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

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
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 m 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) | + a 2 parenleftBig A + A parenrightBig where E (0) n = planckover2pi1 ( n + 1 2 ) , and = radicalBig planckover2pi1 M . The operators A and A satisfy A | n (0) ) = n | ( n 1) (0) ) and A | n (0) ) = n + 1 | ( n + 1) (0) ) . With V = 2 ( A + A ) we find E (1) n = 2 ( n (0) | parenleftBig A + A parenrightBig | n (0) ) = 0 and | n (1) ) = summationdisplay m negationslash = n | m (0) ) ( m (0) | ( A + A ) | n (0) ) 2 planckover2pi1 ( m n ) = 2 planckover2pi1 summationdisplay m negationslash = n | m (0) ) ( m (0) | ( n | ( n 1) (0) ) + n + 1 | ( n + 1) (0) ) ) m n = 2 planckover2pi1 summationdisplay m negationslash = n | m (0) ) n m,n 1 + n + 1 m,n +1 m n = 2 planckover2pi1 parenleftBig n | ( n 1) (0) ) + n + 1 | ( n + 1) (0) ) parenrightBig (1) For the second order energy shift we find E (2) n = 2 2 planckover2pi1 ( n (0) | parenleftBig A + A parenrightBigparenleftBig n | ( n 1) (0) ) + n + 1 | ( n + 1) (0) ) parenrightBig = 2 2 planckover2pi1 ( 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 = 2 2 planckover2pi1 ( n + n + 1) = 2 2 planckover2pi1 1 Putting the pieces together gives E n planckover2pi1 parenleftbigg n + 1 2 parenrightbigg a 2 2 2 planckover2pi1 and | n ) | n (0) ) + a 2 planckover2pi1 parenleftBig n | ( n 1) (0) ) n + 1 | ( n + 1) (0) ) parenrightBig The exact solution for the energy is E n = planckover2pi1 parenleftbigg n + 1 2 parenrightbigg a 2 2 2 planckover2pi1 so that the second-order result is exact. The exact solution for the wavefunction is | n ) = U S parenleftBig a 2 planckover2pi1 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 n ( x ) = ( x | n ) = ( x | U S parenleftBig a 2 planckover2pi1 parenrightBig | n (0) ) = ( x + a 2 planckover2pi1 | n (0) ) = n parenleftBig x + a 2 planckover2pi1 parenrightBig where 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

This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.

Page1 / 11

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online