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Linear perturbation to harmonic oscillator

  1. Mar 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the first-order corrections to energy and the wavefunction, for a 1D harmonic oscillator which is linearly perturbed by ##H'=ax##.

    2. Relevant equations
    First-order correction to the energy is given by, ##E^{(1)}=\langle n|H'|n\rangle##, while first-order correction to the wave-function is, $$|n^{(1)}\rangle=\Sigma_{m\neq n}\frac{\langle m|H'|n\rangle}{E_n-E_m}|m\rangle.$$

    3. The attempt at a solution
    The energy correction is just ##0## by orthogonality arguments. For the corrections to the wavefunction, I have, $$|n^{(1)}\rangle=\frac{a\sqrt{\frac{\hbar}{2m\omega}}}{\hbar\omega}\Sigma_{m\neq n}\frac{\langle m|a+a^+|n\rangle}{n-m}|m\rangle.$$ The first term within the summation is only exists for ##m=n-1##, such that the term becomes, ##\sqrt{n}|n-1\rangle##. Similarly, the second term in the summation only exists for ##m=n+1##, such that the term becomes,##-\sqrt{n+1}|n+1\rangle##. Hence, the first order correction should be, $$\frac{a\sqrt{\frac{\hbar}{2m\omega}}}{\hbar\omega}(\sqrt{n}|n-1\rangle-\sqrt{n+1}|n+1\rangle).$$ I am not sure however, as I'm still getting my feet wet with perturbation theory and would like to know if I'm on the correct track.
     
  2. jcsd
  3. Mar 5, 2017 #2

    blue_leaf77

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    Looks fine, except that the last expression is applicable only for ##n>0##. What about the one for ##n=0##?
     
  4. Mar 5, 2017 #3
    For, ##n=0##, I'm thinking that the term involving the ##|n-1\rangle## would just go to ##0## by definition of the raising/lowering operators. Would that be correct?
     
  5. Mar 5, 2017 #4

    blue_leaf77

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    Yes.
     
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