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2nd order pertubation theory of harmonic oscillator

  1. May 9, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm having some trouble calculating the 2nd order energy shift in a problem.
    I am given the pertubation:
    [tex]\hat{H}'=\alpha \hat{p},[/tex]
    where $\alpha$ is a constant, and [itex]\hat{p}[/itex] is given by:
    [tex]p=i\sqrt{\frac{\hbar m\omega }{2}}\left( {{a}_{+}}-{{a}_{-}} \right),[/tex]
    where [itex]{a}_{+}[/itex] and [itex]{a}_{-}[/itex] are the usual ladder operators.

    2. Relevant equations
    Now, according to my book, the 2nd order energy shift is given by:
    [tex]E_{n}^{2}=\sum\limits_{m\ne n}{\frac{{{\left| \left\langle \psi _{m}^{0} \right|H'\left| \psi _{n}^{0} \right\rangle \right|}^{2}}}{E_{n}^{0}-E_{m}^{0}}}[/tex]

    3. The attempt at a solution
    Now, what I have tried to do is to calculate the term inside the power of 2. And so far I have done this:
    [tex]\begin{align}
    & E_{n}^{1}=\alpha i\sqrt{\frac{\hbar m\omega }{2}}\int{\psi _{m}^{*}\left( {{{\hat{a}}}_{+}}-{{{\hat{a}}}_{-}} \right)}\,{{\psi }_{n}}\,dx=\alpha i\sqrt{\frac{\hbar m\omega }{2}}\left( \int{\psi _{m}^{*}\,{{{\hat{a}}}_{+}}{{\psi }_{n}}\,dx-\int{\psi _{m}^{*}\,{{{\hat{a}}}_{-}}{{\psi }_{n}}\,dx}} \right) \\
    & =\alpha i\sqrt{\frac{\hbar m\omega }{2}}\left( \sqrt{n+1}\int{\psi _{m}^{*}\,{{\psi }_{n+1}}\,dx-\sqrt{n}\int{\psi _{m}^{*}\,{{\psi }_{n-1}}\,dx}} \right)
    \end{align}
    [/tex]
    As you can see, I end up with the two integrals. But I don't know what to do next ? 'Cause if [itex]m > n[/itex], and only by 1, then the first integral will be 1, and the other will be zero. And if [itex]n > m[/itex], only by 1, then the second integral will be 1, and the first will be zero. Otherwise both will be zero.
    And it seems wrong to have to make two expressions for the energy shift for [itex]n > m[/itex] and [itex]m > n[/itex].

    So am I on the right track, or doing it totally wrong ?

    Thanks in advance.


    Regards
     
  2. jcsd
  3. May 9, 2013 #2
    You are on the right track. You will not have two different expressions depending on your n and m values because you sum over the m values. So you will only have two integrals that are not zero but you will sum and square them giving you one second order energy shift.
     
  4. May 10, 2013 #3
    Hmmm, not sure I understand.
    How can the two integrals be not zero at the same time ?

    As far as I know, the integrals above is either zero if [itex]n \neq m[/itex], and 1 if [itex]n = m[/itex].
    And this is, as far as I can see, not possible for both of them at the same time. At least not in the way I have written it down. Or...?
     
  5. May 10, 2013 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    The two integrals do not have to be nonzero at the same time (i.e., for the same m). One integral will be nonzero for one value of m and the other integral will be nonzero for a different value of m.
     
  6. May 10, 2013 #5
    Right so just for an example pretend you had m=5. When summing over n you would keep the n=4 term and the n=6 term because these would be the only two non-zero integrals.
     
  7. May 10, 2013 #6
    I see...
    Thank you very much :)
     
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