{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

852HW4_Solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}