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852HW4_Solutions - PHYS852 Quantum Mechanics II Spring 2009...

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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2009 HOMEWORK ASSIGNMENT 4: 1. [20 points] Consider the shifted harmonic oscillator: H = P 2 2 m + 1 2 mω 2 X 2 + aX. Use perturbation theory to compute the eigenvalues to second order in a and the eigenstates to first- order. Answer: We can rewrite the Hamiltonian as H = ∞ summationdisplay n =0 E (0) n | n (0) )( n (0) | + aλ √ 2 parenleftBig A + A † parenrightBig where E (0) n = planckover2pi1 ω ( n + 1 2 ) , and λ = radicalBig planckover2pi1 Mω . The operators A and A † satisfy A | n (0) ) = √ n | ( n − 1) (0) ) and A † | n (0) ) = √ n + 1 | ( n + 1) (0) ) . With V = λ √ 2 ( A + A † ) we find E (1) n = λ √ 2 ( n (0) | parenleftBig A + A † parenrightBig | n (0) ) = 0 and | n (1) ) = − summationdisplay m negationslash = n | m (0) ) λ ( m (0) | ( A + A † ) | n (0) ) √ 2 planckover2pi1 ω ( m − n ) = − λ √ 2 planckover2pi1 ω summationdisplay m negationslash = n | m (0) ) ( m (0) | ( √ n | ( n − 1) (0) ) + √ n + 1 | ( n + 1) (0) ) ) m − n = − λ √ 2 planckover2pi1 ω summationdisplay m negationslash = n | m (0) ) √ nδ m,n − 1 + √ n + 1 δ m,n +1 m − n = − λ √ 2 planckover2pi1 ω parenleftBig − √ n | ( n − 1) (0) ) + √ n + 1 | ( n + 1) (0) ) parenrightBig (1) For the second order energy shift we find E (2) n = − λ 2 2 planckover2pi1 ω ( n (0) | parenleftBig A + A † parenrightBigparenleftBig − √ n | ( n − 1) (0) ) + √ n + 1 | ( n + 1) (0) ) parenrightBig = − λ 2 2 planckover2pi1 ω ( n (0) | parenleftBig radicalbig n ( n − 1) | ( n − 2) (0) ) − n | n (0) ) + ( n + 1) | n (0) ) + radicalbig ( n + 1)( n + 2) | ( n + 2) (0) ) parenrightBig = − λ 2 2 planckover2pi1 ω ( − n + n + 1) = − λ 2 2 planckover2pi1 ω 1 Putting the pieces together gives E n ≈ planckover2pi1 ω parenleftbigg n + 1 2 parenrightbigg − a 2 λ 2 2 planckover2pi1 ω and | n ) ≈ | n (0) ) + aλ √ 2 planckover2pi1 ω parenleftBig √ n | ( n − 1) (0) ) − √ n + 1 | ( n + 1) (0) ) parenrightBig The exact solution for the energy is E n = planckover2pi1 ω parenleftbigg n + 1 2 parenrightbigg − a 2 λ 2 2 planckover2pi1 ω so that the second-order result is exact. The exact solution for the wavefunction is | n ) = U S parenleftBig − aλ 2 planckover2pi1 ω parenrightBig | n (0) ) , where U S ( d ) = e − idP/ planckover2pi1 is the shift operator, defined by U S ( d ) | x ) = | x + d ) . To verify that this is the solution we can compute the wavefunction via ψ n ( x ) = ( x | n ) = ( x | U S parenleftBig − aλ 2 planckover2pi1 ω parenrightBig | n (0) ) = ( x + aλ 2 planckover2pi1 ω | n (0) ) = φ n parenleftBig x + aλ 2 planckover2pi1 ω parenrightBig where φ n ( x ) is the unperturbed n th harmonic oscillator wavefunction. This gives the correct result: the wavefunction is simply shifted to the left by the introduction of the linear potential....
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