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No friction. No air resistance. Perfect SpringPowerPoint Presentation

No friction. No air resistance. Perfect Spring

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Coupled Pendulums. Weak spring. Two normal modes. No friction. No air resistance. Perfect Spring. Time Dependent Two State Problem. Copyright – Michael D. Fayer, 2007. Start A Swinging. Some time later - . B swings with full amplitude. A stationary. Copyright – Michael D. Fayer, 2007.

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Weak spring

Two normal modes.

No friction. No air resistance. Perfect Spring

Time Dependent Two State Problem

Copyright – Michael D. Fayer, 2007

Some time later -

B swings with full amplitude.

A stationary

Copyright – Michael D. Fayer, 2007

Vibrational TransferVibration moves betweentwo modes of a molecule.

Electronic Excitation TransferElectronic excited state movesbetween two molecules.

Electron TransferElectron moves betweenmetal centers.

Copyright – Michael D. Fayer, 2007

Take molecules to be identical, so later will set

States of System

Consider two molecules and their lowest two energy levels

Copyright – Michael D. Fayer, 2007

H is time independent.Therefore

Time independent kets

Spatial wavefunction

Time dependent

Part of wavefunction

Initially, take there to be NO interaction between them

No “spring”

Copyright – Michael D. Fayer, 2007

(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

If molecules reasonably close together

Intermolecular Interactions

Copyright – Michael D. Fayer, 2007

For molecules that are identical

Pick energy scale so:

Therefore

Coupling strength

Thus:

Time dependent phase factors

Very Important

Copyright – Michael D. Fayer, 2007

Most general state is a superposition

Normalized

May be time dependent

Kets have time dependent parts

For case of identical molecules being considered:

Any time dependence must be in C1 & C2.

Then:

&

Have two kets describing states of the system.

Copyright – Michael D. Fayer, 2007

Left multiply by

Then:

Eq. of motion of coefficients

time independent. All time dependence in C1 & C2

Substitute into time dependent Schrödinger Equation:

Take derivative.

Copyright – Michael D. Fayer, 2007

then:

Second derivative of function equalsnegative constant times function – solutions, sin and cos.

Solving Equations of Motion:

have:

Copyright – Michael D. Fayer, 2007

And:

Copyright – Michael D. Fayer, 2007

This yields

To go further, need initial condition

Take for t = 0

Means: Molecule A excited at t = 0,

B not excited.

Sum of probabilities equals 1.

Copyright – Michael D. Fayer, 2007

probability amplitudes

For t = 0

Means: Molecule A excited at t = 0,

B not excited.

R = 1 & Q = 0

Copyright – Michael D. Fayer, 2007

In general:

Coefficient – Amplitude (for normalized kets)

Projection Operator:

Time dependent coefficients

Copyright – Michael D. Fayer, 2007

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

Consider:

Copyright – Michael D. Fayer, 2007

Projection Ops. Probability of finding system in

given it is in

Total probability is always 1 since cos2 + sin2 = 1

Copyright – Michael D. Fayer, 2007

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

At t = 0 PA = 1 (A excited)PB = 0 (B not excited)

Copyright – Michael D. Fayer, 2007

Consider two superpositions of

Eigenstate, Eigenvalue

Similarly

Eigenstate, Eigenvalue

Observables of Energy Operator

Stationary States

Copyright – Michael D. Fayer, 2007

2g

E0

Dimer splitting

E = 0

Delocalized States

pentacene in

p-terphenyl crystal

1.3 K

Probability of finding either molecule excited is equal

Recall E0 = 0If E0 not 0, splitting still symmetricabout E0 with splitting 2g.

Copyright – Michael D. Fayer, 2007

Make energy measurement equal probability of finding or -

is not an eigenstate

Use projection operators to find probability of being in eigenstate,

given that the system is in

complex conjugate

of previous expression

Copyright – Michael D. Fayer, 2007

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

Expectation Value

Copyright – Michael D. Fayer, 2007

E

D

B

E

E

A

B

As E increases Oscillations Faster

Less Probability Transferred

Non-Degenerate Case

Thermal Fluctuations change E &

Copyright – Michael D. Fayer, 2007

Three levels

…

…

n levels

Crystals

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, 2007

lattice spacing

j-1 j j+1

ground state of the jth molecule in lattice (normalized, orthogonal)

excited state of this 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

set of n-fold degenerate eigenstates

in the absence of intermolecular interactions

because any of the n molecules can be excited.

Excitation of a One Dimensional Lattice

Copyright – Michael D. Fayer, 2007

Translating a lattice by any number of lattice spacings, ,lattice looks identical.

Bloch Theorem – from group theory and symmetry properties of lattices

The exponential is the translationoperator. It moves functionone lattice spacing.

p is integer ranging from 0 to n-1.L = n, size of lattice

k = 2p/L

Bloch Theorem of Solid State Physics – Periodic Lattice Eigenstates

Because lattice is identical, following translation

Potential is unchanged by translation

Hamiltonian unchanged by translation

Eigenvectors unchanged by translation

Copyright – Michael D. Fayer, 2007

Bloch Theorem – eigenstates of lattice

Ket with excited state on jth molecule

Sum over all j possible positions (translations) of excited state.

Normalization so there is only a total of one excited state on entire 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 = 2p/L

Any number of lattice translations produces an equivalent function, result is a superposition of the kets with each of the n possible translations.

Copyright – Michael D. Fayer, 2007

Intermolecular coupling between adjacent molecules.Couples a molecule to molecules on either side.Like coupling in two state (two molecule) problem.

Molecular Hamiltonian

in absence of intermolecular

interactions.

Sum of single molecule Hamiltonians

The jth term gives Ee,the other terms give zero because

the ground state energy is zero.

One dimensional lattice problem with nearest neighbor interactions only

In the absence of intermolecular interactions,the energy of an excitation in the latticeis just the energy of the molecular excited state.

Copyright – Michael D. Fayer, 2007

state with jth molecule excited

Operate on Bloch states – eigenstates.

Inclusion of intermolecular interactions breaks the excited state degeneracy.

Copyright – Michael D. Fayer, 2007

combining

j + 1

j 1

Therefore, the exp. times ket, summed over all sites ( j )

Each of the terms in the square brackets can be multiplied by

In spite of difference in indices, the sum over j is sum over all lattice

sites because of cyclic boundary condition.

Copyright – Michael D. Fayer, 2007

Adding this result to

Gives the energy for the full Hamiltonian.

The nearest neighbor interaction with strength breaks the degeneracy.

Replacing exp. times ket, summed over all sites ( j ) with

Copyright – Michael D. Fayer, 2007

Each state delocalized over entire crystal.

2

1.5

)

g

1

(units of

0.5

0

e

-0.5

E(k) - E

-1

-1.5

-2

p

p

p

p

0

a

a

a

a

2

2

k

Quasi-continuous Range of energies from 2 to -2

Result (one dimension, nearest neighbor interaction only, )

k wave vector – labels levels a lattice spacing

Brillouin zone

Copyright – Michael D. Fayer, 2007

Exciton packet moves with well defined velocity. Coherent Transport.

Thermal fluctuations (phonon scattering)

localization, incoherent transport, hopping

Exciton Transport

Exciton wave packet more or less localized like free particle wave packet

Dispersion Relation:

Copyright – Michael D. Fayer, 2007

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