midterm_solutions

midterm_solutions - PH 216 Midterm 2011 Instructions Your...

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PH 216, Midterm, 2011 Instructions: Your solutions are due Saturday at 9 AM , via email or in my office (I should be present on Saturday morning, but if you want to drop it friday night, or miss me, you can slip it under the door.) Late midterms will not be accepted except in case of natural catastrophe or national emergency. Feel free to consult any written (paper and ink) sources, but not computers or other people. I’ve made every effort to keep ambiguity out of the questions, but if some remains for you, I will try to remove it; please email ([email protected]) or feel free to call (325-6832 between 8 am and 9 pm). 1. An electron moves in Coulomb field centered at the origin of coordinates. With neglect of spin and relativistic corrections, the first excited state ( n = 2) is 4-fold degenerate: l = 0 , m = 0; l = 1 , m = ± 1 , 0. Consider what happens to this level in the presence of an additional non-central potentail V 0 = f ( r ) xy , where f ( r ) is some central function, well-behaved but not otherwise specified (it falls off rapidly enough as r → ∞ .) This perturbation is to be treated to first order. To this order the degenerate n = 2 level splits into several levels of different energies, each characterized by an energy shift Δ E and a degeneracy (perhaps non-degenerate, perhaps multiply degenerate.) (a) How many distinct energy levels are there? (b) What is the degeneracy of each? (c) Given the energy shift, call it A ( A > 0), for one of the levels, what are the values of the shifts for all of the other levels? Solution: With V = f ( r ) xy = f ( r ) r 2 sin 2 θ sin φ cos φ treated as perturbation, the unperturbed wave functions for energy level n = 2 are = 0 , m = 0 , R 20 ( r ) Y 00 , = 1 , m = 1 , R 21 ( r ) Y 11 , = 1 , m = 0 , R 21 ( r ) Y 10 , = 1 , m = - 1 , R 21 ( r ) Y 1 , - 1 . As they all correspond to the same energy, i.e. degeneracy occurs, we have to first calculate H 0 0 m 0 ‘m = h 0 m 0 | V | ‘m i = Z R 2 0 ( r ) R 2 ( r ) r 2 f ( r ) Y * 0 m 0 sin 2 θ sin φ cos φY ‘m dV. The required spherical harmonics are 1
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Y 00 = 1 4 π 1 2 , Y 11 = 3 8 π 1 2 sin θe , Y 10 = 3 4 π 1 2 cos θ, Y 1 , - 1 = 3 8 π 1 2 sin θe - . Considering the factor involving φ in the matrix elements H 0 0 m 0 ‘m we note that all such elements have on of the following factors: 2 π Z 0 sin φ cos φdφ = 0 , 2 π Z 0 e ± sin φ cos φdφ = 0 , except H 0 1 , - 1 , 1 , 1 and H 0 1 , 1 , 1 , - 1 , which have nonzero values H 0 1 , - 1 , 1 , 1 = 3 8 π Z [ R 21 ( r )] 2 r 4 f ( r ) dr π Z 0 sin 5 θdθ 2 π Z 0 sin φ cos φe - 2 = iA, H 0 1 , 1 , 1 , - 1 = - iA, with A = 1 5 Z [ R ( r )] 2 r 4 f ( r ) dr. We then calculate the secular equation 0 0 0 0 0 0 0 iA 0 0 0 0 0 - iA 0 0 - Δ E I = Δ E 0 0 0 0 Δ E 0 iA 0 0 Δ E 0 0 - iA 0 Δ E = 0 , whose solutions are Δ E = 0 , Δ E = 0 , Δ E = A, Δ E = - A .
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