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 2004 Test 3 Solutions April 16, 2004 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) Explain qualitatively the difference between classical hard sphere cross section and quantum mechanical low-energy cross section for hard spheres. σcl = πa 2, σQM = 4πa 2. The quantum result is identical to the result obtained for classical waves, i.e. it accounts for diffraction of waves around the sphere into the shadow region behind the sphere. If you prefer to think of a beam of electrons hitting the hard sphere, then quantum mechanically the wave packets can scatter from the sphere when their impact parameter is greater than a. (c) Optical theorem. State the optical theorem and discuss. σ = 4π k Imf(0). Incident plane wave brings in probability current density in z direction. Some of it gets scattered into various directions. Must give rise to decrease in current density behind target (θ = 0) due to destructive interference of incident plane wave and scattered wave in forward direction. So total flux scattered (vσ) related to forward scattering amplitude f(0). 1 (d) Fermi Golden rule. A beam of protons is scatters off a target containing neutrons. The scattering amplitude for p − n scattering is modelled in a certain energy range by f(θ) = χ†f (A + Bσp · σn)χi (1) where χf and χi are the final and initial spin states of the n − p system. Assume the initial (probe) proton spin is up, the inital (target) neutron spin is unknown, and only the final proton spin is measured. Use the Fermi Golden rule to find the ratio of the rates of spin up and spin down measurements. First note that from the definition of a cross section, the rate of scattering is 1 τ = σv. What we want is 1/τ↑ 1/τ↓ = σ↑ σ↓ = |f↑ f↓ |2, where 1/τ↑ is the rate of detecting scattered ↑ particles, etc. Now the initial state is | ↑ mn〉 (proton first), and final state is 〈m′pm′n|, where 1/2 + mn = m ′ p + m ′ n by a.m. conservation. So for final proton spin up we have |f↑|2 ∝ |〈↑↑ |V | ↑↑〉|2 + |〈↑↓ |V | ↑↓〉|2 = |A + B|2 + |A−B|2, whereas for final proton spin down we get |f↓|2 ∝ |〈↓↑ |V | ↑↓〉|2 = |2B|2, where matrix elements are easy to calculate using Sp · Sn = (SpzSnz + 2[Sp+Sn− + Sp−Sn+]). So ratio of transition rates is (|A + B|2 + |A − B|2)/4|B|2. (e) Spontaneous emission. A hydrogen atom in an excited state in free space decays after a certain time of order 10−9 sec. typically. How can this be? After all, the atom is in an eigenstate, which is a stationary state of the Hamiltonian, i.e. the time evolution of the state does not mix in any other eigenstates. Discuss this paradox. The paradox arises because the excited atom is in free space which, if it were truly empty, would mean that the atom would be in an eigenstate of the full Hamiltonian, i.e. it would have an infinite lifetime. Electromag- netic field turns out to have zero-point oscillations, however, constantly interacting with atom even in ground state. Proper answer requires quan- tum electrodynamics, which calls these processes vacuum polarizations. 2 (c) Calculate perturbatively the probability for the system to make a transition to the upper state after a time T . For sinusoidal perturbations, Pi→f ' |Vfi| 2 h̄2 sin2(ω − ω0)T/2 (ω − ω0)2 , where Vfi = 〈f |V0|i〉, and V0 is the t-independent part of the potential, i.e. B0(gpSpz + gnSnz). Here h̄ω0 = Ef − Ei = J/4 − (−3J/4) = J . We need (1st spin is proton, 2nd neutron): 〈S = 1,ms|gpSpz + gnSnz|S = 0, 0〉 = 〈S = 1,ms|gpSpz + gnSnz| ↑↓ − ↓↑〉/ √ 2 = 〈S = 1,ms| [ gp h̄ 2 √ 2 − gn h̄ 2 √ 2 ] | ↑↓ + ↓↑〉 = 〈S = 1,ms| [ gp h̄ 2 − gn h̄ 2 ] |S = 1,ms = 0〉 = (gp − gn) h̄ 2 δms=0 ⇒ Pi→f = (gp − gn) 2 4 B20 sin2(ω − ω0)T/2 (ω − ω0)2 (d) Sketch the probability of a transition as a function of time for fixed ω. Discuss with this picture the criterion for the breakdown of perturbation theory. Central peak has height |Vfit/2h̄|2 and width 4π/t, getting higher and narrower as time goes on (see fig.) Recall this is perturbative treatment, however: can’t get bigger than 1, so perturbation theory breaks down eventually. 3. Hard sphere. Particles are scattered by a potential V (r)    V0 r < r0 0 r > r0 (2) with V0 > 0. 5 (a) Find the differential cross section in Born approximation. f = − m 2πh̄2 ∫ d3r′V (r′)ei~q·~r ′ = − m 2πh̄2 ∫ 1 −1 d cos θ ∫ r0 0 dr′ r′2 V0eiqr cos θ = − m 2πh̄2 ∫ r0 0 dr′ r′2 2 sin qr′ qr′ = −2mV0 h̄2q ∫ r0 0 dr′ r′ sin qr′ = −2mV0 h̄2q3 (sin qr0 − qr0 cos qr0), and dσ/dΩ = |f |2. (b) What is the limit of your result when the particles are very slow? Very slow means |q| → 0, so ei~q·~r′ ' 1. So you can take limit as q → 0 of part a), or just say f = − m 2πh̄2 ∫ r0 0 4πr′2dr′V0 = −2mV0r 2 0 3h̄2 ⇒ dσ dΩ = 4m2V 20 r 6 0 9h̄4 (c) For this part, take dσ/dΩ = 10−12 cm2. If 1016 particles per cm2 per sec are incident on this target at slow speeds, how many per second are detected in a 1 cm2 detector located at 90◦ from the incident direction 1m (À r0) from the target? dN dt = (nv) · dσ dΩ · δΩ ' 1016 · 10−12 · 10−4 = 1 particle/sec, where I took δΩ ' 10−4 since the detector area 1 cm2 was approx. 10−4 of the total surface area at a distance of 100 cm. Since the particles are slow, the scattering angle is irrelevant. (d) Let V0 → ∞. Write down and solve the radial Schrödinger equation for the radial wave fctn. u0 = rψ|`=0 in the s-wave scattering approximation. Find the s-wave scattering phase shift δ0(k), and use it to calculate the differential scattering cross section. Compare with your answer in b) and explain the difference. Radial Schrödinger eqn. for radial function u = rψ (remember just like 1D eqn. for r): 6 − h̄ 2 2m ∂2 ∂r2 u0 = h̄2k2 2m u0 Outside the scattering potential, the solutions are just sums of 1D plane waves, e.g. u0 = Ae ikr + Be−ikr = rψ0. Boundary conditions u0(r0) = 0 since V0 → ∞, gives you −A/B = exp(−2ikr0). and now recall asymptotic ψ0 in scattering form: ψ ' 1 2i (−e−ikr kr + η0 eikr kr ) , and compare the 2 forms of ψ0 you have, giving η0 = −A B = e−2ikr0 δ0 = −kr0, where I used the definition of η0 = e 2iδ0 . The differential scattering cross section is then (compare with form of plane wave & identify coefficients of outgoing wave) dσ/dΩ = sin2 kr0/k 2. Difference with b) has to do that, although both parts assume particle is slow (` = 0), in part b) we had a weak potential to apply Born approx., whereas in part d) we assumed potential went to ∞. So we don’t expect the same answer. 4. Scattering from molecular hydrogen. A light, neutral spinless probe par- ticle of wave vector k scatters weakly off protons. The asymptotic (r → ∞) wave function has the form ψ(r) = e−k·r + a eikr r , (3) where a is a constant. (a) Find the differential scattering cross section and the total scattering cross section for a single hydrogen atom in this approximation. The coefficient of the outgoing wave eikr/r is a, but this is also the definition of the scattering amplitude f . Therefore |f |2 = dσ/dΩ = a2. Since a is constant the total cross section is σ = ∫ dΩa2 = 4πa2. (b) A hydrogen molecule now consists of two protons separated by a displace- ment d, which is unaltered by the scattering event. For small k, the scat- tering is not exactly isotropic. Find the differential cross section for an 7