1. 2014 International Symposium on Molecular Spectroscopy
Full-Dimensional Franck-Condon Factors for the Acetylene à (1Au)—X̃ (1Sg+) Transition in the Harmonic Normal Mode Basis
Barratt Park, Bob Field
2014 International Symposium on Molecular Spectroscopy

2. Outline Methodology Main gerade FC progressions ungerade Bn polyads
Coordinate transformation
Considerations for bent—linear systems
Method of Generating Functions
Change of basis for v5’’
Calculation of D and d
Main gerade FC progressions
v3’←0
2jm Vkn
Evaluation of new S1 Force Field
ungerade Bn polyads
Trends in IR-UV absorption
IR-UV-DF progressions from 3n61
ZOBS ratios in IR-UV-DF from 3n61
gerade Bn polyads: Fluorescence intensities and local bender plucks

3. Franck-Condon Factors
Transition Intensity:
Often m is approximately constant:
But, the acetylene ×X̃ transition is vibronically allowed, so .
Y’vib and Y’’vib have different vibrational normal coordinates, so a change of basis
must be performed in order to evaluate the integral.

4. Coordinate Transformation
Shift of equilibrium geometry
Normal mode displacement
L = Eckart Rot (Axis Switching)
l = cartesian displacements to normal coord
g = converts to dimensionless q
L depends nonlinearly on q’’ (and parametrically on re’ and re’’). It is defined according to
A first-order Taylor expansion (by Watson) gives

5. Eckart Considerations for the linear—bent problem? Ã X̃
2R + 7V ↔ 3R + 6V
2 Rotations
3 Stretches
+ 2 Bends (Doubly degenerate)
9 rotovibrational
degrees of freedom
3 Rotations
+ 6 Vibrations
9 rotovibrational degrees of freedom
t indexes the bending coordinates

6. Eckart Considerations for the linear—bent problem? Ã X̃
2R + 7V ↔ 3R + 6V
2 Rotations
3 Stretches
+ 2 Bends (Doubly degenerate)
9 rotovibrational
degrees of freedom
3 Rotations
+ 6 Vibrations
9 rotovibrational degrees of freedom z
In the linear state, the (x, y) subblock of I is trivially diagonal.
Therefore, c is not defined.
t indexes the bending coordinates
x and y are not
well defined.
Instead, rotation about the molecular axis is defined by the
bending coordinates,

7. Eckart Considerations for the linear—bent problem? Ã X̃
2R + 7V ↔ 3R + 6V
2 Rotations
3 Stretches
+ 2 Bends (Doubly degenerate)
9 rotovibrational
degrees of freedom
3 Rotations
+ 6 Vibrations
9 rotovibrational degrees of freedom
To restore c, choose a nonlinear reference configuration and locate the x and y axes to diagonalize I.
The natural choice of reference configuration for the problem of the Ã←X̃ transition is the trans-bent à equilibrium.
t indexes the bending coordinates
Only bending modes have amplitude in the x,y plane.
Also, since the trans à equilibrium is centrosymmetric, q5,x = 0.
Therefore, q4 has the only non-zero contribution to Ixy.

8. Eckart Considerations for the linear—bent problem
Diagonalize I in the trans-bent equilibrium
2R + 7V ↔ 3R + 6V
t indexes the bending coordinates

9. Eckart Considerations for the linear—bent problem
2R + 7V ↔ 3R + 6V x
Choose the solution z ,
or, in polar coordinates, .
Constraint: Displacement in q4’’ must lie in the plane of the Ã-state Equilibrium.
This constraint is equivalent to removing the angular factor exp(il4f4) factor from the 2DHO wavefunction and integrating it in the rotational part.

10. The Eckart FC Propensity Rule for Linear-Bent Transition+
v4’ = v5,y’’
(torsion)
(out-of-plane cis-bend)
The v4’ = v5,y’’ rule is very rigid. Out-of-plane trans-bend (n4,y’’) does not participate in the Franck-Condon integral, because it violates the Eckart conditions for the transition by rotating the Ã-state molecule.
The in-plane cis-bend FC propensity is much weaker, primarily because of Duscinsky rotation into the other bu mode (antisymmetric stretch).
v6’ v5,x’’ ~
(in plane cis-bend)

11. Sharp and Rosenstock: Method of Generating Functions
Make use of the exponential generating function for Hermite Polynomials ,
which allows us to obtain wavefunctions from .
In an n-dimensional product basis, the generating function becomes .
We seek overlap integrals between wavefunctions, which are given by

12. Method of Generating Functions: Linear—Bent considerations
The integral is exactly solvable and the solutions are given by Sharp and Rosenstock1-2 (in terms of D and d).
However, to set f4 to zero, it is useful to incorporate the generating function for degenerate bending mode wavefunctions
T.E. Sharp and H.M. Rosenstock, JCP 41, 3453 (1964).
Botter, Dibeler, Walker, Rosenstock JCP 44, 1271 (1966).

13. v3’ ← Origin progression
Tobiason’s Force Field1
(numbers match Watson2)
Jiang’s revised Force Field3 4 1 1 3
Tobiason, Utz, Sibert, Crim, JCP 99, 5762 (1993).
Watson, J. Mol Spec., 207, 276 (2001).
Jiang, Baraban, Park, Clark, Field, J. Phys. Chem. A, (2013).
Ingold and King, J. Chem. Soc. 1953, (1953)

14. FC Progressions in Emission from gerade levels
Lower State
X̃(mv2’’ + 8v4’’)
X̃(kv2’’ + nv4’’)
normal
modes
local
modes
Upper State
k = 0
Tobiason’s Force Field
Ã(origin)
Jiang’s Force Field
k = 1
Ã-state gerade modes
Mode
Description
Symmetry
w/cm-1
n1’
sym stretch ag
3052.1
n2’
CC stretch
1420.9
n3’
trans bend
1098.0
Ã(32)
k = 3
Ã(2131)
X̃-state gerade modes
Mode
Description
Symmetry
w/cm-1
n1’’
sym stretch
sg+
n2’’
CC stretch
n4’’
trans bend pg
608.73
k = 3
Ã(2132) 2 4 6 5 10 15 20 m n

15. IR-UV-DF progressions from Ã(3n61)
Lower State
X̃(mv2’’ + 5v4’’ + v5’’)
X̃(nv4’’ + v5’’)
Upper State
Ã-state ungerade modes
Mode
Description
Symmetry
w/cm-1
n4’
torsion au
787.7
n5’
antisym stretch bu
3032.4
n6’
cis bend
801.6
Ã(3261)
Ã(3361)
X̃-state ungerade modes
Mode
Description
Symmetry
w/cm-1
n3’’
antisym stretch
su+
n5’’
cis bend pu
729.08 1 2 3 5 10 15 m n

16. X̃-state ungerade intermediates for IR-UV Double Resonance
pu intermediate:
n3’’ + n4’’
n1’’ + n5’’
D∞h:
(sg+ ⊗ pu) = pu
(su+ ⊗ pg) = pu
Gives intensity to au levels
Gives intensity to bu levels
pu correllates with au ⊕ bu. Why the strong preference?
Merer, Yamakita, Tsuchiya, Steeves, Bechtel, Field JCP 129, (2008): “ f”

17. IR-UV to ungerade Bn polyads
The calculation correctly reproduces the observation that it matters which pu intermediate you choose. v1’’+v5’’ gives more intensity to au levels while v3’’+v4’’ gives more intensity to bu levels. In fact to zeroth order, v3’’+v4’’ cannot go to au states because the bu component of v4’’ does not participate in FC overlap.
Calculated FC Factors to Bn polyads

18. The search for the perfect “Local Bender Pluck”

19. Change of Basis Normal—Local Mode
Define the coordinates
Derive the ladder operators
Apply the ladder operators to determine
change-of-basis coefficients
Caveat: these wavefunctions do not have
well-defined +/- or g/u symmetry, so they
must be symmetrized.

20. The Perfect Pluck
Intensities from ag members of Bn polyads to the extreme local benders.
Addition of v3’ will help gain intensity.
v3’ will also turn off Darling-Dennison between v4’ and v6’ via x36. NB NB NB

21. Thank You Joshua Baraban Bryan Changala
FIELD GROUP Bob Field Jun Jiang Carrie Womack David Grimes Yan Zhou Tim Barnum Ethan Klein John Muenter Catherine Salidragas Steve Coy
ALSO STARRING
Joshua Baraban
Bryan Changala
Adam Steeves
Prof. Anthony Merer
FUNDING!!
DOE Grant DE-FG02-87ER13671

22. Coordinate Transformation
Shift of equilibrium geometry
Normal mode displacement
L = Eckart Rot (Axis Switching)
l = cartesian to normal coord
g = converts to dimensionless q
L depends nonlinearly on q’’ (and parametrically on re’ and re’’). It is defined according to
A first-order Taylor expansion (by Watson) gives

23. Coordinate Transformation
A first-order Taylor expansion (by Watson) gives

24. Method of Generating Functions: Linear—Bent considerations
The overall integral is simplified if we use bending mode wavefunctions only for q4’’ and not for q5’’ (otherwise we end up with exp(f5), sqrt(r5) out front, and other headaches.)
Use Laplace’s approximation to deal with sqrt(r4).
Deal with q5 by solving in the q5x, q5y basis and doing a basis transformation to v5, l5 afterwards.