# ch08 - CHAPTER 8 1 The solutions are of the form where un(x...

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CHAPTER 8 1. The solutions are of the form ψ n 1 n 2 n 3 ( x , y , z ) = u n 1 ( x ) u n 2 ( y ) u n 3 ( z ) where u n ( x ) = 2 a sin n π x a ,and so on. The eigenvalues are E = E n 1 + E n 2 + E n 3 = h 2 π 2 2 m a 2 ( n 1 2 + n 2 2 + n 3 2 ) 2. (a) The lowest energy state corresponds to the lowest values of the integers { n 1 , n 2 , n 3 }, that is, {1,1,1)Thus E ground = h 2 π 2 2 ma 2 × 3 In units of h 2 π 2 2 ma 2 the energies are {1,1,1} 3 nondegenerate) {1,1,2},(1,2,1},(2,1,1} 6 (triple degeneracy) {1,2,2},{2,1,2}.{2,2,1} 9 (triple degeneracy) {3,1,1},{1,3,1},{1,1,3} 11 (triple degeneracy) {2,2,2} 12 (nondegenrate) {1,2,3},{1,3,2},{2,1,3},{2,3,1},{3,1,2},{3,2,1} 14 (6-fold degenerate) {2,2,3},{2,3,2},{3,2,2} 17 (triple degenerate) {1,1,4},{1,4,1},{4,1,1} 18 (triple degenerate) {1,3,3},{3,1,3},{3,3,1} 19 (triple degenerate) {1,2,4},{1,4,2},{2,1,4},{2,4,1},{4,1,2},{4,2,1} 21 (6-fold degenerate) 3. The problem breaks up into three separate, here identical systems. We know that the energy for a one-dimensional oscillator takes the values h ϖ ( n + 1 /2 ) , so that here the energy eigenvalues are E = h ϖ ( n 1 + n 2 + n 3 + 3 /2 ) The ground state energy correspons to the n values all zero. It is 3 2 h ϖ . 4. The energy eigenvalues in terms of h ϖ with the corresponding integers are (0,0,0) 3/2 degeneracy 1 (0,0,1) etc 5/2 3 (0,1,1) (0,0,2) etc 7/2 6 (1,1,1),(0,0,3),(0,1,2) etc 9/2 10 (1,1,2),(0,0,4),(0,2,2),(0,1,3) 11/2 15 (0,0,5),(0,1,4),(0,2,3)(1,2,2)

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(1,1,3) 13/2 21 (0,0,6),(0,1,5),(0,2,4),(0,3,3) (1,1,4),(1,2,3),(2,2,2), 15/2 28 (0,0,7),(0,1,6),(0,2,5),(0,3,4) (1,1,5),(1,2,4),(1,3,3),(2,2,3) 17/2 36 (0,0,8),(0,1,7),(0,2,6),(0,3,5) (0,4,4),(1,1,6),(1,2,5),(1,3,4) (2,2,4),(2,3,3) 19/2 45 (0,0,9),(0,1,8),(0,2,7),(0,3,6) (0,4,5)(1,1,7),(1,2,6),(1,3,5) (1,4,4),(2,2,5) (2,3,4),(3,3,3) 21/2 55 5. It follows from the relations x = ρ co s φ , y = ρ s in φ that d x = d ρ co s φ - ρ s in φ d φ ; dy=d ρ s in φ + ρ co s φ d φ Solving this we get d ρ = co s φ d x + s in φ d y ; ρ d φ =- s in φ d x + co s φ d y so that x = ρ x ρ + φ x φ = co s φ ρ - s in φ ρ φ and y = ρ y ρ + φ y φ = s in φ ρ + co s φ ρ φ We now need to work out 2 x 2 + 2 y 2
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• Spring '05
• mokhtari
• Energy, ground state energy, ϕ ρϕ, triple degeneracy

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