can be viewed as similar to the condition of two
vectors v1 and v2 being perpendicular, in which case their scalar (sometimes called “dot”)
product vanishes v1 ? v2 = 0. I want you to keep this property in mind because you will
soon see that it is a characteristic not only of these particle-in-a-box wave functions but
of all wave functions obtained from any Schr?dinger equation.
In fact, the orthogonality property is even broader than the above discussion
suggests. It turns out that all quantum mechanical operators formed as discussed earlier
(replacing Cartesian momenta p by the corresponding -i h ?/?q operator and leaving all
Cartesian coordinates as they are) can be shown to be so-called Hermitian operators. This
means that they form Hermitian matrices when they are placed between pairs of functions
and the coordinates are integrated over. For example, the matrix representation of an
operator F when acting on a set of functions denoted {φJ} is:
52
FI,J = <φI | F |φJ> = ∫ φI* F φJ dq.
For all of the operators formed following the rules stated earlier, one finds that these
matrices have the following property:
FI,J = FJ,I*
which makes the matrices what we call Hermitian. If the functions upon which F acts and
F itself have no imaginary parts (i.e., are real), then the matrices turn out to be
symmetric:
FI,J = FJ,I .
The importance of the Hermiticity or symmetry of these matrices lies in the fact that it
can be shown that such matrices have all real (i.e., not complex) eigenvalues and have
eigenvectors that are orthogonal.
So, all quantum mechanical operators, not just the Hamiltonian, have real
eigenvalues (this is good since these eigenvalues are what can be measured in any
experimental observation of that property) and orthogonal eigenfunctions. It is important
to keep these facts in mind because we make use of them many times throughout this
text.
53
F. Time Propagation of Wave functions
For a system that exists in an eigenstate Ψ(x) = (2/L)1/2 sin(npix/L) having an
energy En = n2 pi2h2/(2mL2), the time-dependent wave function is
Ψ(x,t) = (2/L)1/2 sin(npix/L) exp(-itEn/h),
which can be generated by applying the so-called time evolution operator U(t,0) to the
wave function at t = 0:
Ψ(x,t) = U(t,0) Ψ(x,0)
where an explicit form for U(t,t’) is:
U(t,t’) = exp[-i(t-t’)H/ h].
The function Ψ(x,t) has a spatial probability density that does not depend on time because
Ψ*(x,t) Ψ(x,t) = (2/L) sin2(npix/L);
since exp(-itEn/h) exp(itEn/h) = 1. However, it is possible to prepare systems (even in real
laboratory settings) in states that are not single eigenstates; we call such states
superposition states. For example, consider a particle moving along the x- axis within the
“box” potential but in a state whose wave function at some initial time t = 0 is
54
Ψ(x,0) = 2-1/2 (2/L)1/2 sin(1pix/L) – 2-1/2 (2/L)1/2 sin(2pix/L).
This is a superposition of the n =1 and n = 2 eigenstates. The probability density
associated with this function is
|Ψ|2 = 1/2{(2/L) sin2(1pix/L)+ (2/L) sin2(2pix/L) -2(2/L) sin(1pix/L)sin(2pix/L)}.
The n = 1 and n = 2 components, the superposition Ψ, and the probability density at t = 0
|Ψ|2 are shown in the first three panels of Fig. 1.12.
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Figure 1. 12. The n = 1 and n = 2 wave functions, their superposition, and the t = 0 and
time-evolved probability densities of the superposition
It should be noted that the probability density associated with this superposition state is
not symmetric about the x=L/2 midpoint even though the n = 1 and n = 2 component
wave functions and densities are. Such a density describes the particle localized more
strongly in the large-x region of the box than in the small-x region.
Now, let’s consider the superposition wave function and its density at later times.
Applying the time evolution operator exp(-itH/h) to Ψ(x,0) generates this time-evolved
function at time t:
Ψ(x,t) = exp(-itH/h) {2-1/2 (2/L)1/2 sin(1pix/L) – 2-1/2 (2/L)1/2 sin(2pix/L)}
= {2-1/2 (2/L)1/2 sin(1pix/L) ) exp(-itE1/h). – 2-1/2 (2/L)1/2 sin(2pix/L) ) exp(-itE2/h) }.
The spatial probability density associated with this Ψ is:
|Ψ(x,t)|2 = 1/2{(2/L) sin2(1pix/L)+ (2/L) sin2(2pix/L)
-2(2/L) cos(E2-E1)t/h) sin(1pix/L)sin(2pix/L)}.
56
At t = 0, this function clearly reduces to that written earlier for Ψ(x,0). Notice that as time
evolves, this density changes because of the cos(E2-E1)t/h) factor it contains. In particular,
note that as t moves through a period of length δt = pi h/(E2-E1), the cos factor changes
sign. That is, for t = 0, the cos factor is +1; for t = pi h/(E2-E1), the cos factor is –1; for t =
2 pi h/(E2-E1), it returns to +1. The result of this time-variation in the cos factor is that |Ψ|2
changes in form from that shown in the bottom left panel of Fig. B. 12 to that shown in
the bottom right panel (at t = pi h/(E2-E1)) and then back to the form in the bottom left
panel (at t = 2 pi h/(E2-E1)). One can interpret this time variation as describing the
particle’s probability density (not its classical position!), initially localized toward the
right side of the box, moving to the left and then back to the right. Of course, this time
evolution will continue over more and more cycles as time evolves further.
This example illustrates once again the difficulty with attempting to localize
particles that are being described by quantum wave functions. For example, a particle that
is characterized by the eigenstate (2/L)1/2 sin(1pix/L) is more likely to be detected near x
= L/2 than near x = 0 or x = L because the square of this function is large near x = L/2. A
particle in the state (2/L)1/2 sin(2pix/L) is most likely to be found near x = L/4 and x =
3L/4, but not near x = 0, x = L/2, or x =L. The issue of how the particle in the latter state
moves from being near x = L/4 to x = 3L/4 is not something quantum mechanics deals
with. Quantum mechanics does not allow us to follow the particle’s trajectory which is
what we need to know when we ask how it moves from one place to another.
Nevertheless, superposition wave functions can offer, to some extent, the opportunity to
follow the motion of the particle. For example, the superposition state written above as
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2-1/2 (2/L)1/2 sin(1pix/L) – 2-1/2 (2/L)1/2 sin(2pix/L) has a probability amplitude that changes
with time as shown in the figure. Moreover, this amplitude’s major peak does move from
side to side within the box as time evolves. So, in this case, we can say with what
frequency the major peak moves back and forth. In a sense, this allows us to “follow” the
particle’s movements, but only to the extent that we are satisfied with ascribing its
location to the position of the major peak in its probability distribution. That is, we can
not really follow its “precise” location, but we can follow the location of where it is very
likely to be found. This is an important observation that I hope the student will keep fresh
in mind. It is also an important ingredient in modern quantum dynamics in which
localized wave packets, similar to superposed eigenstates, are used to detail the position
and speed of a particle’s main probability density peak.
The above example illustrates how one time-evolves a wave function that can be
expressed as a linear combination (i.e., superposition) of eigenstates of the problem at
hand. As noted above, there is a large amount of current effort in the theoretical
chemistry community aimed at developing efficient approximations to the exp(-itH/h)
evolution operator that do not require Ψ(x,0) to be explicitly written as a sum of
eigenstates. This is important because, for most systems of direct relevance to molecules,
one can not solve for the eigenstates; it is simply too difficult to do so. You can find a
significantly more detailed treatment of the research-level treatment of this subject in my
Theory Page web site and my QMIC text book. However, let’s spend a little time on a
brief introduction to what is involved.
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The problem is to express exp(-itH/ h) Ψ(qj), where Ψ(qj) is some initial wave
function but not an eigenstate, in a manner that does not require one to first find the
eigenstates {ΨJ} of H and to expand Ψ in terms of these eigenstates:
Ψ = ΣJ CJ ΨJ
after which the desired function is written as
exp(-itH/ h) Ψ(qj) = ΣJ CJ ΨJ exp(-itEJ/h).
The basic idea is to break H into its kinetic T and potential V energy components and to
realize that the differential operators appear in T only. The importance of this observation
lies in the fact that T and V do not commute which means that TV is not equal to VT
(n.b., for two quantities to commute means that their order of appearance does not
matter). Why do they not commute? Because T contains second derivatives with respect
to the coordinates {qj} that V depends on, so, for example, d2/dq2(V(q) Ψ(q)) is not equal
to V(q)d2/dq2Ψ(q). The fact that T and V do not commute is important because the most
common approaches to approximating exp(-itH/ h) is to write this single exponential in
terms of exp(-itT/ h) and exp(-itV/ h). However, the identity
exp(-itH/ h) = exp(-itV/ h) exp(-itT/ h)
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is not fully valid as one can see by expanding all three of the above exponential factors as
exp(x) = 1 + x + x2/2! + ..., and noting that the two sides of the above equation only agree
if one can assume that TV = VT, which, as we noted, is not true.
In most modern approaches to time propagation, one divides the time interval t
into many (i.e., P of them) small time “slices” τ = t/P. One then expresses the evolution
operator as a product of P short-time propagators:
exp(-itH/ h) = exp(-iτH/ h) exp(-iτH/ h) exp(-iτH/ h) ... = [exp(-iτH/ h) ]P.
If one can then develop an efficient means of propagating for a short time τ, one can then
do so over and over again P times to achieve the desired full-time propagation.
It can be shown that the exponential operator involving H can better be
approximated in terms of the T and V exponential operators as follows:
exp(-iτH/ h) ≈ exp(-τ2 (TV-VT)/ h2) exp(-iτV/ h) exp(-iτT/ h).
So, if one can be satisfied with propagating for very short time intervals (so that the τ2
term can be neglected), one can indeed use
exp(-iτH/ h) ≈ exp(-iτV/ h) exp(-iτT/ h)
as an approximation for the propagator U(τ,0).
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To progress further, one then expresses exp(-iτT/ h) acting on the initial function
Ψ(q) in terms of the eigenfunctions of the kinetic energy operator T. Note that these
eigenfunctions do not depend on the nature of the potential V, so this step is valid for any
and all potentials. The eigenfunctions of T = - h2/2m d2/dq2 are
ψp(q) = (1/2pi)1/2 exp(ipq/ h)
and they obey the following orthogonality
∫ ψp'*(q) ψp(q) dq = δ(p'-p)
and completeness relations
∫ ψp(q) ψp*(q') dp = δ(q-q').
Writing Ψ(q) as
Ψ(q) = ∫ δ(q-q') Ψ(q'),
and using the above expression for δ(q-q') gives:
Ψ(q) = ∫∫ ψp(q) ψp*(q') Ψ(q') dq' dp.
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Then inserting the explicit expressions for ψp(q) and ψp*(q') in terms of
ψp(q) = (1/2pi)1/2 exp(ipq/ h) gives
Ψ(q) = ∫∫ (1/2pi)1/2 exp(ipq/ h) (1/2pi)1/2 exp(-ipq'/ h) Ψ(q') dq' dp.
Now, allowing exp(-iτT/ h) to act on this form for Ψ(q) produces
exp(-iτT/ h) Ψ(q) = ∫∫ exp(-iτp2h2/2mh) (1/2pi)1/2 exp(ip(q-q')/ h) (1/2pi)1/2 Ψ(q') dq' dp.
The integral over p above can be carried out analytically and gives
exp(-iτT/ h) Ψ(q) = (m/2piτh)1/2 ∫ exp(im(q-q')2/2τh) Ψ(q') dq'.
So, the final expression for the short-time propagated wave function is:
Ψ(q.τ) = exp(-iτV(q)/ h) (m/2piτh)1/2 ∫ exp(im(q-q')2/2τh) Ψ(q') dq',
which is the working equation one uses to compute Ψ(q,τ) knowing Ψ(q). Notice that all
one needs to know to apply this formula is the potential V(q) at each point in space. One
does not to know any of the eigenfunctions of the Hamiltonian to apply this method.
However, one does have to use this formula over and over again to propagate the initial
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wave function through many small time steps τ to achieve full propagation for the desired
time interval t = P τ.
Because this type of time propagation technique is a very active area of research
in the theory community, it is likely to continue to be refined and improved. Further
discussion of it is beyond the scope of this book, so I will not go further into this
direction.
III. Free Particle Motions in More Dimensions
The number of dimensions depends on the number of particles and the number of
spatial (and other) dimensions needed to characterize the position and motion of each
particle
A. The Schr?dinger Equation
Consider an electron of mass m and charge e moving on a two-dimensional
surface that defines the x,y plane (e.g., perhaps an electron is constrained to the surface of
a solid by a potential that binds it tightly to a narrow region in the z-direction), and
assume that the electron experiences a constant and not time-varying potential V0 at all
points in this plane. The pertinent time independent Schr?dinger equation is:
- h2/2m (?2/?x2 +?2/?y2)ψ(x,y) +V0ψ(x,y) = E ψ(x,y).
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The task at hand is to solve the above eigenvalue equation to determine the “allowed”
energy states for this electron. Because there are no terms in this equation that couple
motion in the x and y directions (e.g., no terms of the form xayb or ?/?x ?/?y or x?/?y),
separation of variables can be used to write ψ as a product ψ(x,y)=A(x)B(y). Substitution
of this form into the Schr?dinger equation, followed by collecting together all x-
dependent and all y-dependent terms, gives;
- h2/2m A-1?2A/?x2 - h2/2m B-1?2B/?y2 =E-V0.
Since the first term contains no y-dependence and the second contains no x-dependence,
and because the right side of the equation is independent of both x and y, both terms on
the left must actually be constant (these two constants are denoted Ex and Ey,
respectively). This observation allows two separate Schr?dinger equations to be written:
- h2/2m A-1?2A/?x2 =Ex, and
- h2/2m B-1?2B/?y2 =Ey.
The total energy E can then be expressed in terms of these separate energies Ex and Ey as
Ex + Ey = E-V0. Solutions to the x- and y- Schr?dinger equations are easily seen to be:
A(x) = exp(ix(2mEx/h2)1/2) and exp(-ix(2mEx/h2)1/2) ,
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B(y) = exp(iy(2mEy/h2)1/2) and exp(-iy(2mEy/h2)1/2).
Two independent solutions are obtained for each equation because the x- and y-space
Schr?dinger equations are both second order differential equations (i.e., a second order
differential equation has two independent solutions).
B. Boundary Conditions
The boundary conditions, not the Schr?dinger equation, determine whether the
eigenvalues will be discrete or continuous
If the electron is entirely unconstrained within the x,y plane, the energies Ex and Ey
can assume any values; this means that the experimenter can 'inject' the electron onto the
x,y plane with any total energy E and any components Ex and Ey along the two axes as
long as Ex + Ey = E. In such a situation, one speaks of the energies along both
coordinates as being 'in the continuum' or 'not quantized'.
In contrast, if the electron is constrained to remain within a fixed area in the x,y
plane (e.g., a rectangular or circular region), then the situation is qualitatively different.
Constraining the electron to any such specified area gives rise to boundary conditions that
impose additional requirements on the above A and B functions. These constraints can
arise, for example, if the potential V0(x,y) becomes very large for x,y values outside the
region, in which case, the probability of finding the electron outside the region is very
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small. Such a case might represent, for example, a situation in which the molecular
structure of the solid surface changes outside the enclosed region in a way that is highly
repulsive to the electron (e.g., as in the case of molecular corrals on metal surfaces). This
case could then represent a simple model of so-called “corrals” in which the particle is
constrained to a finite region of space.
For example, if motion is constrained to take place within a rectangular region
defined by 0 ≤ x ≤ Lx; 0 ≤ y ≤ Ly, then the continuity property that all wave functions
must obey (because of their interpretation as probability densities, which must be
continuous) causes A(x) to vanish at 0 and at Lx. That is, because A must vanish for x < 0
and must vanish for x > Lx, and because A is continuous, it must vanish at x = 0 and at x
= Lx. Likewise, B(y) must vanish at 0 and at Ly. To implement these constraints for A(x),
one must linearly combine the above two solutions exp(ix(2mEx/h2)1/2) and
exp(-ix(2mEx/h2)1/2) to achieve a function that vanishes at x=0:
A(x) = exp(ix(2mEx/h2)1/2) - exp(-ix(2mEx/h2)1/2).
One is allowed to linearly combine solutions of the Schr?dinger equation that have the
same energy (i.e., are degenerate) because Schr?dinger equations are linear differential
equations. An analogous process must be applied to B(y) to achieve a function that
vanishes at y=0:
B(y) = exp(iy(2mEy/h2)1/2) - exp(-iy(2mEy/h2)1/2).
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Further requiring A(x) and B(y) to vanish, respectively, at x=Lx and y=Ly, gives
equations that can be obeyed only if Ex and Ey assume particular values:
exp(iLx(2mEx/h2)1/2) - exp(-iLx(2mEx/h2)1/2) = 0, and
exp(iLy(2mEy/h2)1/2) - exp(-iLy(2mEy/h2)1/2) = 0.
These equations are equivalent (i.e., using exp(ix) = cos(x) + i sin(x)) to
sin(Lx(2mEx/h2)1/2) = sin(Ly(2mEy/h2)1/2) = 0.
Knowing that sin(θ) vanishes at θ = npi, for n=1,2,3,..., (although the sin(npi) function
vanishes for n=0, this function vanishes for all x or y, and is therefore unacceptable
because it represents zero probability density at all points in space) one concludes that the
energies Ex and Ey can assume only values that obey:
Lx(2mEx/h2)1/2 =nxpi,
Ly(2mEy/h2)1/2 =nypi, or
Ex = nx2pi2 h2/(2mLx2), and
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Ey = ny2pi2 h2/(2mLy2), with nx and ny =1,2,3, ...
It is important to stress that it is the imposition of boundary conditions, expressing the
fact that the electron is spatially constrained, that gives rise to quantized energies. In the
absence of spatial confinement, or with confinement only at x =0 or Lx or only at y =0 or
Ly, quantized energies would not be realized.
In this example, confinement of the electron to a finite interval along both the x and
y coordinates yields energies that are quantized along both axes. If the electron were
confined along one coordinate (e.g., between 0 ≤ x ≤ Lx) but not along the other (i.e.,
B(y) is either restricted to vanish at y=0 or at y=Ly or at neither point), then the total
energy E lies in the continuum; its Ex component is quantized but Ey is not. Analogs of
such cases arise, for example, when a linear triatomic molecule has more than enough
energy in one of its bonds to rupture it but not much energy in the other bond; the first
bond's energy lies in the continuum, but the second bond's energy is quantized.
Perhaps more interesting is the case in which the bond with the higher dissociation
energy is excited to a level that is not enough to break it but that is in excess of the
dissociation energy of the weaker bond. In this case, one has two degenerate states- i. the
strong bond having high internal energy and the weak bond having low energy (ψ1), and
ii. the strong bond having little energy and the weak bond having more than enough
energy to rupture it (ψ2). Although an experiment may prepare the molecule in a state
that contains only the former component (i.e., ψ= C1ψ1 + C2ψ2 with C1 = 1, C2 = 0),
coupling between the two degenerate functions (induced by terms in the Hamiltonian H
that have been ignored in defining ψ1 and ψ2) usually causes the true wave function
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Ψ = exp(-itH/h) ψ to acquire a component of the second function as time evolves. In such
a case, one speaks of internal vibrational energy relaxation (IVR) giving rise to
unimolecular decomposition of the molecule.
C. Energies and Wave functions for Bound States
For discrete energy levels, the energies are specified functions that depend on
quantum numbers, one for each degree of freedom that is quantized
Returning to the situation in which motion is constrained along both axes, the
resultant total energies and wave functions (obtained by inserting the quantum energy
levels into the expressions for A(x) B(y)) are as follows:
Ex = nx2pi2 h2/(2mLx2), and
Ey = ny2pi2 h2/(2mLy2),
E = Ex + Ey +V0
ψ(x,y) = (1/2Lx)1/2 (1/2Ly)1/2[exp(inxpix/Lx) -exp(-inxpix/Lx)]
[exp(inypiy/Ly) -exp(-inypiy/Ly)], with nx and ny =1,2,3, ... .
The two (1/2L)1/2 factors are included to guarantee that ψ is normalized:
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∫ |ψ(x,y)|2 dx dy = 1.
Normalization allows |ψ(x,y)|2 to be properly identified as a probability density for
finding the electron at a point x, y.
Shown in Fig. 1. 13 are plots of four such two dimensional wave functions for nx and
ny values of (1,1), (2,1), (1.2) and (2,2), respectively.
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Figure 1. 13. Plots of the (1,1), (2,1), (1,2) and (2,2) wave functions
Note that the functions vanish on the boundaries of the box, and notice how the number
of nodes (i.e., zeroes encountered as the wave function oscillates from positive to
negative) is related to the nx and ny quantum numbers and to the energy. This pattern of
more nodes signifying higher energy is one that we encounter again and again in quantum
mechanics and is something the student should be able to use to “guess” the relative
energies of wave functions when their plots are at hand. Finally, you should also notice
that, as in the one-dimensional box case, any attempt to classically interpret the
probabilities P(x,y) corresponding to the above quantum wave functions will result in
failure. As in the one-dimensional case, the classical P(x,y) would be constant along
slices of fixed x and varying y or slices of fixed y and varying x within the box because
the speed is constant there. However, the quantum P(x,y) plots, at least for small quantum
numbers, are not constant. For large nx and ny values, the quantum P(x,y) plots will again,
via the quantum-classical correspondence principle, approach the (constant) classical
P(x,y) form.
D. Quantized Action Can Also be Used to Derive Energy Levels
There is another approach that can be used to find energy levels and is especially
straightforward to use for systems whose Schr?dinger equations are separable. The so-
called classical action (denoted S) of a particle moving with momentum p along a path
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leading from initial coordinate qi at initial time ti to a final coordinate qf at time tf is
defined by:
S = ??
qi;ti
qf;tf
p?dq .
Here, the momentum vector p contains the momenta along all coordinates of the system,
and the coordinate vector q likewise contains the coordinates along all such degrees of
freedom. For example, in the two-dimensional particle in a box problem considered
above, q = (x, y) has two components as does p = (px, py), and the action integral is:
S = ??
xi;yi;ti
xf;yf;tf
(px dx + py dy) .
In computing such actions, it is essential to keep in mind the sign of the momentum as the
particle moves from its initial to its final positions. An example will help clarify these
matters.
For systems such as the above particle in a box example for which the Hamiltonian
is separable, the action integral decomposes into a sum of such integrals, one for each
degree of freedom. In this two-dimensional example, the additivity of H:
H = Hx + Hy = px2/2m + py2/2m + V(x) + V(y)
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= - h2/2m ?2/?x2 + V(x) - h2/2m ?2/?y2 + V(y)
means that px and py can be independently solved for in terms of the potentials V(x) and
V(y) as well as the energies Ex and Ey associated with each separate degree of freedom:
px = ± 2m(Ex - V(x))
py = ± 2m(Ey - V(y)) ;
the signs on px and py must be chosen to properly reflect the motion that the particle is
actually undergoing. Substituting these expressions into the action integral yields:
S = Sx + Sy
= ??
xi;ti
xf;tf
± 2m(Ex - V(x)) dx + ??
yi;ti
yf;tf
± 2m(Ey - V(y)) dy .
The relationship between these classical action integrals and existence of quantized
energy levels has been shown to involve equating the classical action for motion on a
closed path to an integral multiple of Planck's constant:
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Sclosed = ??
qi;ti
qf=qi;tf
p?dq = n h. (n = 1, 2, 3, 4, ...)
Applied to each of the independent coordinates of the two-dimensional particle in a box
problem, this expression reads:
nx h = ??
x=0
x=Lx
2m(Ex - V(x)) dx + ??
x=Lx
x=0
- 2m(Ex - V(x)) dx
ny h = ??
y=0
y=Ly
2m(Ey - V(y)) dy + ??
y=Ly
y=0
- 2m(Ey - V(y)) dy .
Notice that the sign of the momenta are positive in each of the first integrals appearing
above (because the particle is moving from x = 0 to x = Lx, and analogously for y-
motion, and thus has positive momentum) and negative in each of the second integrals
(because the motion is from x = Lx to x = 0 (and analogously for y-motion) and thus the
particle has negative momentum). Within the region bounded by 0 ≤ x ≤ Lx; 0 ≤ y ≤ Ly,
the potential vanishes, so V(x) = V(y) = 0. Using this fact, and reversing the upper and
lower limits, and thus the sign, in the second integrals above, one obtains:
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nx h = 2 ??
x=0
x=Lx
2mEx dx = 2 2mEx Lx
ny h = 2 ??
y=0
y=Ly
2mEy dy = 2 2mEy Ly.
Solving for Ex and Ey, one finds:
Ex = (nxh)
2
8mLx2
Ey = (nyh)
2
8mLy2 .
These are the same quantized energy levels that arose when the wave function boundary
conditions were matched at x = 0, x = Lx and y = 0, y = Ly. In this case, one says that the
Bohr-Sommerfeld quantization condition:
n h = ??
qi;ti
qf=qi;tf
p?dq
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has been used to obtain the result.
The use of action quantization as illustrated above has become a very important tool.
It has allowed scientists to make great progress toward bridging the gap between classical
and quantum descriptions of molecular dynamics. In particular, by using classical
concepts such as trajectories and then appending quantal action conditions, people have
been able to develop so-called semi-classical models of molecular dynamics. In such
models, one is able to retain a great deal of classical understanding while building in
quantum effects such as energy quantization, zero-point energies, and interferences.
E. Quantized Action Does Not Always Work
Unfortunately, the approach of quantizing the action does not always yield the
correct expression for the quantized energies. For example, when applied to the so-called
harmonic oscillator problem that we will study in quantum form later, which serves as the
simplest reasonable model for vibration of a diatomic molecule AB, one expresses the
total energy as
E = p2/2μ + k/2 x2
where μ = mAmB/(mA + mB) is the reduced mass of the AB diatom, k is the force constant
describing the bond between A and B, x is the bond-length displacement, and p is the
momentum along the bond length. The quantized action requirement then reads
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n h = ∫ p dx = ∫ [2μ(E-k/2 x2)]1/2 dx.
This integral is carried out between x = - (2E/k)1/2 and (2E/k)1/2 the left and right turning
points of the oscillatory motion and back again to form a closed path. Carrying out this
integral and equating it to n h gives the following expression for the energy E:
E = n (h/2pi) [k/μ]1/2
where the quantum number n is allowed to assume integer values ranging from n = 0, 1,
2, to infinity. The problem with this result is that it is wrong! As experimental data
clearly show, the lowest-energy level for the vibrations of a molecule do not have E = 0;
they have a “zero-point” energy that is approximately equal to 1/2 (h/2pi) [k/μ]1/2. So,
although the action quantization condition yields energies whose spacings are reasonably
in agreement with laboratory data for low-energy states (e.g., such states have
approximately constant spacings), it fails to predict the zero-point energy content of such
vibrations. As we will see later, a proper quantum mechanical treatment of the harmonic
oscillator yields energies of the form
E = (n + 1/2) (h/2pi) [k/μ]1/2
which differs from the action-based result by the proper zero-point energy.
Even with such difficulties known, much progress has been made in extending the
most elementary action-based methods to more and more systems by introducing, for
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example, rules that allow the quantum number n to assume half-integer as well as integer
values. Clearly, if n were allowed to equal 1/2, 3/2, 5/2, ..., the earlier action integral
would have produced the correct result. However, how does one know when to allow n to
assume only integer or only half-integer or both integer and half-integer values. The
answers to this question are beyond the scope of this text and constitute an active area of
research. For now, it is enough for the student to be aware that one can often find energy
levels by using action integrals, but one must be careful in doing so because sometimes
the answers are wrong.
Before leaving this section, it is worth noting that the appearance of half-integer
quantum numbers does not only occur in the harmonic oscillator case. To illustrate, let us
consider the Lz angular momentum operator discussed earlier. As we showed, this
operator, when computed as the z-component of r x p, can be written in polar (r, θ, φ)
coordinates as
Lz = - i h d/dφ.
The eigenfunctions of this operator have the form exp(iaφ), and the eigenvalues are a h.
Because geometries with azimuthal angles equal to φ or equal to φ + 2pi are exactly the
same geometries, the function exp(iaφ) should be exactly the same as exp(ia(φ+2pi)). This
can only be the case if a is an integer. Thus, one concludes that only integral multiples of
h can be “allowed” values of the z-component of angular momentum. Experimentally,
one measures the z-component of an angular momentum by placing the system
possessing the angular momentum in a magnetic field of strength B and observing how
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many z-component energy states arise. For example, a boron atom with its 2p orbital has
one unit of orbital angular momentum, so one finds three separate z-component values
which are usually denoted m = -1, m=0, and m=1. Another example is offered by the
scandium atom with one unpaired electron in a d orbital; this atom’s states split into five
(m = -2, -1, 0, 1, 2) z-component states. In each case, one finds 2L + 1 values of the m
quantum number, and, because L is an integer, 2L + 1 is an odd integer. Both of these
observations are consistent with the expectation that only integer values can occur for Lz
eigenvalues.
However, it has been observed that some species do not possess 3 or 5 or 7 or 9 z-
component states but an even number of such states. In particular, when electrons,
protons, or neutrons are subjected to the kind of magnetic field experiment mentioned
above, these particles are observed to have only two z-component eigenvalues. Because,
as we discuss later in this text, all angular momenta have z-component eigenvalues that
are separated from one aother by unit multiples of h, one is forced to conclude that these
three fundamental building-block particles have z-component eigenvalues of 1/2 h and
– 1/2 h. The appearance of half-integral angular momenta is not consistent with the
observation made earlier that φ and φ + 2pi correspond to exactly the same physical point
in coordinate space, which, in turn, implies that only full-integer angular momenta are
possible.
The resolution of the above paradox (i.e., how can half-integer angular momenta
exist?) involved realizing that some angular momenta correspond not to the r x p angular
momenta of a physical mass rotating, but, instead, are intrinsic properties of certain
particles. That is, the intrinsic angular momenta of electrons, protons, and neutrons can
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not be viewed as arising from rotation of some mass that comprises these particles.
Instead, such intrinsic angular momenta are fundamental “built in” characteristics of
these particles. For example, the two 1/2 h and – 1/2 h angular momentum states of an
electron, usually denoted α and β, respectively, are two internal states of the electron that
are degenerate in the absence of a magnetic field but which represent two distinct states
of the electron. Analogously, a proton has 1/2 h and – 1/2 h states, as do neutrons. All
such half-integral angular momentum states can not be accounted for using classical
mechanics but are known to arise in quantum mechanics.
Chapter 2. Model Problems That Form Important Starting Points
The model problems discussed in this Section form the basis for chemists’
understanding of the electronic states of atoms, molecules, clusters, and solids as well as
the rotational and vibrational motions of molecules.
I. Free Electron Model of Polyenes
The particle-in-a-box problem provides an important model for several relevant
chemical situations
The 'particle in a box' model for motion in two dimensions discussed earlier can
obviously be extended to three dimensions or to one. For two and three dimensions, it
provides a crude but useful picture for electronic states on surfaces or in metallic crystals,
respectively. I say metallic crystals because it is in such systems that the outermost
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valence electrons are reasonably well treated as moving freely. Free motion within a
spherical volume gives rise to eigenfunctions that are used in nuclear physics to describe
the motions of neutrons and protons in nuclei. In the so-called shell model of nuclei, the
neutrons and protons fill separate s, p, d, etc. orbitals with each type of nucleon forced to
obey the Pauli principle (i.e., to have no more than two nucleons in each orbital because
protons and neutrons are Fermions). To remind you, I display in Fig. 2. 1 the angular
shapes that characterize s, p, and d orbitals.
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Figure 2.1. The angular shapes of s, p, and d functions
This same spherical box model has also been used to describe the orbitals of valence
electrons in clusters of metal atoms such as Csn, Cun, Nan and their positive and negative
ions. Because of the metallic nature of these species, their valence electrons are
essentially free to roam over the entire spherical volume of the cluster, which renders this
simple model rather effective. In this model, one thinks of each electron being free to
roam within a sphere of radius R (i.e., having a potential that is uniform within the sphere
and infinite outside the sphere). Finally, as noted above, this same spherical box model
forms the basis of the so-called shell model of nuclear structure. In this model, one
assumes that the protons and neutrons that make up a nucleus, both of which are
Fermions, occupy spherical-box orbitals (one set of orbitals for protons, another set for
neutrons because they are distinguishable from one another). By placing the protons and
neutrons into these orbitals, two to an orbital, one achieves a description of the energy
levels of the nucleus. Excited states are achieved by promoting a neutron or proton from
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an occupied orbital to a virtual (i.e., previously unoccupied) orbital. In such a model,
especially stable nuclei are achieved when “closed-shell” configurations such as 1s2 or
1s22s22p6 are realized (e.g., 4He has both neutrons and protons in 1s2 configurations).
The orbitals that solve the Schr?dinger equation inside such a spherical box are not
the same in their radial 'shapes' as the s, p, d, etc. orbitals of atoms because, in atoms,
there is an additional radial potential V(r) = -Ze2/r present. However, their angular shapes
are the same as in atomic structure because, in both cases, the potential is independent of
θ and φ. As the orbital plots shown above indicate, the angular shapes of s, p, and d
orbitals display varying number of nodal surfaces. The s orbitals have none, p orbitals
have one, and d orbitals have two. Analogous to how the number of nodes related to the
total energy of the particle constrained to the x, y plane, the number of nodes in the
angular wave functions indicates the amount of angular or rotational energy. Orbitals of s
shape have no angular energy, those of p shape have less then do d orbitals, etc.
One-dimensional free particle motion provides a qualitatively correct picture for pi-
electron motion along the ppi orbitals of delocalized polyenes. The one Cartesian
dimension then corresponds to motion along the delocalized chain. In such a model, the
box length L is related to the carbon-carbon bond length R and the number N of carbon
centers involved in the delocalized network L=(N-1) R. In Fig. 2.2, such a conjugated
network involving nine centers is depicted. In this example, the box length would be
eight times the C-C bond length.
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Figure 2.2. The pi atomic orbitals of a conjugated chain of nine carbon atoms
The eigenstates ψn(x) and their energies En represent orbitals into which electrons are
placed. In the example case, if nine pi electrons are present (e.g., as in the 1,3,5,7-
nonatetraene radical), the ground electronic state would be represented by a total wave
function consisting of a product in which the lowest four ψ's are doubly occupied and the
fifth ψ is singly occupied:
Ψ = ψ1αψ1βψ2αψ2βψ3αψ3βψ4αψ4βψ5α.
The z-component angular momentum states of the electrons are labeled α and β as
discussed earlier.
A product wave function is appropriate because the total Hamiltonian involves the
kinetic plus potential energies of nine electrons. To the extent that this total energy can be
represented as the sum of nine separate energies, one for each electron, the Hamiltonian
allows a separation of variables
H ? Σj H(j)
Conjugated pi Network with 9 Centers Involved
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in which each H(j) describes the kinetic and potential energy of an individual electron.
Recall that when a partial differential equation has no operators that couple its different
independent variables (i.e., when it is separable), one can use separation of variables
methods to decompose its solutions into products. Thus, the (approximate) additivity of H
implies that solutions of H Ψ = E Ψ are products of solutions to
H (j) ψ(rj) = Ej ψ(rj).
The two lowest pi-excited states would correspond to states of the form
Ψ* = ψ1α ψ1β ψ2α ψ2β ψ3α ψ3β ψ4α ψ5β ψ5α , and
Ψ'* = ψ1α ψ1β ψ2α ψ2β ψ3α ψ3β ψ4α ψ4β ψ6α ,
where the spin-orbitals (orbitals multiplied by α or β) appearing in the above products
depend on the coordinates of the various electrons. For example,
ψ1α ψ1β ψ2α ψ2β ψ3α ψ3β ψ4α ψ5β ψ5α
denotes
85
ψ1α(r1) ψ1β (r2) ψ2α (r3) ψ2β (r4) ψ3α (r5) ψ3β (r6) ψ4α (r7)
ψ5β (r8) ψ5α (r9).
The electronic excitation energies from the ground state to each of the above excited
states within this model would be
?E* = pi2 h2/2m [ 52/L2 - 42/L2] and
?E'* = pi2 h2/2m [ 62/L2 - 52/L2].
It turns out that this simple model of pi-electron energies provides a qualitatively correct
picture of such excitation energies. Its simplicity allows one, for example, to easily
suggest how a molecule’s color (as reflected in the complementary color of the light the
molecule absorbs) varies as the conjugation length L of the molecule varies. That is,
longer conjugated molecules have lower-energy orbitals because L2 appears in the
denominator of the energy expression. As a result, longer conjugated molecules absorb
light of lower energy than do shorter molecules.
This simple particle-in-a-box model does not yield orbital energies that relate to
ionization energies unless the potential 'inside the box' is specified. Choosing the value of
this potential V0 such that V0 + pi2 h2/2m [ 52/L2] is equal to minus the lowest ionization
energy of the 1,3,5,7-nonatetraene radical, gives energy levels (as E = V0 + pi2 h2/2m
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[ n2/L2]) which can then be used as approximations to ionization energies.
The individual pi-molecular orbitals
ψn = (2/L)1/2 sin(npix/L)
are depicted in Fig. 2.3 for a model of the 1,3,5 hexatriene pi-orbital system for which the
'box length' L is five times the distance RCC between neighboring pairs of carbon atoms.
The magnitude of the kth C-atom centered atomic orbital in the nth pi-molecular orbital is
given by (2/L)1/2 sin(npikRCC/L).
Figure 2.3. The phases of the six molecular orbitals of a chain containing six atoms.
n = 6
n = 5
n = 4
n = 3
n = 2
n = 1
(2/L)1/2 sin(npix/L); L = 5 x RCC
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In this figure, positive amplitude is denoted by the clear spheres, and negative amplitude
is shown by the darkened spheres. Where two spheres of like shading overlap, the wave
function has enhanced amplitude; where two spheres of different shading overlap, a node
occurs. Once again, we note that the number of nodes increases as one ranges from the
lowest-energy orbital to higher energy orbitals. The reader is once again encouraged to
keep in mind this ubiquitous characteristic of quantum mechanical wave functions.
This simple model allows one to estimate spin densities at each carbon center and
provides insight into which centers should be most amenable to electrophilic or
nucleophilic attack. For example, radical attack at the C5 carbon of the nine-atom
nonatetraene system described earlier would be more facile for the ground state Ψ than
for either Ψ* or Ψ'*. In the former, the unpaired spin density resides in ψ5, which has
non-zero amplitude at the C5 site x=L/2. In Ψ* and Ψ'*, the unpaired density is in ψ4 and
ψ6, respectively, both of which have zero density at C5. These densities reflect the values
(2/L)1/2 sin(npikRCC/L) of the amplitudes for this case in which L = 8 x RCC for n = 5, 4,
and 6, respectively. Plots of the wave functions for n ranging from 1 to 7 are shown in
another format in Fig. 2.4 where the nodal pattern is emphasized.
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Figure 2.4. The nodal pattern for a chain containing seven atoms
I hope that by now the student is not tempted to ask how the electron “gets” from one
region of high amplitude, through a node, to another high-amplitude region. Remember,
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such questions are cast in classical Newtonian language and are not appropriate when
addressing the wave-like properties of quantum mechanics.
II. Bands of Orbitals in Solids
Not only does the particle in a box model offer a useful conceptual representation of
electrons moving in polyenes, but it also is the zeroth-order model of band structures in
solids. Let us consider a simple one-dimensional “crystal” consisting of a large number of
atoms or molecules, each with a single orbital (the blue spheres shown below) that it
contributes to the bonding. Let us arrange these building blocks in a regular “lattice” as
shown in the Fig. 2.5.
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Figure 2.5. The energy levels arising from 1, 2, 3, 4, and an infinite number of orbitals
In the top four rows of this figure we show the case with 1, 2, 3, and 4 building blocks.
To the left of each row, we display the energy splitting pattern into which the building
blocks’ orbitals evolve as they overlap and form delocalized molecular orbitals. Not
surprisingly, for n = 2, one finds a bonding and an antibonding orbital. For n = 3, one has
a bonding, one non-bonding, and one antibonding orbital. Finally, in the bottom row, we
attempt to show what happens for an infinitely long chain. The key point is that the
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discrete number of molecular orbitals appearing in the 1-4 orbital cases evolves into a
continuum of orbitals called a band. This band of orbital energies ranges from its bottom
(whose orbital consists of a fully in-phase bonding combination of the building block
orbitals) to its top (whose orbital is a fully out-of-phase antibonding combination).
In Fig. 2.6 we illustrate these fully bonding and fully antibonding band orbitals for two
cases- the bottom involving s-type building block orbitals, and the top involving p-type
orbitals. Notice that when the energy gap between the building block s and p orbitals is
larger than is the dispersion (spread) in energy within the band or s or band of p orbitals,
a band gap occurs between the highest member of the s band and the lowest member of
the p band. The splitting between the s and p orbitals is a property of the individual atoms
comprising the solid and varies among the elements of the periodic table. The dispersion
in energies that a given band of orbitals is split into as these atomic orbitals combine to
form a band is determined by how strongly the orbitals on neighboring atoms overlap.
Small overlap produces small dispersion, and large overlap yields a broad band.
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Figure 2.6. The bonding through antibonding energies and band orbitals arising from s
and from p orbitals
Depending on how many valence electrons each building block contributes, the various
bands formed by overlapping the building-block orbitals of the constituent atoms will be
filled to various levels. For example, if each orbital shown above has a single valence
electron in an s-orbital (e.g., as in the case of the alkali metals), the s-band will be half
filled in the ground state with α and β -paired electrons. Such systems produce very good
conductors because their partially filled bands allow electrons to move with very little
(e.g., only thermal) excitation among other orbitals in this same band. On the other hand,
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for alkaline earth systems with two s electrons per atom, the s-band will be completely
filled. In such cases, conduction requires excitation to the lowest members of the nearby
p-orbital band. Finally, if each building block were an Al (3s2 3p1) atom, the s-band
would be full and the p-band would be half filled. Systems whose highest energy
occupied band is completely filled and for which the gap in energy to the lowest unfilled
band is large are called insulators because they have no way to easily (i.e., with little
energy requirement) promote some of their higher-energy electrons from orbital to orbital
and thus effect conduction. If the band gap between a filled band and an unfilled band is
small, it may be possible for thermal excitation (i.e., collisions with neighboring atoms or
molecules) to cause excitation of electrons from the former to the latter thereby inducing
conductive behavior. An example of such a case is illustrated in Fig. 2.7.
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Figure 2.7. The valence and conduction bands and the band gap.
In contrast, systems whose highest energy occupied band is partially filled are conductors
because the have little spacing among their occupied and unoccupied orbitals.
95
To form a semiconductor, one starts with an insulator as shown in Fig.2.8 with its
filled (dark) band and a band gap between this band and its empty (clear) upper band.
Figure 2.8. The filled and empty bands, the band gap, and exmpty acceptor or filled
donor bands.
If this insulator material were synthesized with a small amount of “dopant” whose
valence orbitals have energies between the filled and empty bands of the insulator, one
may generate a semiconductor. If the dopant species has no valence electrons (i.e., has an
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empty valence orbital), it gives rise to an empty band lying between the filled and empty
bands of the insulator as shown below in case a. In this case, the dopant band can act as
an electron acceptor for electrons excited (either thermally or by light) from the filled
band into the dopant band. Once electrons enter the dopant band, charge can flow and
the system becomes a conductor. Another case is illustrated in the b part of the figure.
Here, the dopant has its own band filled but lies close to the empty band of the insulator.
Hence excitation of electrons from the dopant band to the empty band can induce current
to flow.
III. Densities of States in 1, 2, and 3 dimensions.
When a large number of neighboring orbitals overlap, band are formed.
However, the nature of these bands is very different in different dimensions.
Before leaving our discussion of bands of orbitals and orbital energies in solids, I
want to address the issue of the density of electronic states and the issue of what
determines the energy range into which orbitals of a given band will split. First, let’s
recall the energy expression for the 1 and 2- dimensional electron in a box case, and let’s
generalize it to three dimensions. The general result is
E = Σj nj2 pi2 h2/(2mLj2)
where the sum over j runs over the number of dimensions (1, 2, or 3), and Lj is the length
of the box along the jth direction. For one dimension, one observes a pattern of energy
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levels that grows with increasing n, and whose spacing between neighboring energy
levels also grows. However, in 2 and 3 dimensions, the pattern of energy level spacing
displays a qualitatively different character at high quantum number.
Consider first the 3-dimensional case and, for simplicity, let’s use a “box” that has
equal length sides L. In this case, the total energy E is (h2pi2/2mL2) times (nx2 + ny2 + nz2).
The latter quantity can be thought of as the square of the length of a vector R having three
components nx, ny, nz. Now think of three Cartesian axes labeled nx, ny, and nz and view a
sphere of radius R in this space. The volume of the 1/8 th sphere having positive values of
nx, ny, and nz and having radius R is 1/8 (4/3 piR3). Because each cube having unit length
along the nx, ny, and nz axes corresponds to a single quantum wave function and its
energy, the total number Ntot(E) of quantum states with positive nx, ny, and nz and with
energy between zero and E = (h2pi2/2mL2)R2 is
Ntot = 1/8 (4/3 piR3) = 1/8 (4/3 pi [2mEL2/( h2pi2)]3/2
The number of quantum states with energies between E and E+dE is (dNtot/dE) dE, which
is the density ?(E) of states near energy E:
?(E) = 1/8 (4/3 pi [2mL2/( h2pi2)]3/2 3/2 E1/2.
Notice that this state density increases as E increases. This means that, in the 3-
dimensional case, the number of quantum states per unit energy grows; in other words,
the spacing between neighboring state energies decreases, very unlike the 1-dimensioal
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case where the spacing between neighboring states grows as n and thus E grows. This
growth in state density in the 3-dimensional case is a result of the degenacies and near-
degenracies that occur. For example, the states with nx, ny, nz = 2,1,1 and 1, 1, 2, and 1, 2,
1 are degenerate, and those with nx, ny, nz = 5, 3, 1 or 5, 1, 3 or 1, 3, 5 or 1, 5, 3 or 3, 1, 5
or 3, 5, 1 are degenerate and nearly degenerate to those having quantum numbers 4, 4, 1
or 1, 4, 4, or 4, 1, 4.
In the 2-dimensional case, degeracies also occur and cause the density of states to
possess an interesting E dependence. In this case, we think of states having energy
E = (h2pi2/2mL2)R2, but with R2 = nx2 + ny2. The total number of states having energy
between zero and E is
Ntotal= 4piR2 = 4pi E(2mL2/ h2pi2)
So, the density of states between E and E+dE is
?(E) = dNtotal/dE = 4pi (2mL2/ h2pi2)
That is, in this 2-dimensional case, the number of states per unit energy is constant for
high E values (where the analysis above applies best).
This kind of analysis for the 1-dimensional case gives
Ntotal= R = (2mEL2/ h2pi2)1/2
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so, the state density between E and E+ dE is:
?(E) = 1/2 (2mL2/ h2pi2)1/2 E-1/2,
which clearly shows the widening spacing, and thus lower density, as one goes to higher
energies.
These findings about densities of states in 1-, 2-, and 3- dimensions are important
because, in various problems one encounters in studying electronic states of extended
systems such as solids and surfaces, one needs to know how the number of states
available at a given total energy E varies with E. Clearly, the answer to this question
depends upon the dimensionality of the problem, and this fact is what I want the students
reading this text to keep in mind.
IV. The Most Elementary Model of Orbital Energy Splittings: Hückel or Tight
Binding Theory
Now, let’s examine what determines the energy range into which orbitals (e.g., ppi
orbitals in polyenes or metal s or p orbitals in a solid) split. To begin, consider two
orbitals, one on an atom labeled A and another on a neighboring atom labeled B; these
orbitals could be, for example, the 1s orbitals of two hydrogen atoms, such as Figure 2.9
illustrates.
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Figure 2.9. Two 1s orbitals combine to produce a σ bonding and a σ* antibonding
molecular orbital
However, the two orbitals could instead be two ppi orbitals on neighboring carbon atoms
such as are shown in Fig. 2.10 as they form bonding and pi* anti-bonding orbitals.
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Figure 2.10. Two atomic ppi orbitals form a bonding pi and antibonding pi* molecular
orbital.
In both of these cases, we think of forming the molecular orbitals (MOs) φk as linear
combinations of the atomic orbitals (AOs) χa on the constituent atoms, and we express
this mathematically as follows:
φK = Σa CK,a χa,
where the CK,a are called linear combination of atomic orbital to form molecular orbital
(LCAO-MO) coefficients. The MOs are supposed to be solutions to the Schr?dinger
equation in which the Hamiltonian H involves the kinetic energy of the electron as well
as the potentials VL and VR detailing its attraction to the left and right atomic centers:
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H = - h2/2m ?2 + VL + VR.
In contrast, the AOs centered on the left atom A are supposed to be solutions of the
Schr?dinger equation whose Hamiltonian is H = - h2/2m ?2 + VL , and the AOs on the
right atom B have H = - h2/2m ?2 + VR. Substituting φK = Σa CK,a χa into the MO’s
Schr?dinger equation HφK = εK φK and then multiplying on the left by the complex
conjugate of χb and integrating over the r, θ and φ coordinates of the electron produces
Σa <χb| - h2/2m ?2 + VL + VR |χa> CK,a = εK Σa <χb|χa> CK,a
Recall that the Dirac notation <a|b> denotes the integral of a* and b, and <a| op| b>
denotes the integral of a* and the operator op acting on b.
In what is known as the Hückel model in organic chemistry or the tight-binding
model in solid-state theory, one approximates the integrals entering into the above set of
linear equations as follows:
i. The diagonal integral <χb| - h2/2m ?2 + VL + VR |χb> involving the AO centered on the
right atom and labeled χb is assumed to be equivalent to <χb| - h2/2m ?2 + VR |χb>, which
means that net attraction of this orbital to the left atomic center is neglected. Moreover,
this integral is approximated in terms of the binding energy (denoted α, not to be
confused with the electron spin function α) for an electron that occupies the χb orbital:
<χb| - h2/2m ?2 + VR |χb> = αb. The physical meaning of αb is the kinetic energy of the
103
electron in χb plus the attraction of this electron to the right atomic center while it resides
in χb. Of course, an analogous approximation is made for the diagonal integral involving
χa; <χa| - h2/2m ?2 + VL |χa> = αa .
ii. The off-diagonal integrals <χb| - h2/2m ?2 + VL + VR |χa> are expressed in terms of a
parameter βa,b which relates to the kinetic and potential energy of the electron while it
resides in the “overlap region” in which both χa and χb are non-vanishing. This region is
shown pictorially above as the region where the left and right orbitals touch or overlap.
The magnitude of β is assumed to be proportional to the overlap Sa,b between the two
AOs : Sa,b = <χa|χb>. It turns out that β is usually a negative quantity, which can be seen
by writing it as <χb| - h2/2m ?2 + VR |χa> + <χb| VL |χa>. Since χa is an eigenfunction of -
h2/2m ?2 + VR having the eigenvalue αa, the first term is equal to αa (a negative quantity)
times <χb|χa>, the overlap S. The second quantity <χb| VL |χa> is equal to the integral of
the overlap density χb(r) χa(r) multiplied by the (negative) Coulomb potential for
attractive interaction of the electron with the left atomic center. So, whenever χb(r) and
χa(r) have positive overlap, β will turn out negative.
iii. Finally, in the most elementary Hückel or tight-binding model, the overlap integrals
<χa|χb> = Sa,b are neglected and set equal to zero on the right side of the matrix
eigenvalue equation. However, in some Hückel models, overlap between neighboring
orbitals is explictly treated, so, in some of the discussion below we will retain Sa,b.
104
With these Hückel approximations, the set of equations that determine the orbital
energies εK and the corresponding LCAO-MO coefficients CK,a are written for the two-
orbital case at hand as in the first 2x2 matrix equations shown below
which is sometimes written as
These equations reduces with the assumption of zero overlap to
The α parameters are identical if the two AOs χa and χb are identical, as would be
the case for bonding between the two 1s orbitals of two H atoms or two 2ppi orbitals of
two C atoms or two 3s orbitals of two Na atoms. If the left and right orbitals were not
identical (e.g., for bonding in HeH+ or for the pi bonding in a C-O group), their α values
would be different and the Hückel matrix problem would look like:
α β
β α
?
?
?
?
?
?
?
?
C
L
C
R
?
?
?
?
?
?
?
?
= ε
S 0
0 S
?
?
?
?
?
?
C
L
C
R
?
?
?
?
?
?
?
?
α ? ε β ? εS
β ? εS α ? ε
?
?
?
?
?
?
?
?
C
L
C
R
?
?
?
?
?
?
?
?
=
0
0
?
?
?
?
?
?
α β
β α
?
?
?
?
?
?
?
?
C
L
C
R
?
?
?
?
?
?
?
?
= ε
1 0
0 1
?
?
?
?
?
?
C
L
C
R
?
?
?
?
?
?
?
?
105
To find the MO energies that result from combining the AOs, one must find the
values of ε for which the above equations are valid. Taking the 2x2 matrix consisting of ε
times the overlap matrix to the left hand side, the above set of equations reduces to the
third set displayed earlier. It is known from matrix algebra that such a set of linear
homogeneous equations (i.e., having zeros on the right hand sides) can have non-trivial
solutions (i.e., values of C that are not simply zero) only if the determinant of the matrix
on the left side vanishes. Setting this determinant equal to zero gives a quadratic equation
in which the ε values are the unknowns:
(α-ε)2 – (β-εS)2 = 0.
This quadratic equation can be factored into a product
(α - β - ε +εS) (α + β - ε -εS) = 0
which has two solutions
ε = (α + β)/(1 + S), and ε = (α -β)/(1 – S).
α β
β α'
?
?
?
?
?
?
C
L
C
R
?
?
?
?
?
?
= ε
1 S
S 1
?
?
?
?
?
?
C
L
C
R
?
?
?
?
?
?
106
As discussed earlier, it turns out that the β values are usually negative, so the
lowest energy such solution is the ε = (α + β)/(1 + S) solution, which gives the energy of
the bonding MO. Notice that the energies of the bonding and anti-bonding MOs are not
symmetrically displaced from the value α within this version of the Hückel model that
retains orbital overlap. In fact, the bonding orbital lies less than β below α, and the
antibonding MO lies more than β above α because of the 1+S and 1-S factors in the
respective denominators. This asymmetric lowering and raising of the MOs relative to the
energies of the constituent AOs is commonly observed in chemical bonds; that is, the
antibonding orbital is more antibonding than the bonding orbital is bonding. This is
another important thing to keep in mind because its effects pervade chemical bonding and
spectroscopy.
Having noted the effect of inclusion of AO overlap effects in the Hückel model, I
should admit that it is far more common to utilize the simplified version of the Hückel
model in which the S factors are ignored. In so doing, one obtains patterns of MO orbital
energies that do not reflect the asymmetric splitting in bonding and antibonding orbitals
noted above. However, this simplified approach is easier to use and offers qualitatively
correct MO energy orderings. So, let’s proceed with our discussion of the Hückel model
in its simplified version.
To obtain the LCAO-MO coefficients corresponding to the bonding and
antibonding MOs, one substitutes the corresponding α values into the linear equations
α ? ε β
β α ?ε
?
?
?
?
?
?
C
L
C
R
?
?
?
?
?
?
=
0
0
?
?
?
?
?
?
107
and solves for the Ca coefficients (acutally, one can solve for all but one Ca, and then use
normalization of the MO to determine the final Ca). For example, for the bonding MO,
we substitute ε = α + β into the above matrix equation and obtain two equations for CL
and CR:
? β CL + β CR = 0
β CL - β CR = 0.
These two equations are clearly not independent; either one can be solved for one C in
terms of the other C to give:
CL = CR,
which means that the bonding MO is
φ = CL (χL + χR).
The final unknown, CL, is obtained by noting that φ is supposed to be a normalized
function <φ|φ> = 1. Within this version of the Hückel model, in which the overlap S is
negleted, the normalization of φ leads to the following condition:
108
1 = <φ|φ> = CL2 (<χL|χL> + <χRχR>) = 2 CL2
with the final result depending on assuming that each χ is itself also normalized. So,
finally, we know that CL = (1/2)1/2, and hence the bonding MO is:
φ = (1/2)1/2 (χL + χR).
Actually, the solution of 1 = 2 CL2 could also have yielded CL = - (1/2)1/2 and then, we
would have
φ = - (1/2)1/2 (χL + χR).
These two solutions are not independent (one is just –1 time the other), so only one
should be included in the list of MOs. However, either one is just as good as the other
because, as shown very early in this text, all of the physical properties that one computes
from a wave function depend not on ψ but on ψ*ψ. So, two wave functions that differ
from one another by an overall sign factor as we have here have exactly the same ψ*ψ
and thus are equivalent.
In like fashion, we can substitute ε = α - β into the matrix equation and solve for
the CL can CR values that are appropriate for the antibonding MO. Doing so, gives us:
φ* = (1/2)1/2 (χL - χR)
109
or, alternatively,
φ* = (1/2)1/2 (χR - χL).
Again, the fact that either expression for φ* is acceptable shows a property of all
solutions to any Scrr?dinger equations; any multiple of a solution is also a solution. In the
above example, the two “answers” for φ* differ by a multiplicative factor of (-1).
Let’s try another example to practice using Hückel or tight-binding theory. In
particular, I’d like you to imagine two possible structures for a cluster three Na atoms
(i.e., pretend that someone came to you and asked what geometry you think such a cluster
would assume in its ground electronic state), one linear and one an equilateral triangle.
Further, assume that the Na-Na distances in both such clusters are equal (i.e., that the
person asking for your theoretical help is willing to assume that variations in bond
lengths are not the crucial factor in determining which structure is favored). In Fig. 2.11,
I shown the two candidate clusters and their 3s orbitals.
Figure 2.11. Linear and equilateral triangle structures of sodium trimer.
Na Na Na Na
Na
Na
110
Numbering the three Na atoms’ valence 3s orbitals χ1, χ2, and χ3, we then set up
the 3x3 Hückel matrix appropriate to the two candidate structures:
for the linear structure (n.b., the zeros arise because χ1 and χ3 do not overlap and thus
have no β coupling matrix element). Alternatively, for the triangular structure, we find’
as the Hückel matrix. Each of these 3x3 matrices will have three eigenvalues that we
obtain by subtracting ε from their diagonals and setting the determinants of the resulting
matrices to zero. For the linear case, doing so generates
(α-ε)3 – 2 β2 (α-ε) = 0,
and for the triangle case it produces
(α-ε)3 –3 β2 (α-ε) + 2 β2 = 0.
α β 0
β α β
0 β α
?
?
?
?
?
?
?
?
α β β
β α β
β β α
?
?
?
?
?
?
?
?
?
?
111
The first cubic equation has three solutions that give the MO energies:
ε = α + (2)1/2 β, ε = α, and ε = α - (2)1/2 β,
for the bonding, non-bonding and antibonding MOs, respectively. The second cubic
equation also has three solutions
ε = α + 2β, ε = α - β , and ε = α - β.
So, for the linear and triangular structures, the MO energy patterns are as shown in Fig.
2.12.
112
Figure 2.12. Energy orderings of molecular orbitals of linear and triangular
sodium trimer
For the neutral Na3 cluster about which you were asked, you have three valence
electrons to distribute among the lowest available orbitals. In the linear case, we place
two electrons into the lowest orbital and one into the second orbital. Doing so produces a
3-electron state with a total energy of E= 2(α+21/2 β) + α = 3α +2 21/2β. Alternatively, for
the triangular species, we put two electrons into the lowest MO and one into either of the
degenerate MOs resulting in a 3-electron state with total energy E = 3 α + 3β. Because β
is a negative quantity, the total energy of the triangular structure is lower than that of the
linear structure since 3 > 2 21/2.
The above example illustrates how we can use Hückel/tight-binding theory to
make qualitative predictions (e.g., which of two “shapes” is likely to be of lower energy).
Notice that all one needs to know to apply such a model to any set of atomic orbitals that
overlap to form MOs is
α
α + (2)
1/2
β
α ? (2)
1/2
β
Linear Na3 Cluster MO Energies
α + 2β
α ? β
Triangle Na3 Cluster MO Energies
113
(i) the individual AO energies α (which relate to the electronegativity of the AOs) and
(ii) the degree to which the AOs couple (the β parameters which relate to AO overlaps).
Let’s see if you can do some of this on your own. Using the above results, would
you expect the cation Na3+ to be linear or triangular? What about the anion Na3-? Next, I
want you to substitute the MO energies back into the 3x3 matrix and find the C1, C2, and
C3 coefficients appropriate to each of the 3 MOs of the linear and of the triangular
structure. See if doing so leads you to solutions that can be depicted as shown in Fig.
2.13, and see if you can place each set of MOs in the proper energy ordering.
Figure 2.13. The molecular orbitals of linear and triangular sodium trimer (note,
they are not energy ordered).
114
Now, I want to show you how to broaden your horizons and use tight-binding
theory to describe all of the bonds in a more complicated molecule such as ethylene
shown in Fig. 2.14. Within the model described above, each pair of orbitals that “touch”
or overlap gives rise to a 2x2 matrix. More correctly, all n of the constituent AOs form an
nxn matrix, but this matrix is broken up into 2x2 blocks whenever each AO touches only
one other AO. Notice that this did not happen in the triangular Na3 case where each AO
touched two other AOs. For the ethlyene case, the valence AO’s consist of (a) four
equivalent C sp2 orbitals that are directed toward the four H atoms, (b) four H 1s orbitals,
(c) two C sp2 orbitals directed toward one another to form the C-C σ bond, and (d) two C
ppi orbitals that will form the C-C pi bond. This total of 12 AOs generates 6 Hückel
matrices as shown below the ethylene molecule.
Figure 2.14 Ethylene molecule with four C-H bonds, one C-C σ bond, and one C-C pi
bond.
115
We obtain one 2x2 matrix for the C-C σ bond of the form
and one 2x2 matrix for the C-C pi bond of the form
Finally, we also obtain four identical 2x2 matrices for the C-H bonds:
The above matrices will then produce (a) four identical C-H bonding MOs having
energies ε = 1/2 {(αH + αC) –[(αH-αC)2 + 4β2]1/2}, (b) four identical C-H antibonding MOs
having energies ε* = 1/2 {(αH + αC) + [(αH - αC)2 + 4β2]1/2}, (c) one bonding C-C
pi orbital with ε = αppi + β , (d) a partner antibonding C-C orbital with ε* = αppi - β, (e) a
C-C σ bonding MO with ε = αsp2 + β , and (f) its antibonding partner with ε* = αsp2 - β. In
all of these expressions, the β parameter is supposed to be that appropriate to the specific
orbitals that overlap as shown in the matrices.
α
sp
2 β
sp
2
, sp
2
β
sp
2
,sp
2 α
sp
2
?
?
?
?
?
?
α
sp
2 β
sp
2
, H
β
sp
2
, H
α
H
?
?
?
?
?
?
α
p
pi
β
p
pi, ppi
β
p
pi
, p
pi
α
p
pi
?
?
?
?
?
?
116
If you wish to practice this exercise of breaking a large molecule down into sets
of interacting AOs, try to see what Hückel matrices you obtain and what bonding and
antibonding MO energies you obtain for the valence orbitals of methane shown in Fig.
2.15.
Figure 2.15. Methane molecule with four C-H bonds.
Before leaving this discussion of the Hückel/tight-binding model, I need to stress
that it has its flaws (because it is based on approximations and involves neglecting certain
terms in the Schr?dinger equation). For example, it predicts (see above) that ethylene has
four energetically identical C-H bonding MOs (and four degenerate C-H antibonding
MOs). However, this is not what is seen when photoelectron spectra are used to probe
the energies of these MOs. Likewise, it suggests that methane has four equivalent C-H
bonding and antibonding orbitals, which, again is not true. It turns out that, in each of
these two cases (ethylene and methane), the experiments indicate a grouping of four
nearly iso-energetic bonding MOs and four nearly iso-energetic antibonding MOs.
However, there is some “splitting” among these clusters of four MOs. The splittings can
117
be interpreted, within the Hückel model, as arising from couplings or interactions among,
for example, one sp2 or sp3 orbital on a given C atom and another such orbital on the
same atom. Such couplings cause the nxn Hückel matrix to not block-partition into
groups of 2x2 sub-matrices because now there exist off-diagonal β factors that couple
one pair of directed AOs to another. When such couplings are included in the analysis,
one finds that the clusters of MOs expected to be degenerate are not but are split just as
the photoelectron data suggest.
V. Hydrogenic Orbitals
The Hydrogenic atom problem forms the basis of much of our thinking about
atomic structure. To solve the corresponding Schr?dinger equation requires separation
of the r, θ, and φ variable.
The Schr?dinger equation for a single particle of mass μ moving in a central
potential (one that depends only on the radial coordinate r) can be written as
- h
?2
2μ ??
?
??
??2
?x2 +
?2
?y2 +
?2
?z2 ψ + V?
? ??x2+y2+z2 ψ = Eψ.
or, introducing the short-hand notation ?2:
- h2/2m ?2 ψ + V ψ = E ψ.
118
This equation is not separable in Cartesian coordinates (x,y,z) because of the way x,y,
and z appear together in the square root. However, it is separable in spherical coordinates
where it has the form
- h
?2
2μr2 ??
?
??
??
?r ??
?
??
?r2 ?ψ
?r +
1
r2Sinθ
?
?θ ??
?
??
?
Sinθ ?ψ?θ
+ 1r2Sin2θ ?
2ψ
?φ2 + V(r)ψ = - h2/2m ?2 ψ + V ψ = Eψ .
Subtracting V(r)ψ from both sides of the equation and multiplying by - 2μr
2
h?2 then
moving the derivatives with respect to r to the right-hand side, one obtains
1
Sinθ
?
?θ ??
?
??
?
Sinθ ?ψ?θ + 1Sin2θ ?
2ψ
?φ2
= -2μr
2
h?2 ( )E-V(r) ψ -
?
?r ??
?
??
?r2 ?ψ
?r .
Notice that, except for ψ itself, the right-hand side of this equation is a function of r only;
119
it contains no θ or φ dependence. Let's call the entire right hand side F(r) ψ to emphasize
this fact.
To further separate the θ and φ dependence, we multiply by Sin2θ and subtract the
θ derivative terms from both sides to obtain
?2ψ
?φ2 = F(r)ψSin2θ - Sinθ
?
?θ ??
?
??
?
Sinθ ?ψ?θ .
Now we have separated the φ dependence from the θ and r dependence. We now
introduce the procedure used to separate variables in differntial equations and assume ψ
can be written as a function of φ times a function of r and θ: ψ = Φ(φ) Q(r,θ). Dividing by
Φ Q, we obtain
1Φ ?
2Φ
?φ2 =
1
Q ??
?
??
?
F(r)Sin2θ Q - Sinθ ??θ ??
?
??
?
Sinθ ?Q?θ .
Now all of the φ dependence is isolated on the left hand side; the right hand side contains
only r and θ dependence.
Whenever one has isolated the entire dependence on one variable as we have done
above for the φ dependence, one can easily see that the left and right hand sides of the
equation must equal a constant. For the above example, the left hand side contains no r
or θ dependence and the right hand side contains no φ dependence. Because the two
120
sides are equal, they both must actually contain no r, θ, or φ dependence; that is, they are
constant.
For the above example, we therefore can set both sides equal to a so-called
separation constant that we call -m2. It will become clear shortly why we have chosen to
express the constant in the form of minus the square of an integer. You may recall that we
studied this same φ - equation earlier and learned how the integer m arises via. the
boundary condition that φ and φ + 2pi represent identical geometries.
1. The Φ Equation
The resulting Φ equation reads (the “ symbol is used to represent second
derivative)
Φ" + m2Φ = 0 .
This equation should be familiar because it is the equation that we treated much earlier
when we discussed z-component of angular momentum. So, its further analysis should
also be familiar, but for completeness, I repeat much of it. The above equation has as its
most general solution
Φ = Αeimφ + Be-imφ .
121
Because the wave functions of quantum mechanics represent probability densities, they
must be continuous and single-valued. The latter condition, applied to our Φ function,
means (n.b., we used this in our earlier discussion of z-component of angular momentum)
that
Φ(φ) = Φ(2pi + φ) or,
Aeimφ( )1 - e2impi + Be-imφ( )1 - e-2impi = 0.
This condition is satisfied only when the separation constant is equal to an integer m = 0,
±1, ± 2, ... . and provides another example of the rule that quantization comes from the
boundary conditions on the wave function. Here m is restricted to certain discrete values
because the wave function must be such that when you rotate through 2pi about the z-axis,
you must get back what you started with.
2. The Θ Equation
Now returning to the equation in which the φ dependence was isolated from the r
and θ dependence.and rearranging the θ terms to the left-hand side, we have
122
1
Sinθ
?
?θ ??
?
??
?
Sinθ ?Q?θ - m
2Q
Sin2θ = F(r)Q.
In this equation we have separated θ and r variations so we can further decompose the
wave function by introducing Q = Θ(θ) R(r) , which yields
1
Θ
1
Sinθ
?
?θ ??
?
??
?
Sinθ ?Θ?θ - m
2
Sin2θ =
F(r)R
R = -λ,
where a second separation constant, -λ, has been introduced once the r and θ dependent
terms have been separated onto the right and left hand sides, respectively.
We now can write the θ equation as
1
Sinθ
?
?θ ??
?
??
?
Sinθ ?Θ?θ - m
2Θ
Sin2θ = -λ Θ,
where m is the integer introduced earlier. To solve this equation for Θ , we make the
substitutions z = Cosθ and P(z) = Θ(θ) , so 1-z2 = Sinθ , and
?
?θ =
?z
?θ
?
?z = - Sinθ
?
?z .
123
The range of values for θ was 0 ≤ θ < pi , so the range for z is -1 < z < 1. The equation
for Θ , when expressed in terms of P and z, becomes
d
dz ??
?
??
?
(1-z2) dPdz - m
2P
1-z2 + λP = 0.
Now we can look for polynomial solutions for P, because z is restricted to be less than
unity in magnitude. If m = 0, we first let
P = ∑
k=0
∞
akzk ,
and substitute into the differential equation to obtain
∑
k=0
∞
(k+2)(k+1) ak+2 zk - ∑
k=0
∞
(k+1) k akzk + λ ∑
k=0
∞
akzk = 0.
Equating like powers of z gives
ak+2 = ak(k(k+1)-λ)(k+2)(k+1) .
124
Note that for large values of k
ak+2
ak →
k2??
?
??
?
1+1k
k2??
?
??
?
1+2k ??
?
??
?
1+1k
= 1.
Since the coefficients do not decrease with k for large k, this series will diverge for z = ±
1 unless it truncates at finite order. This truncation only happens if the separation
constant λ obeys λ = l(l+1), where l is an integer. So, once again, we see that a boundary
condition (i.e., that the wave function not diverge and thus be normalizable in this case)
give rise to quantization. In this case, the values of λ are restricted to l(l+1); before, we
saw that m is restricted to 0, ±1, ± 2, .. .
Since the above recursion relation links every other coefficient, we can choose to
solve for the even and odd functions separately. Choosing a0 and then determining all of
the even ak in terms of this a0, followed by rescaling all of these ak to make the function
normalized generates an even solution. Choosing a1 and determining all of the odd ak in
like manner, generates an odd solution.
For l= 0, the series truncates after one term and results in Po(z) = 1. For l= 1 the
same thing applies and P1(z) = z. For l= 2, a2 = -6 ao2 = -3ao , so one obtains P2 = 3z2-1,
and so on. These polynomials are called Legendre polynomials.
For the more general case where m ≠ 0, one can proceed as above to generate a
polynomial solution for the Θ function. Doing so, results in the following solutions:
125
P lm(z) = (1-z2)
|m|
2 d
|m| P
l
(z)
dz|m| .
These functions are called Associated Legendre polynomials, and they constitute the
solutions to the Θ problem for non-zero m values.
The above P and eimφ functions, when re-expressed in terms of θ and φ, yield the
full angular part of the wave function for any centrosymmetric potential. These solutions
are usually written as Yl,m(θ,φ) = P lm(Cosθ) (2pi)-
1
2 exp(imφ), and are called spherical
harmonics. They provide the angular solution of the r,θ, φ Schr?dinger equation for any
problem in which the potential depends only on the radial coordinate. Such situations
include all one-electron atoms and ions (e.g., H, He+, Li++ , etc.), the rotational motion of
a diatomic molecule (where the potential depends only on bond length r), the motion of a
nucleon in a spherically symmetrical "box" (as occurs in the shell model of nuclei), and
the scattering of two atoms (where the potential depends only on interatomic distance).
The Yl,m functions possess varying number of angular nodes, which, as noted earlier, give
clear signatures of the angular or rotational energy content of the wave function. These
angular nodes originate in the oscillatory nature of the Legendre and associated Legendre
polynomials PlM (cosθ); the higher l is, the more sign changes occur within the
polynomial.
3. The R Equation
Let us now turn our attention to the radial equation, which is the only place that
126
the explicit form of the potential appears. Using our earlier results for the equation
obeyed by the R(r) function and specifying V(r) to be the coulomb potential appropriate
for an electron in the field of a nucleus of charge +Ze, yields:
1
r2
d
dr ??
?
??
?
r2 dRdr + ??
?
??
?2μ
h?2 ?
?? ???E + Ze2r - l(l + 1)r2 R = 0.
We can simplify things considerably if we choose rescaled length and energy units
because doing so removes the factors that depend on μ, h? , and e. We introduce a new
radial coordinate ρ and a quantity σ as follows:
ρ = ??
?
??
?-8μE
h?2
1
2 r, and σ2 = -μZ2e4
2Eh?2 .
Notice that if E is negative, as it will be for bound states (i.e., those states with energy
below that of a free electron infinitely far from the nucleus and with zero kinetic energy),
ρ is real. On the other hand, if E is positive, as it will be for states that lie in the
continuum, ρ will be imaginary. These two cases will give rise to qualitatively different
behavior in the solutions of the radial equation developed below.
We now define a function S such that S(ρ) = R(r) and substitute S for R to obtain:
127
1
ρ2
d
dρ ??
?
??
?ρ2 dS
dρ + ??
?
??
?- 1
4 -
l(l+1)
ρ2 +
σ
ρ S = 0.
The differential operator terms can be recast in several ways using
1
ρ2
d
dρ ??
?
??
?ρ2 dS
dρ =
d2S
dρ2 +
2
ρ
dS
dρ =
1
ρ
d2
dρ2 (ρS) .
The strategy that we now follow is characteristic of solving second order
differential equations. We will examine the equation for S at large and small ρ values.
Having found solutions at these limits, we will use a power series in ρ to "interpolate"
between these two limits.
Let us begin by examining the solution of the above equation at small values of ρ
to see how the radial functions behave at small r. As ρ→0, the second term in the
brackets will dominate. Neglecting the other two terms in the brackets, we find that, for
small values of ρ (or r), the solution should behave like ρL and because the function must
be normalizable, we must have L ≥ 0. Since L can be any non-negative integer, this
suggests the following more general form for S(ρ) :
S(ρ) ≈ ρL e-aρ.
This form will insure that the function is normalizable since S(ρ) → 0 as r → ∞ for all L,
128
as long as ρ is a real quantity. If ρ is imaginary, such a form may not be normalized (see
below for further consequences).
Turning now to the behavior of S for large ρ, we make the substitution of S(ρ)
into the above equation and keep only the terms with the largest power of ρ (e.g., the first
term in brackets). Upon so doing, we obtain the equation
a2ρLe-aρ = 14 ρLe-aρ ,
which leads us to conclude that the exponent in the large-ρ behavior of S is a = 12 .
Having found the small- and large-ρ behaviors of S(ρ), we can take S to have the
following form to interpolate between large and small ρ-values:
S(ρ) = ρLe-
ρ
2 P(ρ),
where the function P is expanded in an infinite power series in ρ as P(ρ) = ∑ak ρk .
Again substituting this expression for S into the above equation we obtain
P"ρ + P'(2L+2-ρ) + P(σ-L-l) = 0,
129
and then substituting the power series expansion of P and solving for the ak's we arrive at
a recursion relation for the ak coefficients:
ak+1 = (k-σ+L+l) ak(k+1)(k+2L+2) .
For large k, the ratio of expansion coefficients reaches the limit ak+1ak = 1k , which has the
same behavior as the power series expansion of eρ. Because the power series expansion
of P describes a function that behaves like eρ for large ρ, the resulting S(ρ) function
would not be normalizable because the e-
ρ
2 factor would be overwhelmed by this eρ
dependence. Hence, the series expansion of P must truncate in order to achieve a
normalizable S function. Notice that if ρ is imaginary, as it will be if E is in the
continuum, the argument that the series must truncate to avoid an exponentially diverging
function no longer applies. Thus, we see a key difference between bound (with ρ real)
and continuum (with ρ imaginary) states. In the former case, the boundary condition of
non-divergence arises; in the latter, it does not because exp(ρ/2) does not diverge if ρ is
imaginary.
To truncate at a polynomial of order n', we must have n' - σ + L+ l= 0. This
implies that the quantity σ introduced previously is restricted to σ = n' + L + l , which is
certainly an integer; let us call this integer n. If we label states in order of increasing n =
1,2,3,... , we see that doing so is consistent with specifying a maximum order (n') in the
P(ρ) polynomial n' = 0,1,2,... after which the L-value can run from L = 0, in steps of unity
130
up to L = n-1.
Substituting the integer n for σ , we find that the energy levels are quantized
because σ is quantized (equal to n):
E = - μZ
2e4
2h?2n2
and the scaled distance turns out to be
ρ = Zraon .
Here, the length ao is the so called Bohr radius ??
?
??
?
ao = h
?2
μe2 ; it appears once the above E-
expression is substituted into the equation for ρ. Using the recursion equation to solve
for the polynomial's coefficients ak for any choice of n and l quantum numbers generates
a so-called Laguerre polynomial; Pn-L-1(ρ). They contain powers of ρ from zero through
n-L-1, and they have n-L-1 sign changes as the radial coordinate ranges from zero to
infinity. It is these sign changes in the Laguerre polynomials that cause the radial parts of
the hydrogenic wave functions to have n-L-1 nodes. For example, 3d orbitals have no
radial nodes, but 4d orbitals have one; and, as shown in Fig. 2.16, 3p orbitals have one
while 3s orbitals have two. Once again, the higher the number of nodes, the higher the
energy in the radial direction.
131
Figure 2.16. Plots of the Radial Parts of the 3s and 3p Orbitals
Let me again remind you about the danger of trying to understand quantum wave
functions or probabilities in tems of classical dynamics. What kind of potential V(r)
would give rise to, for example, the 3s P(r) plot shown above? Classical mechanics
suggests that P should be large where the particle moves slowly and small where it moves
quickly. So, the 3s P(r) plot suggests that the radial speed of the electron has three regions
where it is low (i.e., where the peaks in P are) and two regions where it is very large (i.e.,
where the nodes are). This, in turn, suggests that the radial potential V(r) experienced by
the 3s electron is high in three regions (near peaks in P) and low in two regions (and at
the nucleus). Of course, this conclusion about the form of V(r) is nonsense and again
illustrates how one must not be drawn into trying to think of the classical motion of the
132
particle, especially for quantum states with small quantum number. In fact, the low
quantum number states of such one-electron atoms and ions have their radial P(r) plots
focused in regions of r-space where the potential –Ze2/r is most attractive (i.e., largest in
magnitude).
Finally, we note that the energy quantization does not arise for states lying in the
continuum because the condition that the expansion of P(ρ) terminate does not arise. The
solutions of the radial equation appropriate to these scattering states (which relate to the
scattering motion of an electron in the field of a nucleus of charge Z) are a bit outside the
scope of this text, so we will not treat them further here. For the interested student, they
are treated on p. 90 of the text by Eyring, Walter, and Kimball.
To review, separation of variables has been used to solve the full r,θ,φ
Schr?dinger equation for one electron moving about a nucleus of charge Z. The θ and φ
solutions are the spherical harmonics YL,m (θ,φ). The bound-state radial solutions
Rn,L(r) = S(ρ) = ρLe-
ρ
2 P
n-L-1(ρ)
depend on the n and l quantum numbers and are given in terms of the Laguerre
polynomials.
4. Summary
To summarize, the quantum numbers L and m arise through boundary conditions
requiring that ψ(θ) be normalizable (i.e., not diverge) and ψ(φ) = ψ(φ+2pi). The radial
133
equation, which is the only place the potential energy enters, is found to possess both
bound-states (i.e., states whose energies lie below the asymptote at which the potential
vanishes and the kinetic energy is zero) and continuum states lying energetically above
this asymptote. The resulting hydrogenic wave functions (angular and radial) and
energies are summarized on pp. 133-136 in the text by L. Pauling and E. B. Wilson for n
up to and including 6 and L up to 5.
There are both bound and continuum solutions to the radial Schr?dinger equation
for the attractive coulomb potential because, at energies below the asymptote, the
potential confines the particle between r=0 and an outer turning point, whereas at
energies above the asymptote, the particle is no longer confined by an outer turning point
(see Fig. 2.17).
-Zee/r
r
0.0
Continuum State
Bound
States
134
Figure 2.17. Radial Potential for Hydrogenic Atoms and Bound and Continuum Orbital
Energies.
The solutions of this one-electron problem form the qualitative basis for much of atomic
and molecular orbital theory. For this reason, the reader is encouraged to gain a firmer
understanding of the nature of the radial and angular parts of these wave functions. The
orbitals that result are labeled by n, L, and m quantum numbers for the bound states and
by L and m quantum numbers and the energy E for the continuum states. Much as the
particle-in-a-box orbitals are used to qualitatively describe pi- electrons in conjugated
polyenes, these so-called hydrogen-like orbitals provide qualitative descriptions of
orbitals of atoms with more than a single electron. By introducing the concept of
screening as a way to represent the repulsive interactions among the electrons of an atom,
an effective nuclear charge Zeff can be used in place of Z in the ψn,L,m and En to generate
approximate atomic orbitals to be filled by electrons in a many-electron atom. For
example, in the crudest approximation of a carbon atom, the two 1s electrons experience
the full nuclear attraction so Zeff = 6 for them, whereas the 2s and 2p electrons are
screened by the two 1s electrons, so Zeff = 4 for them. Within this approximation, one then
occupies two 1s orbitals with Z = 6, two 2s orbitals with Z = 4 and two 2p orbitals with
Z=4 in forming the full six-electron wave function of the lowest-energy state of carbon.
VI. Electron Tunneling
135
Tunneling is a phenomenon of quantum mechanics, not classical mechanics. It is
an extremely important subject that occurs in a wide variety of chemical species.
Solutions to the Schr?dinger equation display several properties that are very different
from what one experiences in Newtonian dynamics. One of the most unusual and
important is that the particles one describes using quantum mechanics can move into
regions of space where they would not be “allowed” to go if they obeyed classical
equations. Let us consider an example to illustrate this so-called tunneling phenomenon.
Specifically, we think of an electron (a particle that we likely would use quantum
mechanics to describe) moving in a direction we will call R under the influence of a
potential that is:
a. Infinite for R < 0 (this could, for example, represent a region of space within a solid
material where the electron experiences very repulsive interactions with other electrons);
b. Constant and negative for some range of R between R = 0 and Rmax (this could
represent the attractive interaction of the electrons with those atoms or molecules in a
finite region of a solid);
c. Constant and repulsive by an amount δV + De for another finite region from Rmax to
Rmax +δ (this could represent the repulsive interactions between the electrons and a layer
of molecules of thickness δ lying on the surface of the solid at Rmax);
d. Constant and equal to De from Rmaz +δ to infinity (this could represent the electron
being removed from the solid, but with a “work function energy cost of De, and moving
freely in the vacuum above the surface and the ad-layer). Such a potential is shown in
Fig. 2.18.
136
Figure 2.18. One-dimensional potential showing a well, a barrier, and the asymptotic
region.
The piecewise nature of this potential allows the one-dimensional Schr?dinger equation
to be solved analytically. For energies lying in the range De < E < De +δV, an especially
interesting class of solutions exists. These so-called resonance states occur at energies
δV
R
max
+ δR
max
Electron Position Coordinate R
D
e
0.0
V(R)
137
that are determined by the condition that the amplitude of the wave function within the
barrier (i.e., for 0 ≤ R ≤ Rmax ) be large. Let us now turn our attention to this specific
energy regime, which also serves to introduce the tunneling phenomenon.
The piecewise solutions to the Schr?dinger equation appropriate to the resonance
case are easily written down in terms of sin and cos or exponential functions, using the
following three definitions:
The combination of sin(kR) and cos(kR) that solve the Schr?dinger equation in the inner
region and that vanish at R=0 (because the function must vanish within the region where
V is infinite and because it must be continuous, it must vanish at R=0) is:
Ψ = Asin(kR) (for 0 ≤ R ≤ Rmax ).
Between Rmax and Rmax +δ, there are two solutions that obey the Schr?diger equation, so
the most general solution is a combination of these two:
Ψ = B+ exp(κ'R) + B- exp(-κ'R) (for Rmax ≤ R ≤ Rmax +δ).
Finally, in the region beyond Rmax + δ, we can use a combination of either sin(k’R) and
cos(k’R) or exp(ik’R) and exp(-ik’R) to express the solution. Unlike the region near R=0,
where it was most convenient to use the sin and cos functions because one of them could
k = 2m
e
E / h
2
,k' = 2m
e
(E ? D
e
) / h
2
,κ' = 2m
e
(D
e
+ δV ? E) / h
2
138
be “thrown away” since it could not meet the boundary condition of vanishing at R=0, in
this large-R region, either set is acceptable. We choose to use the exp(ik’R) and
exp(-ik’R) set because each of these functions is an eigenfunction of the momentum
operator –ih?/?R. This allows us to discuss amplitudes for electrons moving with positive
momentum and with negative momentum. So, in this region, the most general solution is
Ψ = C exp(ik'R) + D exp(-ik'R) (for Rmax +δ ≤ R < ∞).
There are four amplitudes (A, B+, B-, and C) that can be expressed in terms of the
specified amplitude D of the incoming flux (e.g., pretend that we know the flux of
electrons that our experimental apparatus “shoots” at the surface). Four equations that can
be used to achieve this goal result when Ψ and dΨ/dR are matched at Rmax and at Rmax +
δ (one of the essential properties of solutions to the Schr?dinger equation is that they and
their first derivative are continuous; these properties relate to Ψ being a probability and
–ih?/?R being a momentum operator). These four equations are:
Asin(kRmax) = B+ exp(κ'Rmax) + B- exp(-κ'Rmax),
Akcos(kRmax) = κ'B+ exp(κ'Rmax) - κ'B- exp(-κ'Rmax),
B+ exp(κ'(Rmax + δ)) + B- exp(-κ'(Rmax + δ))
139
= C exp(ik'(Rmax + δ) + D exp(-ik'(Rmax + δ),
κ'B+ exp(κ'(Rmax + δ)) - κ'B- exp(-κ'(Rmax + δ))
= ik'C exp(ik'(Rmax + δ)) -ik' D exp(-ik'(Rmax + δ)).
It is especially instructive to consider the value of A/D that results from solving this set of
four equations in four unknowns because the modulus of this ratio provides information
about the relative amount of amplitude that exists inside the barrier in the attractive
region of the potential compared to that existing in the asymptotic region as incoming
flux.
The result of solving for A/D is:
A/D = 4 κ'exp(-ik'(Rmax+δ))
{exp(κ'δ)(ik'-κ')(κ'sin(kRmax)+kcos(kRmax))/ik'
+ exp(-κ'δ)(ik'+κ')(κ'sin(kRmax)-kcos(kRmax))/ik' }-1.
Further, it is instructive to consider this result under conditions of a high (large De + δV -
E) and thick (large δ) barrier. In such a case, the "tunneling factor" exp(-κ'δ) will be very
small compared to its counterpart exp(κ'δ), and so
140
A/D = 4 ik'κ'ik'-κ' exp(-ik'(Rmax+δ)) exp(-κ'δ) {κ'sin(kRmax)+kcos(kRmax) }-1.
The exp(-κ'δ) factor in A/D causes the magnitude of the wave function inside the barrier
to be small in most circumstances; we say that incident flux must tunnel through the
barrier to reach the inner region and that exp(-κ'δ) gives the probability of this tunneling.
Keep in mind that, in the energy range we are considering (E < De+δ), a classical
particle could not even enter the region Rmax < R < Rmax + δ; this is why we call this the
classically forbidden or tunneling region. A classical particle starting in the large-R
region can not enter, let alone penetrate, this region, so such a particle could never end up
in the 0 <R < Rmax inner region. Likewise, a classical particle that begins in the inner
region can never penetrate the tunneling region and escape into the large-R region. Were
it not for the fact that electrons obey a Schr?dinger equation rather than Newtonian
dynamics, tunneling would not occur and, for example, scanning tunneling microscopy
(STM), which has proven to be a wonderful and powerful tool for imaging molecules on
and near surfaces, would not exist. Likewise, many of the devices that appear in our
modern electronic tools and games, which depend on currents induced by tunneling
through various junctions, would not be available. But, or course, tunneling does occur
and it can have remarkable effects.
Let us examine an especially important (in chemistry) phenomenon that takes
place because of tunneling and that occurs when the energy E assumes very special
values. The magnitude of the A/D factor in the above solutions of the Schr?dinger
equation can become large if the energy E is such that
141
κ'sin(kRmax)+kcos(kRmax)
is small. In fact, if
tan(kRmax) = - k/κ'
the denominator factor in A/D will vanish and A/D will become infinite. It can be shown
that the above condition is similar to the energy quantization condition
tan(kRmax) = - k/κ
that arises when bound states of a finite potential well are examined. There is, however, a
difference. In the bound-state situation, two energy-related parameters occur
k = 2μE/h2
and
κ = 2μ(De - E)/h2 .
In the case we are now considering, k is the same, but
142
κ' = 2μ(De +δV - E)/h2 )
rather than κ occurs, so the two tan(kRmax) equations are not identical, but they are quite
similar.
Another observation that is useful to make about the situations in which A/D
becomes very large can be made by considering the case of a very high barrier (so that κ'
is much larger than k). In this case, the denominator that appears in A/D
κ'sin(kRmax)+kcos(kRmax) ? κ' sin(kRmax)
can become small if
sin(kRmax) ? 0.
This condition is nothing but the energy quantization condition that occurs for the
particle-in-a-box potential shown in Fig. 2.19.
Rmax
Electron Coordinate R
De
0.0
V(R)
143
Figure 2.19. One-dimensional potential similar to the tunneling potential but without the
barrier and asymptotic region.
This potential is identical to the potential that we were examining for 0 ≤ R ≤ Rmax , but
extends to infinity beyond Rmax ; the barrier and the dissociation asymptote displayed by
our potential are absent.
Let’s consider what this tunneling problem has taught us. First, it showed us that
quantum particles penetrate into classically forbidden regions. It showed that, at certain
so-called resonance energies, tunneling is much more likely that at energies that are “off
resonance”. In our model problem, this means that electrons impinging on the surface
with resonance energies will have a very high probability of tunneling to produce an
electron that is trapped in the 0 < R < Rmax region.
By the way, we could have solved the four equations for the amplitude C of the
outgoing wave in the R > Rmax region in terms of the A amplitude. We might want to take
this approach if wanted to model an experiment in which the electron began in the 0 < R
< Rmax region and we wanted to compute the relative amplitude for the electron to escape.
However, if we were to solve for C/A and then examined under what conditions the
amplitude of this ratio would become small (so the electron can not escape), we would
find the same tan(kRmax) = - k/κ' resonance condition as we found from the other point of
view. This means that the resonance energies tell us for what collision energies the
electron will tunnel inward and produce a trapped electron and, at these same energies, an
electron that is trapped will not escape quickly.
144
Whenever one has a barrier on a potential energy surface, at energies above the
dissociation asymptote De but below the top of the barrier (De + δV here), one can expect
resonance states to occur at "special" scattering energies E. As we illustrated with the
model problem, these so-called resonance energies can often be approximated by the
bound-state energies of a potential that is identical to the potential of interest in the inner
region (0 ≤ R ≤ Rmax ) but that extends to infinity beyond the top of the barrier (i.e.,
beyond the barrier, it does not fall back to values below E).
The chemical significance of resonances is great. Highly rotationally excited
molecules may have more than enough total energy to dissociate (De), but this energy
may be "stored" in the rotational motion, and the vibrational energy may be less than De.
In terms of the above model, high angular momentum may produce a significant
centrifugal barrier in the effective potential that characterizes the molecule’s vibration,
but the system's vibrational energy may lie significantly below De. In such a case, and
when viewed in terms of motion on an angular-momentum-modified effective potential
such as I show in Fig. 2.20 , the lifetime of the molecule with respect to dissociation is
determined by the rate of tunneling through the barrier.
145
Figure 2.20. Radial potential for non-rotating (J = 0) molecule and for rotating molecule.
In that case, one speaks of "rotational predissociation" of the molecule. The
lifetime τ can be estimated by computing the frequency ν at which flux that exists inside
Rmax strikes the barrier at Rmax
ν = hk2μRmax (sec-1)
R
V(R) for non-rotating molecule
V(R) +h2(J(J+1))/8pi2I
Metastable rotational level
146
and then multiplying by the probability P that flux tunnels through the barrier from Rmax
to Rmax + δ:
P = exp(-2κ'δ).
The result is that
τ -1= hk2μRmax exp(-2κ'δ)
with the energy E entering into k and κ' being determined by the resonance condition:
(κ'sin(kRmax)+kcos(kRmax)) = minimum. By looking back at the defintion of κ’, we note
that the probability of tunneling falls of exponentially with a factor depending on the
width δ of the barrier through which the particle must tunnel multiplied by κ’, which
depends on the height of the barrier De + δ above the energy E available. This exponential
dependence on thickness and height of the barriers is something you should keep in mind
because it appears in all tunneling rate expressions.
Another important case in which tunneling occurs is in electronically metastable
states of anions. In so-called shape resonance states, the anion’s “extra” electron
experiences
a. an attractive potential due to its interaction with the underlying neutral molecule’s
dipole, quadrupole, and induced electrostatic moments, as well as
147
b. a centrifugal potential of the form L(L+1)h2/8pi2meR2 whose magnitude depends on
the angular character of the orbital the extra electron occupies.
When combined, the above attractive and centrifugal potentials produce an effective
radial potential of the form shown in Fig. 2.21 for the N2- case in which the added
electron occupies the pi* orbital which has L=2 character when viewed from the center of
the N-N bond. Again, tunneling through the barrier in this potential determines the
lifetimes of such shape resonance states.
Figure 2.21 Effective radial potential for the excess electron in N2- occupying the pi*
orbital which has a dominant L = 2 component.
N N
R
N2 pi* orbitalhaving d angular
symmetry
Effective radial potential
for electron in the pi*
orbital
with the L=2
L(L+1)h2/8pi2mR2
148
Although the examples treated above involved piecewise constant potentials (so
the Schr?dinger equation and the boundary matching conditions could be solved exactly),
many of the characteristics observed carry over to more chemically realistic situations. In
fact, one can often model chemical reaction processes in terms of motion along a
"reaction coordinate" (s) from a region characteristic of reactant materials where the
potential surface is positively curved in all direction and all forces (i.e., gradients of the
potential along all internal coordinates) vanish; to a transition state at which the potential
surface's curvature along s is negative while all other curvatures are positive and all
forces vanish; onward to product materials where again all curvatures are positive and all
forces vanish. A prototypical trace of the energy variation along such a reaction
coordinate is in Fig. 2.22.
Figure 2.22. Energy profile along a reaction path showing the barrier through which
tunneling may occur.
149
Near the transition state at the top of the barrier on this surface, tunneling through the
barrier plays an important role if the masses of the particles moving in this region are
sufficiently light. Specifically, if H or D atoms are involved in the bond breaking and
forming in this region of the energy surface, tunneling must usually be considered in
treating the dynamics.
Within the above "reaction path" point of view, motion transverse to the reaction
coordinate s is often modeled in terms of local harmonic motion although more
sophisticated treatments of the dynamics is possible. This picture leads one to consider
motion along a single degree of freedom (s), with respect to which much of the above
treatment can be carried over, coupled to transverse motion along all other internal
degrees of freedom taking place under an entirely positively curved potential (which
therefore produces restoring forces to movement away from the "streambed" traced out
by the reaction path s). This point of view constitutes one of the most widely used and
successful models of molecular reaction dynamics and is treated in more detail in Chapter
8 of this text.
VII. Angular Momentum
1. Orbital Angular Momentum
A particle moving with momentum p at a position r relative to some coordinate
origin has so-called orbital angular momentum equal to L = r x p . The three components
150
of this angular momentum vector in a Cartesian coordinate system located at the origin
mentioned above are given in terms of the Cartesian coordinates of r and p as follows:
Lz = x py - y px ,
Lx = y pz - z py ,
Ly = z px - x pz .
Using the fundamental commutation relations among the Cartesian coordinates
and the Cartesian momenta:
[qk,pj] = qk pj - pj qk = ih δj,k ( j,k = x,y,z) ,
it can be shown that the above angular momentum operators obey the following set of
commutation relations:
[Lx, Ly] = ih Lz ,
[Ly, Lz] = ih Lx ,
[Lz, Lx] = ih Ly .
151
Although the components of L do not commute with one another, they can be shown to
commute with the operator L2 defined by
L2 = Lx2 + Ly2 + Lz2 .
This new operator is referred to as the square of the total angular momentum operator.
The commutation properties of the components of L allow us to conclude that
complete sets of functions can be found that are eigenfunctions of L2 and of one, but not
more than one, component of L. It is convention to select this one component as Lz, and
to label the resulting simultaneous eigenstates of L2 and Lz as |l,m> according to the
corresponding eigenvalues:
L2 |l,m> = h2 l(l+1) |l,m>, l = 0,1,2,3,....
Lz |l,m> = h m |l,m>, m = ± l, ±(l-1), ±(l-2), ... ±(l-(l-1)), 0.
These eigenfunctions of L2 and of Lz will not, in general, be eigenfunctions of either Lx
or of Ly. This means that any measurement of Lx or Ly will necessarily change the
wavefunction if it begins as an eigenfunction of Lz.
The above expressions for Lx, Ly, and Lz can be mapped into quantum
mechanical operators by substituting x, y, and z as the corresponding coordinate
operators and -ih?/?x, -ih?/?y, and -ih?/?z for px, py, and pz, respectively. The resulting
operators can then be transformed into spherical coordinates the results of which are:
152
Lz =-ih ?/?φ ,
Lx = ih {sinφ ?/?θ + cotθ cosφ ?/?φ} ,
Ly = -ih {cosφ ?/?θ - cotθ sinφ ?/?φ} ,
L2 = - h2 {(1/sinθ) ?/?θ (sinθ ?/?θ) + (1/sin2θ) ?2/?φ2} .
2. Properties of General Angular Momenta
There are many types of angular momenta that one encounters in chemistry.
Orbital angular momenta, such as that introduced above, arise in electronic motion in
atoms, in atom-atom and electron-atom collisions, and in rotational motion in molecules.
Intrinsic spin angular momentum is present in electrons, H1, H2, C13, and many other
nuclei. In this section, we will deal with the behavior of any and all angular momenta and
their corresponding eigenfunctions.
At times, an atom or molecule contains more than one type of angular
momentum. The Hamiltonian's interaction potentials present in a particular species may
or may not cause these individual angular momenta to be coupled to an appreciable
extent (i.e., the Hamiltonian may or may not contain terms that refer simultaneously to
two or more of these angular momenta). For example, the NH- ion, which has a 2Π
153
ground electronic state (its electronic configuration is 1sN22σ23σ22ppix22ppy1) has
electronic spin, electronic orbital, and molecular rotational angular momenta. The full
Hamiltonian H contains terms that couple the electronic spin and orbital angular
momenta, thereby causing them individually to not commute with H.
In such cases, the eigenstates of the system can be labeled rigorously only by
angular momentum quantum numbers j and m belonging to the total angular momentum
J. The total angular momentum of a collection of individual angular momenta is defined,
component-by-component, as follows:
Jk = Σi Jk(i),
where k labels x, y, and z, and i labels the constituents whose angular momenta couple to
produce J.
For the remainder of this Section, we will study eigenfunction-eigenvalue
relationships that are characteristic of all angular momenta and which are consequences
of the commutation relations among the angular momentum vector's three components.
We will also study how one combines eigenfunctions of two or more angular momenta
{J(i)} to produce eigenfunctions of the the total J.
a. Consequences of the Commutation Relations
Any set of three operators that obey
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[Jx, Jy] = ih Jz ,
[Jy, Jz] = ih Jx ,
[Jz, Jx] = ih Jy ,
will be taken to define an angular momentum J, whose square J2= Jx2 + Jy2 + Jz2
commutes with all three of its components. It is useful to also introduce two
combinations of the three fundamental operators:
J± = Jx ± i Jy ,
and to refer to them as raising and lowering operators for reasons that will be made clear
below. These new operators can be shown to obey the following commutation relations:
[J2, J±] = 0,
[Jz, J±] = ± h J± .
Using only the above commutation properties, it is possible to prove important
properties of the eigenfunctions and eigenvalues of J2 and Jz. Let us assume that we have
found a set of simultaneous eigenfunctions of J2 and Jz ; the fact that these two operators
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commute tells us that this is possible. Let us label the eigenvalues belonging to these
functions:
J2 |j,m> = h2 f(j,m) |j,m>,
Jz |j,m> = h m |j,m>,
in terms of the quantities m and f(j,m). Although we certainly "hint" that these quantities
must be related to certain j and m quantum numbers, we have not yet proven this,
although we will soon do so. For now, we view f(j,m) and m simply as symbols that
represent the respective eigenvalues. Because both J2 and Jz are Hermitian,
eigenfunctions belonging to different f(j,m) or m quantum numbers must be orthogonal:
<j,m|j',m'> = δm,m' δj,j' .
We now prove several identities that are needed to discover the information about
the eigenvalues and eigenfunctions of general angular momenta that we are after. Later in
this Section, the essential results are summarized.
i. There is a Maximum and a Minimum Eigenvalue for Jz
Because all of the components of J are Hermitian, and because the scalar product
of any function with itself is positive semi-definite, the following identity holds:
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<j,m|Jx2 + Jy2|j,m> = <Jx<j,m| Jx|j,m> + <Jy<j,m| Jy|j,m> ≥ 0.
However, Jx2 + Jy2 is equal to J2 - Jz2, so this inequality implies that
<j,m| J2 - Jz2 |j,m> = h2 {f(j,m) - m2} ≥ 0,
which, in turn, implies that m2 must be less than or equal to f(j,m). Hence, for any value
of the total angular momentum eigenvalue f, the z-projection eigenvalue (m) must have a
maximum and a minimum value and both of these must be less than or equal to the total
angular momentum squared eigenvalue f.
ii. The Raising and Lowering Operators Change the Jz Eigenvalue but not the J2
Eigenvalue When Acting on |j,m>
Applying the commutation relations obeyed by J± to |j,m> yields another useful
result:
Jz J± |j,m> - J± Jz |j,m> = ± h J± |j,m>,
J2 J± |j,m> - J± J2 |j,m> = 0.
Now, using the fact that |j,m> is an eigenstate of J2 and of Jz, these identities give
Jz J± |j,m> = (mh ± h) J± |j,m> = h (m±1) |j,m>,
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J2 J± |j,m> = h2 f(j,m) J± |j,m>.
These equations prove that the functions J± |j,m> must either themselves be
eigenfunctions of J2 and Jz, with eigenvalues h2 f(j,m) and h (m+1) or J± |j,m> must
equal zero. In the former case, we see that J± acting on |j,m> generates a new eigenstate
with the same J2 eigenvalue as |j,m> but with one unit of h higher or lower in Jz
eigenvalue. It is for this reason that we call J± raising and lowering operators. Notice that,
although J± |j,m> is indeed an eigenfunction of Jz with eigenvalue (m±1) h, J± |j,m> is
not identical to |j,m±1>; it is only proportional to |j,m±1>:
J± |j,m> = C±j,m |j,m±1>.
Explicit expressions for these C±j,m coefficients will be obtained below. Notice also that
because the J± |j,m>, and hence |j,m±1>, have the same J2 eigenvalue as |j,m> (in fact,
sequential application of J± can be used to show that all |j,m'>, for all m', have this same
J2 eigenvalue), the J2 eigenvalue f(j,m) must be independent of m. For this reason, f can
be labeled by one quantum number j.
iii. The J2 Eigenvalues are Related to the Maximum and Minimum Jz Eigenvalues Which
are Related to One Another
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Earlier, we showed that there exists a maximum and a minimum value for m, for
any given total angular momentum. It is when one reaches these limiting cases that J±
|j,m> = 0 applies. In particular,
J+ |j,mmax> = 0,
J- |j,mmin> = 0.
Applying the following identities:
J- J+ = J2 - Jz2 -h Jz ,
J+ J- = J2 - Jz2 +h Jz,
respectively, to |j,mmax> and |j,mmin> gives
h2 { f(j,mmax) - mmax2 - mmax} = 0,
h2 { f(j,mmin) - mmin2 + mmin} = 0,
which immediately gives the J2 eigenvalue f(j,mmax) and f(j,mmin) in terms of mmax or
mmin:
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f(j,mmax) = mmax (mmax+1),
f(j,mmin) = mmin (mmin-1).
So, we now know the J2 eigenvalues for |j,mmax> and |j,mmin>. However, we earlier
showed that |j,m> and |j,m-1> have the same J2 eigenvalue (when we treated the effect of
J± on |j,m>) and that the J2 eigenvalue is independent of m. If we therefore define the
quantum number j to be mmax , we see that the J2 eigenvalues are given by
J2 |j,m> = h2 j(j+1) |j,m>.
We also see that
f(j,m) = j(j+1) = mmax (mmax+1) = mmin (mmin-1),
from which it follows that
mmin = - mmax .
iv. The j Quantum Number Can Be Integer or Half-Integer
160
The fact that the m-values run from j to -j in unit steps (because of the property of
the J± operators), there clearly can be only integer or half-integer values for j. In the
former case, the m quantum number runs over -j, -j+1, -j+2, ..., -j+(j-1), 0, 1, 2, ... j;
in the latter, m runs over -j, -j+1, -j+2, ...-j+(j-1/2), 1/2, 3/2, ...j. Only integer and half-
interger values can range from j to -j in steps of unity. Species with integer spin are
known as Bosons and those with half-integer spin are called Fermions.
v. More on J± |j,m>
Using the above results for the effect of J± acting on |j,m> and the fact that J+ and
J- are adjoints of one another, allows us to write:
<j,m| J- J+ |j,m> = <j,m| (J2 - Jz2 -h Jz ) |j,m>
= h2 {j(j+1)-m(m+1)} = <J+<j,m| J+|j,m> = (C+j,m)2,
where C+j,m is the proportionality constant between J+|j,m> and the normalized function
|j,m+1>. Likewise, the effect of J- can be expressed as
<j,m| J+ J- |j,m> = <j,m| (J2 - Jz2 +h Jz) |j,m>
= h2 {j(j+1)-m(m-1)} = <J-<j,m| J-|j,m> = (C-j,m)2,
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where C-j,m is the proportionality constant between J- |j,m> and the normalized |j,m-1>.
Thus, we can solve for C±j,m after which the effect of J± on |j,m> is given by:
J± |j,m> = h {j(j+1) –m(m-1)}1/2 |j,m±1>.
3. Summary
The above results apply to any angular momentum operators. The essential
findings can be summarized as follows:
(i) J2 and Jz have complete sets of simultaneous eigenfunctions. We label these
eigenfunctions |j,m>; they are orthonormal in both their m- and j-type indices:
<j,m| j',m'> = δm,m' δj,j' .
(ii) These |j,m> eigenfunctions obey:
J2 |j,m> = h2 j(j+1) |j,m>, { j= integer or half-integer},
Jz |j,m> = h m |j,m>, { m = -j, in steps of 1 to +j}.
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(iii) The raising and lowering operators J± act on |j,m> to yield functions that are
eigenfunctions of J2 with the same eigenvalue as |j,m> and eigenfunctions of Jz with
eigenvalue of (m±1) h :
J± |j,m> = h {j(j+1) - m(m±1)}1/2 |j,m±1>.
(iv) When J± acts on the "extremal" states |j,j> or |j,-j>, respectively, the result is zero.
The results given above are, as stated, general. Any and all angular momenta have
quantum mechanical operators that obey these equations. It is convention to designate
specific kinds of angular momenta by specific letters; however, it should be kept in mind
that no matter what letters are used, there are operators corresponding to J2, Jz, and J±
that obey relations as specified above, and there are eigenfunctions and eigenvalues that
have all of the properties obtained above. For electronic or collisional orbital angular
momenta, it is common to use L2 and Lz ; for electron spin, S2 and Sz are used; for
nuclear spin I2 and Iz are most common; and for molecular rotational angular momentum,
N2 and Nz are most common (although sometimes J2 and Jz may be used). Whenever two
or more angular momenta are combined or coupled to produce a "total" angular
momentum, the latter is designated by J2 and Jz.
4. Coupling of Angular Momenta
163
If the Hamiltonian under study contains terms that couple two or more angular
momenta J(i), then only the components of the total angular momentum J = Σi J(i) and
J2 will commute with H. It is therefore essential to label the quantum states of the system
by the eigenvalues of Jz and J2 and to construct variational trial or model wavefunctions
that are eigenfunctions of these total angular momentum operators. The problem of
angular momentum coupling has to do with how to combine eigenfunctions of the
uncoupled angular momentum operators, which are given as simple products of the
eigenfunctions of the individual angular momenta Πi |ji,mi>, to form eigenfunctions of J2
and Jz.
a. Eigenfunctions of Jz
Because the individual elements of J are formed additively, but J2 is not , it is
straightforward to form eigenstates of
Jz = Σi Jz(i);
simple products of the form Πi |ji,mi> are eigenfunctions of Jz:
Jz Πi |ji,mi> = Σk Jz(k) Πi |ji,mi> = Σk h mk Πi |ji,mi>,
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and have Jz eigenvalues equal to the sum of the individual mk h eigenvalues. Hence, to
form an eigenfunction with specified J and M eigenvalues, one must combine only those
product states Πi |ji,mi> whose mih sum is equal to the specified M value.
b. Eigenfunctions of J2; the Clebsch-Gordon Series
The task is then reduced to forming eigenfunctions |J,M>, given particular values
for the {ji} quantum numbers. When coupling pairs of angular momenta { |j,m> and
|j',m'>}, the total angular momentum states can be written, according to what we
determined above, as
|J,M> = Σm,m' CJ,Mj,m;j',m' |j,m> |j',m'>,
where the coefficients CJ,Mj,m;j',m' are called vector coupling coefficients (because
angular momentum coupling is viewed much like adding two vectors j and j' to produce
another vector J), and where the sum over m and m' is restricted to those terms for which
m+m' = M. It is more common to express the vector coupling or so-called Clebsch-
Gordon (CG) coefficients as <j,m;j'm'|J,M> and to view them as elements of a "matrix"
whose columns are labeled by the coupled-state J,M quantum numbers and whose rows
are labeled by the quantum numbers characterizing the uncoupled "product basis"
j,m;j',m'. It turns out that this matrix can be shown to be unitary so that the CG
coefficients obey:
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Σm,m' <j,m;j'm'|J,M>* <j,m;j'm'|J',M'> = δJ,J' δM,M'
and
ΣJ,M <j,n;j'n'|J,M> <j,m;j'm'|J,M>* = δn,m δn',m'.
This unitarity of the CG coefficient matrix allows the inverse of the relation
giving coupled functions in terms of the product functions:
|J,M> = Σm,m' <j,m;j'm'|J,M> |j,m> |j',m'>
to be written as:
|j,m> |j',m'> = ΣJ,M <j,m;j'm'|J,M>* |J,M>
= ΣJ,M <J,M|j,m;j'm'> |J,M>.
This result expresses the product functions in terms of the coupled angular momentum
functions.
c. Generation of the CG Coefficients
The CG coefficients can be generated in a systematic manner; however, they can
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also be looked up in books where they have been tabulated (e.g., see Table 2.4 of Dick
Zare's book on angular momentum). Here, we will demonstrate the technique by which
the CG coefficients can be obtained, but we will do so for rather limited cases and refer
the reader to more extensive tabulations.
The strategy we take is to generate the |J,J> state (i.e., the state with maximum M-
value) and to then use J- to generate |J,J-1>, after which the state |J-1,J-1> (i.e., the state
with one lower J-value) is constructed by finding a combination of the product states in
terms of which |J,J-1> is expressed (because both |J,J-1> and |J-1,J-1> have the same M-
value M=J-1) which is orthogonal to |J,J-1> (because |J-1,J-1> and |J,J-1> are
eigenfunctions of the Hermitian operator J2 corresponding to different eigenvalues, they
must be orthogonal). This same process is then used to generate |J,J-2> |J-1,J-2> and (by
orthogonality construction) |J-2,J-2>, and so on.
i. The States With Maximum and Minimum M-Values
We begin with the state |J,J> having the highest M-value. This state must be
formed by taking the highest m and the highest m' values (i.e., m=j and m'=j'), and is
given by:
|J,J> = |j,j> |j'j'>.
Only this one product is needed because only the one term with m=j and m'=j' contributes
to the sum in the above CG series. The state
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|J,-J> = |j,-j> |j',-j'>
with the minimum M-value is also given as a single product state.
Notice that these states have M-values given as ±(j+j'); since this is the maximum M-
value, it must be that the J-value corresponding to this state is J= j+j'.
ii. States With One Lower M-Value But the Same J-Value
Applying J- to |J,J> , and expressing J- as the sum of lowering operators for the
two individual angular momenta:
J- = J-(1) + J-(2)
gives
J-|J,J> = h{J(J+1) -J(J-1)}1/2 |J,J-1>
= (J-(1) + J-(2)) |j,j> |j'j'>
= h{j(j+1) - j(j-1)}1/2 |j,j-1> |j',j'> + h{j'(j'+1)-j'(j'-1)}1/2 |j,j> |j',j'-1>.
This result expresses |J,J-1> as follows:
|J,J-1>= [{j(j+1)-j(j-1)}1/2 |j,j-1> |j',j'>
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+ {j'(j'+1)-j'(j'-1)}1/2 |j,j> |j',j'-1>] {J(J+1) -J(J-1)}-1/2;
that is, the |J,J-1> state, which has M=J-1, is formed from the two product states |j,j-1>
|j',j'> and |j,j> |j',j'-1> that have this same M-value.
iii. States With One Lower J-Value
To find the state |J-1,J-1> that has the same M-value as the one found above but
one lower J-value, we must construct another combination of the two product states with
M=J-1 (i.e., |j,j-1> |j',j'> and |j,j> |j',j'-1>) that is orthogonal to the combination
representing |J,J-1>; after doing so, we must scale the resulting function so it is properly
normalized. In this case, the desired function is:
|J-1,J-1>= [{j(j+1)-j(j-1)}1/2 |j,j> |j',j'-1>
- {j'(j'+1)-j'(j'-1)}1/2 |j,j-1> |j',j'>] {J(J+1) -J(J-1)}-1/2 .
It is straightforward to show that this function is indeed orthogonal to |J,J-1>.
iv. States With Even One Lower J-Value
Having expressed |J,J-1> and |J-1,J-1> in terms of |j,j-1> |j',j'> and |j,j> |j',j'-1>,
we are now prepared to carry on with this stepwise process to generate the states |J,J-2>,
|J-1,J-2> and |J-2,J-2> as combinations of the product states with M=J-2. These product
states are |j,j-2> |j',j'>, |j,j> |j',j'-2>, and |j,j-1> |j',j'-1>. Notice that there are precisely as
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many product states whose m+m' values add up to the desired M-value as there are total
angular momentum states that must be constructed (there are three of each in this case).
The steps needed to find the state |J-2,J-2> are analogous to those taken above:
a. One first applies J- to |J-1,J-1> and to |J,J-1> to obtain |J-1,J-2> and |J,J-2>,
respectively as combinations of |j,j-2> |j',j'>, |j,j> |j',j'-2>, and |j,j-1> |j',j'-1>.
b. One then constructs |J-2,J-2> as a linear combination of the |j,j-2> |j',j'>, |j,j> |j',j'-2>,
and |j,j-1> |j',j'-1> that is orthogonal to the combinations found for |J-1,J-2> and |J,J-2>.
Once |J-2,J-2> is obtained, it is then possible to move on to form |J,J-3>, |J-1,J-3>,
and |J-2,J-3> by applying J- to the three states obtained in the preceding application of the
process, and to then form |J-3,J-3> as the combination of |j,j-3> |j',j'>, |j,j> |j',j'-3>,
|j,j-2> |j',j'-1>, |j,j-1> |j',j'-2> that is orthogonal to the combinations obtained for |J,J-3>,
|J-1,J-3>, and |J-2,J-3>.
Again notice that there are precisely the correct number of product states (four
here) as there are total angular momentum states to be formed. In fact, the product states
and the total angular momentum states are equal in number and are both members of
orthonormal function sets (because J2(1), Jz(1), J2(2), and Jz(2) as well as J2 and Jz are
Hermitian operators). This is why the CG coefficient matrix is unitary; because it maps
one set of orthonormal functions to another, with both sets containing the same number
of functions.
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d. An Example
Let us consider an example in which the spin and orbital angular momenta of the
Si atom in its 3P ground state can be coupled to produce various 3PJ states. In this case,
the specific values for j and j' are j=S=1 and j'=L=1. We could, of course take j=L=1 and
j'=S=1, but the final wavefunctions obtained would span the same space as those we are
about to determine.
The state with highest M-value is the 3P(Ms=1, ML=1) state, which can be
represented by the product of an αα spin function (representing S=1, Ms=1) and a
3p13p0 spatial function (representing L=1, ML=1), where the first function corresponds to
the first open-shell orbital and the second function to the second open-shell orbital. Thus,
the maximum M-value is M= 2 and corresponds to a state with J=2:
|J=2,M=2> = |2,2> = αα 3p13p0 .
Clearly, the state |2,-2> would be given as ββ 3p-13p0.
The states |2,1> and |1,1> with one lower M-value are obtained by applying J- =
S- + L- to |2,2> as follows:
J- |2,2> = h{2(3)-2(1)}1/2 |2,1>
= (S- + L-) αα 3p13p0 .
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To apply S- or L- to αα 3p13p0, one must realize that each of these operators is, in turn,
a sum of lowering operators for each of the two open-shell electrons:
S- = S-(1) + S-(2),
L- = L-(1) + L-(2).
The result above can therefore be continued as
(S- + L-) αα 3p13p0 = h{1/2(3/2)-1/2(-1/2)}1/2 βα 3p13p0
+ h{1/2(3/2)-1/2(-1/2)}1/2 αβ 3p13p0
+ h{1(2)-1(0)}1/2 αα 3p03p0
+ h{1(2)-0(-1)}1/2 αα 3p13p-1.
So, the function |2,1> is given by
|2,1> = [βα 3p13p0 + αβ 3p13p0 + {2}1/2 αα 3p03p0
+ {2}1/2 αα 3p13p-1]/2,
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which can be rewritten as:
|2,1> = [(ba + ab)3p13p0 + {2}1/2 aa (3p03p0 + 3p13p-1)]/2.
Writing the result in this way makes it clear that |2,1> is a combination of the product
states |S=1,MS=0> |L=1,ML=1> (the terms containing |S=1,MS=0> = 2-1/2(ab+ba)) and
|S=1,MS=1> |L=1,ML=0> (the terms containing |S=1,MS=1> = aa).
To form the other function with M=1, the |1,1> state, we must find another
combination of |S=1,MS=0> |L=1,ML=1> and |S=1,MS=1> |L=1,ML=0> that is
orthogonal to |2,1> and is normalized. Since
|2,1> = 2-1/2 [|S=1,MS=0> |L=1,ML=1> + |S=1,MS=1> |L=1,ML=0>],
we immediately see that the requisite function is
|1,1> = 2-1/2 [|S=1,MS=0> |L=1,ML=1> - |S=1,MS=1> |L=1,ML=0>].
In the spin-orbital notation used above, this state is:
|1,1> = [(βα + αβ)3p13p0 - {2}1/2 αα (3p03p0 + 3p13p-1)]/2.
Thus far, we have found the 3PJ states with J=2, M=2; J=2, M=1; and J=1, M=1.
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To find the 3PJ states with J=2, M=0; J=1, M=0; and J=0, M=0, we must once
again apply the J- tool. In particular, we apply J- to |2,1> to obtain |2,0> and we apply J-
to |1,1> to obtain |1,0>, each of which will be expressed in terms of |S=1,MS=0>
|L=1,ML=0>, |S=1,MS=1> |L=1,ML=-1>, and |S=1,MS=-1> |L=1,ML=1>. The |0,0>
state is then constructed to be a combination of these same product states which is
orthogonal to |2,0> and to |1,0>. The results are as follows:
|J=2,M=0> = 6-1/2[2 |1,0> |1,0> + |1,1> |1,-1> + |1,-1> |1,1>],
|J=1,M=0> = 2-1/2[|1,1> |1,-1> - |1,-1> |1,1>],
|J=0, M=0> = 3-1/2[|1,0> |1,0> - |1,1> |1,-1> - |1,-1> |1,1>],
where, in all cases, a short hand notation has been used in which the |S,MS> |L,ML>
product stated have been represented by their quantum numbers with the spin function
always appearing first in the product. To finally express all three of these new functions
in terms of spin-orbital products it is necessary to give the |S,MS> |L,ML> products with
M=0 in terms of these products. For the spin functions, we have:
|S=1,MS=1> = αα,
|S=1,MS=0> = 2-1/2(αβ+βα).
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|S=1,MS=-1> = ββ.
For the orbital product function, we have:
|L=1, ML=1> = 3p13p0 ,
|L=1,ML=0> = 2-1/2(3p03p0 + 3p13p-1),
|L=1, ML=-1> = 3p03p-1.
e. Coupling Angular Momenta of Equivalent Electrons
If equivalent angular momenta are coupled (e.g., to couple the orbital angular
momenta of a p2 or d3 configuration), one must use the following "box" method to
determine which of the term symbols violate the Pauli principle. To carry out this step,
one forms all possible unique (determinental) product states with non-negative ML and
MS values and arranges them into groups according to their ML and MS values. For
example, the boxes appropriate to the p2 orbital occupancy are shown below:
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ML 2 1 0
---------------------------------------------------------
MS 1 |p1αp0α| |p1αp-1α|
0 |p1αp1β| |p1αp0β|, |p0αp1β| |p1αp-1β|,
|p-1αp1β|,
|p0αp0β|
There is no need to form the corresponding states with negative ML or negative MS
values because they are simply "mirror images" of those listed above. For example, the
state with ML= -1 and MS = -1 is |p-1βp0β|, which can be obtained from the ML = 1, MS
= 1 state |p1αp0α| by replacing α by β and replacing p1 by p-1.
Given the box entries, one can identify those term symbols that arise by applying
the following procedure over and over until all entries have been accounted for:
i. One identifies the highest MS value (this gives a value of the total spin quantum
number that arises, S) in the box. For the above example, the answer is S = 1.
ii. For all product states of this MS value, one identifies the highest ML value (this gives a
value of the total orbital angular momentum, L, that can arise for this S). For the above
example, the highest ML within the MS =1 states is ML = 1 (not ML = 2), hence L=1.
iii. Knowing an S, L combination, one knows the first term symbol that arises from this
configuration. In the p2 example, this is 3P.
iv. Because the level with this L and S quantum numbers contains (2L+1)(2S+1) states
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with ML and MS quantum numbers running from -L to L and from -S to S, respectively,
one must remove from the original box this number of product states. To do so, one
simply erases from the box one entry with each such ML and MS value. Actually, since
the box need only show those entries with non-negative ML and MS values, only these
entries need be explicitly deleted. In the 3P example, this amounts to deleting nine
product states with ML, MS values of 1,1; 1,0; 1,-1; 0,1; 0,0; 0,-1; -1,1; -1,0; -1,-1.
v. After deleting these entries, one returns to step 1 and carries out the process again. For
the p2 example, the box after deleting the first nine product states looks as follows (those
that appear in italics should be viewed as already cancelled in counting all of the 3P
states):
ML 2 1 0
---------------------------------------------------------
MS 1 |p1αp0α| |p1αp-1α|
0 |p1αp1β| |p1αp0β|, |p0αp1β| |p1αp-1β|,
|p-1αp1β|,
|p0αp0β|
It should be emphasized that the process of deleting or crossing off entries in various ML,
MS boxes involves only counting how many states there are; by no means do we identify
the particular L,S,ML,MS wavefunctions when we cross out any particular entry in a box.
For example, when the |p1αp0β| product is deleted from the ML= 1, MS=0 box in
177
accounting for the states in the 3P level, we do not claim that |p1αp0β| itself is a member
of the 3P level; the |p0αp1β| product state could just as well been eliminated when
accounting for the 3P states.
Returning to the p2 example at hand, after the 3P term symbol's states have been
accounted for, the highest MS value is 0 (hence there is an S=0 state), and within this MS
value, the highest ML value is 2 (hence there is an L=2 state). This means there is a 1D
level with five states having ML = 2,1,0,-1,-2. Deleting five appropriate entries from the
above box (again denoting deletions by italics) leaves the following box:
ML 2 1 0
---------------------------------------------------------
MS 1 |p1αp0α| |p1αp-1α|
0 |p1αp1β| |p1αp0β|, |p0αp1β| |p1αp-1β|,
|p-1αp1β|,
|p0αp0β|
The only remaining entry, which thus has the highest MS and ML values, has MS = 0 and
ML = 0. Thus there is also a 1S level in the p2 configuration.
178
Thus, unlike the non-equivalent 2p13p1 case, in which 3P, 1P, 3D, 1D, 3S, and 1S
levels arise, only the 3P, 1D, and 1S arise in the p2 situation. This "box method" is
necessary to carry out whenever one is dealing with equivalent angular momenta.
If one has mixed equivalent and non-equivalent angular momenta, one can
determine all possible couplings of the equivalent angular momenta using this method
and then use the simpler vector coupling method to add the non-equivalent angular
momenta to each of these coupled angular momenta. For example, the p2d1 configuration
can be handled by vector coupling (using the straightforward non-equivalent procedure)
L=2 (the d orbital) and S=1/2 (the third electron's spin) to each of 3P, 1D, and 1S. The
result is 4F, 4D, 4P, 2F, 2D, 2P, 2G, 2F, 2D, 2P, 2S, and 2D.
VIII. Rotations of Molecules
1. Rotational Motion For Rigid Diatomic and Linear Polyatomic Molecules
This Schr?dinger equation relates to the rotation of diatomic and linear
polyatomic molecules. It also arises when treating the angular motions of electrons in
any spherically symmetric potential.
A diatomic molecule with fixed bond length R rotating in the absence of any
external potential is described by the following Schr?dinger equation:
h2/2μ {(R2sinθ)-1?/?θ (sinθ ?/?θ) + (R2sin2θ)-1 ?2/?φ2 } ψ = E ψ
179
or
L2ψ/2μR2 = E ψ,
where L2 is the square of the total angular momentum operator Lx2 + Ly2 + Lz2 expressed
in polar coordinates above. The angles θ and φ describe the orientation of the diatomic
molecule's axis relative to a laboratory-fixed coordinate system, and μ is the reduced
mass of the diatomic molecule μ=m1m2/(m1+m2). The differential operators can be seen
to be exactly the same as those that arose in the hydrogen-like-atom caseas discussed
above. Therefore, the same spherical harmonics that served as the angular parts of the
wave function in the hydrogen-atom case now serve as the entire wave function for the
so-called rigid rotor: ψ = YJ,M(θ,φ). These are exactly the same functions as we plotted
earlier when we graphed the s (L=0), p (L=1), and d (L=2) orbitals. As detailed in
Chapter 6 of this text, the energy eigenvalues corresponding to each such eigenfunction
are given as:
EJ = h2 J(J+1)/(2μR2) = B J(J+1)
and are independent of M. Thus each energy level is labeled by J and is 2J+1-fold
degenerate (because M ranges from -J to J). Again, this is just like we saw when we
looked at the hydrogen orbitals; the p orbitals are 3-fold degenerate and the d orbitals are
5-fold degenerate. The so-called rotational constant B (defined as h2/2μR2) depends on
the molecule's bond length and reduced mass. Spacings between successive rotational
180
levels (which are of spectroscopic relevance because, as shown in Chapter 6, angular
momentum selection rules often restrict the changes ?J in J that can occur upon photon
absorption to 1,0, and -1) are given by
?E = B (J+1)(J+2) - B J(J+1) = 2B(J+1).
These energy spacings are of relevance to microwave spectroscopy which probes the
rotational energy levels of molecules. In fact, microwave spectroscopy offers the most
direct way to determine molecular rotational constants and hence molecular bond lengths.
The rigid rotor provides the most commonly employed approximation to the
rotational energies and wave functions of linear molecules. As presented above, the
model restricts the bond length to be fixed. Vibrational motion of the molecule gives rise
to changes in R which are then reflected in changes in the rotational energy levels. The
coupling between rotational and vibrational motion gives rise to rotational B constants
that depend on vibrational state as well as dynamical couplings, called centrifugal
distortions, that cause the total ro-vibrational energy of the molecule to depend on
rotational and vibrational quantum numbers in a non-separable manner.
Within this "rigid rotor" model, the absorption spectrum of a rigid diatomic
molecule should display a series of peaks, each of which corresponds to a specific J ==>
J + 1 transition. The energies at which these peaks occur should grow linearly with J. An
example of such a progression of rotational lines is shown in the Fig. 2.23.
181
Figure 2.23. Typical rotational absorption profile showing intensity vs. J value of the
absorbing level
The energies at which the rotational transitions occur appear to fit the ?E = 2B (J+1)
formula rather well. The intensities of transitions from level J to level J+1 vary strongly
with J primarily because the population of molecules in the absorbing level varies with J.
These populations PJ are given, when the system is at equilibrium at temperature T, in
terms of the degeneracy (2J+1) of the Jth level and the energy of this level B J(J+1) by the
Boltzmann formula:
PJ = Q-1 (2J+1) exp(-BJ(J+1)/kT),
where Q is the rotational partition function:
Q = ΣJ (2J+1) exp(-BJ(J+1)/kT).
182
For low values of J, the degeneracy is low and the exp(-BJ(J+1)/kT) factor is near unity.
As J increases, the degeracy grows linearly but the exp(-BJ(J+1)/kT) factor decreases
more rapidly. As a result, there is a value of J, given by taking the derivative of (2J+1)
exp(-BJ(J+1)/kT) with respect to J and setting it equal to zero,
2Jmax + 1 = 2kT/B
at which the intensity of the rotational transition is expected to reach its maximum. This
behavior is clearly displayed in the above figure.
The eigenfunctions belonging to these energy levels are the spherical harmonics
YL,M(θ,φ) which are normalized according to
?�
�?
0
pi
( ??
0
2pi
(Y*L,M(θ,φ) YL',M'(θ,φ) sinθ dθ dφ)) = δL,L' δM,M' .
As noted above, these functions are identical to those that appear in the solution of the
angular part of Hydrogen-like atoms. The above energy levels and eigenfunctions also
apply to the rotation of rigid linear polyatomic molecules; the only difference is that the
moment of inertia I entering into the rotational energy expression is given by
I = Σa ma Ra2
183
where ma is the mass of the ath atom and Ra is its distance from the center of mass of the
molecule. This moment of inertia replaces μR2 in the earlier rotational energy level
expressions.
2. Rotational Motions of Rigid Non-Linear Molecules
a. The Rotational Kinetic Energy
The rotational kinetic energy operator for a rigid polyatomic molecule is
Hrot = Ja2/2Ia + Jb2/2Ib + Jc2/2Ic
where the Ik (k = a, b, c) are the three principal moments of inertia of the molecule (the
eigenvalues of the moment of inertia tensor). This tensor has elements in a Cartesian
coordinate system (K, K' = X, Y, Z), whose origin is located at the center of mass of the
molecule, that can be computed as:
IK,K = Σj mj (Rj2 - R2K,j) (for K = K')
IK,K' = - Σj mj RK,j RK',j (for K ≠ K').
184
As discussed in more detail in Chapter 6, the components of the quantum mechanical
angular momentum operators along the three principal axes are:
Ja = -ih cosχ [cotθ ?/?χ - (sinθ)-1?/?φ ] - -ih sinχ ?/?θ
Jb = ih sinχ [cotθ ?/?χ - (sinθ)-1?/?φ ] - -ih cosχ ?/?θ
Jc = - ih ?/?χ.
The angles θ, φ, and χ are the Euler angles needed to specify the orientation of the rigid
molecule relative to a laboratory-fixed coordinate system. The corresponding square of
the total angular momentum operator J2 can be obtained as
J2 = Ja2 + Jb2 + Jc2
= - ?2/?θ2 - cotθ ?/?θ
+ (1/sinθ) (?2/?φ2 + ?2/?χ2 - 2 cosθ?2/?φ?χ),
and the component along the lab-fixed Z axis JZ is - ih ?/?φ as we saw much earlier in
this text.
b. The Eigenfunctions and Eigenvalues for Special Cases
185
i. Spherical Tops
When the three principal moment of inertia values are identical, the molecule is
termed a spherical top. In this case, the total rotational energy can be expressed in terms
of the total angular momentum operator J2
Hrot = J2/2I.
As a result, the eigenfunctions of Hrot are those of J2 and Ja as well as JZ both of which
commute with J2 and with one another. JZ is the component of J along the lab-fixed Z-
axis and commutes with Ja because JZ = - ih ?/?φ and Ja = - ih ?/?χ act on different
angles. The energies associated with such eigenfunctions are
E(J,K,M) = h2 J(J+1)/2I2,
for all K (i.e., Ja quantum numbers) ranging from -J to J in unit steps and for all M (i.e.,
JZ quantum numbers) ranging from -J to J. Each energy level is therefore (2J + 1)2
degenarate because there are 2J + 1 possible K values and 2J + 1 possible M values for
each J.
The eigenfunctions |J,M,K> of J2, JZ and Ja , are given in terms of the set of so-
called rotation matrices DJ,M,K:
|J,M,K> = 2J + 18 pi2 D*J,M,K(θ,φ,χ)
186
which obey
J2 |J,M,K> = h2 J(J+1) |J,M,K>,
Ja |J,M,K> = h K |J,M,K>,
JZ |J,M,K> = h M |J,M,K>.
These DJ,M,K functions are propotional to the spherical harmonics YJ,M(θ,φ) multiplied by
exp(iKχ), which reflects its χ-dependence.
ii. Symmetric Tops
Molecules for which two of the three principal moments of inertia are equal are
called symmetric tops. Those for which the unique moment of inertia is smaller than the
other two are termed prolate symmetric tops; if the unique moment of inertia is larger
than the others, the molecule is an oblate symmetric top. An American football is prolate,
and a frisbee is oblate.
Again, the rotational kinetic energy, which is the full rotational Hamiltonian, can
be written in terms of the total rotational angular momentum operator J2 and the
component of angular momentum along the axis with the unique principal moment of
inertia:
187
Hrot = J2/2I + Ja2{1/2Ia - 1/2I}, for prolate tops
Hrot = J2/2I + Jc2{1/2Ic - 1/2I}, for oblate tops.
Here, the moment of inertia I denotes that moment that is common to two directions; that
is, I is the non-unique moment of inertia. As a result, the eigenfunctions of Hrot are those
of J2 and Ja or Jc (and of JZ), and the corresponding energy levels are:
E(J,K,M) = h2 J(J+1)/2I2 + h2 K2 {1/2Ia - 1/2I},
for prolate tops
E(J,K,M) = h2 J(J+1)/2I2 + h2 K2 {1/2Ic - 1/2I},
for oblate tops, again for K and M (i.e., Ja or Jc and JZ quantum numbers, respectively)
ranging from -J to J in unit steps. Since the energy now depends on K, these levels are
only 2J + 1 degenerate due to the 2J + 1 different M values that arise for each J value.
Notice that for prolate tops, because Ia is smaller than I, the energies increase with
increasing K for given J. In contrast, for oblate tops, since Ic is larger than I, the energies
decrease with K for given J. The eigenfunctions |J, M,K> are the same rotation matrix
functions as arise for the spherical-top case, so they do not require any further discussion
at this time.
188
iii. Asymmetric Tops
The rotational eigenfunctions and energy levels of a molecule for which all three
principal moments of inertia are distinct (a so-called asymmetric top) can not easily be
expressed in terms of the angular momentum eigenstates and the J, M, and K quantum
numbers. In fact, no one has ever solved the corresponding Schr?dinger equation for this
case. However, given the three principal moments of inertia Ia, Ib, and Ic, a matrix
representation of each of the three contributions to the rotational Hamiltonian
Hrot = Ja2/2Ia + Jb2/2Ib + Jc2/2Ic
can be formed within a basis set of the {|J, M, K>} rotation-matrix functions discussed
earlier. This matrix will not be diagonal because the |J, M, K> functions are not
eigenfunctions of the asymmetric top Hrot. However, the matrix can be formed in this
basis and subsequently brought to diagonal form by finding its eigenvectors {Cn, J,M,K}
and its eigenvalues {En}. The vector coefficients express the asymmetric top eigenstates
as
Ψn (θ, φ, χ) = ΣJ, M, K Cn, J,M,K |J, M, K>.
Because the total angular momentum J2 still commutes with Hrot, each such eigenstate
will contain only one J-value, and hence Ψn can also be labeled by a J quantum number:
189
Ψn,J (θ, φ, χ) = Σ M, K Cn, J,M,K |J, M, K>.
To form the only non-zero matrix elements of Hrot within the |J, M, K> basis, one
can use the following properties of the rotation-matrix functions (see, for example, R. N.
Zare, Angular Momentum, John Wiley, New York (1988)):
<J, M, K| Ja2| J, M, K> = <J, M, K| Jb 2| J, M, K>
= 1/2 <J, M, K| J2 - Jc2 | J, M, K> = h2 [ J(J+1) - K2 ],
<J, M, K| Jc2| J, M, K> = h2 K2,
<J, M, K| Ja2| J, M, K ± 2> = - <J, M, K| Jb 2| J, M, K ± 2>
= h2 [J(J+1) - K(K± 1)]1/2 [J(J+1) -(K± 1)(K± 2)]1/2
<J, M, K| Jc2| J, M, K ± 2> = 0.
Each of the elements of Jc2, Ja2, and Jb2 must, of course, be multiplied, respectively, by
1/2Ic, 1/2Ia, and 1/2Ib and summed together to form the matrix representation of Hrot.
The diagonalization of this matrix then provides the asymmetric top energies and wave
functions.
190
IX. Vibrations of Molecules
This Schr?dinger equation forms the basis for our thinking about bond stretching and
angle bending vibrations as well as collective vibrations called phonons in solids.
The radial motion of a diatomic molecule in its lowest (J=0) rotational level can
be described by the following Schr?dinger equation:
- (h2/2μ) r-2?/?r (r2?/?r) ψ +V(r) ψ = E ψ,
where μ is the reduced mass μ = m1m2/(m1+m2) of the two atoms. If the molecule is
rotating, then the above Schr?dinger equation has an additional term J(J+1) h2/2μ r-2 ψ
on its left-hand side. Thus, each rotational state (labeled by the rotational quantum
number J) has its own vibrational Schr?dinger equantion and thus its own set of
vibrational energy levels and wave functions. It is common to examine the J=0
vibrational problem and then to use the vibrational levels of this state as approximations
to the vibrational levels of states with non-zero J values (treating the vibration-rotation
coupling via perturbation theory introduced in Sec. VII.). Let us thus focus on the J=0
situation.
By substituting ψ= F(r)/r into this equation, one obtains an equation for F(r) in
which the differential operators appear to be less complicated:
- h2/2μ d2F/dr2 + V(r) F = E F.
191
This equation is exactly the same as the equation seen earlier in this text for the radial
motion of the electron in the hydrogen-like atoms except that the reduced mass μ replaces
the electron mass m and the potential V(r) is not the Coulomb potential.
If the vibrational potential is approximated as a quadratic function of the bond
displacement x = r-re expanded about the equilibrium bond length re where V has its
minimum:
V = 1/2 k(r-re)2,
the resulting harmonic-oscillator equation can be solved exactly. Because the potential V
grows without bound as x approaches ∞ or -∞, only bound-state solutions exist for this
model problem. That is, the motion is confined by the nature of the potential, so no
continuum states exist in which the two atoms bound together by the potential are
dissociated into two separate atoms.
In solving the radial differential equation for this potential, the large-r behavior is
first examined. For large-r, the equation reads:
d2F/dx2 = 1/2 k x2 (2μ/h2) F,
where x = r-re is the bond displacement away from equilibrium. Defining ξ= (μk/h2)1/4 x
as a new scaled radial coordinate allows the solution of the large-r equation to be written
as:
192
Flarge-r = exp(- ξ2/2).
The general solution to the radial equation is then expressed as this large-r
solution multiplied by a power series in the ζ variable:
F = exp(- ξ2/2) ∑
n=0
∞
ξn Cn ,
where the Cn are coefficients to be determined. Substituting this expression into the full
radial equation generates a set of recursion equations for the Cn amplitudes. As in the
solution of the hydrogen-like radial equation, the series described by these coefficients is
divergent unless the energy E happens to equal specific values. It is this requirement that
the wave function not diverge so it can be normalized that yields energy quantization.
The energies of the states that arise are given by:
En = h (k/μ)1/2 (n+1/2),
and the eigenfunctions are given in terms of the so-called Hermite polynomials Hn(y) as
follows:
ψn(x) = (n! 2n)-1/2 (α/pi)1/4 exp(- αx2/2) Hn(α1/2 x),
193
where α =(kμ/h2)1/2. Within this harmonic approximation to the potential, the vibrational
energy levels are evenly spaced:
?E = En+1 - En = h (k/μ)1/2 .
In experimental data such evenly spaced energy level patterns are seldom seen; most
commonly, one finds spacings En+1 - En that decrease as the quantum number n
increases. In such cases, one says that the progression of vibrational levels displays
anharmonicity.
Because the Hermite functions Hn are odd or even functions of x (depending on
whether n is odd or even), the wave functions ψn(x) are odd or even. This splitting of the
solutions into two distinct classes is an example of the effect of symmetry; in this case,
the symmetry is caused by the symmetry of the harmonic potential with respect to
reflection through the origin along the x-axis (i.e., changing x to –x). Throughout this
text, many symmetries arise; in each case, symmetry properties of the potential cause the
solutions of the Schr?dinger equation to be decomposed into various symmetry
groupings. Such symmetry decompositions are of great use because they provide
additional quantum numbers (i.e., symmetry labels) by which the wave functions and
energies can be labeled.
The basic idea underlying how such symmetries split the solutions of the
Schr?dinger equation into different classes relates to the fact that a symmetry operator
(e.g., the reflection plane in the above example) commutes with the Hamiltonian. That is,
the symmetry operator S obeys
194
S H Ψ = H S Ψ.
So S leaves H unchanged as it acts on H (this allows us to pass S through H in the above
equation). Any operator that leaves the Hamiltonian (i.e., the energy) unchanged is called
a symmetry operator.
If you have never learned about how point group symmetry can be used to hjelp
simplify the solution of the Schr?dinger equation, this would be a good time to interrupt
your reading and go to Sec. VIII and read the material there.
The harmonic oscillator energies and wave functions comprise the simplest
reasonable model for vibrational motion. Vibrations of a polyatomic molecule are often
characterized in terms of individual bond-stretching and angle-bending motions, each of
which is, in turn, approximated harmonically. This results in a total vibrational wave
function that is written as a product of functions, one for each of the vibrational
coordinates.
Two of the most severe limitations of the harmonic oscillator model, the lack of
anharmonicity (i.e., non-uniform energy level spacings) and lack of bond dissociation,
result from the quadratic nature of its potential. By introducing model potentials that
allow for proper bond dissociation (i.e., that do not increase without bound as x=>∞), the
major shortcomings of the harmonic oscillator picture can be overcome. The so-called
Morse potential (see Fig. 2.24)
V(r) = De (1-exp(-a(r-re)))2,
195
is often used in this regard.
0 1 2 3 4
-6
-4
-2
0
2
4
Internuclear distance
Energy
Figure 2.24. Morse potential energy as a function of bond length
In the Morse potential function, De is the bond dissociation energy, re is the equilibrium
bond length, and a is a constant that characterizes the 'steepness' of the potential and thus
affects the vibrational frequencies. The advantage of using the Morse potential to
improve upon harmonic-oscillator-level predictions is that its energy levels and wave
functions are also known exactly. The energies are given in terms of the parameters of the
potential as follows:
En = h(k/μ)1/2 { (n+1/2) - (n+1/2)2 h(k/μ)1/2/4De },
196
where the force constant is given by k=2De a2. The Morse potential supports both bound
states (those lying below the dissociation threshold for which vibration is confined by an
outer turning point) and continuum states lying above the dissociation threshold. Its
degree of anharmonicity is governed by the ratio of the harmonic energy h(k/μ)1/2 to the
dissociation energy De.
The eigenfunctions of the harmonic and Morse potentials display nodal character
analogous to what we have seen earlier in the particle-in-boxes model problems. Namely,
as the energy of the vibrational state increases, the number of nodes in the vibrational
wave function also increases. The state having vibrational quantum number v has v
nodes. I hope that by now the student is getting used to seeing the number of nodes
increase as the quantum number and hence the energy grows.
Chapter 3. Characteristics of Energy Surfaces
Born-Oppenheimer energy surfaces (or the empirical functions often used to
represent them) possess important critical points that detail the properties of stable
molecular structures, transition states, and reaction paths, all of which play central roles
in the theoretical description of molecular properties. In this Chapter, you will learn
about these special points on the surfaces, how to find them, and what to do with them
once you know them.
I. Strategies for Geometry Optimization
197
The extension of the harmonic and Morse vibrational models to polyatomic
molecules requires that the multidimensional energy surface be analyzed in a manner
that allows one to approximate the molecule’s motions in terms of many nearly
independent vibrations. In this Section, we will explore the tools that one uses to carry
out such an analysis of the surface.
Many strategies that attempt to locate minima on molecular potential energy
landscapes begin by approximating the potential energy V for geometries (collectively
denoted in terms of 3N Cartesian coordinates {qj}) in a Taylor series expansion about
some “starting point” geometry (i.e., the current molecular geometry in an iterative
process):
V (qk) = V(0) + Σk (?V/?qk) qk + 1/2 Σj,k qj Hj,k qk + ... .
Here, V(0) is the energy at the current geometry, (?V/?qk) = gk is the gradient of the
energy along the qk coordinate, Hj,k = (?2V/?qj?qk) is the second-derivative or Hessian
matrix, and qk is the length of the “step” to be taken along this Cartesian direction. An
example of an energy surface in only two dimensions is given in the Figure 3.1 where
various special aspects are illustrated. For example, minima corresponding to stable
molecular structures, transition states (first order saddle points) connecting such minima,
and higher order saddle points are displayed.
198
Figure 3.1. Two-dimensional potential surface showing minima, transition states, and
paths connecting them.
If the only knowledge that is available is V(0) and the gradient components (e.g.,
computation of the second derivatives is usually much more computationally taxing than
is evaluation of the gradient), the linear approximation
V (qk) = V(0) + Σk gK qk
suggests that one should choose “step” elements qk that are opposite in sign from that of
the corresponding gradient elements gk = (?V/?qk). The magnitude of the step elements is
usually kept small in order to remain within the “trust radius” within which the linear
approximation to V is valid to some predetermined desired precision.
When second derivative data is available, there are different approaches to
199
predicting what step {qk} to take in search of a minimum. We first write the quadratic
Taylor expansion
V (qk) = V(0) + Σk gk qk + 1/2 Σj,k qj Hj,k qk
in matrix-vector notation
V(q) = V(0) + qT ? g + 1/2 qT ? H ? q
with the elements{qk} collected into the column vector q whose transpose is denoted qT.
Introducing the unitary matrix U that diagonalizes the symmetric H matrix, the above
equation becomes
V(q) = V(0) + gT U UT q + 1/2 qT U UTHU UT q.
Because UTHU is diagonal
(UTHU)k,l = δk,l λk
and has eigenvalues λk. For non-linear molecules, 3N-6 of these eigenvalues will be non-
zero; for linear molecules, 3N-5 will be non-zero. The 5 or 6 zero eigenvalues of H have
eigenvectors that describe translation and rotation of the entire molecule; they are zero
200
because the energy surface V does not change if the molecule is rotated or translated.
The eigenvectors of H form the columns of the array U that brings H to diagonal form:
Σλ Hk,l Ul,m = λm Uk,m
Therefore, if we define
Qm = Σk UTm,k qk and Gm = Σk UYm,k gk
to be the component of the step {qk} and of the gradeint along the mth eigenvector of H,
the quadratic expansion of V can be written in terms of steps along the 3N-5 or 3N-6
{Qm} directions that correspond to non-zero Hessian eigenvalues:
V (qk) = V(0) + Σm GTm Qm + 1/2 Σm Qm λm Qm.
The advantage to transforming the gradient, step, and Hessian to the eigenmode basis is
that each such mode (labeled m) appears in an independent uncoupled form in the
expansion of V. This allows us to take steps along each of the Qm directions in an
independent manner with each step designed to lower the potential energy (as we search
for minima).
For each eigenmode direction, one can ask for what step Q would the quantity GQ
+ 1/2 λ Q2 be a minimum. Differentiating this quadratic form with respect to Q and
setting the result equal to zero gives
201
Qm = - Gm/λm ;
that is, one should take a step opposite the gradient but with a magnitude given by the
gradient divided by the eigenvalue of the Hessian matrix. If the current molecular
geometry is one that has all positive λm values, this indicates that one may be “close” to a
minimum on the energy surface (because all λm are positive at minima). In such case, the
step Qm = - Gm/λm is opposed to the gradient along all 3N-5 or 3N-6 directions. The
energy change that is expected to occur if the step {Qm} is taken can be computed by
substituting Qm = - Gm/λm into the quadratic equation for V:
V(after step) = V(0) + Σm GTm (- Gm/λm) + 1/2 Σm λm (- Gm/λm)2
= V(0) - 1/2 Σm λm (- Gm/λm)2.
This clearly suggests that the step will lead “downhill” in energy as long as all of the λm
values are positive.
However, if one or more of the λm are negative at the current geometry, one is in a
region of the energy surface that is not close to a minimum. In fact, if only one λm is
negative, one anticipates being near a transition state (at which all gradient components
vanish and all but one λm are positive with one λm negative). In such a case, the above
analysis suggests taking a step Qm = - Gm/λm along all of the modes having positive λm,
but taking a step of opposite direction Qn = + Gn/λn along the direction having negative
202
λn.
In any event, once a step has been suggested within the eigenmode basis, one
needs to express that step in terms of the original Cartesian coordinates qk so that these
Cartesian values can be altered within the software program to effect the predicted step.
Given values for the 3N-5 or 3N-6 step components Qm (n.b., the step components Qm
along the 5 or 6 modes having zero Hessian eigenvalues can be taken to be zero because
the would simply translate or rotate the molecule), one must compute the {qk}. To do so,
we use the relationship
Qm = Σk UTm,k qk
and write its inverse (using the unitary nature of the U matrix):
qk = Σm Uk,m Qm
to compute the desired Cartesian step components.
In using the Hessian-based approaches outlined above, one has to take special
care when one or more of the Hessian eigenvalues is small. This often happens when
i. one has a molecule containing “soft modes” (i.e., degrees of freedom along which the
energy varies little), or
ii. as one moves from a region of negative curvature into a region of positive curvature
(or vice versa)- in such cases, the the curvature must move through or near zero.
203
For these situations, the expression Qm = - Gm/λm can produce a very large step along the
mode having small curvature. Care must be taken to not allow such incorrect artificially
large steps to be taken.
Before closing this Section, I should note that there are other important regions of
potential energy surfaces that one must be able to locate and characterize. Above, we
focused on local minima and transition states. In Chapter 8, we will discuss how to
follow so-called reaction paths that connect these two kinds of stationary points using the
type of gradient and Hessian information that we introduced earlier in this Chapter.
Finally, it is sometimes important to find geometries at which two Born-
Oppenheimer energy surfaces V1(q) and V2(q) intersect. First, let’s spend a few minutes
thinking about whether such surfaces can indeed intersect because students often hear
that surfaces do not intersect but, instead, undergo “avoided crossings”. To understand
the issue, let us assume that we have two wave functions Φ1 and Φ2 both of which depend
on 3N-6 coordinates {q}. These two functions are not assumed to be exact eigenfunctions
of the Hamiltonian H, but likely are chosen to approximate such eigenfunctions. To find
the improved functions Ψ1 and Ψ2 that more accurately represent the eigenstates, one
usually forms linear combinations of Φ1 and Φ2,
ΨK = CK,1 Φ1 + CK,2 Φ2
from which a 2x2 matrix eigenvalue problem arises:
H
1,1
? E H
1,2
H
2,1
H
2,2
? E
= 0 .
204
This quadratic equation has two solutions
2E
±
= (H
1,1
+ H
2,2
) ± (H
1,1
? H
2,2
)
2
+ 4H
1,2
2 .
These two solutions can be equal (i.e., the two state energies can cross) only if the square
root factor vanishes. Because this factor is a sum of two squares (each thus being positive
quantities), this can only happen if two identities hold:
H1,1 = H2,2
and
H1,2 = 0.
The main point then is that in the 3N-6 dimensional space, the two states will generally
not have equal energy. However, in a space of two lower dimensions (because there are
two conditions that must simultaneously be obeyed- H1,1 = H2,2 and H1,2 = 0), their
energies may be equal. They do not have to be equal, but it is possible that they are. It is
based upon such an analysis that one usually says that potential energy surfaces in 3N-6
dimensions may undergo intersections in spaces of dimension 3N-8. If the two states are
of different symmetry, the off-diagonal element H1,2 vanishes automatically, so only one
205
other condition is needed to realize crossing. So, we say that two states of different
symmetry can cross in a space of dimension 3N-7.
To find the lower-dimensional space in which two surfaces cross, one must have
available information about the gradients and Hessians of both functions V1 and V2. One
then uses this information to locate a geometry at which the difference function F = [V1
–V2]2 passes through zero by using conventional “root finding” methods designed to
locate where F = 0. Once one such geometry (q0) has been located, one subsequently
tries to follow the “seam” along which the function F remains zero. This is done by
parameterizng steps away from (q0) in a manner that constrains such steps to have no
component along the gradient of F (i.e., to lie in the tangent plane where F is constant).
For a system with 3N-6 geometrical degrees of freedom, this seam will be a subsurface of
lower dimension (3N-8 or 3N-7 as noted earlier). Such intersection seam location
procedures are becoming more commonly employed, but are still under very active
development. Locating these intersections is an important ingredient when one is
interested in studying, for example, photochemical reactions in which the reactants and
products may move from one electronic surface to another.
II. Normal Modes of Vibration
Having seen how one can use information about the gradients and Hessians on a
Born-Oppenheimer surface to locate geometries corresponding to stable species, let us
now move on to see how this same data is used to treat vibrations on this surface.
For a polyatomic molecule whose electronic energy's dependence on the 3N
206
Cartesian coordinates of its N atoms, the potential energy V can be expressed
(approximately) in terms of a Taylor series expansion about any of the local minima. Of
course, different local minima (i.e., different isomers) will have different values for the
equilibrium coordinates and for the derivatives of the energy with respect to these
coordinates. The Taylor series expansion of the electronic energy is written as:
V (qk) = V(0) + Σk (?V/?qk) qk + 1/2 Σj,k qj Hj,k qk + ... ,
where V(0) is the value of the electronic energy at the stable geometry under study, qk is
the displacement of the kth Cartesian coordinate away from this starting position,
(?V/?qk) is the gradient of the electronic energy along this direction, and the Hj,k are the
second derivative or Hessian matrix elements along these directions Hj,k = (?2V/?qj?qk).
If the geometry corresponds to a stable species, the gradient terms will all vanish
(meaning this geometry corresponds to a minimum, maximum, or saddle point), and the
Hessian matrix will possess 3N - 5 (for linear species) or 3N -6 (for non-linear
molecules) positive eigenvalues and 5 or 6 zero eigenvalues (corresponding to 3
translational and 2 or 3 rotational motions of the molecule). If the Hessian has one
negative eigenvalue, the geometry corresponds to a transition state. From now on, we
assume that the geometry under study corresponds to that of a stable minimum about
which vibrational motion occurs. The treatment of unstable geometries is of great
importance to chemistry, but this material will be limited to vibrations of stable species.
A. The Newton Equations of Motion for Vibration
207
1. The Kinetic and Potential Energy Matrices
Truncating the Taylor series at the quadratic terms (assuming these terms
dominate because only small displacements from the equilibrium geometry are of
interest), one has the so-called harmonic potential:
V (qk) = V(0) + 1/2 Σj,k qj Hj,k qk.
The classical mechanical equations of motion for the 3N {qk} coordinates can be written
in terms of the above potential energy and the following kinetic energy function:
T = 1/2 Σj mj q? j2,
where q? j denotes the time rate of change of the coordinate qj and mj is the mass of the
atom on which the jth Cartesian coordinate resides. The Newton equations thus obtained
are:
mj q??j = - Σk Hj,k qk
where the force along the jth coordinate is given by minus the derivative of the potential
V along this coordinate (?V/?qj) = Σk Hj,k qk within the harmonic approximation.
208
These classical equations can more compactly be expressed in terms of the time
evolution of a set of so-called mass weighted Cartesian coordinates defined as:
xj = qj (mj)1/2,
in terms of which the above Newton equations become
x??j = - Σk H'j,k xk
and the mass-weighted Hessian matrix elements are
H'j,k = Hj,k (mjmk)-1/2.
2. The Harmonic Vibrational Energies and Normal Mode Eigenvectors
Assuming that the xj undergo some form of sinusoidal time evolution:
xj(t) = xj (0) cos(ωt),
and substituting this into the Newton equations produces a matrix eigenvalue equation:
ω2 xj = Σk H'j,k xk
209
in which the eigenvalues are the squares of the so-called normal mode vibrational
frequencies and the eigenvectors give the amplitudes of motion along each of the 3N
mass weighted Cartesian coordinates that belong to each mode. Hence, to perform a
normal-mode analysis of a molecule, one forms the mass-weighted Hessian matrix and
then finds the 3N-5 or 3N-6 non-zero eigenvalues ωj2 as well as the corresponding
eigenvectors xk(j).
Within this harmonic treatment of vibrational motion, the total vibrational energy
of the molecule is given as
E(v1, v2, ··· v3N-5 or 6) = ∑
j=1
3N-5or6
hωj (vj + 1/2)
a sum of 3N-5 or 3N-6 independent contributions one for each normal mode. The
corresponding total vibrational wave function
Ψ = Πj=1,3N-5or 6 ψvj (x(j))
is a product of 3N-5 or 3N-6 harmonic oscillator functions ψvj (x(j)) one for each normal
mode. The energy gap between one vibrational level and another in which one of the vj
quantum numbers is increased by unity (i.e., for fundamental vibrational transitions) is
?Evj → vj + 1 = h ωj
210
The harmonic model thus predicts that the "fundamental" (v=0 → v = 1) and "hot band"
(v=1 → v = 2) transitions should occur at the same energy, and the overtone (v=0 → v=2)
transitions should occur at exactly twice this energy.
B. The Use of Symmetry
1. Symmetry Adapted Modes
It is often possible to simplify the calculation of the normal mode frequencies and
eigenvectors by exploiting molecular point group symmetry. For molecules that possess
symmetry at a particular stable geometry, the electronic potential V(qj) displays
symmetry with respect to displacements of symmetry equivalent Cartesian coordinates.
For example, consider the water molecule at its C2v equilibrium geometry as illustrated in
Fig. 3.2. A very small movement of the H2O molecule's left H atom in the positive x
direction (?xL) produces the same change in the potential V as a correspondingly small
displacement of the right H atom in the negative x direction
(-?xR). Similarly, movement of the left H in the positive y direction (?yL) produces an
energy change identical to movement of the right H in the positive y direction (?yR).
H
O
Hθ
r
2
r
1
y
x
211
Figure 3.2. Water molecule showing its two bond lengths and angle
The equivalence of the pairs of Cartesian coordinate displacements is a result of
the fact that the displacement vectors are connected by the point group operations of the
C2v group. In particular, reflection of ?xL through the yz plane (the two planes are
depicted in Fig. 3.3) produces - ?xR, and reflection of ?yL through this same plane yields
?yR.
Figure 3.3. Two planes of symmetry of the water molecule.
More generally, it is possible to combine sets of Cartesian displacement
coordinates {qk} into so-called symmetry adapted coordinates {QΓ,j}, where the index Γ
labels the irreducible representation in the appropriate point group and j labels the
212
particular combination of that symmetry. These symmetry adapted coordinates can be
formed by applying the point group projection operators (that are treated in detail in Sec.
VIII) to the individual Cartesian displacement coordinates.
To illustrate, again consider the H2O molecule in the coordinate system described
above. The 3N = 9 mass weighted Cartesian displacement coordinates (XL, YL, ZL, XO,
YO, ZO, XR, YR, ZR) can be symmetry adapted by applying the following four projection
operators:
PA1 = 1 + σyz + σxy + C2
Pb1 = 1 + σyz - σxy - C2
Pb2 = 1 - σyz + σxy - C2
Pa2 = 1 - σyz - σxy + C2
to each of the 9 original coordinates (the symbol σ denotes reflection through a plane and
C2 means rotation about the molecule’s C2 axis). Of course, one will not obtain 9 x 4 =
36 independent symmetry adapted coordinates in this manner; many identical
combinations will arise, and only 9 will be independent.
The independent combination of a1 symmetry (normalized to produce vectors of
unit length) are
Qa1,1 = 2-1/2 [XL - XR]
Qa1,2 = 2-1/2 [YL + YR]
213
Qa1,3 = [YO]
Those of b2 symmetry are
Qb2,1 = 2-1/2 [XL + XR]
Qb2,2 = 2-1/2 [YL - YR]
Qb2,3 = [XO],
and the combinations
Qb1,1 = 2-1/2 [ZL + ZR]
Qb1,2 = [ZO]
are of b1 symmetry, whereas
Qa2,1 = 2-1/2 [ZL - ZR]
is of a2 symmetry.
2. Point Group Symmetry of the Harmonic Potential
214
These nine QΓ,j are expressed as unitary transformations of the original mass
weighted Cartessian coordinates:
QΓ,j = ∑
k
CΓ,j,k Xk
These transformation coefficients {CΓ,j,k} can be used to carry out a unitary
transformation of the 9x9 mass-weighted Hessian matrix. In so doing, we need only form
blocks
HΓj,l =
∑k k'
CΓ,j,k Hk,k' (mk mk')-1/2 CΓ,l,k'
within which the symmetries of the two modes are identical. The off-diagonal elements
H Γ Γ'j l = Σk,k’ CΓ,j,k Hk,k' (mk mk')-1/2 CΓ',l,k'
vanish because the potential V (qj) (and the full vibrational Hamiltonian H = T + V)
commutes with the C2V point group symmetry operations.
As a result, the 9x9 mass-weighted Hessian eigenvalue problem can be sub
divided into two 3x3 matrix problems (of a1 and b2 symmetry), one 2x2 matrix of b1
symmetry and one 1x1 matrix of a2 symmetry. For example, the a1 symmetry block Ha1j l
is formed as follows:
215
?
?
?
?
?
?
?
?1
2 -
1
2 0
1
2
1
2 0
0 0 1
??
?
?
?
?m-1H ?2v?x
L2
m-1H ?
2v
?xL ?xR (mH mO)-1/2
?2v
?xL ?xO
m-1H ?
2v
?xR?xL m-1H
?2v
?X2R (mH mO)-1/2
?2v
?xR ?xO
(mH mO)-1/2 ?
2v
?xO ?xL (mH mO)-1/2
?2v
?xO ?xR m-1O
?2v
?xO2
?
?
?
?
?
?
?
?1
2
1
2 0
- 12 12 0
0 0 1
The b2, b1 and a2 blocks are formed in a similar manner. The eigenvalues of each of
these blocks provide the squares of the harmonic vibrational frequencies, the eigenvectors
provide the normal mode displacements as linear combinations of the symmetry adapted
{QΓj}.
Regardless of whether symmetry is used to block diagonalize the mass-weighted
Hessian, six (for non-linear molecules) or five (for linear species) of the eigenvalues will
equal zero. The eigenvectors belonging to these zero eigenvalues describe the 3
translations and 2 or 3 rotations of the molecule. For example,
1
3 [XL + XR + XO]
1
3 [YL + YR + YO]
1
3 [ZL +ZR + ZO]
are three translation eigenvectors of b2, a1 and b1 symmetry, and
216
1
2 (ZL - ZR)
is a rotation (about the Y-axis in the figure shown above) of a2 symmetry. This rotation
vector can be generated by applying the a2 projection operator to ZL or to ZR. The other
two rotations are of b1 and b2 symmetry and involve spinning of the molecule about the
X- and Z- axes of the Fig. B.39, respectively.
So, of the 9 Cartesian displacements, 3 are of a1 symmetry, 3 of b2 , 2 of b1, and 1
of a2. Of these, there are three translations (a1, b2, and b1) and three rotations (b2, b1, and
a2). This leaves two vibrations of a1 and one of b2 symmetry. For the H2O example
treated here, the three non zero eigenvalues of the mass-weighted Hessian are therefore of
a1 b2 , and a1 symmetry. They describe the symmetric and asymmetric stretch vibrations
and the bending mode, respectively as illustrated in Fig. 3.4.
H
O
H H
O
H H
O
H
Figure 3.4. Symmetric and asymmetric stretch modes and bending mode of water
The method of vibrational analysis presented here can work for any polyatomic
molecule. One knows the mass-weighted Hessian and then computes the non-zero
217
eigenvalues, which then provide the squares of the normal mode vibrational frequencies.
Point group symmetry can be used to block diagonalize this Hessian and to label the
vibrational modes according to symmetry as we show in Fig. 3.5 for the CF4 molecule in
tetrahedral symmetry.
Figure 3.5. Symmetries of vibrations of methane
Chapter 4. Some Important Tools of Theory
218
For all but the most elementary problems, many of which serve as fundamental
approximations to the real behavior of molecules (e.g., the Hydrogenic atom, the
harmonic oscillator, the rigid rotor, particles in boxes), the Schr?dinger equation can not
be solved exactly. It is therefore extremely useful to have tools that allow one to
approach these insoluble problems by solving other Schr?dinger equations that can be
trusted to reasonably describe the solutions of the impossible problem. The tools
discussed in this Chapter are the most important tools of this type.
I. Perturbation Theory and the Variational Method
In most practical applications of quantum mechanics to molecular problems, one
is faced with the harsh reality that the Schr?dinger equation pertinent to the problem at
hand can not be solved exactly. To illustrate how desperate this situation is, I note that
neither of the following two Schr?dinger equations have ever been solved exactly
(meaning analytically):
1. The Schr?dinger equation for the two electrons moving about the He nucleus:
[- h2/2me ?l2 - h2/2me ?22 – 2e2/r1 – 2e2/r2 + e2/r1,2]ψ = E ψ,
2. The Schr?dinger equation for the two electrons moving in an H2 molecule even if the
locations of the two nuclei (labeled A and B) are held clamped:
[- h2/2me ?l2 - h2/2me ?22 – e2/r1,A – e2/r2,A – e2/r1,B – e2/r2,B + e2/r1,2]ψ = E ψ.
219
These two problems are examples of what is called the “three-body problem” meaning
solving for the behavior of three bodies moving relative to one another. Motions of the
sun, earth, and moon (even neglecting all the other planets and their moons) constitute
another three-body problem. None of these problems, even the classical Newton’s
equation for the sun, earth, and moon, have ever been solved exactly. So, what does one
do when faced with trying to study real molecules using quantum mechanics?
There are two very powerful tools that one can use to “sneak up” on the solutions
to the desired equations by first solving an easier “model” problem and then using the
solutions to this problem to approximate the solutions to the real Schr?dinger problem of
interest. For example, to solve for the energies and wave functions of a boron atom, one
could use hydrogenic 1s orbitals (but with Z = 5) and hydrogenic 2s and 2p orbitals with
Z = 3 to account for the screening of the full nuclear charge by the two 1s electrons as a
starting point. To solve for the vibrational energies of a diatomic molecule whose energy
vs. bond length E(R) is known, one could use the Morse oscillator wave functions as
starting points. But, once one has decided on a reasonable “starting point” model to use,
how does one connect this model to the real system of interest? Perturbation theory and
the variational method are the two tools that are most commonly used for this purpose.
A. Perturbation Theory
In this method, one has available a set of equations for generating a sequence of
approximations to the true energy E and true wave function ψ. I will now briefly outline
220
the derivation of these working equations for you. First, one decomposes the true
Hamiltonian H into a so-called zeroth-order part H0 (this is the Hamiltonian of the model
problem one has chosen to use to represent the real system) and the difference (H-H0)
which is called the perturbation and often denoted V:
H = H0 + V.
The fundamental assumption of perturbation theory is that the wave functions and
energies can be expanded in a Taylor series involving various powers of the perturbation.
That is, one expands the energy E and the wave function ψ into zeroth-, first-, second,
etc, order pieces which form the unknowns in this method:
E = E0 + E1 +E2 + E3 + ...
ψ = ψ0 + ψ1 + ψ2 + ψ3 + ...
Next, one substitutes these expansions of E of H and of ψ into Hψ = Eψ. This produces
one equation whose right and left hand sides both contain terms of various “powers” in
the perturbation. For example, terms of the form E1 ψ2 and V ψ2 and E0 ψ3 are all of third
power (also called third order). Next, one equates the terms on the left and right sides that
are of the same order. This produces a set of equations, each containing all the terms of a
given order. The zeroth, first, and second order such equations are given below:
221
H0 ψ0 = E0 ψ0,
H0 ψ1 + V ψ0 = E0 ψ1 + E1 ψ0
H0 ψ2 + V ψ1 = E0 ψ2 + E1 ψ1 + E2 ψ0.
The zeroth order equation simply instructs us to solve the zeroth-order Schr?dinger
equation to obtain the zeroth-order wave function ψ0 and its zeroth-order energy E0.
In the first-order equation, the unknowns are ψ1 and E1 (recall that V is assumed to be
known because it is the difference between the Hamiltonian one wants to solve and the
model Hamiltonian H0).
To solve the first-order and higher-order equations, one expands each of the
corrections to the wave function ψK in terms of the complete set of wave functions of the
zeroth-order problem {ψ0J}. This means that one must be able to solve H0 ψ0J = E0J ψ0J not
just for the zeroth-order state one is interested in (denoted ψ0 above) but for all of the
other (e.g., excited states if ψ0 is the ground state) zeroth-order states {ψ0J}. For example,
expanding ψ1 in this manner gives:
ψ1 = ΣJ C1J ψ0J.
Now, the unknowns in the first-order equation become E1 and the C1J expansion
coefficients. Substituting this expansion into H0 ψ1 + V ψ0 = E0 ψ1 + E1 ψ0 and solving for
these unknows produces the following final first-order working equations:
222
E1 = <ψ0 | V | ψ0>
ψ1 = ΣJ ψ0J {<ψ0 | V | ψ0J>/(E0 – E0J)},
where the index J is restricted such that ψJ0 not equal the state ψ0 you are interested in.
These are the fundamental working equations of first-order perturbation theory. They
instruct us to compute the average value of the perturbation taken over a probability
distrubution equal to ψ0* ψ0 to obtain the first-order correction to the energy E1. They also
tell us how to compute the first-order correction to the wave function in terms of
coefficients multiplying various other zeroth-order wave functions ψJ0.
An analogous approach is used to solve the second- and higher-order equations.
Although modern quantum mechanics does indeed use high-order perturbation theory in
some cases, much of what the student needs to know is contained in the first- and second-
order results to which I will therefore restrict our attention. The expression for the
second- order energy correction is found to be:
E2 = ΣJ <ψ0 | V | ψ0J>2/(E0 – E0J),
where again, the index J is restricted as noted above. Let’s now consider an example
problem that illustrates how perturbation theory is used.
Example Problem for Perturbation Theory
223
As we discussed earlier, an electron moving in a conjugated bond framework can
be modeled as a particle in a box. An externally applied electric field of strength ε
interacts with the electron in a fashion that can described by adding the perturbation V =
eε??
?
??
?
x - L2 to the zeroth-order Hamiltonian. Here, x is the position of the electron in the
box, e is the electron's charge, and L is the length of the box.
First, we will compute the first order correction to the energy of the n=1 state and
the first order wave function for the n=1 state. In the wave function calculation, we will
only compute the contribution to ψ made by ψ02 (this is just an approximation to keep
things simple in this example) . Let me now do all the steps needed to solve this part of
the problem. Try to make sure you can do the algebra but also make sure you understand
how we are using the first-order perturbation equations.
V = eε??
?
??
?
x - L2 , Ψ(0)n = ??
?
??
?2
L
1
2 Sin??? ???npix
L , and
E(0)n = h
?2pi2n2
2mL2 .
E (1)n=1 = < >Ψ (0)n=1|V|Ψ (0)n=1 = < >Ψ (0)n=1|eε?? ??x - L2 |Ψ (0)n=1
= ??
?
??
?2
L ?
?
0
L
Sin2?? ??pixL eε?? ??x - L2 dx
224
= ??
?
??
?2eε
L ?�
?
0
L
Sin2??
?
??
?pix
L xdx - ??
?
??
?2eε
L
L
2?�
?
0
L
Sin2??
?
??
?pix
L dx
The first integral can be evaluated using the following identity with a = piL :
??
0
L
Sin2( )ax xdx = x
2
4 -
x Sin(2ax)
4a -
Cos(2ax)
8a2 ??
?L
0 =
L2
4
The second integral can be evaluated using the following identity with θ = pixL
and dθ = piL dx :
?�
?
0
L
Sin2??
?
??
?pix
L dx =
L
pi??0
pi
Sin2θdθ
??
0
pi
Sin2θdθ = -14 Sin(2θ) + θ2 ??
?pi
0 =
pi
2 .
Making all of these appropriate substitutions we obtain:
E (1)n=1 = ??
?
??
?2eε
L ??
?
??
?L2
4 -
L
2
L
pi
pi
2 = 0
225
Ψ (1)n=1 =
< >Ψ (0)n=2|eε?? ??x - L2 |Ψ (0)n=1 Ψ (0)n=2
E (0)n=1 - E (0)n=2
Ψ (1)n=1 =
??
?
??
?2
L ?�
?
0
L
Sin??
?
??
?2pix
L eε??
?
??
?
x - L2 Sin??
?
??
?pix
L dx
h?2pi2
2mL2( )12 - 22
??
?
??
?2
L
1
2 Sin??? ???2pix
L
The two integrals in the numerator need to be evaluated:
?�
?
0
L
xSin??
?
??
?2pix
L Sin??
?
??
?pix
L dx , and ?�
?
0
L
Sin??
?
??
?2pix
L Sin??
?
??
?pix
L dx .
Using the integral identities ??xCos(ax)dx = 1a2 Cos(ax) + xa Sin(ax), and ??Cos(ax)dx
= 1a Sin(ax), we obtain the following:
?�
?
0
L
Sin??
?
??
?2pix
L Sin??
?
??
?pix
L dx =
1
2??
??
??
??
?�
?
0
L
Cos??
?
??
?pix
L dx - ?�
?
0
L
Cos??
?
??
?3pix
L dx
= 12??
?
??
?L
piSin??
?
??
?pix
L ??
?L
0 -
L
3piSin??
?
??
?3pix
L ??
?L
0 = 0
226
?�
?
0
L
xSin??
?
??
?2pix
L Sin??
?
??
?pix
L dx =
1
2??
??
??
??
?�
?
0
L
xCos??
?
??
?pix
L dx - ?�
?
0
L
xCos??
?
??
?3pix
L dx
= 12??
?
??
?
??
?
??
?L2
pi2Cos??
?
??
?pix
L +
Lx
pi Sin??
?
??
?pix
L ??
?L
0 - ?
?? ???L29pi2Cos??? ???3pixL + Lx3piSin??? ???3pixL
??
?L
0
= -2L
2
2pi2 -
-2L2
18pi2 =
L2
9pi2 -
L2
pi2 = -
8L2
9pi2
Making all of these appropriate substitutions we obtain:
Ψ (1)n=1 =
??
?
??
?2
L (eε)??
?
??
?
-8L
2
9pi2 -
L
2(0)
-3h?2pi2
2mL2
??
?
??
?2
L
1
2 Sin??? ???2pix
L
Ψ (1)n=1 = 32mL
3eε
27h?2pi4 ??
?
??
?2
L
1
2 Sin??? ???2pix
L
Now, let’s compute the induced dipole moment caused by the polarization of the electron
density due to the electric field effect using the equation μinduced = - e?�
?
Ψ*??
?
??
?
x - L2 Ψdx
with Ψ now being the sum of our zeroth- and first-order wave functions. In computing
this integral, we neglect the term proportional to ε2 because we are interested in only the
term linear in ε because this is what gives the dipole moment. Again, allow me to do the
algebra and see if you can follow.
227
μinduced = - e?�
?
Ψ*??
?
??
?
x - L2 Ψdx , where, Ψ = ?? ??Ψ(0)1 + Ψ(1)1 .
μinduced = - e?�
?
0
L
?? ??Ψ(0)1 + Ψ(1)1 *??? ???x - L2 ?? ??Ψ(0)1 + Ψ(1)1 dx
= -e?�
?
0
L
Ψ(0)1 *??
?
??
?
x - L2 Ψ(0)1 dx - e?�
?
0
L
Ψ(0)1 *??
?
??
?
x - L2 Ψ(1)1 dx
- e?�
?
0
L
Ψ(1)1 *??
?
??
?
x - L2 Ψ(0)1 dx - e?�
?
0
L
Ψ(1)1 *??
?
??
?
x - L2 Ψ(1)1 dx
The first integral is zero (see the evaluation of this integral for E(1)1 above). The fourth
integral is neglected since it is proportional to ε2. The second and third integrals are the
same and are combined to give:
μinduced = -2e?�
?
0
L
Ψ(0)1 *??
?
??
?
x - L2 Ψ(1)1 dx
Substituting Ψ(0)1 = ??
?
??
?2
L
1
2 Sin??? ???pix
L and Ψ
(1)
1 =
32mL3eε
27h?2pi4 ??
?
??
?2
L
1
2 Sin??? ???2pix
L , we obtain:
μinduced = -2e32mL
3eε
27h?2pi4 ??
?
??
?2
L ?�
?
0
L
Sin??
?
??
?pix
L ??
?
??
?
x - L2 Sin??
?
??
?2pix
L dx
228
These integrals are familiar from what we did to compute Ψ1; doing them we finally
obtain:
μinduced = -2e32mL
3eε
27h?2pi4 ??
?
??
?2
L ??
?
??
?
-8L
2
9pi2
μinduced = mL
4e2ε
h?2pi6
210
35
Now. Let’s compute the polarizability, α, of the electron in the n=1 state of the
box, and try to undestand physically why α should depend as it does upon the length of
the box L. To compute the polarizability, we need to know that α = ?μ?ε ??
?
ε=0
.Using our
induced moment result above, we then find
α = ??
?
??
??μ
?ε ε=0 =
mL4e2
h?2pi6
210
35
Notice that this finding suggests that the larger the box (molecule), the more polarizable
the electron density. This result also suggests that the polarizability of conjugated
polyenes should vary non-linearly with the length of the congugated chain.
B. The Variational Method
Let us now turn to the other method that is used to solve Schr?dinger equations
229
approximately, the variational method. In this approach, one must again have some
reasonable wave function ψ0 that is used to approximate the true wave function. Within
this approximate wave function, one imbeds one or more variables {αJ} that one
subsequently varies to achieve a minimum in the energy of ψ0 computed as an
expectation value of the true Hamiltonian H:
E({αJ}) = <ψ0| H | ψ0>/<ψ0 | ψ0>.
The optimal values of the αJ parameters are determined by making
dE/dαJ = 0
To achieve the desired energy minimum (n.b., we also should verify that the second
derivative matrix (?2E/?αJ?αL) has all positive eigenvalues).
The theoretical basis underlying the variational method can be understood through
the following derivation. Suppose the someone knew the exact eigenstates (i.e., true ΨK
and true EK) of the true Hamiltonian H. These states obey
H ΨK = EK ΨK.
Because these true states form a complete set (it can be shown that the eigenfunctions of
230
all the Hamiltonian operators we ever encounter have this property), our so-called “trial
wave function” ψ0 can, in principle, be expanded in terms of these ΨK:
ψ0 = ΣK CK ΨK.
Before proceeding further, allow me to overcome one likely misconception. What I am
going through now is only a derivation of the working formula of the variational method.
The final formula will not require us to ever know the exact ΨK or the exact EK, but we
are allowed to use them as tools in our derivation because we know they exist even if we
never know them.
With the above expansion of our trial function in terms of the exact
eigenfunctions, let us now substitute this into the quantity <ψ0| H | ψ0>/<ψ0 | ψ0> that the
varitational method instructs us to compute:
E = <ψ0| H | ψ0>/<ψ0 | ψ0> = <ΣK CK ΨK | H | ΣL CL ΨL>/<ΣK CK ΨK|ΣL CL ΨL>.
Using the fact that the ΨK obey HΨK = EKΨK and that the ΨK are orthonormal (I hope you
remember this property of solutions to all Schr?dinger equations that we discussed
earlier)
<ΨK|ΨL> = δK.L
231
the above expression reduces to
E = ΣΚ < CK ΨK | H | CK ΨK>/(ΣK< CK ΨK| CK ΨK>) = ΣK |CK|2 EK/ΣK|CK|2.
One of the basic properties of the kind of Hamiltonia we encounter is that they have a
lowest-energy state. Sometimes we say they are bounded from below, which means their
energy states do not continue all the way to minus infinity. There are systems for which
this is not the case, but we will now assume that we are not dealing with such systems.
This allows us to introduce the inequality EK ≥ E0 which says that all of the energies are
higher than or equal to the energy of the lowest state which we denote E0. Introducing
this inequality into the above expression gives
E ≥ ΣK |CK|2 E0 /ΣK|CK|2 = E0.
This means that the variational energy, computed as <ψ0| H | ψ0>/<ψ0 | ψ0> will lie above
the true ground-state energy no matter what trial function ψ0 we use.
The significance of the above result that E ≥ E0 is as follows. We are allowed to
imbed into our trial wave function ψ0 parameters that we can vary to make E, computed
as <ψ0| H | ψ0>/<ψ0 | ψ0> as low as possible because we know that we can never make
<ψ0| H | ψ0>/<ψ0 | ψ0> lower than the true ground-state energy. The philosophy then is to
vary the parameters in ψ0 to render E as low as possible, because the closer E is to E0 the
232
“better” is our variational wave function. Let me now demonstrate how the variational
method is used in such a manner by solving an example problem.
Example Variational Problem
Suppose you are given a trial wave function of the form:
φ = Ze
3
pia03 exp??
?
??
?-Zer1
a0 exp??
?
??
?-Zer2
a0
to represent a two-electron ion of nuclear charge Z and suppose that you are lucky
enough that I have already evaluated the <ψ0| H | ψ0>/<ψ0 | ψ0> integral, which I’ll call
W, for you and found
W = ??
?
??
?
Ze2 - 2ZZe + 58 Ze e
2
a0 .
Now, let’s find the optimum value of the variational parameter Ze for an arbitrary nuclear
charge Z by setting dWdZe = 0 . After finding the optimal value of Ze, we’ll then find the
optimal energy by plugging this Ze into the above W expression. I’ll do the algebra and
see if you can follow.
W = ??
?
??
?
Ze2 - 2ZZe + 58 Ze e
2
a0
233
dW
dZe = ??
?
??
?
2Ze - 2Z + 58 e
2
a0 = 0
2Ze - 2Z + 58 = 0
2Ze = 2Z - 58
Ze = Z - 516 = Z - 0.3125
(n.b., 0.3125 represents the shielding factor of one 1s electron to the other).
Now, using this optimal Ze in our energy expression gives
W = Ze??
?
??
?
Ze - 2Z + 58 e
2
a0
W = ??
?
??
?
Z - 516 ??
?
??
?
??
?
??
?
Z - 516 - 2Z + 58 e
2
a0
W = ??
?
??
?
Z - 516 ??
?
??
?
-Z + 516 e
2
a0
W = -??
?
??
?
Z - 516 ??
?
??
?
Z - 516 e
2
a0 = -??
?
??
?
Z - 516 2 e
2
a0
= - (Z - 0.3125)2(27.21) eV
(n.b., since a0 is the Bohr radius 0.529 ?, e2/a0 = 27.21 eV, or one atomic unit of energy).
Is this energy “any good”? The total energies of some two-electron atoms and ions have
been experimentally determined to be:
234
Z = 1 H- -14.35 eV
Z = 2 He -78.98 eV
Z = 3 Li+ -198.02 eV
Z = 4 Be+2 -371.5 eV
Z = 5 B+3 -599.3 eV
Z = 6 C+4 -881.6 eV
Z = 7 N+5 -1218.3 eV
Z = 8 O+6 -1609.5 eV
Using our optimized expression for W, let’s now calculate the estimated total energies of
each of these atoms and ions as well as the percent error in our estimate for each ion.
Z Atom Experimental Calculated % Error
Z = 1 H- -14.35 eV -12.86 eV 10.38%
235
Z = 2 He -78.98 eV -77.46 eV 1.92%
Z = 3 Li+ -198.02 eV -196.46 eV 0.79%
Z = 4 Be+2 -371.5 eV -369.86 eV 0.44%
Z = 5 B+3 -599.3 eV -597.66 eV 0.27%
Z = 6 C+4 -881.6 eV -879.86 eV 0.19%
Z = 7 N+5 -1218.3 eV -1216.48 eV 0.15%
Z = 8 O+6 -1609.5 eV -1607.46 eV 0.13%
The energy errors are essentially constant over the range of Z, but produce a larger
percentage error at small Z.
In 1928, when quantum mechanics was quite young, it was not known whether
the isolated, gas-phase hydride ion, H-, was stable with respect to loss of an electrn ot
form a hydrogen atom. Let’s compare our estimated total energy for H- to the ground
state energy of a hydrogen atom and an isolated electron (which is known to be
-13.60 eV). When we use our expression for W and take Z = 1, we obtain W = -12.86 eV,
which is greater than -13.6 eV (H + e-), so this simple variational calculation erroneously
predicts H- to be unstable. More complicated variational treatments give a ground state
energy of H- of -14.35 eV, in agreement with experiment.
II. Point Group Symmetry
236
It is assumed that the reader has previously learned, in undergraduate inorganic or
physical chemistry classes, how symmetry arises in molecular shapes and structures and
what symmetry elements are (e.g., planes, axes of rotation, centers of inversion, etc.). For
the reader who feels, after reading this appendix, that additional background is needed,
the text by Eyring, Walter, and Kimball or by Atkins and Friedman can be consulted. We
review and teach here only that material that is of direct application to symmetry analysis
of molecular orbitals and vibrations and rotations of molecules. We use a specific
example, the ammonia molecule, to introduce and illustrate the important aspects of point
group symmetry.
A. The C3v Symmetry Group of Ammonia - An Example
The ammonia molecule NH3 belongs, in its ground-state equilibrium geometry, to
the C3v point group. Its symmetry operations consist of two C3 rotations, C3, C32
(rotations by 120° and 240°, respectively about an axis passing through the nitrogen atom
and lying perpendicular to the plane formed by the three hydrogen atoms), three vertical
reflections, σv, σv', σv", and the identity operation. Corresponding to these six operations
are symmetry elements: the three-fold rotation axis, C3 and the three symmetry planes
σv, σv' and σv" that contain the three NH bonds and the z-axis (see Fig. 4.1).
237
N
σ
v'
σ
v
σ
xz
is σ
v'' H3
H
2
H
1
C
3
axis (z)
x- axis
y- axis
Figure 4.1 Ammonia Molecule and its Symmetry Elements
These six symmetry operations form a mathematical group. A group is defined as
a set of objects satisfying four properties.
1. A combination rule is defined through which two group elements are combined to
give a result which we call the product. The product of two elements in the group
must also be a member of the group (i.e., the group is closed under the
combination rule).
2. One special member of the group, when combined with any other member of the
group, must leave the group member unchanged (i.e., the group contains an
identity element).
238
3. Every group member must have a reciprocal in the group. When any group
member is combined with its reciprocal, the product is the identity element.
4. The associative law must hold when combining three group members (i.e., (AB)C
must equal A(BC)).
The members of symmetry groups are symmetry operations; the combination rule
is successive operation. The identity element is the operation of doing nothing at all.
The group properties can be demonstrated by forming a multiplication table. Let us label
the rows of the table by the first operation and the columns by the second operation.
Note that this order is important because most groups are not commutative. The C3v
group multiplication table is as follows:
E C3 C32 σv σv' σv" second operation
E E C3 C32 σv σv' σv"
C3 C3 C32 E σv' σv" σv
C32 C32 E C3 σv" σv σv'
σv σv σv" σv' E C32 C3
σv' σv' σv σv" C3 E C32
σv" σv" σv' σv C32 C3 E
First
operation
239
Note the reflection plane labels do not move. That is, although we start with H1 in the σv
plane, H2 in σv'', and H3 in σv", if H1 moves due to the first symmetry operation, σv
remains fixed and a different H atom lies in the σv plane.
B. Matrices as Group Representations
In using symmetry to help simplify molecular orbital (mo) or vibration/rotation
energy level identifications, the following strategy is followed:
1. A set of M objects belonging to the constituent atoms (or molecular fragments, in a
more general case) is introduced. These objects are the orbitals of the individual atoms
(or of the fragments) in the m.o. case; they are unit vectors along the x, y, and z directions
located on each of the atoms, and representing displacements along each of these
directions, in the vibration/rotation case.
2. Symmetry tools are used to combine these M objects into M new objects each of which
belongs to a specific symmetry of the point group. Because the Hamiltonian (electronic in
the m.o. case and vibration/rotation in the latter case) commutes with the symmetry
operations of the point group, the matrix representation of H within the symmetry
adapted basis will be "block diagonal". That is, objects of different symmetry will not
interact; only interactions among those of the same symmetry need be considered.
To illustrate such symmetry adaptation, consider symmetry adapting the 2s orbital
of N and the three 1s orbitals of the three H atoms. We begin by determining how these
240
orbitals transform under the symmetry operations of the C3v point group. The act of each
of the six symmetry operations on the four atomic orbitals can be denoted as follows:
(SN,S1,S2,S3) →E (SN,S1,S2,S3)
→C3 (SN,S3,S1,S2)
→C3
2
(SN,S2,S3,S1)
→σv (SN,S1,S3,S2)
→σv" (SN,S3,S2,S1)
→σv' (SN,S2,S1,S3)
Here we are using the active view that a C3 rotation rotates the molecule by 120°. The
equivalent passive view is that the 1s basis functions are rotated -120°. In the C3
rotation, S3 ends up where S1 began, S1, ends up where S2 began and S2 ends up where
S3 began.
These transformations can be thought of in terms of a matrix multiplying a
vector with elements (SN,S1,S2,S3). For example, if D(4) (C3) is the representation
matrix giving the C3 transformation, then the above action of C3 on the four basis orbitals
can be expressed as:
241
D(4)(C3)
??
??
??
??
SN
S1
S2
S3
=
??
??
??
??1 0 0 00 0 0 1
0 1 0 0
0 0 1 0
??
??
??
??
SN
S1
S2
S3
=
??
??
??
??
SN
S3
S1
S2
We can likewise write matrix representations for each of the symmetry operations of the
C3v point group:
D(4)(C32) =
??
??
??
??1 0 0 00 0 1 0
0 0 0 1
0 1 0 0
, D(4)(E) =
??
??
??
??1 0 0 00 1 0 0
0 0 1 0
0 0 0 1
D(4)(σv) =
??
??
??
??1 0 0 00 1 0 0
0 0 0 1
0 0 1 0
, D(4)(σv') =
??
??
??
??1 0 0 00 0 0 1
0 0 1 0
0 1 0 0
D(4)(σv") =
??
??
??
??1 0 0 00 0 1 0
0 1 0 0
0 0 0 1
242
It is easy to verify that a C3 rotation followed by a σv reflection is equivalent to a σv'
reflection alone. In other words
σv C3 = σv' , or,
S1
S2 S3
→C3
S3
S1 S2
→σv
S3
S2 S1
Note that this same relationship is carried by the matrices:
D(4)(σv) D(4)(C3) = ??
?
?
?
?1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0
??
?
?
?
?1 0 0 0
0 0 0 1
0 1 0 0
0 0 1 0
= ??
?
?
?
?1 0 0 0
0 0 0 1
0 0 1 0
0 1 0 0
= D(4)(σv')
Likewise we can verify that C3 σv = σv" directly and we can notice that the matrices also
show the same identity:
D(4)(C3) D(4)(σv) = ??
?
?
?
?1 0 0 0
0 0 0 1
0 1 0 0
0 0 1 0
??
?
?
?
?1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0
= ??
?
?
?
?1 0 0 0
0 0 1 0
0 1 0 0
0 0 0 1
= D(4)(σv").
In fact, one finds that the six matrices, D(4)(R), when multiplied together in all 36
possible ways obey the same multiplication table as did the six symmetry operations. We
243
say the matrices form a representation of the group because the matrices have all the
properties of the group.
1. Characters of Representations
One important property of a matrix is the sum of its diagonal elements which is
called the trace of the matrix D and is denoted Tr(D):
Tr(D) = ∑
i
Dii = χ.
So, χ is called the trace or character of the matrix. In the above example
χ(E) = 4
χ(C3) = χ(C32) = 1
χ(σv) = χ(σv') = χ(σv") = 2.
The importance of the characters of the symmetry operations lies in the fact that they do
not depend on the specific basis used to form them. That is, they are invariant to a unitary
or orthorgonal transformation of the objects used to define the matrices. As a result, they
contain information about the symmetry operation itself and about the space spanned by
the set of objects. The significance of this observation for our symmetry adaptation
process will become clear later.
244
Note that the characters of both rotations are the same as are those of all three
reflections. Collections of operations having identical characters are called classes. Each
operation in a class of operations has the same character as other members of the class.
The character of a class depends on the space spanned by the basis of functions on which
the symmetry operations act.
2. Another Basis and Another Representation
Above we used (SN,S1,S2,S3) as a basis. If, alternatively, we use the one-
dimensional basis consisting of the 1s orbital on the N-atom, we obtain different
characters, as we now demonstrate.
The act of the six symmetry operations on this SN can be represented as follows:
SN →E SN; SN →C3 SN; SN →C3
2
SN;
SN →σv SN; SN →σv' SN; SN →σv" SN.
We can represent this group of operations in this basis by the one-dimensional set of
matrices:
D(1) (E) = 1; D(1) (C3) = 1; D(1) (C32) = 1,
D(1) (σv) = 1; D(1)(σv") = 1; D(1) (σv') = 1.
245
Again we have
D(1) (σv) D(1) (C3) = 1 ? 1 = D(1) (σv"), and
D(1) (C3) D(1) (σv) = 1 ? 1 = D(1) (σv').
These six matrices form another representation of the group. In this basis, each character
is equal to unity. The representation formed by allowing the six symmetry operations to
act on the 1s N-atom orbital is clearly not the same as that formed when the same six
operations acted on the (SN,S1,S2,S3) basis. We now need to learn how to further analyze
the information content of a specific representation of the group formed when the
symmetry operations act on any specific set of objects.
C. Reducible and Irreducible Representations
1. A Reducible Representation
Note that every matrix in the four dimensional group representation labeled D(4)
has the so-called block diagonal form
246
1 0 0 0
0
0 3 x 3 matrix
0
This means that these D(4) matrices are really a combination of two separate group
representations (mathematically, it is called a direct sum representation). We say that D(4)
is reducible into a one-dimensional representation D(1) and a three-dimensional
representation formed by the 3x3 submatrices that we will call D(3).
D(3)(E) = ??
?
??
?1 0 0
0 1 0
0 0 1
; D(3)(C3) = ??
?
??
?0 0 1
1 0 0
0 1 0
; D(3)(C32) = ??
?
??
?0 1 0
0 0 1
1 0 0
D(3)(σv) = ??
?
??
?1 0 0
0 0 1
0 1 0
; D(3)(σv') = ??
?
??
?0 0 1
0 1 0
1 0 0
; D(3)(σv") = ??
?
??
?0 1 0
1 0 0
0 0 1
247
The characters of D(3) are χ(E) = 3, χ(2C3) = 0, χ(3σv) = 1. Note that we would have
obtained this D(3) representation directly if we had originally chosen to examine the basis
(S1,S2,S3); also note that these characters are equal to those of D(4) minus those of D(1).
2. A Change in Basis
Now let us convert to a new basis that is a linear combination of the original
S1,S2,S3 basis:
T1 = S1 + S2 + S3
T2 = 2S1 - S2 - S3
T3 = S2 - S3
(Don't worry about how we construct T1, T2, and T3 yet. As will be demonstrated later,
we form them by using symmetry projection operators defined below). We determine
how the "T" basis functions behave under the group operations by allowing the
operations to act on the Sj and interpreting the results in terms of the Ti. In particular,
(T1,T2 ,T3) →σv (T1,T2,-T3); (T1,T2,T3) →E (T1,T2,T3) ;
(T1,T2,T3) →σv' (S3+S2+S1,2S3-S2-S1,S2-S1) = (T1, -12 T2 - 32 T3, -12 T2 + 12 T3);
248
(T1,T2,T3) →σv" (S2+S1+S3,2S2-S1-S3,S1-S3) = (T1, -12 T2 + 32 T3, 12 T2 + 12 T3);
(T1,T2,T3) →C3 (S3+S1+S2,2S3-S1-S2,S1-S2) = (T1, -12 T2 - 32 T3, 12 T2 - 12 T3);
(T1,T2,T3) →C3
2
(S2+S3+S1,2S2-S3-S1,S3-S1) = (T1, -12 T2 + 32 T3, -12 T2 - 12 T3).
So the matrix representations in the new Ti basis are:
D(3)(E) = ??
??
??
??1 0 00 1 0
0 0 1
;D(3)(C3) = ??
?
?
?
?1 0 0
0 -12 -32
0 +12 -12
;
D(3)(C32) = ??
?
?
?
?1 0 0
0 -12 +32
0 -12 -12
; D(3)(σv) = ??
??
??
??1 0 00 1 0
0 0 -1
;
D(3)(σv') = ??
?
?
?
?1 0 0
0 -12 -32
0 -12 +12
; D(3)(σv") =??
?
?
?
?1 0 0
0 -12 +32
0 +12 +12
.
249
3. Reduction of the Reducible Representation
These six matrices can be verified to multiply just as the symmetry operations
do; thus they form another three-dimensional representation of the group. We see that in
the Ti basis the matrices are block diagonal. This means that the space spanned by the Ti
functions, which is the same space as the Sj span, forms a reducible representation that
can be decomposed into a one dimensional space and a two dimensional space (via
formation of the Ti functions). Note that the characters (traces) of the matrices are not
changed by the change in bases.
The one-dimensional part of the above reducible three-dimensional
representation is seen to be the same as the totally symmetric representation we arrived at
before, D(1). The two-dimensional representation that is left can be shown to be
irreducible ; it has the following matrix representations:
D(2)(E) = ???
?
??
??1 0
0 1 ; D
(2)(C3) = ??
?
??
?-12 -32
+12 -12
; D(2)(C32) = ??
?
??
?-12 +32
-12 -12
;
D(2)(σv) = ???
?
??
??1 0
0 -1 ; D
(2)(σv') = ??
?
??
?-12 -32
-12 +12
; D(2)(σv'') = ??
?
??
?-12 -32
-12 +12
250
The characters can be obtained by summing diagonal elements:
χ(E) = 2, χ (2C3) = -1, χ (3σv) = 0.
4. Rotations as a Basis
Another one-dimensional representation of the group can be obtained by taking
rotation about the Z-axis (the C3 axis) as the object on which the symmetry operations
act:
Rz →E Rz; Rz →C3 Rz; Rz →C3
2
Rz;
Rz →σv -Rz; Rz →σv" -Rz; Rz →σv' -Rz.
In writing these relations, we use the fact that reflection reverses the sense of a rotation.
The matrix representations corresponding to this one-dimensional basis are:
D(1)(E) = 1; D(1)(C3) = 1; D(1)(C32) = 1;
D(1)(σv) = -1;D(1)(σv") = -1; D(1) (σv') = -1.
These one-dimensional matrices can be shown to multiply together just like the symmetry
operations of the C3v group. They form an irreducible representation of the group
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(because it is one-dimensional, it can not be further reduced). Note that this one-
dimensional representation is not identical to that found above for the 1s N-atom orbital,
or the T1 function.
5. Overview
We have found three distinct irreducible representations for the C3v symmetry
group; two different one-dimensional and one two dimensional representations. Are
there any more? An important theorem of group theory shows that the number of
irreducible representations of a group is equal to the number of classes. Since there are
three classes of operation (i.e., E, C3 and σv), we have found all the irreducible
representations of the C3v point group. There are no more.
The irreducible representations have standard names; the first D(1) (that arising
from the T1 and 1sN orbitals) is called A1, the D(1) arising from Rz is called A2 and D(2)
is called E (not to be confused with the identity operation E). We will see shortly where
to find and identify these names.
Thus, our original D(4) representation was a combination of two A1
representations and one E representation. We say that D(4) is a direct sum representation:
D(4) = 2A1 ⊕ E. A consequence is that the characters of the combination representation
D(4) can be obtained by adding the characters of its constituent irreducible
representations.
252
E 2C3 3σv
A1 1 1 1
A1 1 1 1
E 2 -1 0
2A1 ⊕ E 4 1 2
6. How to Decompose Reducible Representations in General
Suppose you were given only the characters (4,1,2). How can you find out
how many times A1, E, and A2 appear when you reduce D(4) to its irreducible parts?
You want to find a linear combination of the characters of A1, A2 and E that add up
(4,1,2). You can treat the characters of matrices as vectors and take the dot product of A1
with D(4)
??
?
??
?1 1 1 1 1 1
E C3 σv ?
??
?
??
??
?
??
4 E
1 C3
1
2 σv
2
2
= 4 + 1 + 1 + 2 + 2 + 2 = 12.
The vector (1,1,1,1,1,1) is not normalized; hence to obtain the component of (4,1,1,2,2,2)
along a unit vector in the (1,1,1,1,1,1) direction, one must divide by the norm of
253
(1,1,1,1,1,1); this norm is 6. The result is that the reducible representation contains 126 =
2 A1 components. Analogous projections in the E and A2 directions give components of 1
and 0, respectively. In general, to determine the number nΓ of times irreducible
representation Γ appears in the reducible representation with characters χred, one
calculates
nΓ = 1g ∑
R
χΓ(R)χred(R) ,
where g is the order of the group and χΓ(R) are the characters of the Γth irreducible
representation.
7. Commonly Used Bases
We could take any set of functions as a basis for a group representation.
Commonly used sets include: coordinates (x,y,z) located on the atoms of a polyatomic
molecule (their symmetry treatment is equivalent to that involved in treating a set of p
orbitals on the same atoms), quadratic functions such as d orbitals - xy,yz,xz,x2-y2,z2, as
well as rotations about the x, y and z axes. The transformation properties of these very
commonly used bases are listed in the character tables shown at the end of this Section.
8. Summary
254
The basic idea of symmetry analysis is that any basis of orbitals,
displacements, rotations, etc. transforms either as one of the irreducible representations or
as a direct sum (reducible) representation. Symmetry tools are used to first determine
how the basis transforms under action of the symmetry operations. They are then used to
decompose the resultant representations into their irreducible components.
D. Another Example
1. The 2p Orbitals of Nitrogen
For a function to transform according to a specific irreducible representation
means that the function, when operated upon by a point-group symmetry operator, yields
a linear combination of the functions that transform according to that irreducible
representation. For example, a 2pz orbital (z is the C3 axis of NH3) on the nitrogen atom
belongs to the A1 representation because it yields unity times itself when C3, C32, σv ,
σv',σv" or the identity operation act on it. The factor of 1 means that 2pz has A1
symmetry since the characters (the numbers listed opposite A1 and below E, 2C3, and
3σv in the C3v character table shown at the end of this Section) of all six symmetry
operations are 1 for the A1 irreducible representation.
The 2px and 2py orbitals on the nitrogen atom transform as the E representation
since C3, C32, σv, σv', σv" and the identity operation map 2px and 2py among one
another. Specifically,
255
C3 ???
?
??
??2px
2py = ??
??
??
??Cos120° -Sin120°
Sin120° Cos120° ??
??
??
??2px
2py ;
C32 ???
?
??
??2px
2py
=
??
??
??
??Cos240° -Sin240°
Sin240° Cos240° ??
??
??
??2px
2py ;
E ???
?
??
??2px
2py = ??
??
??
??1 0
0 1 ??
??
??
??2px
2py ;
σv ???
?
??
??2px
2py = ??
??
??
??-1 0
0 1 ??
??
??
??2px
2py ;
σv' ???
?
??
??2px
2py = ??
??
??
??+12 +
3
2
+ 32 -12
???
?
??
??2px
2py ;
σv" ???
?
??
??2px
2py = ??
??
??
??+12 -
3
2
- 32 -12
???
?
??
??2px
2py .
256
The 2 x 2 matrices, which indicate how each symmetry operation maps 2px and 2py into
some combinations of 2px and 2py, are the representation matrices ( D(IR)) for that
particular operation and for this particular irreducible representation (IR). For example,
??
??
??
??+12 +
3
2
+ 32 -12
= D(E)(σv')
This set of matrices have the same characters as the D(2) matrices obtained earlier when
the Ti displacement vectors were analyzed, but the individual matrix elements are
different because we used a different basis set (here 2px and 2py ; above it was T2 and
T3). This illustrates the invariance of the trace to the specific representation; the trace
only depends on the space spanned, not on the specific manner in which it is spanned.
2. A Short-Cut
A short-cut device exists for evaluating the trace of such representation
matrices (that is, for computing the characters). The diagonal elements of the
representation matrices are the projections along each orbital of the effect of the
symmetry operation acting on that orbital. For example, a diagonal element of the C3
matrix is the component of C32py along the 2py direction. More rigorously, it
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is ∫ 2py* C3 2py dτ. Thus, the character of the C3 matrix is the sum of ∫ 2py* C3 2py dτ
and ∫ 2px* C3 2px dτ. In general, the character χ of any symmetry operation S can be
computed by allowing S to operate on each orbital φi, then projecting Sφi along φi (i.e.,
forming ??φi*Sφidτ ), and summing these terms,
∑
i
??φi*Sφidτ = χ(S).
If these rules are applied to the 2px and 2py orbitals of nitrogen within the C3v
point group, one obtains
χ(E) = 2, χ(C3) = χ(C32) = -1, χ(σv) = χ(σv") = χ(σv') = 0.
This set of characters is the same as D(2) above and agrees with those of the E
representation for the C3v point group. Hence, 2px and 2py belong to or transform as the
E representation. This is why (x,y) is to the right of the row of characters for the E
representation in the C3v character table shown at the end of this Section. In similar
fashion, the C3v character table (please refer to this table now) states that dx2?y2 and dxy
orbitals on nitrogen transform as E, as do dxy and dyz, but dz2 transforms as A1.
Earlier, we considered in some detail how the three 1sH orbitals on the
hydrogen atoms transform. Repeating this analysis using the short-cut rule just
described, the traces (characters) of the 3 x 3 representation matrices are computed by
258
allowing E, 2C3, and 3σv to operate on 1sH1, 1sH2, and 1sH3 and then computing the
component of the resulting function along the original function. The resulting characters
are χ(E) = 3, χ(C3) = χ(C32) = 0, and χ(σv) = χ(σv') = χ(σv") = 1, in agreement with
what we calculated before.
Using the orthogonality of characters taken as vectors we can reduce the above
set of characters to A1 + E. Hence, we say that our orbital set of three 1sH orbitals forms
a reducible representation consisting of the sum of A1 and E IR's. This means that the
three 1sH orbitals can be combined to yield one orbital of A1 symmetry and a pair that
transform according to the E representation.
E. Projector Operators: Symmetry Adapted Linear Combinations of Atomic
Orbitals
To generate the above A1 and E symmetry-adapted orbitals, we make use of
so-called symmetry projection operators PE and PA1. These operators are given in terms
of linear combinations of products of characters times elementary symmetry operations
as follows:
PA1 = ∑
S
χA(S) S
259
PE = ∑
S
χE(S) S
where S ranges over C3, C32, σv, σv' and σv" and the identity operation. The result of
applying PA1 to say 1sH1 is
PA1 1sH1 = 1sH1 + 1sH2 + 1sH3 + 1sH2 + 1sH3 + 1sH1
= 2(1sH1 + 1sH2 + 1sH3) = φA1,
which is an (unnormalized) orbital having A1 symmetry. Clearly, this same φA1 would
be generated by PA1 acting on 1sH2 or 1sH3. Hence, only one A1 orbital exists.
Likewise,
PE1sH1 = 2 ? 1sH1 - 1sH2 - 1sH3 ≡ φE,1
which is one of the symmetry adapted orbitals having E symmetry. The other E orbital
can be obtained by allowing PE to act on 1sH2 or 1sH3:
PE1sH2 = 2 ? 1sH2 - 1sH1 - 1sH3 ≡ φE,2
260
PE1sH3 = 2 ? 1sH3 - 1sH1 - 1sH2 = φE,3 .
It might seem as though three orbitals having E symmetry were generated, but only two
of these are really independent functions. For example, φE,3 is related to φE,1 and φE,2
as follows:
φE,3 = -(φE,1 + φE,2).
Thus, only φE,1 and φE,2 are needed to span the two-dimensional space of the E
representation. If we include φE,1 in our set of orbitals and require our orbitals to be
orthogonal, then we must find numbers a and b such that φ'E = aφE,2 + bφE,3 is
orthogonal to φE,1: ??φ'E φE,1dτ = 0. A straightforward calculation gives a = -b or φ'E =
a (1sH2 - 1sH3) which agrees with what we used earlier to construct the Ti functions in
terms of the Sj functions.
F. Summary
Let us now summarize what we have learned. Any given set of atomic
orbitals {φi}, atom-centered displacements or rotations can be used as a basis for the
symmetry operations of the point group of the molecule. The characters χ(S) belonging
to the operations S of this point group within any such space can be found by summing
261
the integrals ??φi*Sφidτ over all the atomic orbitals (or corresponding unit vector
atomic displacements). The resultant characters will, in general, be reducible to a
combination of the characters of the irreducible representations χi(S). To decompose
the characters χ(S) of the reducible representation to a sum of characters χi(S) of the
irreducible representation
χ(S) = ∑
i
ni χi(S) ,
it is necessary to determine how many times, ni, the i-th irreducible representation
occurs in the reducible representation. The expression for ni is
ni = 1g ∑
S
χ(S) χi(S)
in which g is the order of the point group- the total number of symmetry operations in
the group (e.g., g = 6 for C3v).
For example, the reducible representation χ(E) = 3, χ(C3) = 0, and χ(σv) = 1
formed by the three 1sH orbitals discussed above can be decomposed as follows:
nA1 = 16 (3 ? 1 + 2 ? 0 ? 1 + 3 ? 1 ? 1) = 1,
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nA2 = 16 (3 ? 1 + 2 ? 0 ? 1 + 3 ? 1 ? (-1)) = 0,
nE = 16 (3 ? 2 + 2 ? 0 ? (-1) + 3 ? 1 ? 0) = 1.
These equations state that the three 1sH orbitals can be combined to give one A1 orbital
and, since E is degenerate, one pair of E orbitals, as established above. With
knowledge of the ni, the symmetry-adapted orbitals can be formed by allowing the
projectors
Pi = ∑
i
χi(S) S
to operate on each of the primitive atomic orbitals. How this is carried out was
illustrated for the 1sH orbitals in our earlier discussion. These tools allow a symmetry
decomposition of any set of atomic orbitals into appropriate symmetry-adapted orbitals.
Before considering other concepts and group-theoretical machinery, it should
once again be stressed that these same tools can be used in symmetry analysis of the
translational, vibrational and rotational motions of a molecule. The twelve motions of
NH3 (three translations, three rotations, six vibrations) can be described in terms of
combinations of displacements of each of the four atoms in each of three (x,y,z)
directions. Hence, unit vectors placed on each atom directed in the x, y, and z
directions form a basis for action by the operations {S} of the point group. In the case
263
of NH3, the characters of the resultant 12 x 12 representation matrices form a reducible
representation in the C2v point group: χ(E) = 12, χ(C3) = χ(C32) = 0, χ(σv) = χ(σv') =
χ (σv") = 2. For example under σv, the H2 and H3 atoms are interchanged, so unit
vectors on either one will not contribute to the trace. Unit z-vectors on N and H1
remain unchanged as well as the corresponding y-vectors. However, the x-vectors on N
and H1 are reversed in sign. The total character for σv' the H2 and H3 atoms are
interchanged, so unit vectors on either one will not contribute to the trace. Unit z-
vectors on N and H1 remain unchanged as well as the corresponding y-vectors.
However, the x-vectors on N and H1 are reversed in sign. The total character for σv is
thus 4 - 2 = 2. This representation can be decomposed as follows:
nA1 = 16 [1?1?12 + 2?1?0 + 3?1?2] = 3,
nA2 = 16 [1?1?12 + 2?1?0 + 3?(-1)?2] = 1,
nE = 16 [1?2?12 + 2?(-1)?0 + 3?0?2] = 4.
From the information on the right side of the C3v character table, translations of all four
atoms in the z, x and y directions transform as A1(z) and E(x,y), respectively, whereas
rotations about the z(Rz), x(Rx), and y(Ry) axes transform as A2 and E. Hence, of the
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twelve motions, three translations have A1 and E symmetry and three rotations have A2
and E symmetry. This leaves six vibrations, of which two have A1 symmetry, none
have A2 symmetry, and two (pairs) have E symmetry. We could obtain symmetry-
adapted vibrational and rotational bases by allowing symmetry projection operators of
the irreducible representation symmetries to operate on various elementary cartesian
(x,y,z) atomic displacement vectors.
G. Direct Product Representations
1. Direct Products in N-Electron Wave functions
We now turn to the symmetry analysis of orbital products. Such knowledge
is important because one is routinely faced with constructing symmetry-adapted N-
electron configurations that consist of products of N individual spin orbitals, one for
each electron. A point-group symmetry operator S, when acting on such a product of
orbitals, gives the product of S acting on each of the individual orbitals
S(φ1φ2φ3...φN) = (Sφ1) (Sφ2) (Sφ3) ... (SφN).
For example, reflection of an N-orbital product through the σv plane in NH3 applies the
reflection operation to all N electrons.
Just as the individual orbitals formed a basis for action of the point-group
operators, the configurations (N-orbital products) form a basis for the action of these
265
same point-group operators. Hence, the various electronic configurations can be treated
as functions on which S operates, and the machinery illustrated earlier for decomposing
orbital symmetry can then be used to carry out a symmetry analysis of configurations.
Another shortcut makes this task easier. Since the symmetry adapted
individual orbitals {φi, i = 1, ..., M} transform according to irreducible representations,
the representation matrices for the N-term products shown above consist of products of
the matrices belonging to each φi. This matrix product is not a simple product but what
is called a direct product. To compute the characters of the direct product matrices, one
multiplies the characters of the individual matrices of the irreducible representations of
the N orbitals that appear in the electron configuration. The direct-product
representation formed by the orbital products can therefore be symmetry-analyzed
(reduced) using the same tools as we used earlier.
For example, if one is interested in knowing the symmetry of an orbital
product of the form a12a22e2 (note: lower case letters are used to denote the symmetry
of electronic orbitals, whereas capital letters are reserved to label the overall
configuration’s symmetry) in C3v symmetry, the following procedure is used. For each
of the six symmetry operations in the C2v point group, the product of the characters
associated with each of the six spin orbitals (orbital multiplied by α or β spin) is formed
χ(S) = ∏
i
χi(S) = (χA1(S))2 (χA2(S))2 (χE(S))2.
266
In the specific case considered here, χ(E) = 4, χ(2C3) = 1, and χ(3σv) = 0. Notice that
the contributions of any doubly occupied nondegenerate orbitals (e.g., a12, and a22) to
these direct product characters χ(S) are unity because for all operators (χk(S))2 = 1 for
any one-dimensional irreducible representation. As a result, only the singly occupied
or degenerate orbitals need to be considered when forming the characters of the
reducible direct-product representation χ(S). For this example this means that the
direct-product characters can be determined from the characters χE(S) of the two active
(i.e., nonclosed-shell) orbitals - the e2 orbitals. That is, χ(S) = χE(S) ? χE(S).
From the direct-product characters χ(S) belonging to a particular electronic
configuration (e.g., a12a22e2), one must still decompose this list of characters into a
sum of irreducible characters. For the example at hand, the direct-product characters
χ(S) decompose into one A1, one A2, and one E representation. This means that the e2
configuration contains A1, A2, and E symmetry elements. Projection operators
analogous to those introduced earlier for orbitals can be used to form symmetry-
adapted orbital products from the individual basis orbital products of the form
a12a22exmeym' , where m and m' denote the occupation (1 or 0) of the two degenerate
orbitals ex and ey. When dealing with indistinguishable particles such as electrons, it is
also necessary to further project the resulting orbital products to make them
antisymmetric (for Fermions) or symmetric (for Bosons) with respect to interchange of
any pair of particles. This step reduces the set of N-electron states that can arise. For
example, in the above e2 configuration case, only 3A2, 1A1, and 1E states arise; the 3E,
3A1, and 1A2 possibilities disappear when the antisymmetry projector is applied. In
267
contrast, for an e1e'1 configuration, all states arise even after the wave function has
been made antisymmetric. The steps involved in combining the point group symmetry
with permutational antisymmetry are illustrated in Chapter 10 of my QMIC text. In
Appendix III of Electronic Spectra and Electronic Structure of Polyatomic Molecules ,
G. Herzberg, Van Nostrand Reinhold Co., New York, N.Y. (1966) the resolution of
direct products among various representations within many point groups are tabulated.
2. Direct Products in Selection Rules
Two states ψa and ψb that are eigenfunctions of a Hamiltonian Ho in the
absence of some external perturbation (e.g., electromagnetic field or static electric field
or potential due to surrounding ligands) can be "coupled" by the perturbation V only if
the symmetries of V and of the two wave functions obey a so-called selection rule. In
particular, only if the coupling integral
?? ψa* V ψb dτ = Va,b
is non-vanishing will the two states be coupled by V .
The role of symmetry in determining whether such integrals are non-zero can
be demonstrated by noting that the integrand, considered as a whole, must contain a
component that is invariant under all of the group operations (i.e., belongs to the totally
symmetric representation of the group) if the integral is to not vanish. In terms of the
projectors introduced above we must have
268
∑
S
χA(S) S ψa* V ψb
not vanish. Here the subscript A denotes the totally symmetric representation of
whatever point group applies. The symmetry of the product ψa* V ψb is, according to
what was covered earlier, given by the direct product of the symmetries of ψa* of V
and of ψb. So, the conclusion is that the integral will vanish unless this triple direct
product contains, when it is reduced to its irreducible components, a component of the
totally symmetric representation.
To see how this result is used, consider the integral that arises in formulating
the interaction of electromagnetic radiation with a molecule within the electric-dipole
approximation:
??ψa* r ψb dτ .
Here, r is the vector giving, together with e, the unit charge, the quantum mechanical
dipole moment operator
r = e∑
n
Zn Rn - e∑
j
rj ,
where Zn and Rn are the charge and position of the nth nucleus and rj is the position of
the jth electron. Now, consider evaluating this integral for the singlet n→pi* transition in
269
formaldehyde. Here, the closed-shell ground state is of 1A1 symmetry and the singlet
excited state, which involves promoting an electron from the non-bonding b2 lone pair
orbital on the Oxygen into the pi* b1 orbital on the CO moiety, is of 1A2 symmetry (b1x
b2 = a2). The direct product of the two wave function symmetries thus contains only a2
symmetry. The three components (x, y, and z) of the dipole operator have, respectively,
b1, b2, and a1 symmetry. Thus, the triple direct products give rise to the following
possibilities:
a2 x b1 = b2,
a2 x b2 = b1,
a2 x a1 = a2 .
There is no component of a1 symmetry in the triple direct product, so the integral
vanishes. This allows us to conclude that the n→pi* excitation in formaldehyde is
electric dipole forbidden.
H. Overview
We have shown how to make a symmetry decomposition of a basis of atomic
orbitals (or cartesian displacements or orbital products) into irreducible representation
components. This tool is very helpful when studying spectroscopy and when
270
constructing the orbital correlation diagrams that form the basis of the Woodward-
Hoffmann rules. We also learned how to form the direct-product symmetries that arise
when considering configurations consisting of products of symmetry-adapted spin
orbitals. Finally, we learned how the direct product analysis allows one to determine
whether or not integrals of products of wave functions with operators between them
vanish. This tool is of utmost importance in determining selection rules in spectroscopy
and for determining the effects of external perturbations on the states of the species under
investigation.
