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Unformatted text preview: Physics 137B, Fall 2004 Quantum Mechanics II Final exam 12/20/2004, 8:10-11:00 a.m. Solution outlines The mean on this exam was 60/100. I. (a) l ( l + 1)¯ h 2 = 6¯ h 2 . (b) The values of total j are 3 / 2 and 5 / 2, with one multiplet of each. Hence there are four states with value 15¯ h 2 / 4, and 6 states with value 35¯ h 2 / 4. (c) Apply L · S = 1 2 (( L + S ) 2- L 2 + S 2 ) to the result from (b). (d) The first rule is Δ l = ± 1, so we need to find either l = 1 or l = 3 states, of energy lower than the initial state. Only the 2p orbitals satisfy these conditions. (e) Adding five spin-half variables, the maximum possible total s quantum number is 5/2. No lesser value of s would allow the required eigenvalue of S z , so the eigenvalue of S 2 is s ( s + 1)¯ h 2 = 35¯ h 2 / 4. II. The rate of transitions for a single initial state and single final state is ¯ W ab = πI ( ω ba ) ¯ h 2 c (cos 2 θ ) | D ba | 2 . (1) Because the different i values- 1 , , 1 differ only by rotations, and the initial light is unpolarized, we can calculate the transition rate for one particular initial state, say i = 0. (Of course, computing a different initial state or summing all initial states is acceptable, as long as the final answer is correct.) Then only the z component of the dipole operator gives a nonzero result, which we write as h ψ 210 | - ez | ψ 100 i . The average of cos 2 θ over unpolarized light gives an additional factor of 1/3....
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This note was uploaded on 08/01/2008 for the course PHYSICS 137B taught by Professor Moore during the Fall '07 term at Berkeley.
- Fall '07