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

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

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

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

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

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

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

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

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

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

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

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

 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

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

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

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

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

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

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

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

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

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

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

… 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

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

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

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

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

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

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

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

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