PHY 389K Homework Set 11 Solutions: Problems and Energy Eigenvalues in Quantum Mechanics, Assignments of Health sciences

Download PHY 389K Homework Set 11 Solutions: Problems and Energy Eigenvalues in Quantum Mechanics and more Assignments Health sciences in PDF only on Docsity! PHY 389K QM1, Homework Set 11 Solutions Matthias Ihl 12/06/2006 Note: I will post updated versions of the homework solutions on my home- page: http://zippy.ph.utexas.edu/~msihl/PHY389K/ We will frequently work in God-given units c = ~ = 1. The casual reader may also want to set 1 = 2 = π = −1. 1 Problem 1 (a) The necessary steps are outlined in the textbook (chapter V.3) leading up to equations (3.14) on page 179. (b) This is essentially the content of the mathematical note on pages 180 and 181. (c) For Hermitian Qκ = Q † κ, it is shown in the Appendix to section V.3 of the textbook that the functions cj = −cj, (1) are purely imaginary and the functions aj = aj , (2) are real. 2 Problem 2 An account of stationary perturbation theory can be found in the hand-out “Perturbation theory”. As usual, the zeroeth order energy is given by E (0) j,m = ~ 2 2I j(j + 1) − gqB3 2m ~m. (3) 1 The first order energy eigenvalues can be calculated as follows E (1) j,m = E (0) j,m + α〈j, m|J 2 2 |j, m〉 = E (0) j,m + α 4 〈j, m|(J+J− + J−J+)|j, m〉 = E (0) j,m + α 2 (j(j + 1) − m2). The calculation of the first-order eigenvectors involves a sum over all the zeroth order eigenstates except for the diagonal ones. Therefore only the off-diagonal parts of J22 , namely J 2 + and J 2 −, contribute and we are left with |j, m〉(1) = |j, m〉(0)+α gq~B3 m ( √ j(j + 1) − m(m + 1) √ j(j + 1) − (m + 1)(m + 2)|j, m + 2〉(0) − √ j(j + 1) − m(m − 1) √ j(j + 1) − (m − 1)(m − 2)|j, m − 2〉(0) ) . 3 Problem 3 Here we consider a samll perturbation to the harmonic oscillator Hamiltonian H(0) = P 2 2µ + 1 2 KQ2, H = H(0) + H ′ = H(0) + λQ. (4) Since Q = √ ~ 2µω (a† + a), (5) does not have a diagonal piece, it is immediately evident that the first order correction to the eigenvalues vanishes, i.e., E(1)n = E (0) n + λ〈n (0)|Q|n(0)〉 = E(0)n . (6)