Download Introduction to Quantum Mechanics II - Problem Set 6 with Solutions | PHY 4605 and more Assignments Physics in PDF only on Docsity! 1 PHY4605 Problem Set 6 Solutions 1. Atomic configurations. Griffiths Problem 5.12. (a) H: 1s; He: 1s2; Li: 1s22s; Be: 1s22s2; B: 1s22s22p; C: 1s22s22p2; N: 1s22s22p3; O: 1s22s22p4; F: 1s22s22p5; Ne: 1s22s22p6 (b) H: 2S1/2; He: 1S0; Li: 2S1/2. There is only one possibility in each of these cases for the term symbol, because ` = 0, so the triangle rule allows only one value for j, i.e. s. For the other cases one must be careful to determine first the possible total `, then the possible total s, then the possible j’s which can result. B: the one extra p electron means ` = 1, s = 1/2, so the possibilities for j are 1/2, 3/2: 2P1/2, 2P3/2. C: two 2p electrons can have spins which add to 1 or 0, while their orbital a.m. can be 2, 1, or 0. This makes ten possible combinations: 1S0, 3S1, 1P1, 3P0, 3P1, 3P2, 1D2, 3D1, 3D2, 3D3. For N there are 3 ` = 1 momenta which can add to 0,1,2, or 3, and 3 spins 1/2 which can add to 1/2, or 3/2: 4S3/2, 2S1/2, 4P1/2, 4P3/2, 4P5/2, 2P1/2, 2P3/2, 4D1/2, 4D3/2, 4D5/2, 4D7/2, 2D3/2, 2D5/2, 4F3/2, 4F5/2, 4F7/2, 4F9/2, 2F7/2, 2F5/2. 2. Yukawa Hydrogen Griffiths Problem 7.14. Take φ = √ 1 πb3 e−r/b, b a variational parameter. The expectation value of the kinetic energy 〈T 〉 has the same form as for usual Hydrogen, but with a0 → b, i.e. 〈T 〉 → ~2 2mb2 . The expectation value of the Yukawa potential in φ is 〈V 〉 = − e 2 4π²0 4π πb3 ∫ ∞ 0 dr r2 e−2r/b e−µr r = = − e 2 4π²0 4 b3 ∫ ∞ 0 dr r e−(µ+2/b)r = − e 2 4π²0 4 b3 1 (µ + 2/b)2 = − e 2 4π²0 1 b(1 + µb/2)2 ≈ − e 2 4π²0 1 b (1− µb + 3µ2b2/4), 2 where the last expansion follows for sufficiently small µ. The expectation value of H is then 〈H〉 = ~ 2 2mb2 − e 2 4π²0 (1− µb + 3(µb)2/4)/b ⇒ ∂〈H〉 ∂b = 0 = − ~ 2 mb3 + e2 4π²0 ( 1 b2 − 3µ2/4 ) This is a cubic equation, but we only need the perturbative solution. The leading order solution, if we neglect the µ2 term, gives b ≈ 4π²0~ 2 me2 = a0, so write b = a0 + δ and expand for small δ: 0 = − 1 (a0 + δ)3 + 1 a0 ( 1 (a0 + δ)2 − 3µ2/4 ) 0 ≈ −1 + 1 + δ a0 − 3µ 2 4 1 a0 (a30 + ...) ⇒ δ a0 ≈ 3µ2a20/4 So b ≈ a0(1 + 3µ2a20/4). Inserting into the expression for 〈H〉, we find 〈H〉 = −R + e2µ/(4π²0) +O(µ2). 3. Rubber band Helium Griffiths Problem 7.17. (a) First invert variable change, r1 = (u + v)/ √ 2; r2 = (u− v)/ √ 2; r21 + r 2 2 = u 2 + v2. Now I want to show that ( ∂2f ∂x21 + ∂ 2f ∂x22 ) = ( ∂2f ∂u2x + ∂ 2f ∂v2x ) , also for
