Past Test 3 with Solutions - Introduction to Quantum Mechanics II | PHY 4605, Exams of Physics

Download Past Test 3 with Solutions - Introduction to Quantum Mechanics II | PHY 4605 and more Exams Physics in PDF only on Docsity! PHY4605–Introduction to Quantum Mechanics II Spring 2005 Test 3 SOLUTIONS April 15, 2005 1. Short Answer. Must attempt (only) 4 of 6. Circle answers to be graded. (a) Using the figure sketching the wave function of a particle with the scat- tering potential a) turned off and b) turned on, state whether the po- tential was most likely attractive or repulsive, and estimate (quantita- tively!) the cross-section of the scat- tering process in the s-wave approx- imation in terms of the wave vector k of the scattering particle. Potential must be attractive in lower figure since wave function has been “pulled in” to well (δ0 > 0). From the figure, δ0 ' π/4, so dσ dΩ = sin2 δ0 k2 → σ = 4π sin 2 δ0 k2 = 2π k2 (b) State the electric dipole selection rules. What are the most likely de- cay paths to the ground state for a) an H atom starting in a 3d state? Draw them on the picture, and ex- plain why the other possibilities are “forbidden”. Do the same starting from b) the 2s state. All electric dipole transitions should have ∆` = ±1, and ∆m = ±1, 0. Therefore the only state to which the 322 state can decay is the 211. The 200 state is infinitely long lived in this approximation since it has nowhere to decay to which satisfies the selection rule. (c) Describe the three kinds of radiation processes related statistically by Ein- stein’s detailed balance argument. Discuss the conclusions of this argu- ment, or alternatively state how any two of the rates in the argument are related. 1 Einstein’s argument relates the rates of spontaneous emission, stimulated emission, and absorption of radiation of a population of atoms in equilib- rium with a bath of photons with frequency ω0. You were supposed to say what each process is. The rate of spontaneous emission is N2A, rate of stimulated emission is BnN2, and absorption is nN1C, where N2, 1 is population of upper, lower state and n the number of photons in the cavity. Einstein showed that A = B = C. (d) State the physical meaning of the optical theorem. Optical theorem: σ = Im(4πf(0))/k. The amount of flux scattered out of the incoming beam is σvcl. The for- ward part of the outgoing wave must be diminished by this amount, by conservation of probability. (e) State the Fermi Golden rule. Be sure to identify each quantity occurring in the expression you give. Rf = dPf dt ' Pf t = π|〈f |V̂0|i〉|2 2h̄2 ρ(ω0) (1) In other words, the rate of making transitions from state i → f is propor- tional to the matrix element of the perturbing potential V squared, times the density of states at an energy h̄ω0 equal to the difference between the two levels Ef − Ei. This form of the Fermi Golden Rule is for a bunch of two-level atoms exposed to incoherent radiation characterized by some distribution of energies ρ(ω), e.g. the distribution of radiation in a cavity. (f) Explain why the quantum total cross-section σ for hard-sphere scattering is 4 times the geometrical cross-section. Like the classical cross section for low-energy classical waves, the QM cross- section for particles scattering from a hard sphere is a factor of 4 larger than the classical cross-section for pariticles πa2, since in QM the particles can “diffract” around the sphere, filling in the classical shadow zone and leading to enhanced scattering. 2 3. Born approximation. A particle of mass m is scattered by a potential V (r) = V0 exp(−r/a). (a) Find the differential scattering cross section in the first Born approxima- tion. f(θ) = − m 2πh̄2 ∫ d3r′V (r′)eiq·r ′ = −mV0 2πh̄2 (2π) ∫ r′2 dr e−r ′/a ∫ π 0 dθ sin θ eiqr ′ cos θ = −2mV0 h̄2q ∫ ∞ 0 dr′ r′ sin(qr′)e−r ′/a = − 4mV0a 3 h̄2(1 + q2a2)2 , where q = k|k̂ − r̂| = 2k sin(θ/2). So dσ dΩ = |f |2 = 16m 2V 20 a 6 h̄4(1 + 4k2a2 sin2 θ/2)4 (b) Sketch the angular dependence for small and large k, where k is the wave number of the particle being scattered. At what k value does the scattering start to become significantly non-isotropic (estimate)? If ka ¿ 1, the sin θ/2 in the denominator won’t matter. So roughly speak- ing if k ¿ a−1, there will be no angular dependence of the cross section. One could have guessed this without doing the calculation for part a), be- cause as we discussed, the scattering is always isotropic when k is much smaller than the inverse range of the scattering (only length in the prob- lem). 5 (c) The criterion for the validity of the first Born approximation is that the 1st-order correction to the incident plane wave be small, |ψ(1)| ¿ |eik·r|. Using the integral form of the Schrödinger equation, translate this con- dition into a condition on V0, and simplify it for small k, ka ¿ 1. Hint: you may compare the incoming wave function and the first Born scattering correction to it at r = 0. Integral form of the S.-eqn is (see helpful formulae) ψ(r) = ψ0(r)− m 2πh̄2 ∫ eik|r−r ′| |r− r′|V (r ′)ψ(r′)d3r′, and the first Born correction is obtained by inserting ψ0 = e ik·r for ψ, so ∣∣∣∣ ∆ψ(1)(0) ψ(0)(0) ∣∣∣∣ = ∣∣∣∣ V0m 2πh̄2 ∫ eikr ′ r′ e−r ′/aeik·r ′ d3r′ ∣∣∣∣ = ∣∣∣∣ V0m h̄2 ∫ ∞ 0 r′dr′ ∫ π 0 sin θ′ dθ′eikr ′ e−r ′/aeik·r ′ ∣∣∣∣ = ∣∣∣∣ V0m h̄2 ∫ ∞ 0 r′dr′ ∫ π 0 sin θ′ dθ′eikr ′−r′/a+ikr′ cos θ′ ∣∣∣∣ = ∣∣∣∣ V0m h̄2 ∫ ∞ 0 r′dr′ ( −ie−r′/a e 2ikr′ − 1 kr′ )∣∣∣∣ = 2m|V0|a2 h̄2 √ 1 + 4k2a2 , so the criterion that the Born approximation be valid is 2m|V0|a2 h̄2 √ 1 + 4k2a2 ¿ 1 and in the limit ka ¿ 1, |V0| ¿ h̄ 2 2ma2 (d) What is the limit on V0 which comes from this criterion in the high-k limit, i.e. ka >> 1? Compare with answer to c). In limit ka À 1, find |V0| ¿ h̄2k/(ma) = h̄2(ka)/(ma2). So that if ka À 1, the Born approximation is actually valid over a much larger regime. 6