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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2008 HOMEWORK ASSIGNMENT 9: SOLUTIONS 1. Prove [ A, ¯ H ] = A and [ A † , ¯ H ] = − A † , for the harmonic oscillator. Answer: [ A, ¯ H ] = [ A,A † A + 1 / 2] = [ A,A † A ] = AA † A − A † AA = ( A † A + 1) A − A † AA = A [ A † , ¯ H ] = [ A † ,A † A ] = A † A † A − A † AA † = A † A † A − A † ( A † A + 1) = − A † 1 2. Verify that the transformations X → λ ¯ X and P → planckover2pi1 λ ¯ P transforms the Hamiltonian as ¯ H = 1 2 ( ¯ X 2 + ¯ P 2 ) . Derive the required length scale λ , and show that H = planckover2pi1 ω ¯ H . Answer: Start from H = P 2 2 m + 1 2 mω 2 X 2 let X = λ ¯ X and P = ( planckover2pi1 /λ ) ¯ P , to give H = planckover2pi1 2 2 mλ 2 ¯ P 2 + 1 2 mω 2 λ 2 ¯ X 2 let mω 2 λ 2 = planckover2pi1 2 mλ 2 , which requires λ = radicalbigg planckover2pi1 mω , to get mλ 2 planckover2pi1 2 H = 1 2 ( ¯ X 2 + ¯ P 2 ) Since mλ 2 planckover2pi1 2 = planckover2pi1 ω , this gives ¯ H = 1 2 ( ¯ X 2 + ¯ P 2 ) where we have defined H = planckover2pi1 ω ¯ H 2 3. Use the recursion relation ψ n ( x ) = radicalbigg 2 n x λ ψ n 1 ( x ) − radicalbigg n − 1 n ψ n 2 ( x ) to compute ψ 2 ( x ), ψ 3 ( x ), and ψ 4 ( x ). Make plots of the probability densities for the first five eigen states. Answer: ψ 2 ( x ) = x λ ψ 1 ( x ) − radicalbigg 1 2 ψ ( x ) = √ 2 [ √ πλ ] 1 / 2 x 2 λ 2 e x 2 2 λ 2 − 1 [2 √ πλ ] 1 / 2 e x 2 2 λ 2 = 1 [8 √ πλ ] 1 / 2 parenleftbigg 4 x 2 λ 2 − 2 parenrightbigg e x 2 2 λ 2 ψ 3 ( x ) = radicalbigg 2 3 x λ ψ 2 ( x ) − radicalbigg 2 3 ψ 1 ( x ) = radicalBigg 2 24 √ πλ parenleftbigg 4 x 2 λ 2 − 2 parenrightbigg x λ e x 2 2 λ 2 − radicalBigg 4 3 √ πλ x λ e x 2 2 λ 2 = 1 [48 √ πλ ] 1 / 2 parenleftbigg 8 x 3 λ 3 − 12 x λ parenrightbigg e...
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 Fall '08
 MichaelMoore
 mechanics, Work, Boundary value problem, λ, ground state, recursion relation

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