1. Time Dependent Two State Problem
Coupled Pendulums
Weak spring
No friction. No air resistance. Perfect Spring
Two normal modes.
Copyright – Michael D. Fayer, 2017 1 1 1

2. B swings with full amplitude. A stationary
Start A Swinging
Some time later -
B swings with full amplitude.
A stationary
Copyright – Michael D. Fayer, 2017 2 2 2

3. Electron Transfer Electron moves between metal centers.
Vibrational Transfer Vibration moves between two modes of a molecule.
Electronic Excitation Transfer Electronic excited state moves between two molecules.
Copyright – Michael D. Fayer, 2017 3 3 3

4. Consider two molecules and their lowest two energy levels
Take molecules to be identical, so later will set
States of System
Copyright – Michael D. Fayer, 2017 4 4 4

5. Initially, take there to be NO interaction between them No “spring”
H is time independent. Therefore
Time dependent
Part of wavefunction
Time independent kets 
Spatial wavefunction
Copyright – Michael D. Fayer, 2017 5 5 5

6. If molecules reasonably close together Intermolecular Interactions
(Like spring in pendulum problem)
Then energy of molecule A is influenced by B.
Energy of determined by both
Will no longer be eigenket of H
Then:
Coupling strength
Energy of interaction
Copyright – Michael D. Fayer, 2017 6 6 6

7. Thus: Coupling strength Time dependent phase factors Very Important
For molecules that are identical
Pick energy scale so:
Therefore
Copyright – Michael D. Fayer, 2017 7 7 7

8. For case of identical molecules being considered:
Have two kets describing states of the system.
Most general state is a superposition
Normalized
May be time dependent
Kets have time dependent parts
For case of identical molecules being considered:
Then:
&
Any time dependence must be in C1 & C2.
Copyright – Michael D. Fayer, 2017 8 8 8

9. Substitute into time dependent Schrödinger Equation:
Take derivative.
Left multiply by
Left multiply by
Then:
Eq. of motion of coefficients
time independent. All time dependence in C1 & C2
Copyright – Michael D. Fayer, 2017 9 9 9

10. Solving Equations of Motion:
have:
but:
then:
Second derivative of function equals negative constant times function – solutions, sin and cos.
Copyright – Michael D. Fayer, 2017 10 10 10

11. And:
Copyright – Michael D. Fayer, 2017 11 11 11

12. Sum of probabilities equals 1.
normalized
Sum of probabilities equals 1.
This yields
To go further, need initial condition
Take for t = 0
Means: Molecule A excited at t = 0,
B not excited.
Copyright – Michael D. Fayer, 2017 12 12 12

13.  R = 1 & Q = 0 For t = 0 Means: Molecule A excited at t = 0,
B not excited. 
R = 1 & Q = 0
For these initial conditions:
probability amplitudes
Copyright – Michael D. Fayer, 2017 13 13 13

14. Projection Operator: Time dependent coefficients Projection Operator
In general:
Coefficient – Amplitude (for normalized kets)
Copyright – Michael D. Fayer, 2017 14 14 14

15. Closed Brackets  Number
Consider:
Closed Brackets  Number
Absolute value squared of amplitude of particular ket in superposition
Probability of finding system in state given
that it is in superposition of states
Copyright – Michael D. Fayer, 2017 15 15 15

16. Projection Ops.  Probability of finding system in given it is in
Total probability is always 1 since cos2 + sin2 = 1
Copyright – Michael D. Fayer, 2017 16 16 16

17. At t = 0 PA = 1 (A excited) PB = 0 (B not excited)
When
PA = 0 (A not excited)
PB = 1 (B excited)
Excitation has transferred from A to B in time
(A excited again)
(B not excited)
In between times  Probability intermediate
Copyright – Michael D. Fayer, 2017 17 17 17

18. Stationary States Consider two superpositions of
Eigenstate,  Eigenvalue
Similarly
Observables of Energy Operator
Eigenstate,  Eigenvalue
Copyright – Michael D. Fayer, 2017 18 18 18

19. 2g E0 E = 0 Delocalized States
Recall E0 = 0 If E0 not 0, splitting still symmetric about E0 with splitting 2g.
E = 0 E0 2g
Dimer splitting 
Delocalized States
Probability of finding either molecule excited is equal
Copyright – Michael D. Fayer, 2017 19 19 19

20. Use projection operators to find probability of being in eigenstate,
given that the system is in
complex conjugate
of previous expression
Also
Make energy measurement  equal probability of finding  or -
is not an eigenstate
Copyright – Michael D. Fayer, 2017 20 20 20

21. If E0  0, get E0 Expectation Value
Half of measurements yield +; half 
One measurement on many systems  Expectation Value – should be 0.
Using
Expectation Value - Time independent
If E0  0, get E0
Copyright – Michael D. Fayer, 2017 21 21 21

22. As E increases  Oscillations Faster Less Probability Transferred
Non-Degenerate Case E A B D
As E increases  Oscillations Faster
Less Probability Transferred
Thermal Fluctuations change E & 
Copyright – Michael D. Fayer, 2017 22 22 22

23. Crystals 
Dimer splitting 
Three levels …
n levels
Using Bloch Theorem of Solid State Physics  can solve problem of n molecules or atoms where n is very large, e.g., 1020, a crystal lattice.
Copyright – Michael D. Fayer, 2017 23 23 23

24. … Excitation of a One Dimensional Lattice  j-1 j j+1
lattice spacing
ground state of the jth molecule in lattice (normalized, orthogonal)
excited state of jth molecule
Ground state of crystal with n molecules
Take ground state to be zero of energy.
Excited state of lattice, jth molecule excited, all other molecules in ground states
energy of single molecule in excited state, Ee
in the absence of intermolecular interactions
set of n-fold degenerate eigenstates
because any of the n molecules can be excited.
Copyright – Michael D. Fayer, 2017 24 24 24

25. Bloch Theorem of Solid State Physics – Periodic Lattice Eigenstates
Lattice spacing - 
Translating a lattice by any number of lattice spacings, , lattice looks identical.
Because lattice is identical, following translation
Potential is unchanged by translation
Hamiltonian unchanged by translation
Eigenvectors unchanged by translation
Bloch Theorem – from group theory and symmetry properties of lattices
p is integer ranging from 0 to n-1. L = n, size of lattice
k = 2p/L
The exponential is the translation operator. It moves function one lattice spacing.
Copyright – Michael D. Fayer, 2017 25 25 25

26. Different number of half wavelengths on lattice.
Any number of lattice translations produces an equivalent function, result is a superposition of the kets with each of the n possible translations.
Ket with excited state on jth molecule (single site function)
Sum over all j possible positions (translations) of excited state.
Normalization so there is only a total of one excited state on entire lattice.
Bloch Theorem – eigenstates of lattice
k is a wave vector. Different values of k
give different wavelengths.
Different number of half wavelengths on lattice.
In two state problem, there were two molecules and two eigenstates. For a lattice, there are n molecules, and n eigenstates.
There are n different orthonormal
arising from the n different values of the integer p, which give n different values of k.
k = 2p/L
Copyright – Michael D. Fayer, 2017 26 26 26

27. One dimensional lattice problem with nearest neighbor interactions only
Molecular Hamiltonian
in absence of intermolecular
interactions.
Intermolecular coupling between adjacent molecules. Couples a molecule to molecules on either side. Like coupling in two state (two molecule) problem.
Sum of single molecule Hamiltonians
The jth term gives Ee, the other terms give zero because
the ground state energy is zero.
In the absence of intermolecular interactions, the energy of an excitation in the lattice is just the energy of the molecular excited state.
Copyright – Michael D. Fayer, 2017 27 27 27

28. Operate on Bloch states – eigenstates.
Inclusion of intermolecular interactions breaks the excited state degeneracy.
state with jth molecule excited
coupling strength
Operate on Bloch states – eigenstates.
Copyright – Michael D. Fayer, 2017 28 28 28

29. Each of the terms in the square brackets can be multiplied by
combining
j + 1
j  1
In spite of difference in indices, the sum over j is sum over all lattice
sites because of cyclic boundary condition.
Therefore, the exp. times ket, summed over all sites ( j )
Copyright – Michael D. Fayer, 2017 29 29 29

30. Replacing exp. times ket, summed over all sites ( j ) with
factor out
Adding this result to
Gives the energy for the full Hamiltonian.
The nearest neighbor interaction with strength  breaks the degeneracy.
Copyright – Michael D. Fayer, 2017 30 30 30

31. Result (one dimension, nearest neighbor interaction only, )
k  wave vector – labels levels a  lattice spacing
Exciton Band
Each state delocalized over entire crystal.
Quasi-continuous Range of energies from 2 to -2 k -2
-1.5 -1
-0.5
0.5 1
1.5 2 a p
E(k) - E e
(units of g )
Brillouin zone
Copyright – Michael D. Fayer, 2017 31 31 31

32. Exciton packet moves with well defined velocity. Coherent Transport.
Exciton Transport
Exciton wave packet  more or less localized like free particle wave packet
Dispersion Relation:
Group Velocity:
Exciton packet moves with well defined velocity. Coherent Transport.
Thermal fluctuations (phonon scattering)
 localization, incoherent transport, hopping
Copyright – Michael D. Fayer, 2017 32 32 32