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No friction. No air resistance. Perfect Spring

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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No friction. No air resistance. Perfect Spring

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  1. Coupled Pendulums Weak spring Two normal modes. No friction. No air resistance. Perfect Spring Time Dependent Two State Problem Copyright – Michael D. Fayer, 2007

  2. Start A Swinging Some time later - B swings with full amplitude. A stationary Copyright – Michael D. Fayer, 2007

  3. 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

  4. 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

  5. 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

  6. (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

  7. For molecules that are identical Pick energy scale so: Therefore Coupling strength Thus: Time dependent phase factors Very Important Copyright – Michael D. Fayer, 2007

  8. 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

  9. Left multiply by 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

  10. but: then: Second derivative of function equalsnegative constant times function – solutions, sin and cos. Solving Equations of Motion: have: Copyright – Michael D. Fayer, 2007

  11. And: Copyright – Michael D. Fayer, 2007

  12. normalized 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

  13. For these initial conditions: probability amplitudes For t = 0 Means: Molecule A excited at t = 0, B not excited. R = 1 & Q = 0  Copyright – Michael D. Fayer, 2007

  14. Projection Operator In general: Coefficient – Amplitude (for normalized kets) Projection Operator: Time dependent coefficients Copyright – Michael D. Fayer, 2007

  15. 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

  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, 2007

  17. 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 At t = 0 PA = 1 (A excited)PB = 0 (B not excited) Copyright – Michael D. Fayer, 2007

  18. Consider two superpositions of Eigenstate,  Eigenvalue Similarly Eigenstate, Eigenvalue Observables of Energy Operator Stationary States Copyright – Michael D. Fayer, 2007

  19. 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

  20. Also 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

  21. 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

  22. A 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

  23. Dimer splitting    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

  24. 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

  25. Lattice spacing -  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 = 2p/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

  26. 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 = 2p/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

  27. 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

  28. coupling strength state with jth molecule excited Operate on Bloch states – eigenstates. Inclusion of intermolecular interactions breaks the excited state degeneracy. Copyright – Michael D. Fayer, 2007

  29. combining 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

  30. factor out 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

  31. Exciton Band 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

  32. Group Velocity: 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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