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5.80 Small-Molecule Spectroscopy and Dynamics
Fall 2008
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5.80 Lecture #33
Fall, 2008
Page 1 of 10 pages
Lecture #33: Vibronic Coupling
Last time:
 1A – X
 1A
H2CO A
2
1
 -state is planar
Electronically forbidden if A
 -state is planar
vibronically allowed to alternate v′4 vibrational levels if A
 -state is not planar
inertial defect says A
expect to see all v′4 if not planar
staggering of v′4 level spacings⇒inversion through low barrier to planarity
dynamic vs. rigid molecule symmetry classification: molecular symmetry group
How does vibronic coupling really work?
What are the vibrational intensity factors analogous to Franck-Condon factors in the case of
vibronically allowed rather than electronically allowed transition?
See T. Azumi and K. Matsuzaki, Photochemistry and Photobiology 25, 315 (1977) for an
extremely readable review article.
Outline:
Crude Adiabatic Approximation
Correction of ψ for effect of neglected off-diagonal matrix elements
 1A example
H CO A
2
2
What happens to Franck-Condon factors for a “vibronically allowed” transition?
Two electronic basis sets — prediagonalize “symmetry-breaking” vibronic
interaction
Changes in shapes of potential curves (deperturb to a simpler, “more natural”
shape)
K. K. Innes’ model for vibrational band intensities and level staggering
Recall Born-Oppenheimer or “clamped nuclei” approximation.
We use this procedure to define complete sets of electronic and nuclear motion wavefunctions with
which we can FORMALLY expand exact ψ’s and compute (or parametrize) all properties of exact
eigenstates.
The simplest basis set is called “CRUDE ADIABATIC” (CA)
o
CA
(
)
ψCA
r,Q
=
ψ
r,Q
χ
(
)
jt
j
0
jt ( Q)
vibrational state
electronic state
fixed nuclear locations!
Q0 is a convenient reference structure (usually the equilibrium geometry or a high-symmetry potential
energy maximum or saddle point).
ψoj is the electronic wavefunction in the j-th electronic state computed at the chosen and explicitly
specified set of fixed nuclear coordinates Q0.
5.80 Lecture #33
Fall, 2008
Page 2 of 10 pages
χCA
jt (Q) is the vibration-rotation wavefunction computed from an approximate nuclear Schrödinger
Equation.
eigenvalue of
clamped nuclei
nuclear potential
electronic
kinetic energy of Schrödinger
energy bare nuclei Equation at Q0
∆U(r,Q) = U(r,Q) – U(r,Q0)
change in e–↔nuclear and e–↔e–
Coulomb energy
CA CA
⎡ TN (Q) + V(Q) + ε oj ( Q0 ) + ψ oj ( r,Q0 ) ∆ U(r,Q) ψ oj ( r,Q0 ) ⎤ χCA
⎣
⎦ jt ( Q ) = E jt χ jt ( Q )
effective potentialenergy surface
Note that the ∆U integral is evaluated using ψoj (r, Q0) thus cannot contain the exact effect of distortion
of molecule from Q0. To get a better representation of the distortion from Q0, we must use perturbation
theory.
We have explicitly excluded the effects of off-diagonal matrix elements. In order to get a better
approximation to the exact ψ, we must use perturbation theory to correct ψjt.
ψ jt (r,Q) = ψCA
jt (r,Q) +
∑
kr ≠ jt
CA
∆
U
ψ
{ψCA
jt
kr } CA
ψ (r,Q)
CA
E CA
−
E
jt
kr
= ψoj (r,Q0 )χCA
jt (Q) + ∑
∑
k≠ j
o
o
CA
χCA
ψ
∆
U
ψ
χ
kr
k
j
jt
CA
E CA
jt − E kr
r
× ψok (r,Q0 )χCA
kr (Q)
{ } means integrate over r and Q
( ) means integrate over Q
〈 〉 means integrate over r
(
kr
call this a vibronic
mixing coefficient
(both electronic and nuclear)
(nuclear)
(electronic)
This form of ψjt(r,Q) is called the Herzberg-Teller
expansion.
Now expand ∆U(r,Q) in power series about Q0 in each of the normal coordinates.
)
5.80 Lecture #33
Fall, 2008
∆ U = ∆ U(r,Q0 ) + ∑
= 0 by definition
of ∆U
n
Page 3 of 10 pages
⎡ ∂2 U ⎤
1
⎡ ∂U(r,Q) ⎤
Qn + ∑ ⎢
⎥ QnQm etc.
⎢ ∂Q ⎥ goes
2
∂Q
∂Q
into
n,m ⎣
⎣
n
⎦0 nuclear
n
m ⎦0


electronic
matrix
factor
to be initially
neglected
element
Now define the mixing coefficient.
γ nkr, jt
⎡ ∂U ⎤ o
ψ ok ( r,Q0 ) ⎢
ψ j ( r,Q0 )
⎥
⎣ ∂Qn ⎦0
≡
CA
E CA
jt − E kt
CA
χCA
kt Q n χ jt
everything collected into single parameter
n
o
CA
ψ jt (r,Q) = ψ oj (r,Q0 )χCA
jt (Q) + ∑∑∑ γ kr, jt ψ k (r,Q 0 )χ kr (Q)
k≠ j r
n
Electronic
Promoting
states vibrational mode
states
states
note vibrational
wavefunction for
k-th,
NOT j-th
electronic state!
But we can see that γ nkr, jt must vanish if
Γ k ⊗ Γ j ⊄ Γ Qn
Γ r ⊗ Γ t ⊄ Γ Qn
OR
(required by definition of ∆U above)
electronic selection
vibrational selection
rule
rule
which is equivalent to requiring that
Γ kr ⊗ Γ jt ⊂ Γ totally symmetric (and Qn is not totally symmetric).
 1A state.
So now we are ready to consider the specific case of the H2CO A
2
Out-of-plane Bending mode as promoter
b1 ⊗ A2 = B2
b1 vibration
vibronic symmetry
So we are considering vibronic coupling to the 1B2 state.
Non-Lecture
This is a simplified version of Innes’ model, to be discussed later.
Let’s make a really crude model for the out-of-plane bending levels of both 1A2 and 1B2 states.
This is an example when nature is too careless. Deperturb back to a simpler picture.
 state is NOT a double minimum non-planar state!!)
* both are harmonic (NB assume that the A
5.80 Lecture #33
Fall, 2008
Page 4 of 10 pages
* both have same frequency ω
∂U
* coupling is exclusively via Q n term.
∂Q n
O atom π in-plane
⇓
nσ
π
n0
π∗
↓
↓
↓
↓
σ∗
↓
3a12 4a12 1b 22 5a12 1b12 2b 22 2b10 6a10
elect. forbidden
elect. allowed (b-type)
elect. allowed (a-type)
elect. allowed (c-type)
2b1 ← 2b2
6a1 ← 2b2
2b1 ← 1b1
2b1 ← 5a1
 1A
X
1
 1A π*←n
A
2
0
1

B B2 σ*←n0
1
A1 π*←π
1
B1 π*←nσ
0 eV
3.5 eV
7.1 eV
8.0 eV
9.45 eV
 –X
 transition can borrow oscillator strength by “vibronic coupling” with
A
1
B2 via b1 vibration because A2 ⊗ b1 = B2
1
A1 via a2 vibration because A2 ⊗ a2 = A1
(a2 vibration doesn’t exist)
1
B1 via b2 vibration because A2 ⊗ b2 = B1
I will now show, via a simple model, that vibronic coupling accounts for both the oscillator strength of
 –X
 transition and the staggering of ν vibrational levels in A
 -state.
the vibrational bands in the A
4
 and B
 states is
Assume ν4 in A
⎧harmonic - not a double minimum non-planar state
⎪same ω and not displaced (necessarily not displaced if minimum or maximum is
convenient for
⎪⎪
calculating
planar — the high-symmetry point)
vibrational matrix ⎨
⎪
elements
∂U
⎪coupling is exclusively via
Q term
∂Q n n
⎩⎪
5.80 Lecture #33
Fall, 2008
1
Page 5 of 10 pages
B2
2
1
v4 = 0
1
3
A2
2
1
v4 = 0
mode #4, not 4th power
4
o CA
ψAv = ψoA χCA
+
γ
ψ
∑ Bv′, Av B χBv′
v
A
v′
Similarly for ψ Bv′
γ
4
v
 v ′, A
B
(perturbation theory)
⎡ ∂U ⎤ o v′4 Q 4 v4
≡ ψ ⎢
ψA
⎥
CA
E CA
−
E
⎣ ∂Q 4 ⎦0
v
v′
B
A
4
4

o

B
retaining only

levels of B
state in the
Herzberg-Teller
expansion
a mass-independent
electronic factor
∝
actually Q is mass
weighted
≡ β 4B A
1
⎡ v + 1δ v′ ,v +1 + v4 δ v′ ,v −1 ⎤
4 4
4 4
⎦
( µω )1/2 ⎣ 4
B− A
− ∆ T0 − ω 4 ( v′4 − v4 )

( T0 A + v4 ω 4 ) − ( T0 B + v4′ ω 4 )
comes from
modeling ν4 as
the same in

 and A
both B
states.
5.80 Lecture #33
Fall, 2008
Page 6 of 10 pages
Summary of non-zero matrix elements
4
3
2
1
v′4 = 0
1
1
2
2
3
3
4 matrix element
3
2
1
v′4 = 0
So we have
lump everything into this adjustable constant
o
CA
CA
⎡
⎤
ψ Av = ψ oA χ CA
+
βψ
v
χ
+
v
+
1χ
v

v −1
v +1
4
4
B
B
B
A
⎣
4
4
⎦
4
4
o
CA
CA
⎡
⎤
ψ Bv4′ = ψ oB χCA
−
βψ
v′
χ
+
v′
+
1χ
v ′

v ′ −1
v ′ +1
4
4
B
A
A
A
4
⎣
4
4
⎦
same constant but opposite sign because of
the energy denominator.
5.80 Lecture #33
Fall, 2008
Page 7 of 10 pages
 states have same potential surface but X
 is different).
v′ ← X
v′′ ( B
 and A
Transition probability for A
4
4
IAv′ , Xv′′ = ψ Av′ µ ψ Xv′′
4
4
4
2
4
 state is assumed to contribute
Only the B
IAv′ , Xv′′ = β
4
4
2
ψ µb ψ
o

B
o

X
1/2
CA
CA
CA
CA
⎡ v1/2
⎤
χ
χ
+
v
+
1
χ
χ
(
)
4
Bv
−1
Xv
4
Bv
+1
Xv4 ⎦
4
4
4
⎣
2
1/2
X
B− X
⎡ v4 q Bv −−1,v
⎤
+
v
+
1
q
+
2
v
v
+
1
⎡
⎤
(
)
(
)
4
v4 + 1,v4
4
4
2
⎣
⎦
4
4
2
⎥
positive
= β M b,B− X ⎢
squared
v B
v ⎥
v − 1
X
v + 1 X
⎢
× B
4
4
4
4 ⎦
⎣
terms
F-C factor
either sign
cross terms
Note that this is more complicated than usual FRANCK-CONDON expression for allowed transitions.
It is expressed in terms of Franck-Condon factors for B–X NOT A–X!!!! We still have a symmetry
selection rule for the ν4 vibrational mode because it is non-totally symmetric.
 v′ =
 v4 = even (because ν4 is not totally symmetric) which requires A
From v″4 = 0 we can only reach B
4
odd. Note that the intensity expression above vanishes for v′4 = 0 and v″4 = 0 because
B− X
q1,0
≡0
(by symmetry).
 –X
 system that is made
This can be expressed more generally, for any vibrational band in the A
 1B2 state promoted by ν′4.
allowed by vibronic coupling to the B
individual mode F-C factors (product over all modes except the promoting mode)
IAV′, XV′′ = β 2 M b,B − X
b-type
2
∏
vi ≠ v4
q vB vX
i
i
∑
vB4
2
vA4 Q 4 vB4 vB4 v4X
symmetry selection B X
v4 − v4 = even
rule
∴vA4 − v4X = odd
vA4 − vB4 = odd
K. K. Innes J. Mol. Spectrosc. 99, 294 (1983) performed a vibronic coupling calculation which not only
reproduced the mode-4 intensity promotion factors, but also “explained” the level staggering in the
 -state.
A
5.80 Lecture #33
Fall, 2008
Page 8 of 10 pages
In order to define complete basis sets, we solve an approximate Schrödinger equation by neglecting
specified terms in the exact H, or by ignoring off-diagonal elements of these terms.
In the crude adiabatic approximation, we define potential curves by ignoring terms of the form
CA
ψCA
.
jt ∆ H(r,Q) ψkr
We showed that, by expanding ∆U as power series in Q (the normal mode displacements), we get
(H′electronic ) jk = ∑
n
⎡ ∂U ⎤ o
ψoj ( r,Q 0 ) ⎢
ψk ( r,Q 0 ) Q n = ∑ γ njk Q n .
⎥
⎣ ∂Q n ⎦0
n
We can go to a new electronic basis set by diagonalizing H 0 + ∑ γ njk Q n .
n
Suppose we have two harmonic zero-order potential curves, V0(Qn), for mode n of electronic states j and
k. Then we have the following zero-order and diagonalized potential curves.
Vk(Qn)
Vk0 (Qn)
H′ = γnjkQ n
Vj0 (Qn)
Qn = 0
Vj(Qn)
5.80 Lecture #33
Fall, 2008
Page 9 of 10 pages
Upper curve gets narrower.
Lower curve turns into a double minimum curve.
Qn = 0 points of both curves do not shift because H′ → 0 at Q = Q0 by definition.
Vibrational eigenstates of lower curve will exhibit the pattern of a symmetric double minimum potential.
Vk0 ( Q n ) = ω 2k Q 2n ⎫⎪
⎬
Vj0 ( Q n ) = ω 2jQ 2n ⎪⎭
H′ij ( Q n ) = γ njk Q1n
here we are allowing harmonic frequencies to be different
Second-order perturbation theory:
Vk = ω Q
2
k
2
n
(γ ) Q
+
(ω − ω ) Q + T
n 2
jk
2
2
j
n
2
k
2
n
ek
− Tej
= ω′k2Q 2n + αQ n4
(We do power series expansion of the second term about Qn = 0. There can be no
constant term because Vk vanishes at Qn = 0. Note that the coefficient of Q 2n changes
because it is ω 2k plus a Q 2n term from the power series.)
Vj = ω′j2Q 2n − αQ n4
(same α because it is the same expansion but with opposite sign energy denominator)
ω′ = ω +
2
k
(ω
2
k
α≡
2
k
− ω′
2
k
(γ )
n 2
jk
Tek − Tej
) = − (ω
(γ )
n 2
jk
( Tek − Tej )
2
2
j
(ω
from power series expansion
− ω′j
2
(γ )
)=−T
ek
2
k
n 2
jk
get opposite sign shifts in the effective harmonic frequency
− Tej
− ω2j )
get a quartic term that depends on difference in ω's for j and k.
The quartic term vanishes if ωj = ωk.
This shows that upper state ω increases and lower state ω decreases.
“Exact” 2 × 2 deperturbation treatment for the potential curves
⎛ ∆V
−E
⎜
2
⎜
⎜⎜ H
12
⎝
(
⎞
1/2
⎟
⎤
Vk + Vj ⎡ ∆ V2
2
⎟ ⇒ V± =
±⎢
+ H12 ⎥
⎣ 4
⎦
−∆V
2
⎟
− E⎟
⎠
2
H12
)
∆ V = ω 2k − ω 2j Q 2n + Tek − Tej (includes difference between minima of Vk and Vj)
(
2
2
Vk + Vj ω 2k + ω 2j 2 ⎡ ω k − ω j
⎢
V± =
+
Qn ±
2
2
⎣
4
)
2
1/2
( )
Q n4 + γ njk
⎤
2
2⎥
Qn
⎦
5.80 Lecture #33
Fall, 2008
Page 10 of 10 pages
1/2
(the ∆Te/2 term seems to have been omitted from the [ ] term)
For large γn, second term in [ ]1/2 will dominate at small |Qn| but first term will eventually
dominate at large |Qn|. We obtain two perturbed potential energy curves.
Now go back to the original vibronic Hamiltonian and get a degenerate perturbation expression for the
energy levels.
A second-order perturbation treatment of this kind of 2-state interaction in the CA picture cannot give
this type of level stagger. It is necessary to set up and diagonalize two large dimension matrices
⎧odd quanta of upper state
HI ⎨
⎩
even quanta of lower state
⎧even quanta of upper state
H II ⎨
⎩
odd quanta of lower state
because of odd-even symmetry of a symmetric (not necessarily harmonic) potential, there can be no
coupling matrix elements between these two matrices.
The level shifts are larger for the lower states in HII than those in HI. For example, the lower state v = 1
level is pushed down by v = 0 and 2 of the upper state, but the lower state v = 0 level is only pushed
down by v = 1.
This produces level staggering.
−X
 intensity and A
 -state level
K. K. Innes [J. Mol. Spectrosc. 99, 294-301 (1983)] reproduced A
pattern with
 
∆ T0B− A = 28035 cm −1


ω B4 = ω A4 = 1125 cm −1
H′Av  ,Bv  = v  +1 = β ( vA + 1)
1/2
A
B
A
H′Av  ,Bv  = v  −1 = βv1/2

A
A
B
A
β = 3138 cm −1