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p137bp4sol

# p137bp4sol - Physics 137B Fall 2007 Moore Problem Set 4...

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Physics 137B, Fall 2007, Moore Problem Set 4 Solutions 1. We start with Eq. (8.10) in Bransden: H 0 ψ (1) n + H ψ n (0) = E (0) n ψ (1) n + E (1) n ψ (0) n . In Dirac ket notation this reads H 0 | ψ (1) n + H | ψ (0) n = E (0) n | ψ (1) n + E (1) n | ψ (0) n . If we hit this equation with the bra ψ (1) n | we obtain ψ (1) n | H 0 | ψ (1) n + ψ (1) n | H | ψ (0) n = E (0) n ψ (1) n | ψ (1) n + E (1) n ψ (1) n | ψ (0) n . (Note that the first order corrections | ψ (1) n are not normalized, so we can’t simplify the first term on the right-hand side.) Rearranging, this reads ψ (1) n | H - E (1) n | ψ (0) n = - ψ (1) n | H 0 - E (0) n | ψ (1) n . Now taking the Hermitian conjugate, and using the fact that H 0 and H are Hermitian, we have ψ (0) n | H - E (1) n | ψ (1) n = - ψ (1) n | H 0 - E (0) n | ψ (1) n . By Eq. (8.17), the left-hand side is precisely E (2) n . 1

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2. For reference, we write the first two eigenfunctions for the one-dimensional harmonic oscillator, using Eq. (4.168) and α = ( mω/ ¯ h ) 1 / 2 : ψ 0 ( x ) = α π 1 / 2 e - α 2 x 2 / 2 , ψ 1 ( x ) = α 2 π 1 / 2 2 αx e - α 2 x 2 / 2 . The eigenfunctions of the unperturbed two-dimensional harmonic oscillator H 0 = - ¯ h 2 2 m 2 ∂x 2 + 2 ∂y 2 + 1 2 k ( x 2 + y 2 ) are just products of the one-dimensional eigenfunctions, i.e. Ψ n x n y ( x, y ) = ψ n x ( x ) ψ n y ( y ) , E n x n y = ¯ ( n x + n y + 1) .
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p137bp4sol - Physics 137B Fall 2007 Moore Problem Set 4...

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