Download Lecture Notes for Chemistry 543, Part III

Revised 5/2/2014
Lecture Notes for Chemistry 543, Part III
LECTURE 30: SPECTROSCOPY OF RIGID ROTORS AND HARMONIC
OSCILLATORS.
Reading: Chapters 6 and 7 of Bernath; Levine pp 142-174.
As in any branch of spectroscopy, we need to determine (1) the eigenvalues of the Hamiltonian in order to know the frequencies of the possible transitions, (2) the selection rules to
determine what transitions actually occur, (3) the transition dipole moments to determine
the strengths of the transitions, (4) degeneracies, and (5) relative populations in order to
determine the intensity of transitions. The simplest case beyond an atom is a diatomic
molecule. To the lowest order of approximation, we may treat such a molecule as a rigid
rotor and a harmonic oscillator and consider its rotational, vibrational, and electronic degrees of freedom as completely independent quantities. At the next level of approximation,
we consider the interactions of these degrees of freedom as perturbations.
We start with the simplest problem of pure rotations. The Hamiltonian for a rigid rotor
is
Ĥrot =
Jˆ2
Jˆ2
=
,
2µRe2
2I
(743)
where Jˆ is the rotational angular momentum operator, Re is the equilibrium internuclear
distance, and I is the moment of inertia. The eigenfunctions are the well known spherical
harmonics,
1
|M |
YJM (θ, ϕ) = √ PJ (cos θ)eiM ϕ ,
2π
|M |
where PJ
(744)
is a modified Legendre Polynomial. The eigenvalues are
EJM =
J(J + 1)~2
,
2I
(745)
and the degeneracy is
gJ = 2J + 1.
(746)
2
The energy in wave numbers is given by
EJ
= BJ(J + 1),
hc
(747)
where the rotational constant is
Be =
h
.
8π 2 cI
(748)
The selection rule is obtained by evaluating the matrix element of the transition dipole
operator. The integral over ϕ is elementary. For µx , which is proportional to cos ϕ,
∫ 2π
∫
1 2π i(M1 −M2 )ϕ iϕ
−iM2 ϕ
iM1 ϕ
e
cos ϕe
dϕ =
e
(e + e−iϕ )dϕ = πδM1 ,M2 ±1
2
0
0
(749)
vanishes unless
M2 − M1 = ±1.
(750)
The same selection rule results from the integral over ϕ for µy , which is proportional to sin ϕ:
∫
∫ 2π
1 2π i(M1 −M2 )ϕ iϕ
−iM2 ϕ
iM1 ϕ
e
(e − e−iϕ )dϕ.
(751)
e
sin ϕe
dϕ =
2i 0
0
In the case of µz , which does not depend on ϕ,
∫ 2π
ei(M1 −M2 )ϕ dϕ = δM1 M2 ,
(752)
0
which yields M2 = M1 . Thus the selection rule for M is therefore
∆M = 0, ±1.
(753)
For randomly aligned molecules, all three transitions are equally likely, whereas for molecules
aligned along the axes of the electric field, the various allowed transitions correspond to
different polarizations of the electric field.
The selection rule for J follows from the recursion relations for Legendre Polynomials,
]
1 [ |M |+1
|M |
|M |+1
sin θPJ (cos θ) =
PJ+1 (cos θ) − PJ−1 (cos θ)
2J + 1
]
1 [
|M |−1
|M |−1
=
(J + |M |)(J + |M | − 1)PJ−1 (cos θ) − (J − |M | + 1)(J − |M | + 2)PJ+1 (cos θ) ,
2J + 1
(754)
and the orthogonality property,
∫ 1
|M |
|M |
PJ (x)PJ ′ (x)dx =
−1
2(J + M )!
δJJ ′ .
(2J + 1)!(J − M )!
(755)
3
For µz we found that M2 = M1 . Since the θ integral is the same as Eq. (755), we find that
J2 = J1 . For a pure (i.e., non-vibrational and non-electronic) rotational transition, this is
not a transition at all. For µx and µy we found that M2 = M1 ± 1. For this case, Eqs. (754)
and (755) imply that
J2 − J1 = ∆J = ±1.
(756)
From the selection rule we deduce that the pure rotational spectrum is a series of evenly
spaced lines with frequencies
ν̃J = Be [J + 1)(J + 2) − J(J + 1)] = 2Be (J + 1)
(757)
∆ν̃ = 2Be .
(758)
and spacing of
To get the total transition probability (the “line strength”), we need to sum over all
2J + 1 values of M and all three polarization directions. The result is called the HönlLondon factor, which is a function of J (and more generally of K and Λ, which we will learn
about later). These factors are tabulated in Herzberg I, page 208 [2] and Zare, page 286 [7].
Using the same methods may be used to derive the selection rules for rotational Raman
transitions. In this case we evaluate the matrix elements of the polarizability operator,
which is proportional to products of the Cartesian coordinates.The result is:
∆J = ±2.
(759)
The Raman frequency shift for ∆J = 2 is
ν̃J = Be (J + 2)(J + 3) − Be J(J + 1) = 4Be J + 6Be = 4Be (J + 3/2),
(760)
with an equivalent expression for ∆J = −2. The spacing between lines is therefore 4Be .
Next we consider pure vibrational transitions. We will start with the case of a harmonic
oscillator. The Hamiltonian in the center of mass coordinate system is
Ĥvib = −
or
~2 d2
1 2
+
kx ,
2µ dx2 2
Ĥvib
~ d2
1 k 2
=−
+
x,
2
~ω
2µω dx
2 ~ω
(761)
(762)
4
where
x = R − Re
(763)
is the deviation of the bond length from its equilibrium value, and the vibrational frequency
in radians/s is
ω=
Defining the reciprocal scale length,
√
k/µ.
(764)
√
β, by
β 2 = µk/~2 ,
β = µω/~ = k/~ω,
and the dimensionless distance,
y=
√
βx,
(765)
(766)
the reduced Hamiltonian becomes
Ĥvib
1 d2
1
=−
+ y2.
2
~ω
2 dy
2
(767)
Note that y = 1 corresponds to the turning point of a classical oscillator having an energy
of ~ω/2. The harmonic oscillator eigenfunctions are
ψv = |v⟩ = Nv e− 2 y Hv (y),
1 2
(768)
where Hv (y) is a hermite polynomial, and
[√
Nv =
β/π
v
2 v!
]1/2
We derive the selection rule for v by evaluating the integral
∫ ∞
⟨n|ey|m⟩ = e
ψn yψm dy.
(769)
(770)
−∞
We evaluate this and any integral of a polynomial of y using the recursion relation
1
yHm = mHm−1 + Hm+1 ,
2
(771)
which is equivalent to
1 Nm
Nm
|m − 1⟩ +
|m + 1⟩
Nm−1
2 Nm+1
√
√
= m/2 |m − 1⟩ + (m + 1)/2 |m + 1⟩.
y|m⟩ = m
(772)
5
In deriving the previous equation, we used the relations
2
ey /2
Hm =
|m⟩,
Nm
(773)
Nm
1
=√ .
Nm−1
2m
(774)
⟨m|n⟩ = δmn
(775)
v2 − v1 = ∆v = ±1.
(776)
and
Applying the orthogonality condition,
we deduce that
It follows that the infrared for a pure vibrational spectrum of a harmonic oscillator consists
of a single line.
LECTURE 31: THE ANHARMONIC OSCILLATOR AND NON-RIGID ROTOR
Now let’s consider the anharmonic oscillator. We replace the parabolic potential by a
Taylor series in x = R − Re ,
1
1
1
V (R) = V (Re ) + V ′ (Re )x + V ′′ (Re )x2 + V ′′′ (Re )x3 + V (IV ) (Re )x4 + · · · .
2
3!
4!
(777)
We set V (Re ) = 0, and from the definition of Re the first derivative is also zero. We are left
with
1
V (R) = kx2 + cx3 + dx4 + · · ·
2
(778)
V (R)
1
= y 2 + c′ y 3 + d ′ y 4 + · · ·
~ω
2
(779)
or equivalently
We next use time-independent perturbation theory to calculate the effect of these terms on
the eigenvalues and eigenfunctions. To first order,
(1)
Ev
= ⟨v|c′ y 3 + d′ y 4 |v⟩ = d′ ⟨v|y 4 |v⟩.
~ω
(780)
6
The y 3 term vanishes because the integral of the product of an even and odd functions is
zero. We apply the recursion relation twice to both the bra and the ket:
√
√
v
v+1
|v − 1⟩ +
|v + 1⟩
y|v⟩ =
2
2
√
√
v
v+1
2
y |v⟩ =
y|v − 1⟩ +
y|v + 1⟩
2
2
[
[√
] √
]
√
√ √
√
v
v−1
v
v+1
v+1
v+2
=
|v − 2⟩ +
|v⟩ +
|v⟩ +
|v + 2⟩
2
2
2
2
2
2
1
1√
1√
v(v − 1) |v − 2⟩ + (2v + 1) |v⟩ +
(v + 1)(v + 2) |v + 2⟩.
=
2
2
2
(781)
We deduce that
Ev
d′
d′
= [v(v − 1) + (2v + 1)2 + (v + 1)(v + 2)] = [2v 2 + 6v + 3].
~ω
4
4
(1)
(782)
To evaluate the c′ term, we need to go to second order perturbation theory,
En(2)
=
∑
|Hmn |2
m̸=n
En − Em
(1)
(0)
(0)
∑ [c3 ⟨m|y 3 |n⟩]2
=
(m − n)~ω
m̸=n
(783)
From the recursion relation we realize that this sum has four terms,
[
]
(2)
Ev
1
′ 2 1
3
2
3
2
3
2
3
2
= (c )
⟨v|y |v + 3⟩ + ⟨v|y |v + 1⟩ − ⟨v|y |v − 1⟩ − ⟨v|y |v − 3⟩ .
(784)
~ω
3
3
∫∞
Problem 69. Evaluate the integral −∞ ϕ0 x3 ϕ3 dx, where ϕn (x) is the nth harmonic oscillator eigenfunction.
The matrix elements may be evaluated using the methods developed previously. We first
evaluate ⟨v|y 3 and project out terms with v ± 1 and v ± 3. Putting all the pieces together
we find
Ev
= G(v) = ωe (v + 1/2) − ωe xe (v + 1/2)2 + ωe ye (v + 1/2)3 + ωe ze (v + 1/2)4 · · ·
hc
(785)
Terms in ωe ye , ωe ze etc. require going to higher order perturbation theory. The quantity
G(v) is called the term value, where ωe = ν̃ is the frequency in wave numbers measured with
respect to the minimum of the potential energy. Note the minus sign before the ωe xe term.
The zero point energy is given by
1
1
1
= G(0) = ωe − ωe xe + ωe ye .
2
4
8
(786)
7
We may also define the energy with respect to the ground state (v = 0),
G(v) = ω0 v − ω0 x0 v 2 + ω0 y0 v 3 + · · ·
(787)
Until now we were interested only in the shifts of the energy levels produced by an
anharmonic potential. Our larger goal is to determine which transitions are allowed and
to calculate the transition strengths. Do do this, we use first order perturbation theory to
calculate the change in the wave function caused by the lowest order perturbation, namely
c′ y 3 .
|n⟩(1) = |n⟩(0) +
= |n⟩(0) + c′
[1
∑
(1)
|Hmn |
(0)
m̸=n En
−
(0)
Em
|m⟩(0)
⟨n|y 3 |n + 3⟩|n + 3⟩(0) + ⟨n|y 3 |n + 1⟩|n + 1⟩(0)
]
1
− ⟨n|y 3 |n − 1⟩|n − 1⟩(0) − ⟨n|y 3 |n − 3⟩|n − 3⟩(0)
3
3
(788)
The superscript (0) is there to remind you that the expansion is in the unperturbed basis
set.
Once we know what the wave function looks like, we may calculate the transition dipole
moments for different final states: 0 → 1 transitions are called fundamentals; 0 → 2 terms are
called overtones or second harmonics, etc. The calculation of µ12 using the perturbed wave
function also involves use of the recursion relation. For example, |4⟩ contains contributions
from |1⟩(0) ,|1⟩(0) ,|3⟩(0) ,|4⟩(0) , |5⟩(0) , and |7⟩(0) . The a transition |0⟩(0) → |4⟩(0) is allowed
through the |1⟩(0) component of |4⟩.
Problem 70. Use first order perturbation theory to calculate the strength of the first
overtone of an anharmonic oscillator and compare it with the strength of the fundamental.
Express your answer in terms of the single parameter c′ . (This is a long problem and will
count double.)
From the line spacings we can determine the anharmonicity constants. Using the ωe term
8
values and neglecting cubic and higher contributions,
(
)
(
)
1
1
2
G(v) = ωe v +
− ωe xe v + v +
2
4
(
)
(
)
3
9
2
G(v + 1) = ωe v +
− ωe xe v + 3v +
2
4
∆Gv+1/2 = G(v + 1) − G(v) = ωe − 2ωe xe v − 2ωe xe
(789)
∆Gv+3/2 = G(v + 2) − G(v + 1) = ωe − 2ωe xe v − 4ωe xe
∆2 Gv+1 = ∆Gv+3/2 − ∆Gv+1/2 = −2ωe xe
Higher order anharmonicity constants may be obtained by looking at higher overtones.
Where does the anharmonic potential come from? The short answer is that it is obtained
either experimentally or by solving the Schrodinger equation for the electronic energy as a
function of R. It is also convenient to fit this potential energy curve to a model function
such as the Morse function, which has the form
(
)2
V (R) = De 1 − e−α(R−Re ) .
(790)
Here De is the dissociation energy and has nothing to do with the centrifugal distortion
constant. It is straight forward to extract the force constant and therefore ωe from a Taylor
series expansion of the potential,
V (R) = De [(1 − 1 + α(R − Re )]2 =
1
De kx2 ,
2
(791)
or by simply taking the second derivative of V (R) and evaluating it at R = Re . The result
is
k = 2α2 De .
(792)
It turns out that the Schrodinger equation can be solved exactly for this potential, yielding
eigenvalues
~ω
Ev
= v + 1/2 −
(v + 1/2)2 .
~ω
4De
(793)
Some books and data bases define the Morse potential by the equation [6]
(
)2
V = 1 − e−x ,
(794)
x = 2β(R − Re )/Re ,
(795)
where
9
and
ωe
β= √
.
4 B e De
(796)
In this expression, β is dimensionless, ωe is the vibrational frequency, Be is the rotational
constant, and De is the dissociation energy. The latter three quantities are all expressed in
wave numbers.
Problem 71. Derive Eq. (796).
Next, let’s look at vibration and rotation together. The Hamiltonian becomes
Ĥvib−rot =
~2 d2
1
Jˆ2
−
+ kx2 ,
2
2
2µ(Re + x)
2µ dx
2
(797)
plus the anharmonicity terms. We expand the rotational term in a Taylor series. Consider
the function
f (ϵ) =
1
.
(1 + ϵ)2
(798)
Its Taylor series expansion is
1
f (ϵ) = f (0) + f ′ (0)ϵ + f ′′ (0)ϵ2 + · · ·
2
2
f ′ (ϵ) = −
(1 + ϵ)3
6
f ′′ (ϵ) =
(1 + ϵ)4
(799)
f (ϵ) = 1 − 2ϵ + 3ϵ2 + · · ·
The rotational term in the Hamiltonian is therefore
)
(
Jˆ2
Jˆ2
2x 3x2
Ĥrot =
=
1−
+ 2 + ··· .
2µRe2 (1 + x/Re )2
2µRe2
Re
Re
(800)
We see that the effect of vibrational-rotational coupling is to add the term 3Be J(J +1)x2 /Re2
to the first order perturbation calculation and the term −2Be J(J + 1)x/Re to the second
order perturbation calculation. The detailed calculation is straight forward, though tedious.
The linear term makes the rotational constant depend on the vibrational quantum number,
Bv = Be − αe (v + 1/2).
(801)
This is called “vibrational-rotational” coupling. The quadratic term adds an extra term to
the rotational energy,
Erot /hc = Bv J(J + 1) − Dv J 2 (J + 1)2 .
(802)
10
This effect is called “centrifugal distortion.” The centrifugal distortion constant is weakly
dependent on the vibrational level because the average bond length increases with vibrational
energy. When this effect is ignored, the constant is written as De , which is unrelated to the
De in the Morse potential.
Transitions in which J increases by 1 are called R-branch transitions, and those in which J
decreases by 1 are called P-branch transitions. Letting ν0 indicate the change in vibrational
energy, the frequencies of these two types of transitions are given by
νR = ν0 + Bv′ (J + 1)(J + 2) − Bv′′ J(J + 1) = ν0 + 2Bv′ + (3Bv′ − Bv′′ )J + (Bv′ − Bv′′ )J 2 ; J = 0, 1, 2, · · ·
νP = ν0 + Bv′ J(J − 1) − Bv′′ J(J + 1) = ν0 − (Bv′ + Bv′′ )J + (Bv′ − Bv′′ )J 2 ; J = 1, 2, · · ·
(803)
Primes indicate the upper state and double primes the lower state. The contribution from
Dv was neglected. The lines are labeled according to greater of J ′ and J ′′ . For example, a
P3 line corresponds to J ′′ = 3 → J ′ = 2, and an R3 line corresponds to J ′′ = 2 → J ′ = 3.
Transitions with ∆J = 0 are called Q-branch transitions. We will learn later that a Q-branch
can only exist when there is a component of the electronic orbital angular momentum parallel
to the bond axis. A similar derivation yields O and S branches for Raman transitions, with
∆J = ±2.
Problem 72.
Look up the rotational constants Be , αe , De for
1
H 35 Cl in the
NIST database http://physics.nist.gov/PhysRefData/MolSpec/Diatomic/Html/sec6.html
and compare them with the values in Huber and Herzberg.
Problem 73. Use the constants in Huber and Herzberg to calculate as accurately as possible
the frequency of the transition v ′′ = 0, J ′′ = 0 → v ′ = 1, J ′ = 1 of 1 H 35 Cl in the ground
electronic state.
LECTURE 32: SYMMETRY EFFECTS IN ROTATIONAL SPECTRA
Reading: Levine [1], pages 176-185 and Herzberg [2], pages 128-140.
Our goal in this lecture is to discover how the symmetry of a diatomic molecule affects
its rotational spectrum. We saw examples of this already in the requirement that a molecule
must have a permanent dipole moment in order to have a pure rotational transition and
that a homonuclear molecule does not have a pure vibrational spectrum because the dipole
11
moment does not change with stretching. We also found a rotational selection rule that ∆J =
±1. It is interesting that this rule does not require the molecule to be homonuclear. All that
matters is the symmetry of the wave function with respect to the alignment (not orientation)
of the molecular axis. Here we will investigate the consequences of different symmetries and
show that the previous results are special cases of more fundamental principles. The two
primary symmetries that we will be concerned with are parity and nuclear exchange.
In cataloging different types of symmetry of a linear molecule, we must distinguish between molecular and laboratory (i.e., space-fixed) coordinates and also between properties
of just electrons, just nuclei, and both. We summarize in the following table four types of
symmetry that affect the wave function of a diatomic molecule.
Symmetry Frame
Particles
Dipole rule Raman
operation
rule
+ ←→ −
Parity
space
Π̂
θ → π − θ, and
(−1)J
ϕ → ϕ + π (all diatomic
does
electrons
+ ←→ +
− ←→ −
nuclei
not molecules)
affect nuclear
spin
Exchange
space,
nuclei
P̂ab
including
(charges and
nuclear spin
masses must
be
16
Inversion
molecule
equal)
electrons
(only
charges must
be
16
equal)
O18 O, HD
Reflection molecule Σ± , electrons only
σ̂v
Hund’s cases
(a) and (b)
for Λ > 0
s ←→ s, a ←→ a
O2 , H2
only
î
only
g ←→ u
g ←→ g
u ←→ u
12
Parity refers to inversion of the spatial coordinates of all particles (electrons and
nuclei). We wish to discover the parity of a particular rotational level. If we write the total
wave function as a product of a nuclear and an electronic wave function, the parity of the
nuclear part is (−1)J , and the parity of the electronic part is ±1. The latter depends on the
reflection symmetry of the electron wave function in a plane containing the bond axis.
To explain this idea, we need to digress to define the term symbol for an electronic state,
2S+1
Λ±
g,u . In this symbol, Λ is the component of the electronic orbital electronic momentum
along the bond axis. A state with Λ = 0 is called a Σ state; a state with Λ = 1 is called
a Π state; etc. States with Λ > 0 have electrons orbiting around the bond axis (like a
doughnut), and are therefore symmetric tops. The 2S + 1 superscript refers to electron spin.
The ± superscript refers to reflection symmetry of the electronic wave function. Although
it applies to any value of Λ, it only appears in Σ states because the electronic eigenfunctions
for Λ > 0 are actually superpositions of Λ+ and Λ− states. The ± superscript should not
be confused with the ± parity, which unfortunately uses the same sign notation.
Inversion of the nuclear coordinates is equivalent to replacing θ by π − θ and ϕ by ϕ + π.
Problem 74. Prove this last statement.
That is to say that Π̂YJM = (−1)J YJM . This is just the result we found when we
determined the selection rule ∆J = ±1 for rotational transitions. An additional factor of
−1 is introduced for Σ− states. This is because inversion of all three coordinates is equivalent
to a rotation about the bond axis, followed by a reflection in a plane containing the bond;
i.e. σ̂v (yz)Ĉ2 (x). The C2 rotation multiplies the nuclear wave function by (−1)J , while the
reflection multiplies the electronic wave function by ±1. That is,
Π̂Ψtot = Π̂Ψtot ψel = Ĉ2 (x)YJM σ̂v (yz)ψel = ±ψΣ± (−1)J YJM .
(804)
Let’s digress to give examples of reflection symmetry. An example of a state with +
symmetry is the ground state of the hydrogen molecules, which has the term symbol 1 Σ+
g.
Since the atomic orbitals are 1s, the molecular orbital must have + symmetry. An example of − symmetry is the 3 Σ−
g ground state of oxygen. The electronic configuration is
(1σg )2 (1σu )2 (2σg )2 (πu )4 (πgx )1 (πgy )1 , which yields a σ bond from the two electrons πz orbitals
pointing along bond axis and one π bond from the 4 electrons in bonding orbitals and 2
electrons in anti-bonding orbitals perpendicular to the axis. Reflection in either the xz or
yz planes changes the sign of the orbitals perpendicular to that plane. We may express this
13
FIG. 1: Molecular orbital energies of O2 .
14
analytically by looking at the Slater determinants for the unpaired electrons,
sin θ1 eiϕ1 α(1) sin θ1 e−iϕ1 α(1) .
Ψ=
iϕ2
−iϕ2
sin θ2 e α(2) sin θ2 e
α(2) (805)
The minus sign in the right column is there because the projection of the L1 + L2 along the
bond axis is zero for a Σ state. Expanding the determinant gives
(
)
Ψ = sin θ1 sin θ2 ei(ϕ1 −ϕ2 ) − e−i(ϕ1 −ϕ2 ) α(1)α(2) = 2i sin θ1 sin θ2 sin(ϕ1 − ϕ2 )α(1)α(2).
(806)
Reflection in the xz plane leaves sin θ1 sin θ2 unchanged but flips the sign of sin(ϕ1 − ϕ2 ).
Note that if we were to increase the dimension of the determinant include the sigma π bond,
the overall reflection symmetry would not change because the additional pair of electrons
would appear in each term as a multiplicative factor with positive symmetry.
The bottom line is that Σ+ states with even J have + parity, and those with odd J have
− parity. The rule is reversed for Σ− states. The molecule does not have to be homonuclear
to possess parity. The selection rule for a dipole transition is that the parity must change,
while for a Raman transition it must remain the same. More generally, if the orbital angular
momentum quantum numbers of the two electrons are L1 , M1 and L2 , M2 , then the orbital
wave functions for the Σ states are
|Σ+ ; M ⟩ = |L1 , M, L2 , −M ⟩ + |L1 , −M, L2 , M ⟩
−
|Σ ; M ⟩ = |L1 , M, L2 , −M ⟩ − |L1 , −M, L2 , M ⟩
(807)
The reason these equations are true is that reflection flips the sign of M .
Problem 75. The example given above is for the MS = 1 component of the triplet state
(i.e., for α(1)α(2)). Clearly we would get the same result for MS = −1 (i.e., β(1)β(2)) of the
triplet state.. Show that the MS = 0 component of the triplet state also has − symmetry,
whereas the MS = 0 component of the singlet state has + symmetry. (Hint: For MS = 0
you need to take the sum of two Slater determinants.)
A second type of symmetry that affects rotational transitions is the exchange of two
identical nuclei, P̂ab . The exchange operation also occurs in the laboratory frame
and includes nuclear spin. The result of the exchange operation is to change the sign
of the total wave function, with plus and minus signs corresponding to symmetric and antisymmetric wave functions. This process is related to the Pauli Exclusion Principle, which
15
refers to exchange of two electrons (in general, any two Fermions) and always causes a sign
change of the electronic wave function.
Nuclear exchange is important for two reasons. First, there is a selection rule that prohibits the interconversion between symmetric and antisymmetric states by any process (radiative, collisional, etc.). The rule applies to any operator, M (such as the dipole operator),
∫
that is unaffected by exchanging identical nuclei. It follows that the integral ψ1∗ M ψ2 dτ
must vanish unless ψ1 and ψ2 have the same exchange symmetry. This rule is rigorous if the
nuclei have zero spin. If they have non-zero spin, the rule is not rigorous (although it still is
approximately correct) because the spins of the nuclei may point in different directions so
that the nuclei are not truly identical.
The second consequence of nuclear exchange symmetry stems from the Pauli principle.
Particles with half integer spins are called Fermions, and particles with integer spin (including zero) are called Bosons. The operating principle is that the total wave function is
antisymmetric with respect to the exchange of a pair of identical Fermions and is symmetric
with respect to the exchange of a pair of identical Bosons. The total wave function for a
molecule may be written as the product of vibrational, rotational, electronic, and nuclear
factors. The vibrational part is always symmetric because it depends only on internuclear
distances. We will assume for now that the electronic part is also symmetric with respect
to nuclear exchange. Rotational states that change sign upon nuclear exchange are called
asymmetric (denoted by the symbol a), and rotational states that do not change sign upon
nuclear exchange are called symmetric (denoted by the symbol s.) In the special case that
the nuclear spin is zero (e.g. for
16
O2 ,12 C2 , and C 16 O2 ), only the rotational wave func-
tion determines the exchange symmetry. Since nuclei with zero spin are Bosons, it follows
that the rotational wave function must be even, and hence odd rotational states for such
homonuclear molecules do not exist. Lines for such states are are simply absent in the
spectrum.
If the nuclei do have spin, we must include the symmetry of the nuclear spin wave function
in determining the symmetry of the total wave function. If the nuclei are distinguishable, the
Pauli principle does not apply. If the molecule is homonuclear, there are four possibilities: (1)
The nuclei are Fermions and the spin functions are symmetric; (2) the nuclei are Fermions
and the nuclear spin function is antisymmetric; (3) the nuclei are Bosons and the spin
functions are symmetric; (4) the nuclei are Bosons and the spin functions are antisymmetric.
16
An example of cases (1) and (2) is the H2 molecule. The spin of each nucleus is I = 1/2,
and the total nuclear spin of the molecule is T = I1 + I2 = 0 or 1. T = 0 is a singlet
and is antisymmetric; T = 1 is a triplet and is symmetric. The ground nuclear state
“modification” is a singlet. For H2 this modification is called para hydrogen or p − H2 ,
and only even rotational states exist. (Odd states would violate the Pauli principle because
they would make the total wave function symmetric, which is forbidden for Fermion nuclei.)
The excited modification has a triplet nuclear spin wave function, which is symmetric, and
accordingly the rotational quantum numbers are all odd. This case is called ortho hydrogen
or o − H2 . In general, the modification with a symmetrical nuclear spin wave function is
always called ortho.
It is easy to figure out which states are symmetric by noticing that the highest spin state
has all the spins lined up and is therefore symmetric. For H atoms, the nuclei are Fermions
with I = 1/2, resulting in triplet and singlet spin states for the diatomic molecule. The
triplet has s symmetry, and the singlet therefore has a symmetry. Since the total wave
function for Fermionic nuclei must have a symmetry, it follows that the even rotational
states are paired with the a singlet nuclear state. For H2 molecules, even J states are called
para-hydrogen or p − H2 , and odd J states are called ortho hydrogen or o − H2 .
For deuterium atoms, the nuclei are Bosons with I = 1, resulting in singlet, triplet, and
quintet spin states for the diatomic molecule. The quintet has s symmetry. The other two
states alternate in symmetry, with the triplet being a and the singlet s. Since the total wave
function for Bosonic nuclei must have s symmetry, it follows that the even rotational states
are paired with the s singlet and quintet nuclear states. It follows that for deuterium even
J is ortho and odd J is para.
In the case of the H atom, the nuclear spin wave functions are
1
ψns = √ [α(1)β(2) − α(2)β(2)]
2
(808)
for the singlet state and
ψns = α(1)α(2)
1
= √ [α(1)β(2) + α(2)β(2)]
2
(809)
= β(1)β(2)
for the triplet state. The nuclear degeneracies, g = 2T + 1, are 1 for the singlet and 3 for
17
the triplet. It follows that for a thermal equilibrium mixture of para and ortho states, the
intensities of the rotational transitions alternate between factors of 1 and 3. The partition
function for an equilibrated mixture of hydrogen molecules is given by
∑
Q(T ) =
=
∑
(2J + 1)e−BhcJ(J+1)/kT + 3
evenJ
∞
∑
(2J + 1)e−BhcJ(J+1)/kT
oddJ
(810)
−BhcJ(J+1)/kT
[2 − (−1) ](2J + 1)e
J
.
J
Examples of cases (3) and (4) are D2 and N2 . The D nucleus has a spin of I = 1, making
it a Boson. The possible values of T are 0, 1, and 2. The T = 0 and 2 states are symmetric
with degeneracies 1 and 5, while the T = 1 state is antisymmetric and has a degeneracy of
3. The even rotational levels are now ortho. The partition function is given by
Q(T ) = 6
∑
(2J + 1)e−BhcJ(J+1)/kT + 3
evenJ
∑
(2J + 1)e−BhcJ(J+1)/kT .
(811)
oddJ
Problem 76. Calculate the rotational population distribution for H2 in thermal equilibrium
at 100K. The rotational constant is B = 60.8 cm−1 .
Now let’s take into account the exchange symmetry of the electronic wave function. When
we exchange the nuclei in space-fixed coordinates, we also invert the axis that defines the
molecular frame. This has the effect of inverting the coordinates of the electrons in the
molecular frame. If the two nuclei have the same charge (although not necessarily the same
mass, as for example 16 O18 O or HD), the electronic wave function will be either even (gerade,
g) or odd (ungerade, u) with respect to inversion of electronic coordinates. This means that
the rotational wave functions for u electronic states have the opposite exchange symmetry
of the g states. The effects of changing both g → u and Σ+ → Σ− on nuclear exchange
symmetry cancel each other out. Changing just Σ+ → Σ− flips both the exchange symmetry
(s vs. a) and the inversion symmetry (+ vs.− parity).
Putting all the pieces together, the lowest rotational level (J = 0) is s+ for a 1 Σ+
g state,
1 −
1 +
a− for a 1 Σ−
g state, a+ for a Σu state, and s− for a Σu state. See figure 4.14 in Levine
and 6.4 in Herzberg I. The inter-relations between reflection and exchange symmetry are
treated systematically in what are known as the Wigner-Witmer rules, which we will discuss
in a later lecture. These rules give a general recipe for determining all possible term symbols
arising from a pair of electrons with given L and M quantum numbers.
18
LECTURE 33: ROTATIONAL SPECTRA OF POLYATOMIC MOLECULES
Reading: This lecture is based on Chapters 3 and 6 of Zare [7].
We begin by examining the classical treatment of rotation of a rigid body about a fixed
axis. We then construct the Hamiltonian and solve for its eigenvalues and eigenfunctions.
Consider first the very simple case of a single point mass at the end of a massless rod.
The rod rotates about an axis perpendicular to it. The kinetic energy of the mass is
1
1
1
1
T = mv 2 = m(ωR)2 = mR2 ω 2 = Iω 2 .
2
2
2
2
(812)
In these equations, R is the length of the rod, ω is the angular frequency, m is the mass of
the particle, and I is its moment of inertia. Likewise, the rotational angular momentum is
J = mvR = m(Rω)R = Iω.
(813)
We may extend these results to a collection of N particles of mass mi ,
1∑
1 ∑
1
mi vi2 = ω 2
mi ri2 = Iω 2 ,
2 i=1
2 i=1
2
N
T =
N
and
J =ω
N
∑
mi ri2 = Iω,
(814)
(815)
i=1
where
I=
N
∑
mi ri2 .
(816)
i=1
We may generalize this result further to rotation about an arbitrary axis,
J=
N
∑
mi (ri × vi ).
(817)
i=1
But
vi = ω × r i ,
so that
J=
N
∑
mi [ri × (ω × ri )].
(818)
(819)
i=1
Expanding the cross product, using the vector identity,
A × (B × C) = B(A · C) − C(A · B).
(820)
19
gives
J=
=
N
∑
i=1
N
∑
mi [ωri2 − ri (ri · ω)]
(821)
mi [ri2 (ωx î + ωy ĵ + ωz k̂) − (xi ωx + yi ωy + zi ωz )(xi î + yi ĵ + zi k̂)].
i=1
The Cartesian components of J are
Jx = ωx
N
∑
mi (ri2
−
x2i )
− ωy
i=1
N
∑
mi xi yi − ωz
N
∑
i=1
mi x i z i ,
(822)
i=1
with analogous expressions for Jy and Jz . We define coefficients Ipq such that
Jx = Ixx ωx + Ixy ωy + Ixz ωz
(823)
Jy = Iyx ωx + Iyy ωy + Iyz ωz
Jz = Izx ωx + Izy ωy + Izz ωz .
The diagonal terms are called the moments of inertia coefficients; e.g.,
Ixx =
N
∑
mi (ri2 − x2i ),
(824)
i=1
and the off-diagonal terms are called the products of inertia; e.g.,
Ixy = −
N
∑
mi xi yi .
(825)
i=1
The angular momentum is now written as a tensor product,
J = I · ω,
or



I I I
J
 x   xx xy xz
  
 Jy  =  Iyx Iyy Iyz
  
Izx Izy Izz
Jz
(826)


ω
 x 


  ωy  .


ωz
(827)
We see that I is a symmetric, Hermitian matrix, which we may diagonalize. That is to say
that we may find an orthogonal matrix, U such that such that


I
0 0

 aa


−1
Id = U · I · U =  0 Ibb 0 


0 0 Icc
(828)
20
The eigenvectors of the original inertia tensor are the principle axes of rotation along directions (n̂a , n̂b , n̂c ), such that
J = Iaa ωa n̂a + Ibb ωb n̂b + Icc ωc n̂c ,
(829)
and the eigenvalues are the moments of inertia about those axes. In what follows we will
drop the subscript d from the diagonalized intertia tensor. We may likewise write the kinetic
energy as a tensor product,
T =
N
N
N
1∑
1∑
1 ∑
mi vi2 =
mi (vi · vi ) =
mi vi · (ω × ri ).
2 i=1
2 i=1
2 i=1
(830)
We invoke another vector identity
A · (B × C) = B · (C × A)
(831)
to get
∑
1
1
1
1
T = ω·
mi (ri × vi ) = ω · J = ω · I · ω = (ωa2 Iaa + ωb2 Ibb + ωc2 Icc ).
2
2
2
2
i=1
N
(832)
Expanding this last result, we get
T =
Ja2
J2
J2
+ b + c ,
2Iaa 2Ibb 2Icc
(833)
where Ja = Iaa ωa , etc. The equation
Ja2
J2
J2
+ b + c =1
2T Iaa 2T Ibb 2T Icc
(834)
describes a solid known as the “inertia ellipsoid.”
By convention, we order the moments of inertia as Iaa ≤ Ibb ≤ Icc . These equations
describe the motion of a classical top. Different types of tops are defined according to the
equality of some of the moments of inertia:
Iaa = 0, Ibb = Icc ,
linear
Iaa < Ibb = Icc ,
prolate symmetric top
Iaa = Ibb < Icc ,
oblate symmetric top
Iaa = Ibb = Icc ,
spherical top
Iaa < Ibb < Icc ,
asymmetric top
21
A prolate symmetric top is football shaped, such as CH3 I. Think of a cylinder, with two
axes perpendicular to the curved surface and one axis along the cylinder axis. The moments
of inertia about the axis perpendicular to the curved surface are equal and also larger then
the moment with respect to the long axis, The latter statement follows because atoms at
the ends of the cylinder are far from these axes and therefore have large moment arms.
An oblate symmetric top is frisbee-shaped, such as benzene. In this case the two smallest
moments correspond to the axes in the plane of the molecule because some of the atoms lie
on those axes. Examples of spherical tops are methane and SF6 . Symmetric tops must have
a rotational axis of at least third order.
We may now construct the quantum mechanical Hamiltonian by replacing the J 2 terms
by their corresponding operators,
Ĥrot =
Jˆa2
Jˆ2
Jˆ2
+ b + c .
2Iaa 2Ibb 2Icc
(835)
We note that the subscripts a, b, c refer to body-fixed coordinates. Normally they are written
as x, y, z, while the space-fixed coordinate are labeled X, Y, Z. We need three angles to
describe the molecular orientation; θ and ϕ determine the direction of the principle axis of
the molecule in the lab, and χ is the rotational angle about that axis. The wave functions
for symmetric and spherical tops have three quantum numbers, J for the magnitude of the
angular momentum, M for its projection along the lab Z axis, and K for its projection along
the molecular z axis. The Schrodinger equation is therefore
Ĥrot ψJKM (θ, ϕ, χ) = EψJKM (θ, ϕ, χ).
(836)
It is convenient to define three rotational constants, similar to that used for the diatomic
molecule:
(
~2
hc
)
Ĥrot = AJˆa2 + B Jˆb2 + C Jˆc2 ,
(837)
where A ≥ B ≥ C. For a spherical top, where A = B = C, the Schrodinger equation
simplifies to
Ĥrot ψJKM (θ, ϕ, χ) =
Jˆ2
ψJKM (θ, ϕ, χ) = EψJKM (θ, ϕ, χ).
2I
(838)
The eigenvalues are the same as for the diatomic molecule,
EJ
= BJ(J + 1),
hc
(839)
22
only that the degeneracy is now (2J + 1)2 . The extra factor of 2J + 1 comes from the
projection of the angular momentum along the molecular axis.
We consider next symmetric tops. In the oblate case (A = B > C),
(
)
Jˆa2 + Jˆb2
Jˆc2
Jˆa2 + Jˆb2 + Jˆc2
Jˆc2
Jˆc2
Jˆ2
1
1
2
ˆ
Ĥrot =
+
=
+
−
=
+ Jc
−
. (840)
2Iaa
2Icc
2Iaa
2Icc 2Iaa
2Iaa
2Icc 2Iaa
The eigenvalues in wave numbers are
E
= AJ(J + 1) − (A − C)K 2
hc
(841)
where 0 ≤ K ≤ J. The degeneracy is 2J + 1 for K = 0 and 2(2J + 1) for K > 0.
Problem 77. The C − C bond length in benzene is 0.14 nm. Calculate the three moments
of inertia. You may add the mass of H to the nearest C and otherwise ignore their presence.
Problem 78. Use the result of the previous problem to calculate all the rotational energy
levels of benzene for J = 0, 1, 2, 3.
For a prolate top (A > B = C), the same reasoning leads to
E
= CJ(J + 1) + (A − C)K 2
hc
(842)
Figure 6.8 in Zare [7] is a drawing of the ordering of the energy levels of prolate and oblate
tops.
Problem 79. Derive Eq. (842).
The energy levels of an asymmetric top may be obtained by expanding the wave functions
of such a molecule, ΨJM (ϕ, θ, χ), in the symmetric top basis set. We will find out what the
latter are in the next lecture. For now we will simply designate them by the ket |JKM ⟩.We
may choose to use either the prolate or oblate basis set. Since most molecules are closer to
the prolate limit, that is usually the basis of choice. The expansion is
ΨJM (ϕ, θ, χ) =
J
∑
AK |JKM ⟩.
(843)
K=−J
We substitute this wave function into the rotational Schrodinger equation,
Ĥrot ΨJM (ϕ, θ, χ) = EΨJM (ϕ, θ, χ)
to get
J
∑
K=−J
AK Ĥrot |JKM ⟩ = E
J
∑
K=−J
AK |JKM ⟩.
(844)
(845)
23
Left multiplying by ⟨JK ′ M | and integrating over all angles gives a set of algebraic equations,
which may be solved by finding the eigenvalues and eigenfunctions of the secular determinant,
|HK ′ K − EδK ′ K | = 0,
(846)
HK ′ K = ⟨JK ′ M |Ĥrot |JKM ⟩.
(847)
where the matrix elements are
Zare (see section 6.3 in [7]) develops an elegant way of block-diagonalizing the secular determinant using the D2 symmetry representation of the inertia ellipsoid. Here we will just
quote the result that the energy eigenvalue lies on a monotonically smooth curve connecting
the prolate and oblate limits, parameterized by the quantity
κ=
2B − A − C
,
A−C
(848)
where κ = −1 corresponds to the prolate limit (B = C), and κ = +1 corresponds to the
oblate limit A = B). Take a look at Figure 6.9 in Zare. The various curves are labeled
JK−1 ,K1 , where the subscripts refer to the limiting values of K for τ = ±1.
LECTURE 34: PROPAGATORS AND CONSERVATION LAWS
The rotational wave function of a symmetric top is an eigenfunction of the total angular
momentum operator as well as of the projections along both the space-fixed and bodyfixed z-axis. To find these eigenfunctions, we need to learn about the transformation from
space-fixed to body-fixed coordinates.
We will first discuss transformations in general, following closely the material in chapter
3 of Zare. A key concept is that a coordinate transformation is a unitary operation, which
transforms one basis set into another,
ψj′ = Û ψj .
(849)
An operator Q̂ is also transformed by Û according to the rule
Q̂′ = Û Q̂Û −1 = Û Q̂Û † .
(850)
For a microscopic transformation of some variable (e.g., a coordinate or time) by an amount
ϵ, Û may be written as
Û = I + ϵV̂ .
(851)
24
The requirement that Û Û † = I is equivalent to
(I + ϵV̂ )(I + ϵV̂ † ) = I + ϵ(V̂ + V̂ † ) = I
(852)
to first order in ϵ. This condition is satisfied if V̂ = iŜ † , where Ŝ = Ŝ † is Hermitian.
We may subdivide this transformation into n smaller steps, so that
Û = (1 + iδ Ŝ)n ,
(853)
where δ = ϵ/n. We may now consider the case of a finite transformation in which δ → 0
and n → ∞, so that
[
( ϵ )]n
Û = lim 1 + i
Ŝ
= eiϵŜ .
n→∞
n
This last step may be proved from the binomial theorem, which states that
( )
( )
n 2
n k
n
(1 + x) = 1 + nx +
x + ··· +
x ···
2
k
and choosing x = iϵŜ/n = iδ Ŝ. Recall that
( )
n
n!
n(n − 1)(n − 2) · · · (n − k + 1)
=
=
≈ nk /k!
k
k!(n − k)!
k!
(854)
(855)
(856)
for n ≫ k. Since we let n → ∞, this approximation is always satisfied.
Recall next that transforming a system in the forward direction is equivalent to transforming the coordinate system backwards by the same amount. Let’s examine this statement
graphically, using as examples rotation, translation in space, and translation in time. We
showed previously that rotating a vector clockwise by some angle, α, (see drawing (a)) is
equivalent to rotating the coordinate system counterclockwise by the same angle (drawing
(b)). We get the same result for a coordinate displacement by either moving a point to the
right (drawing (c)) or moving the origin to the left (drawing (d)).
Let’s consider next the somewhat trickier example of a translation in time. Suppose we
look a sinusoidal pattern on an oscilloscope and take an old-fashioned polaroid snapshot at
time t = 0. The image we get is labeled ψ0 ,
ψ0 (t) = sin ωt,
(857)
shown in drawing (e). Next, suppose we trigger the camera to take a photo at some later
time t′ > 0. The resulting image is shown in drawing (f),
ψt′ (t) = sin ω(t − t′ ).
(858)
25
FIG. 2: Operators for rotational, translational, and temporal transformation.
26
Comparing the two photos, we see that
ψt′ (t) = ψ0 (t − t′ ) = ψ0 (τ ),
(859)
where τ = t − t′ . But we could have gotten the same result by simply relabeling the axes in
the first photo so as to shift the origin of time backwards by an amount t′ instead of moving
the wave forwards by the same amount in the second photo (see drawing (g)).
Now let’s find the unitary operators that do these two tasks. First let’s transform the
time variable, not the wave function. We need to change the time origin by an amount −t′ ,
as follows:
ψ0 (τ ) = ψ0 (t) + (−t′ )
∂
1
∂2
′ ∂
ψ0 (t) + (−t′ )2 2 ψ0 (t) + · · · = e−t ∂t ψ0 (t) = T̂ (t′ )ψ0 (t). (860)
∂t
2!
∂t
From the time-dependent Schrodinger equation,
Ĥ = i~
∂
,
∂t
(861)
we recognize the unitary operator T̂ (t′ ) to be
′
T̂ (t′ ) = eit Ĥ/~ .
(862)
ψ0 (τ ) = T̂ (t′ )ψ0 (t) = ψ0 (t − t′ ).
(863)
such that
Instead of propagating time backwards by t′ , we may propagate the wave function forwards
in time by the same amount,
ψ(t) = T̂ (−t)ψ(0) = e−itĤ/~ ψ(0) = Û (t)ψ(0).
(864)
where the time evolution operator Û (t) is given by
Û (t) = e−itĤ/~ .
(865)
In the last two equations we dropped the primes on t and the subscript on the wave function.
If ψ is an eigenfunction of Ĥ, we get the familiar result
ψ(t) = e−iE0 t/~ ψ(t = 0).
(866)
27
This result shows that the invariance of a wave function in time (apart from a phase change)
is equivalent to conservation of energy. In other words, if ψ is an eigenstate of Ĥ, then operating on it with e−itĤ/~ replaces multiplies the wave function by a simple phase factor.
Saying that the system is in an eigenstate of the Hamiltonian is equivalent to saying that energy is conserved, and saying that the time propagator reduces to a phase factor is equivalent
to saying that the system is invariant with time.
The same analysis applies to other types of transformations. For example, a spatial
displacement way be written as
|z ′ ⟩ = |z0 ⟩ + (−z ′ )
∂
1 ∂2
′ ∂
|z0 ⟩ + (−z ′ )2
|z0 ⟩ + · · · = e−z ∂z |z0 ⟩.
2
∂z
2! ∂z
(867)
Recognizing that p̂z = −i~∂/∂z, we obtain for the unitary displacement operator,
′
D̂(z) = e−iz p̂z /~ .
(868)
Invariance of the wave function with spatial displacement (apart from a phase factor) is
equivalent to conservation of linear momentum.
The same reasoning applies to rotations. Rotation of a wave function counterclockwise
about the z-axis by an angle α is given by
ψ(ϕ0 + α) = e−α ∂ϕ ψ(ϕ)ϕ=ϕ0 .
∂
(869)
Recalling the angular momentum operator,
∂
Jˆz = −i~ ,
∂ϕ
(870)
we obtain the rotational evolution operator
R̂z (α) = e−iαJz /~ .
ˆ
(871)
Invariance of the wave function with rotation (apart from a phase factor) is equivalent to
conservation of angular momentum. Finally, for rotation by an angle α about an arbitrary
axis n̂,
R̂n (α) = e−iαĴ·n̂/~ .
(872)
28
LECTURE 35: ROTATIONAL WAVE FUNCTIONS OF POLYATOMIC
MOLECULES
We are now in a position to use rotation operators to develop an expression for the
rotation of a molecule about an arbitrary axis, and from that to derive the rotational wave
function. To describe such a motion, we need three angles, two angles (θ, ϕ) to locate the
orientation of the molecular axis in the lab frame and a third angle for rotation about that
axis. The first two angles describe the transformation from the XY Z frame to the xyz frame,
and the third angle (χ) describes rotation about the body-fixed z-axis. Note that the angles
θ and ϕ define the direction of the principle axis of inertia with respect to the lab frame.
For linear molecules and symmetric tops, this axis coincides with the angular momentum
vector. These angles are called the Euler angles. We carry our the transformation in three
steps,
XY Z −−−→ x′ y ′ z ′ −−−−−→ x′′ y ′′ z ′′ −−−−−→ xyz,
(873)
R̂z′′ =z (χ)
R̂y′ =N (θ)
R̂Z (ϕ)
as shown in Fig. 3.2 of Zare. At each stage one axis is the same in two frames: Z = z ′ in
the first rotation, y ′ = y ′′ = N in the second, and z ′′ = z ′ in the third. The intermediate y
axes are called the line of nodes. Note that ϕ is the azimuthal angle about the Z axis, and
χ is the azimuthal angle about the z axis. Note also that this is a passive transformation of
the coordinate system, taking 0 → ϕ, 0 → θ, 0 → χ.
The total transformation operator is a product of three counter-clockwise rotational operators,
R̂(θ, ϕ, χ) = e−iχJz /~ e−iθJN /~ e−iϕJZ /~ .
ˆ
ˆ
ˆ
(874)
This result is exact but awkward to use because it involves rotations about axes in two
coordinate systems. But now the transformation Q̂′ = Û Q̂Û −1 comes to the rescue. We
recognize that the rotation by angle θ about the N = y ′ axis is equivalent to a rotation of
θ about the Y axis if we first carry out the transformation y ′ → Y by the operation eiϕJZ /~
ˆ
and afterwards undo that transformation. In other words,
e−iθJN /~ = e−iϕJZ /~ e−iθJY /~ eiϕJZ /~ .
ˆ
ˆ
ˆ
ˆ
(875)
Likewise, the rotation by angle χ about the z axis is equivalent to a rotation of χ about the
ˆ
Z axis after the transformation of eiθJN /~ . That is,
e−iχJz /~ = e−iθJN /~ e−iχJZ /~ eiθJN /~ .
ˆ
ˆ
ˆ
ˆ
(876)
29
Putting the pieces together,
e
−iχJˆz /~ −iθJˆN /~ −iϕJˆZ /~
e
e
[
]
ˆ
ˆ
−iθJˆN /~ −iχJˆZ /~ iθJˆN /~
= e
e
e
e−iθJN /~ e−iϕJZ /~
. = e−iθJN /~ e−iχJZ /~ e−iϕJZ /~
[
]
ˆ
ˆ
ˆ
ˆ
ˆ
= e−iϕJZ /~ e−iθJY /~ eiϕJZ /~ e−iχJZ /~ e−iϕJZ /~
ˆ
ˆ
ˆ
(877)
= e−iϕJZ /~ e−iθJY /~ e−iχJZ /~ .
ˆ
ˆ
ˆ
In the last step, the rotations about the Z axis are permuted. This remarkable result shows
that the Euler transformation may be carried out entirely in the lab frame by just reversing
the order of the rotations. Next we observe that R̂n (α) (a rotation by angle α about an
arbitrary axis) commutes with Ĵ2 because
2
−αĴn /~
[Rn (α), Ĵ ] = [e
∞
∑
1
, Ĵ ] =
(−iα)s [Ĵsn , Ĵ2 ]/~s = 0.
s!
s=0
2
(878)
If |JM ⟩ is an eigenfunction of Ĵ2 and ĴZ , rotation by R̂n (α) can change M but not J. In
other words, it transforms |JM ⟩ into a superposition of M states,
J
∑
R̂(θ, ϕ, χ)|JM ⟩ =
J
′
DM
′ M (θ, ϕ, χ)|JM ⟩,
(879)
M ′ =−J
with coefficients
J
′
DM
′ M (θ, ϕ, χ) = ⟨JM |R̂(θ, ϕ, χ)|JM ⟩.
(880)
J
The array of DM
′ M coefficients is called the rotation matrix.
An important property of the rotation operator is that it can transform the rotational
wave function from one orientation to another. Recall that the rotation matrix was defined
for a passive transformation. That is,
ψJM (θ′ , ϕ′ , χ′ ) =
∑
J
DM
K (θ1 , ϕ1 , χ1 )ψJM (θ, ϕ, χ).
(881)
K
In this expression, θ, ϕ, χ are the angles of the basis functions before the transformation,
θ′ , ϕ′ , χ′ are the angles after the transformation, and θ1 , ϕ1 , χ1 are the angles of the passive
operator. If we choose the angles of the D matrix to be identical to those of the wave
function, then the transformation simply “unwinds” the angles back down to zero. That is,
ψJM (0, 0, 0) =
∑
K
J
DM
K (θ, ϕ, χ)ψJM (θ, ϕ, χ).
(882)
30
If we now invert the transformation, which is equivalent to carrying out an active transformation,
ψJM (θ, ϕ, χ) =
∑
J∗
DM
K (θ, ϕ, χ)ψJM (0, 0, 0).
(883)
K
This discussion is for a general asymmetric top. For a symmetric top, the wave function
is invariant with respect to rotation by χ (apart from a phase factor). This property means
that only one term may be present in the sum, giving rise to
[
2J + 1
|JKM ⟩ =
8π 2
]1/2
J∗
DM
K (θ, ϕ, χ)
(884)
Noting that
e−iαJZ /~ |JM ⟩ = e−iαM |JM ⟩,
ˆ
(885)
we may reduce the rotational matrix element to a simpler form,
J
−iM ϕ J
DM
dM K (θ)e−iKχ .
K (θ, ϕ, χ) = e
(886)
Zare derives an analytic formula for dJM ′ M (θ). (See his Eq.(3.66) and Table 3.1). Two
limiting cases are
L
D00
(θ, ϕ, χ) = PL (cos θ)
and
[
L
D0M
(θ, ϕ, χ)
4π
=
2L + 1
(887)
]1/2
∗
YL,−M
,
(888)
which reduce to the wave functions for a linear rotator.
Knowing the wave functions, we may determine the selection rules and line strengths.
There are nice closed form expressions in Zare Table 3.1 for the product of three rotation
matrix elements (e.g., two rotational wave functions and the dipole transition operator). For
a pure rotational transition with the permanent dipole moment pointing along the highest
order symmetry axis, Herzberg [3] derives the selection rule ∆J = 0, ±1, ∆K = 0.
For a rotational-vibrational transition, we need to compare the symmetry representation
of the vibration x, y, z with the direction of the symmetry axis. For the transition dipole
moment parallel to the symmetry axis, ∆J = 0, ±1, ∆K = 0 for K ̸= 0 and ∆J ± 1, ∆K = 0
for K = 0. For a perpendicular transition, ∆J ± 1, ∆K = ±1.
31
LECTURE 36: INTRODUCTION TO ELECTRONIC SPECTROSCOPY: THE
BORN-OPPENHEIMER APPROXIMATION
Reading: Chapter 9 of Bernath [8]. Excellent reviews of this topic may be found in
references [10, 11].
In most molecular electronic structure calculations, the electrons move much more rapidly
than the nuclei because of their much lighter mass. A tremendous simplification results if
we assume that the Schrodinger equation may be solved separately for the electrons and
nuclei. In the following notes we show what approximations are involved in this separation.
The Hamiltonian for the one-electron hydrogen molecule ion is given by
Ĥ = −
~2 2
~2 2 e2 e2
e2
~2 2
∇a −
∇b −
∇e −
− +
.
2Ma
2Mb
2me
ra rb Rab
(889)
This is the only molecular problem that can be solved exactly. For an n-electron diatomic,
the Hamiltonian is
Ĥ = −
n
n
n
n
~2 2
~2 2
~2 ∑ 2 ∑ Za e2 ∑ Zb e2 ∑ e2
Za Zb e2
∇a −
∇b −
∇i −
−
+
+
, (890)
2Ma
2Mb
2me i=1
r
r
r
R
ia
ib
ij
ab
i=1
i=1
i̸=j
and for N nuclei and n electrons,
N
n
N
n
n
N
∑
~2 2
~2 ∑ 2 ∑ ∑ Zs e2 ∑ e2 ∑ Zs Zs′ e2
Ĥ = −
∇ −
∇ −
+
+
.
2Ms s 2me i=1 i s=1 i=1 ris
rij s̸=s′ Rss′
s=1
i̸=j
(891)
We will use a shorthand to collect all the nuclear and electronic terms together,
Ĥ = T̂N + T̂e + U (r, R),
(892)
where T̂N is the nuclear kinetic energy operator, T̂e is the electronic kinetic energy operator,
U is the total potential energy, and r and R refer to all the electronic and nuclear coordinates,
respectively. For later reference, we define the electronic Hamiltonian by
Ĥe = T̂e + U (r, R).
(893)
Next, we write the wave function as a product
Ψ(r, R) = ψe (r, R)ϕN (R).
(894)
So far we have not made any approximations. The complete Schrodinger equation becomes
Ĥ(ψe ϕN ) = T̂N (ψe ϕN ) + T̂e (ψe ϕN ) + U (r, R)ψe ϕN = Eψe ϕN .
(895)
32
We next expand the kinetic energy derivatives in T̂N and T̂e ,
∇2N (ψe ϕN ) = ψe ∇2N ϕN + ϕN ∇2N ψe + 2∇N ψe · ∇N ϕN
(896)
∇2e (ψe ϕN ) = ϕN ∇2e ψe .
(897)
and
We are now ready to invoke the Born-Oppenheimer (BO) approximation. We clamp the
nuclei at some geometry, R. This allows us to discard the nuclear kinetic energy term, ∇2N .
This is justified because the terms in Eq. (896) are all of the same order of magnitude, and
when we divide them by M they become much smaller than the term
1
∇2 ψ
m e e
that appears
in the electronic Hamiltonian. This is the essence of the BO approximation.
We may now cancel out ϕN on both sides of Eq. (895) , leaving us with the electronic
Schrodinger equation,
(
)
Ĥe ψe (r; R) = T̂e (r; R) + U (r; R) ψe (r; R) = V (R)ψe (r; R).
(898)
Here, R is treated as a parameter. This is the starting point for all quantum calculation
programs (Gaussian, etc). We solve this equation numerically to get ψe (r; R) and V (R).
V (R) is the energy eigenvalue of Ĥe and is called the adiabatic potential energy surface (PES),
with dimension 3N-5 for linear molecules and 3N-6 for non-linear molecules. Each PES
corresponds to a different electronic configuration. We note, however, that the configuration
may change as R varies, and in general a multi-configurational basis is needed.
If we regard ψe (r; R) as the eigenfunction of a truncated Hamiltonian ψe (r; R), we have
not really made any approximations yet. At this point, however, we use the electronic
Schrodinger equation to replace the second and third terms in the complete Schrodinger
equation, T̂e (r; R) + U (r; R)), by V (R) to yield
T̂N (ψe ϕN ) + V (R)ψe ϕN = Eψe ϕN .
(899)
Now here comes the mathematical content of the BO approximation. We assume that
∇2N (ψe ϕN ) ≈ ψe ∇2N ϕN .
(900)
Ĥ ′ (ψe ϕN ) = ϕN ∇2N ψe + 2∇N ψe · ∇N ϕN = 0.
(901)
This is equivalent to setting
33
In a more complete treatment we could treat Ĥ ′ as a perturbation. What remains of the
complete Schrodinger equation is the nuclear Schrodinger equation,
T̂N ϕN + V (R)ϕN = EϕN ,
(902)
from which we obtain the vibrational and rotational wave functions and energies. In the
familiar harmonic oscillator - rigid rotor approximation, we may further factor ϕN into
separate vibrational and rotational wave functions,
ϕN (R) = χvib (Q)ψ rot (θ, ϕ, χ),
(903)
where χv (Q) is the harmonic oscillator wave function for quantum number v, Q = R − Re is
the displacement of the oscillator coordinate from its equilibrium value, and ψrot is the rigid
rotor wave function. For a polyatomic molecule, χ is a product on 3N-5 or 3N-6 independent
harmonic oscillator functions, one for each normal coordinate, Qs . The rotational wave
function is a 3D function of the Euler angles. We may now partition the total energy into
electronic, vibrational, and rotational term values,
E
= Te + G(v) + F (J, K).
hc
(904)
Since different electronic potential energy functions have different equilibrium distances
and different curvatures, they also have different moments of inertia and vibrational frequencies. The vibrational selection rules that we derived for pure vibrational transitions no
longer hold. This is true because the transition dipole moment may be written as
∫
∫
∫
∫
∗ ∗
∗
3
rot∗ rot
2
vib
µ12 = e ψe2 ϕN 2 rψe1 ϕN 1 dτ = e ψe2 rψe1 d τe ψJ ′ K ′ ψJ ′′ K ′′ d Ω χvib
v ′ χv ′′ dR,
(905)
vib
where ϕN = ϕrot∗
JK ϕv . The square of the last integral is called a Franck-Condon factor,
qv′ v′′ = |⟨v ′ |v ′′ ⟩|2 .
In general, closure requires that
∞
∑
qv′ v′′ = 1.
(906)
(907)
v ′ =0
If the equilibrium constants of the lower and upper states are nearly equal (i.e. if B ′ ≈ B ′′ ),
qv′ v′′ will be largest for v ′ = v ′′ = 0. If the equilibrium constants are very different, the
transition probability is greatest for a vertical transition that maximizes the overlap of the
34
vibrational wave functions. For v ′′ = 0, the transition connects the equilibrium position of
the lower potential energy curve to the turning point of the upper curve. In the limiting
case that oscillator is harmonic and that the frequencies of both electronic states are equal,
one may show that
qv0
1
=
v!
(
(∆R̄e )2
2
where
∆R̄e =
)v
e−(∆R̄e )
2 /2
,
√
β(R̄e′ − R̄e′′ ).
(908)
(909)
Problem 80. Derive Eq. (908). Hint: use the recursion property of the harmonic oscillator
eigenfunctions.
Problem 81. Show that Eq. (908) satisfies the closure condition.
An interesting thing also happens to the rotational spectrum. Consider for example the
P-branch. The transition energy in wave numbers is given
ν̃P = ν̃0 + B ′ J(J − 1) − B ′′ J(J + 1) = ν̃0 − (B ′ + B ′′ )J + (B ′ − B ′′ )J 2 ,
(910)
where J is the rotational quantum number of the lower electronic state (J ′′ ). We see that
if B ′ − B ′′ is positive, the frequency passes through a minimum. In this case the rotational
band turns back on itself and is said to have a “band head.” A similar thing may happen
for an R-branch. In this case,
ν̃R = ν̃0 + B ′ (J + 1)(J + 2) − B ′′ J(J + 1) = ν̃0 + 2B ′ + (3B ′ − B ′′ )J + (B ′ − B ′′ )J 2 . (911)
Here, if B ′ − B ′′ is negative, the frequency passes through a maximum.
We may derive a uniform expression by defining m = J ′′ + 1 for the R-branch and
m = −J ′′ for the P-branch. The transition wave number is then
ν̃P,R = ν̃0 + (B ′ + B ′′ )m + (B ′ − B ′′ )m2 .
(912)
When plotted as ν̃P,R vs. m, the function is known as a Fortrat parabola, with m ≤ 0
corresponding to the P-branch and m > 0 corresponding to the R-branch. The function has
a turning point at
dν̃
= 0 = (B ′ + B ′′ ) + 2m(B ′ − B ′′ ),
dm
producing a band head at
mH = −
B ′ + B ′′
.
2(B ′ − B ′′ )
(913)
(914)
35
Substituting mH into Eq. (912) gives for the distance of the turning point
ν̃H − ν̃0 = −
(B ′ + B ′′ )2
.
4(B ′ − B ′′ )
(915)
Problem 82. Verify Eqs. (912) and (915).
If B ′ > B ′′ the band head will be in the P-branch, and the band is said to blue-degraded
because the spectrum has a sharp edge at longer wavelengths (νH < ν0 ) and tails off at short
wavelengths. If B ′ < B ′′ the band head will be in the R-branch, and the band is said to
red-degraded because the spectrum has a sharp edge at shorter wavelengths (νH > ν0 ) and
tails off at long wavelengths.
Problem 83. Calculate the first few transition frequencies (in wave numbers) for the Pand R-branches of the B(v ′ = 40) ← X(v ′′ = 0) transition of I2 . Calculate enough lines
to construct the Fortrat parabola and find the location of the band head. Use Herzberg’s
spectroscopic parameters for I2 .
Problem 84. Use the Table from Huber and Herzberg [5] to calculate the wavelengths of
the transitions X 1 Σ+ v ′′ = 0, J ′′ = 1 → B 1 Σ+ v ′ = 1, J ′ = 2 and X 1 Σ+ v ′′ = 0, J ′′ = 1 →
B 1 Σ+ v ′ = 1, J ′ = 0 of the HF molecule. Include the effects of anharmonicity, vibrationalrotation interaction, and centrifugal distortion.
LECTURE 37: BREAKDOWN OF THE BORN-OPPENHEIMER
APPROXIMATION
We now turn to the interesting question of what happens when the adiabatic approximation fails. When we wrote down the adiabatic Schrodinger equation, we ignored the nuclear
kinetic energy operator, T̂N . This is a reasonable approximation so long as the energies
of the electronic states are far apart. If two states are degenerate (i.e. if they cross), this
approximation fails completely. Let us see how this can happen.
There is a famous principle in quantum mechanics called the “non-crossing” rule, which
states that the potential energy curves of two electronic states of the same symmetry cannot
cross. It is an example of a general result known as the adiabatic theorem. (Caution: As we
shall see shortly, this is a rigorous result only for diatomic molecules, although it has more
wide spread applications.) In simple terms, it says that if we have two approximate wave
functions and we take a linear combination of them to get a better estimate of the energy,
36
the resulting energies will repel each other; i.e., the lower of the two energies gets still lower
and the higher one gets even higher. The proof, derived by von Neumann and Wigner, is as
follows. Suppose we know all but two of the energies and their eigenfunctions. We choose
two additional wave functions, ψ1 and ψ2 , that are orthogonal to each other and to all the
other wave functions. We construct the missing eigenfunctions as linear combinations of
these two,
ψ = c1 ψ1 + c2 ψ2 ,
(916)
and solve for the roots of the resulting secular equation (which comes from the variational
principle). In other words, we diagonalize the Hamiltonian,

 
H −E
H12
c
 11
 1  = 0
H21
H22 − E
c2
(917)
Each of the matrix elements is a function of R. The roots (eigenvalues) of the equation are
(
)
√
1
2
2
E± =
H11 + H22 ± (H11 − H22 ) + 4H12 .
(918)
2
Clearly, E+ > H22 (the energy of the unperturbed upper state), and E− < H11 . The only
way the two energies could be equal is if simultaneously
H11 (R) = H22 (R)
(919)
H12 (R) = 0.
(920)
and
In the case of a diatomic molecule, there is only one independent degree of freedom (R), and
it is impossible to satisfy both conditions at the same location. Hence the potential energy
curves repel each other. (This does not hold if the two states are of different symmetry,
because then H12 = 0.)
Let us take as an example the ionic and covalent states of NaCl, both of which have 1 Σ+
symmetry. If the neutral atoms approach each other very slowly (“adiabatically”), at some
point their interaction switches from covalent to ionic. What happens physically is that at
some distance an electron jumps from the Na atom, flies through space, and attaches itself
to the Cl atom.
The potential energy curves calculated for the ionic and covalent states, based on electrostatics and short range repulsion, intersect at some point, Rx . In the adiabatic approximation, if the neutral atoms approach each other sufficiently slowly, the electrons adjust
37
their configuration continuously until R = Rx , at which point an electron flies through space
and lands on the more electronegative atom. As R gets still smaller, the two ions approach
each other along the same adiabatic potential energy curve, which is now ionic. This is
called a “harpoon reaction” [12] The crossing point occurs where the attractive Coulomb
potential energy matches the the difference between the ionization potential of the Na atom
the electron affinity of the Cl atom,
e2
= IP − EA
Rx
(921)
or
Rx (Å) =
14.4
.
(IP − EA) (eV )
(922)
The reaction cross section is πRx2 . For the reaction
N a + Cl → N a+ + Cl− ,
(923)
IP (N a) = 5.13 eV , EA(Cl) = 3.6 eV , Rx = 9.6Å, and σ = 290 Å2 . Other examples of
harpoon reactions are
K + CH3 I → KI + CH3
(924)
Xe + Cl2 → XeCl + Cl.
(925)
and
Noting that the non-crossing rule predicts that the two lowest adiabatic potential energy
curves repel each other, we expect the adiabatic path to involve a switch from covalent to
ionic character at some point. On the other hand, we also expect that if the atoms collide
with very high velocity, they will not have time to transfer an electron, and the molecule
remains in its initial (covalent) configuration. In other words, it “jumps the track.” Another
way of saying this is that it is possible to calculate the ionic and covalent potential energy
curves separately without invoking configuration interaction. The resulting states, which are
not eigenstates of the electronic Hamiltonian, cross at R = Rx . These are called “diabatic
states.”
What is the probability of the system jumping across the gap (i.e. to remain on the same
diabatic state)? The answer is given by the famous Landau-Zener formula, which is based
on the assumptions that the diabatic states cross at t = 0 and that the energy difference
between the states varies linearly in time. This is equivalent to saying that the potential
38
energies vary linearly in the region of their intersection and that the atoms move at constant
velocity in the interaction region (say in a collision). If the system is initially in adiabatic
state 1, the probability of it jumping to adiabatic state 2 (i.e., remaining on the same
diabatic state) after passing the interaction region may be calculated using time-dependent
perturbation theory. We write the wave function as a superposition of the diabatic wave
functions, ψ d ,
ψ(t) = A(t)ψ1d e−iϕ1 (t) + B(t)ψ2d e−iϕ2 (t) ,
where
i
ϕs (t) = −
~
∫
t
Es (t′ )dt′ .
(926)
(927)
0
The phases may be derived from the time-dependent Schrodinger Eq. for the individual
states,
i~
∂ψs
= H̄ψs = Es (t)ψs .
∂t
(928)
The diabatic energies of the two states are E1 and E2 , and the energy difference is
E2 − E1 = E12 = v(F1 − F2 )t = αt,
(929)
where v is the velocity, and forces F1 and F2 are the slopes of the diabatic potential energy
curves. This last expression comes from the definition of the force,
dE
dE dx
=
= −F v
dt
dx dt
(930)
and the assumption that the potential varies linearly with distance in the vicinity of the
crossing.
Time-dependent perturbation theory gives for the evolution of the coefficients [13]
H12 B(t) ϕ(t)
e
~
(931)
H21 A(t) −ϕ(t)
e
~
(932)
Ȧ = −
and
Ḃ = −
where
∫
ϕ(t) = ϕ1 (t) − ϕ2 (t) =
t
E12 dt/~ =
0
αt2
2~
(933)
The quantities H12 = H21 are matrix elements of the Hamiltonian in the diabatic representation. We can eliminate one of the unknowns, say A(t), by differentiating Eq. (932) and
39
using Eq. (931) to replace Ȧ by B, giving [13]
B̈ = −
2
2
|H12
|B(t) iE12 Ḃ(t)
|H12
|B(t) iαtḂ(t)
−
=
−
−
.
~2
~
~2
~
(934)
This equation can be solved with some difficulty [14], and the result as t → ∞ is the
Landau-Zener transition probability of a diabatic (i.e, non-adiabatic) transition,
P = |B(t → ∞)|2 = e−π ξ ,
2
(935)
where ξ is called the adiabaticity parameter [12]
ξ = t∆E/h,
(936)
and ∆E(Rx ) is the change in nuclear kinetic energy at the distance where the electron jumps.
A large value of ξ means that the system remains on one adiabatic curve. In this case the
nuclei do not jump from diabatic curve to another. This is equivalent to saying that the
electron jumps in order to follow the adiabatically changing potential.
Suppose the electron jumps in a localized region of width b. Then the time the atoms
spend in this region is t = b/v. ∆E(Rx ) has the value
dV1 dV2 = b|∆Fx |.
∆E(Rx ) = b −
dR
dR Rx
(937)
It follows that the adiabaticity parameter has the value
ξ=
∆E 2
.
hv|∆Fx |
(938)
A large value of χ favoring the electron jump (i.e., remaining an a single adiabatic potential
curve) is the result of a large ∆E, a small v, and a small |∆F |. The last condition is
equivalent to having the two “tracks” nearly parallel. A net reaction occurs if a non-adiabatic
transition occurs as the atoms approach and not as they recede or the reverse. The total
reaction cross section is therefore given by
σ(E) = 2P (1 − P )σx ,
(939)
σx = πRx2 .
(940)
where
The non-adiabatic Harpoon reaction is a kinetic example of non-BO behavior. Much
more famous (and ubiquitous) are spectroscopic consequences of non-BO behavior [4, 8].
40
For example, the Ã1 B2u − X̃ 1 A1g transition of benzene is forbidden because x, y, z have E1u
and A2u symmetry, so that µ12 = 0. Nevertheless, the transition is observed with vibrational
transitions ∆v6 = 1. (The subscript 6 refers to a particular normal mode of the molecule.)
How does this happen?
When two (or more) electronic states are close in energy, the electronic Hamiltonian picks
up extra terms that depend on nuclear normal coordinates. We can write this perturbation
as a Taylor series,
Ĥe =
Ĥe0
+
∑
(
Qs
s
∂ Ĥe
∂Qs
)
+ ...
(941)
Qs =0
Truncating after the linear term gives a perturbation operator,
(
)
∑
∂
Ĥ
e
Ĥ ′ =
Qs
.
∂Qs
s
(942)
Qs =0
First order (time-independent) perturbation theory shows that the wave function of the
perturbed electronic state (in the case of benzene, ψà ) picks up a component from the
perturbing state. If we write the wave function of the unperturbed state as ψf0 , the perturbed
state becomes
ψf = ψf0 +
∑
ckf ψk0 ,
(943)
k
where the sum is over all (usually one) perturbing electronic states. The amount of mixing
(also called “intensity borrowing”) is given by the square of the coefficient,
∫ 0 ′ 0 3
ψk Ĥ ψf d r
ckf =
.
Ef0 − Ek0
(944)
In order for the integral to be non-zero, its integrand must contain a totally symmetric
component. In the benzene example, the perturbing state is C̃(1 E1u ). Since the X̃ → C̃
transition is dipole allowed (because x, y has the same symmetry as the C̃), mixing of the
C̃ and à state lends oscillator strength to the latter. The ν6 vibration, which induces the
perturbation, has e2g symmetry. The symmetry of the matrix element in Eq. (944) is
therefore (B2u ⊗ E2g ⊗ E1u ), which contains a component of A1g symmetry. (Verify this with
a character table.) This effect is called a “vibronic transition,” and the non-totally vibration
that allows it to happen is called the “coupling mode.”
Vibronic coupling leads to many important phenomena. Possibly the most interesting is
the conical intersection between potential energy surfaces. Once thought to be impossible,
41
then later to be very rare, it is now known to be very wide spread and to play a central role
in processes such as vision and combustion, and indeed it is essential for life.
A conical intersection is a point where two electronic states are degenerate. This degeneracy may be visualized by looking at a 2D subspace of the PES spanned by two normal
coordinates. (It is conventional to use the normal coordinates of the ground state.) One
of these coordinates is chosen to be totally symmetric. Variation along this coordinate has
the effect of sliding the energies of the two states up and down. For this reason it is called
a “tuning coordinate,” Qt . The other coordinate, Qc , corresponds to a vibronic coupling
mode. We will use pyrazine as an example [15, 16] and investigate the intersection of the
first two excited singlet states, S1 (B3u ) and S2 (B2u ). S1 is reached from the ground state
by an n → π ∗ transition and S2 by a π → π ∗ transition.
The pyrazine molecule as 24 normal modes, 5ag , 4b1u , 4b2u , 4b3g , 2au , b1g , 2b2g , 2b3u (Students should verify this). For this molecule there are five possible tuning modes of symmetry
ag . For the S1 and S2 states there is only one coupling mode, which must have b1g symmetry
for the coupling matrix element ⟨S1 |Qc |S2 ⟩ not to vanish (Students should verify this). This
mode is an out-of-plane deformation of the ring. Calculation of the two PESs shows that
they intersect for Qc = 0. We note that for Qc ̸= 0 the symmetry of the molecule is lowered
from D2h to C2h . When that happens, the S1 and S2 states have the same Bu symmetry
and do not cross, as predicted by the non-crossing rule. Although this rule applies only to
diatomics, it gives the correct result for this 2D subspace. For Qc ̸= 0, however, the states
have different symmetry, and the coupling matrix element is not zero. For the proper values
of the tuning coordinate the surfaces are able to cross.
A laser tuned to excite a vibronic level in the S2 manifold populates what is called a
“bright state,” namely one which has oscillator strength for the S0 → S2 transition. If there
was no coupling between S2 and S1 , the bright states would be eigenstates of He . Such a
state would live for a very long time, until the molecule either fluoresces or interacts with
its environment. In fact, there is very strong vibronic coupling, resulting is extremely rapid
(10 fs) internal conversion. This happens because the initially excited bright state is not an
eigenstate of the total Hamiltonian, but rather is a superposition of many eigenstates that
are embedded in the quasicontinuum of S1 . This process of internal conversion allows the
molecule to survive intact (without fragmenting) in the presence of UV radiation. Such UV
photostability is critical for survival of biological systems and plays a key role in evolution.
42
LECTURE 38: ANGULAR MOMENTUM COUPLING: THE WIGNER-WITMER
RULES
Reference: Chapter 2 of Zare
In atoms and molecules there are usually many sources of angular momentum, including
electron spin, nuclear spin, orbital angular momentum, and rotation. Just as it is impossible
to know all three components of a single angular momentum vector, similarly there are
constraints on what we may know (i.e., measure simultaneously) about sums of angular
momentum vectors. To give a concrete example, suppose L is the orbital angular momentum,
S is the spin, and J = L + S is the vector sum. It is possible to measure either the set
of observables L̂2 , Ŝ 2 , Jˆ2 , Jˆz or the set L̂2 , L̂z , Ŝ 2 , Ŝz . We must choose between them. In
other words, we may construct a complete basis set out of the eigenfunctions of one set of
operators or the other.
The defining characteristic of an angular momentum vector operator is its commutation
property,
[ĵx , ĵy ] = i~ĵz .
(945)
It is easy to show that if ĵ1 and ĵ2 are each angular momentum vectors, then their sum
ĵ = ĵ1 + ĵ2 is also an angular momentum vector. The individual operators have known
eigenfunctions and eigenvalues:
ĵ12 |j1 m1 ⟩ = j1 (j1 + 1)~2 |j1 m1 ⟩
ĵ22 |j2 m2 ⟩ = j2 (j2 + 1)~2 |j2 m2 ⟩
ĵ 2 |jm⟩ = j1 (j + 1)~2 |jm⟩
ĵ1z |j1 m1 ⟩ = m1 ~|j1 m1 ⟩
(946)
ĵ2z |j2 m2 ⟩ = m2 ~|j2 m2 ⟩
ĵz |jm⟩ = m~|jm⟩
Out of these eigenfunctions we may construct two representations:
1) Uncoupled representation: |j1 m1 ⟩|j2 m2 ⟩ which is also written as |j1 m1 j2 m2 ⟩
2) Coupled representation: |j1 j2 jm⟩ which is also written as |jm⟩
In the uncoupled representation, ĵ1 and ĵ2 are each allowed to precess about the z-axis.
By this is meant that ĵ1x , ĵ1y , ĵ2x , ĵ2y are unknown. This in turn means that the length of the
ĵ vector is unknown. If it were known precisely, it would put constraints on the components
43
ĵ1x , ĵ1y , ĵ2x , ĵ2y that violate the Uncertainty Principle. In fact we can show that j is restricted
by the triangle rule,
j1 − j2 ≤ j ≤ j2 + j1 .
(947)
In the coupled representation, ĵ1 and ĵ2 precess about ĵ. This means that m1 and m2 are
not known individually; only their sum is known:
m = m1 + m2 .
(948)
For given values of j1 and j2 we therefore have a choice of two representations:
In the uncoupled representation basis |j1 m1 ⟩|j2 m2 ⟩ there are (2j1 +1)(2j2 +1) eigenstates.
∑ 2 +j1
In the coupled representation basis |j1 j2 jm⟩ there are jk=j
(2k+1) = (2j1 +1)(2j2 +1)
2 −j1
eigenstates. In this sum, k is the number of projections of each value of j sum.
We may transform one basis set into the other using ladder operators,
ĵ± |jm⟩ =
√
j(j + 1) − m(m ± 1)~|j, m ± 1⟩.
(949)
That these operators are defined by
ĵ± = ĵx ± iĵy ,
(950)
ĵ± = ĵ1± + ĵ2± .
(951)
from which we deduce that
It is easiest to see how this is done by working out a specific example. Let’s use the case
of j1 = 3, j2 = 1, j = 3. We will use the shorthand |3m⟩ for the coupled representation and
|2m1 ⟩|1m2 ⟩ for the uncoupled representation. Clearly,
|33⟩ = |22⟩|11⟩.
(952)
Using the lowering operator for the coupled representation,
ĵ− |33⟩ =
√
6~|32⟩.
(953)
Using the lowering operator for the uncoupled representation,
[ĵ1− + ĵ2− ]|22⟩|11⟩ =
Comparing these two results gives
|32⟩ =
√
√
√
4~|21⟩|11⟩ + 2~|22⟩|10⟩.
2
|21⟩|11⟩ +
3
√
1
|22⟩|10⟩.
3
(954)
(955)
44
Note that the sum of squares of the coefficients is 1, as expected for a unitary matrix.
We repeat this process as far as possible:
|31⟩ = c1 |20⟩|11⟩ + c2 |21⟩|10⟩ + c3 |22⟩|1, −1⟩
|30⟩ = c4 |2, −1⟩|11⟩ + c5 |20⟩|10⟩ + c6 |21⟩|1, −1⟩
|3, −1⟩ = c7 |2 − 2⟩|11⟩ + c8 |2, −1⟩|10⟩ + c9 |20⟩|1, −1⟩
(956)
|3, −2⟩ = c10 |2, −2⟩|10⟩ + c11 |2, −1⟩|1, −1⟩
|3, −3⟩ = |2, −2⟩|1, −1⟩
The coefficients in this transformation are called Clebsch-Gordan coefficients, which are
calculated by using the lowering operator repeatedly. They form a unitary matrix:
|jm⟩ =
∑
C(j1 j2 j; m1 m2 m)|j1 m1 , j2 m2 ⟩
(957)
m1 ,m2
and the inverse
|j1 m1 , j2 m2 ⟩ =
∑
C(j1 j2 j; m1 m2 m)|jm⟩.
(958)
j,m
These are actually single sums because m = m1 + m2 .
The coefficients are defined by
C(j1 j2 j; m1 m2 m) = ⟨j1 m1 , j2 m2 |jm⟩ = ⟨jm|j1 m1 , j2 m2 ⟩.
They are related to the Wigner 3 − j symbols by
⟨j1 m1 , j2 m2 |j3 m3 ⟩ = (−1)j1 −j2 +m3
√

2j3 + 1 
(959)

j1 j2
j3
m1 m2 −m3

(960)
Our first application of angular momentum coupling is to figure out what are all the
possible terms formed when two atoms combine to form a diatomic molecule. Suppose the
atoms have spins S1 , S2 and orbital angular momenta L1 , L2 . We will assume that the spins
are coupled to give a total of S, while the orbital angular momenta are described by the
uncoupled representation, |L1 M1 ⟩|L2 M2 ⟩. Our goal is to discover all possible term symbols,
2S+1
Λ, including ± reflection symmetry and g, u inversion symmetry, when they exist. Recall
that Λ = |M1 + M2 |. Note that in the uncoupled representation, the length of the resultant
vector L is not well defined (i.e., is not known precisely).
45
The first thing we need to do is count the number of possible terms, with degenerate
states counted only once. The possible values of Λ are (L> − L< ) ≤ Λ ≤ (L> + L< ), where
L> is the larger of L1 , L2 and L< is the smaller of L1 , L2 . The states with Λ > 0 are all
doubly degenerate because the energy is unchanged upon changing the sign of M1 + M2 .
Note that a change in the sign of M is equivalent to a reflection in a plane containing the
bond axis because that reverses the direction of the precession. A complication is associated
with the Σ states, which are not degenerate and are characterized by a reflection parity:
|Σ+ ; M ⟩ = |L1 , M, L2 , −M ⟩ + |L1 , −M, L2 , M ⟩
|Σ− ; M ⟩ = |L1 , M, L2 , −M ⟩ − |L1 , −M, L2 , M ⟩
(961)
Our first task is to count the number of Σ states (or terms) and the number of Λ > 0 terms.
It is easy to show that the former is just (2L< + 1)(2S< + 1). For each value of S, a Σ state
state contains components of the type |L1 , −M2 , L2 , M2 ⟩, where L2 < L1 . But the number
of spin states is 2S< + 1, giving the claimed result. It takes a bit more work to show that
the total number of Σ and non-Σ states is (2S< + 1)(2L< + 1)(L> + 1).
The remaining job is to find the number of Σ± states and the number of g, u states; the
latter occurs if the molecule is homonuclear. The answer is given by the so-called WignerWitmer rules, which we shall only cite here. For reflection symmetry, recall that
σ(xz) = c2 (y)i.
(962)
The parity of an atom having electrons with orbital angular quantum numbers li is given by
p = (−1)q
where
q=
∑
li .
(963)
(964)
i
The total inversion parity for both atoms is p1 p2 . The parity associated with C2 (y) is (−1)L1
for atom 1 and (−1)L2 for atom 2. Hence the total reflection parity is p1 p2 (−1)L1 +L2 . If the
number of Σ states is even, half are Σ+ and half are Σ− . If this number is odd, the reflection
symmetry of the extra state is p1 p2 (−1)L1 +L2 . This applies for each spin multiplicity.
Homonuclear molecules also have exchange symmetry. The rules for g, u assignment
depend on the value of S, as follows:
46
For Σ states, if S is even, all Σ+ are g and Σ− are u. If S is odd, all Σ+ are u and Σ−
are g.
For Λ > 0, if Λ is odd, are an even number of states with the same Λ value, half of which
have g symmetry and half u symmetry. If Λ is even, there are an odd number of states, with
even values of S for a given Λ lambda value having g symmetry and odd S values having u
symmetry.
Example: Find all the terms for an oxygen molecule built up from two O(3 P ) atoms.
The possible values of Λ are 0, 1, 2, denoted as Σ, Π, ∆.
The possible values of S are 0,1,2.
The total number of terms is (2 × 1 + 1)(2 × 1 + 1)(1 + 1) = 18.
Of these, the number of Σ states is 9. These fall into three sets of single, triplet, and
quintets. In each set, there is one extra Σ+ state because p1 p2 (−1)L1 +L2 = 1.
The Π states are half g and half u because Λ = 1. The ∆ states are either g or u,
depending on the value of S. The extra Σ± states are also g or u, depending on the value of
S.
The final result is
1
1 + 1 − 1
1
1
Σ+
g , Σg , Σu , Πg , Πu , ∆g
3
3 + 3 − 3
3
3
Σ+
u , Σu , Σg , Πg , Πu , ∆u
5
5 + 5 − 5
5
5
Σ+
g , Σg , Σu , Πg , Πu , ∆g
(965)
LECTURE 39: ANGULAR MOMENTUM COUPLING: HUND’S CASES
In the previous lecture we discussed the effects of electronic angular momentum on the
electronic states of a diatomic molecule. We are now in a position to ask how rotational
angular momentum couples to the electronic part and explore its effect on the rotational
spectrum. The various types of angular momentum (excluding nuclear spin) are:
L = electronic orbital angular momentum
S = electronic spin angular momentum
R = nuclear rotational angular momentum
J = R + L + S = total angular momentum
It is also useful to identify the total angular momentum excluding spin, N = J − S.
47
We denote the projections of these operators on the molecular axis by
⟨L̂z ⟩ = Λ~
⟨Ŝz ⟩ = Σ~
⟨R̂z ⟩ = 0
(966)
⟨Jˆz ⟩ = Ω~ = (Λ + Σ)~
In addition, the projection of the total angular momentum operator on the space-fixed axis
is
⟨JˆZ ⟩ = M ~.
(967)
Note that Herzberg uses N instead of R, and K instead of N.
We cannot know the expectation values all of the angular momentum operators. Fortunately, real molecules have several limiting types of behavior known as the Hund cases. You
may think of these different cases as distinct representations or basis sets. When we say
that a state corresponds to a particular case, we mean that its total wave function may be
well described by one of the eigenfunctions is a particular basis.
The most common case is Hund’s case (a). In this case, the interaction of nuclear rotation
with electronic angular momentum is very weak, while the electronic motion is strongly
coupled to the nuclear axis. We may think of the molecule as a dumbbell with a flywheel
attached, with the inertial axis of the electrons along the bond axis and perpendicular to
R, forming a prolate symmetric top. The good quantum numbers are Λ, Σ, ΩJ, S. We
may denote the eigenfunctions as |ΛΣΩJSn⟩, where Ω = Λ + Σ, and n refers to all other
quantum numbers such as vibration. The rotational term energy is the same as for a prolate
symmetric top (A > B = C), with quantum number K replaced by Ω,
Fv (J) = Bv J(J + 1) + (A − Bv )Ω2 .
(968)
Because of the very light mass of the electron, A ≫ Bv . By convention, this term is absorbed
into the electronic term energy, Te , so that the rotation term becomes
Fv (J) = Bv [J(J + 1) − Ω2 ].
(969)
with J = Ω, Ω + 1, Ω + 2, · · · Examples illustrating the rotational levels of 2 Π1/2 ,2 Π3/2 and
3
∆1 ,3 ∆2 ,3 ∆2 are given in Figure 98 of Herzberg, Vol. I [2].
48
Since the energy depends only on the absolute values of Λ and Ω, the wave function is a
linear combination of eigenfunctions with ±Λ and ∓Σ, of the form |nΛ⟩|SΣ⟩|JΩM ⟩, giving
rise to states of definite parity, p± . For example, for a Π state with S = 1/2,
1
|n2 Π1/2 JM p± ⟩ = √ [|n, 1⟩|1/2, −1/2⟩|J, 1/2, M ⟩ ± |n, −1⟩|1/2, 1/2⟩|J, −1/2, M ⟩]
2
(970)
1
2
±
|n Π3/2 JM p ⟩ = √ [|n, 1⟩|1/2, 1/2⟩|J, 3/2, M ⟩ ± |n, −1⟩|1/2, −1/2⟩|J, −3/2, M ⟩]
2
Each state is doubly degenerate, with the ± sign corresponding to the two parities.
The rotational Hamiltonian for case (a) is given by
ĤRot =
1
(Jˆ − L̂ − Ŝ)2 .
2
2µRe
(971)
We may simplify this expression by noting that R is perpendicular to the molecular axis, so
that in the molecular frame
ĤRot =
1
[(Jˆx − L̂x − Ŝx )2 ] + [(Jˆy − L̂y − Ŝy )2 ]
2µRe2
1 ˆ2
[J + L̂2x + Ŝx2 + Jˆy2 + L̂2y + Ŝy2 − 2Jˆx L̂x − 2Jˆy L̂y − 2Jˆx Ŝx − 2Jˆy Ŝy + 2L̂x Ŝx + 2L̂y Ŝy ]
2µRe2 x
1
=
[(Jˆ2 − Jˆz2 ) + (L̂2 − L̂2z ) + (Ŝ 2 − Ŝz2 ) + (L̂+ Ŝ− + L̂− Ŝ+ ) − (Jˆ+ L̂− + Jˆ− L̂+ ) − (Jˆ+ Ŝ− + Jˆ− Ŝ+ )].
2µRe2
=
(972)
Since the wave function is not an eigenfunction of L̂2 , the expectation value of L̂2 − L̂2z is
lumped into the electronic term energy. The ladder operator terms have only off-diagonal
matrix elements, which couple different electronic states. These are perturbations, which
are omitted in the pure case (a). The origin of the ladder operators in the Hamiltonian may
be understood by examining a typical pair of terms:
L̂+ Ŝ− + L̂− Ŝ+ = (L̂x + iL̂y )(Ŝx − iŜy ) + (L̂x − iL̂y )(Ŝx + iŜy ) = 2L̂x Ŝx + 2L̂y Ŝy .
(973)
These terms are part of the spin-orbit operator
HSO = Av L̂ · Ŝ,
(974)
where Av is the spin-orbit coupling constant for vibrational level v. The effect of these
terms is to mix in states with ∆Λ = ±1 and ∆Σ = ∓1. The resulting perturbation is called
“homogeneous” because Ω is unchanged. In a similar fashion, the terms Jˆ+ L̂− + Jˆ− L̂+
49
are called the “L-uncoupling operator,” which induces an inhomogeneous perturbation that
mixes in states with ∆Λ = ±1 and ∆Ω = ∓1. Likewise, the terms Jˆ+ Ŝ− + Jˆ− Ŝ+ is the
“S-uncoupling operator,” which induces an inhomogeneous perturbation that mixes in states
with ∆Σ = ±1 and ∆Ω = ∓1.
Hund’s case (a) is built on the assumption that L and S are uncoupled and precess about
the molecular axis. This assumption breaks down either if L = 0 or if the molecule rotates
very rapidly. The criterion of rapid rotation is that Bv J ≫ Av Λ. In the limit of high J or for
the case of Σ states (i.e., for S = 0), the molecule is described by Hund’s case (b). In this
case, L still precesses about the molecular axis. Its projection, Λẑ couples to R to produce
an angular momentum vector, N. This vector than adds to S to produce the total angular
momentum, J. The quantum numbers that describe case (b) are contained in |nJSN Λ⟩.
The rotational lines are then split by the coupling of S with N. For example, for S = 1/2,
every J state is split into states labeled F1 and F2 . We can see the transition from case
(a) to case (b) by diagonalizing Ĥrot in a suitable basis set. Using the case (a) basis, the
eigenvectors of the Hamiltonian for a 2 Π state are
|F2 ⟩ = aJ |2 Π1/2 vJ⟩ − bJ |2 Π3/2 vJ⟩
|F1 ⟩ = bJ |2 Π1/2 vJ⟩ + aJ |2 Π3/2 vJ⟩
(975)
and the eigenvalues are
ERot (2 Π, v, J) = Bv [(J − 1/2)(J + 3/2) ± X/2],
(976)
where
X = [4(J − 1/2)(J + 3/2) + (Y − 2)2 ]1/2
Y = Av /Bv
√
(977)
X + (Y − 2)
aJ =
2X
√
X − (Y − 2)
bJ =
2X
2
In the case (a) limit, (Y − 2) ≫ (J − 1/2)(J + 3/2), bJ = 0, and the energy is split by
2ESO . In the case (b) limit, (Y − 2)2 ≪ (J − 1/2)(J + 3/2), aJ = bJ =, and the energy is
split by (2J − 1)(2J + 3).
[1] I. N. Levine, Molecular Spectroscopy, (Wiley-Interscience, New York)
50
[2] G. Herzberg, Molecular Spectra and Molecular Structure. I. Spectra of Diatomic Molecules,
(Van Nostrand, New York)
[3] G. Herzberg, Molecular Spectra and Molecular Structure. II. Infrared and Raman Spectra of
Polyatomic Molecules, (Van Nostrand, New York)
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[5] K. P. Huber and G. Herzberg, Molecular Spectra and Molecular Structure. IV. Constants of
Diatomic Molecules, (Van Nostrand, New York)
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[11] W. Domcke and G. Stock, Adv. Chem. Phys. 100, 1 (1997).
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