Physics 74 Graduate Quantum Mechanics Solutions to Second Exam Spring 17 The points for each question are marked. Each question is worth points. Some possibly useful formulas appear at the end of the test. 1. A particle lies in a 1d moving harmonic oscillator 1 with potential V x t m xa t where a(t) is a smoothly increasing function from a t A. It is initially in at to the ground state with wave function 1/4 m x x m e. What is the probability that it is still in the ground state at t if the increase is (a) adiabatic (b) sudden? In the adiabatic approximation the lowest energy state of the initial Hamiltonian with probability 1. Hence there is no calculation to do. In the sudden approximation the probability is given by P. The ground state for the harmonic oscillator is always the same but we must replace x by the position relative to the minimum a(t) of the potentail at each moment so the ground state at time t is given by 1/4 xtm m xat exp We simply replace a(t) by A at the final time so we have * m m m xxdx exp xa exp x dx m m exp x Ax A x dx m ma m ma exp exp x x dx m ma 1 ma exp exp m 4 m ma ma ma exp exp exp. 4 4 We then square this to get the probability so we have adiabatic: P 1 sudden: P exp m A.
. A particle has time-dependent Hamiltonian H H W t where in some basis t H 1 and W t e where the perturbation W(t) applies only for t >. At t = the system is in the ground state of H. Find the probability as t that it goes to each of the excited states to leading order in the perturbation W. We use the probability for the non-initial state 1 i t leading order scattering matrix S i dtw te have energy n We therefore have for n = 1 so the frequency differences are E E F F F I P I F S together with the. The three eigenstates of H n 1 1 S1 dt W t e dt e e i i i i 1 1 S dt W t e dt e e i i i i It is then straightforward to get the probabilities it 1 t i t it t i t. 1 P 1 S i i P S i i 4
3. Show that the following wave function is an eigenstate of the free Dirac Hamiltonian: E cp ip 1 e where mc with energy E and find a relationship for E in terms of p and m. To check that it is an eigenstate of the Hamiltonian with energy E we have to simply let the Hamiltonian act on it. We find Ecp ip Ecp ip H cα P mc e c pmc e mc mc 4 cp mc E cp ip cp E cp m c ip e e 3 mc cp mc mc E cp mc p 4 cp E cp m c ip e mc E where we used the fact that is an eigenstate of with eigenvalue +1 to simplify. Now if this is an eigenstate with eigenvalue E we must have H E or in other words 4 cp E cp m c ip EE cp ip e e. mc E Emc The lower of the pair of these equation is obviously true and the upper will be true provided 4 cp Ecp m c E Ecp This is the correct relativistic formula 4 cpe c p m c E Ecp 4 c p m c E 4 E c p m c..
4. A single photon is detected by absorbing photons so that the probability of finding a photon is proportional to Er. Calculate this quantity first for a single photon in the state 1 q 1 with wave number q q x ˆ ˆ x q y and polariation y ε ˆ 1. Then compute it again for a single photon in the superposition state 1 1 q 1 1 q 1. We note that in every case we are going from one photon to ero and hence we must annihilate the photon; indeed we must annihilate only the photon that is present. We therefore have k ikr * ikr Er 1 q1 i akεke ak εke 1 q1 k V cq iqxxiqyy iε e a 1 q 1 iˆ e. V q iq r q1 q1 V Therefore for a single photon we would have cq cq Er 1 q1 ˆ ˆ V V where q q q. For the superposition state we have x y cq 1 iqxxiqyy iqxxiqyy ˆ i e e Er Er 1 q1 Er 1 q1 V iqyy iqyy cq iqxx y 1 cq iqxx ie ˆ e e ie ˆ cos qy V V We therefore have in this case cq Er cos qy y. V We note in this case that we have the periodic changes in intensity that one would expect from two waves interfering whereas a single wave yields just a smooth probability distribution.
5. An electron of mass m is in the state 111 of the 3D asymmetric harmonic oscillator 1 with potential V R m X 4Y 9Z. Find the decay rate to each of the possible final states with lower energy due to emission of a single photon in the dipole approximation. The unperturbed Hamiltonian is just three harmonic oscillators with angular frequencies and 3 in the x y and directions respectively. The eigenstates will be of the form 1 1 1 n p q with energies Enpq n p 3q np3q3. The three operators R can only raise or lower one of the three labels in n p q and only by one unit so since we want to decrease the energy the only possible final states will be 11 1 1 and 11 and the only non-ero matrix elements of R will be the x y and - components of these respectively. We then use our formula for the operator X in one dimension in each case but modify the frequency to the appropriate multiple of in each case. We therefore have: r R xˆ xˆ 11111 11111 11111 11 111 11 ax ax 111 m m r R yˆ yˆ 11 111 11 ay ay 111 m 4m r R ˆ ˆ 11 111 11 a a 111. m3 6m The frequency differences in each case are simply given by IF and 3 respectively. We then simply substitute these into the decay rate formulas to yield 4 111 11 3c m 3mc 3 3 8 3 6 4 111 1 1 3c 4m 3mc 4 111 11 3. 3c 6m mc
1D H.O.: 1 V x m x X aa m an nn1 a n n n 1 1 Electric field operator k ikr * Er i ak εke ak εke V ikr k Possibly Helpful Formulas: Free Dirac Hamiltonian Spontaneous H cα P mc icα mc Decay: 4 1 σ 3 IF r α 3c 1 σ 1 i 1 x 1 y i 1 Time-dependent Perturbation Theory 1 T S i dtw t e i t Possibly Helpful Integrals: Ax Bx B 4 A n x n 1 e dx e x e dxn!. A