PHY 389K Homework Set 8 Solutions: Vibrations and Rotations of Diatomic Molecules, Assignments of Quantum Mechanics

Download PHY 389K Homework Set 8 Solutions: Vibrations and Rotations of Diatomic Molecules and more Assignments Quantum Mechanics in PDF only on Docsity! PHY 389K QM1, Homework Set 8 Solutions Matthias Ihl 11/08/2006 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 This problem is an extension of the problem that was on the midterm. (a) The diatomic molecule performs both vibrations and rotations (cf. text- book page 127 and 146 ff.) (b) The energy difference ∆En = En+1 − En can be gleaned from the wave- length of the missing peak λ0 = 3.460 × 10 −6m, (1) because it corresponds to the (forbidden) transition with ∆l = 0. We obtain ∆En = ~ω = h c λ0 = 0.3586eV. (2) (c) We will use λ0 = 3.460 × 10 −6m corresponding to l = 0 and λ1 = 3.485 × 10−6m corresponding to l = 1. Then we have ∆El=0→1 = hc ( 1 λ0 − 1 λ1 ) = 2.573 × 10−3eV. (3) (d) This is due to selection rules that do not allow for transitions with ∆l = 0 (cf. textbook page 141), because the corresponding matrix element 1 vanishes identically. (e) The frequency of oscillation is given by ω = 2π c λ0 = 5.45 × 1014s. (4) As for the moment of inertia, one should determine values of I for all peaks and average over them to get the best result, but for simplicity, let us look at the first peak only: ∆El=0→1 = ~ 2 I =⇒ I = 2.7 × 10−47kgm2. (5) 2 Problem 2 Here, I will omit the vector arrows to avoid unnecessary cluttering. Let us introduce the following definitions: Q = Q2 − Q1, (6) P = m1P1 − m2P2 m1 + m2 , (7) Qcm = m1Q1 + m2Q2 m1 + m2 , (8) Pcm = P1 + P2. (9) The standard Heisenberg commutation relations are [Qi, Pj] = i~δij, [Qi, Qj ] = 0, [Pi, Pj] = 0. (10) Using these relations, it can be easily checked that [Q, P ] = m1 m1 + m2 [Q2, P2] + m2 m1 + m2 [Q1, P1] = m1 + m2 m1 + m2 i~ = i~, (11) [Qcm, Pcm] = m1 + m2 m1 + m2 i~ = i~, (12) [Q, Pcm] = [Q2, P2] − [Q1, P1] = i~ − i~ = 0, (13) [Qcm, P ] = − m1m2 m1 + m2 [Q1, P1] + m2m1 m1 + m2 [Q2, P2] = 0. (14) All other commutators are trivially zero.