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Perturbed Harmonic oscil

  1. Sep 23, 2007 #1

    malawi_glenn

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    1. The problem statement, all variables and given/known data

    Sakural modern quantum.. ch 5 problem 1

    A simple one dimensional harmonic oscillator is subject to a perturbation:

    V = bx, where b is a real constant.

    Calculate the energy shift in ground state to lowest non vanishing order.

    2. Relevant equations

    You may use:

    [tex] \langle k \vert x \vert n \rangle = \sqrt{\dfrac{\hbar}{2m\omega}}\left( \sqrt{n+1}\delta_{k,n+1} + \sqrt{n}\delta_{k,n-1} \right) [/tex]

    where |n> is eigentkets to unperturbed harm. osc

    Energy shift:

    [tex]
    \Delta _{n} \equiv E_n - E^{(0)}_n = \lambda V_{nn} + \lambda^{2} \sum _{k\neq n} \dfrac{\vert V_{nk}\vert^{2}}{E^{(0)}_n - E^{(0)}_k} + . . . [/tex]

    Lamda is order, V_nn is matrix elements.

    Energy levels for harm osc

    [tex] E_N^{(0)} = \hbar \omega (1/2 + N) [/tex]

    3. The attempt at a solution


    I first do the matrix representation of V = bx

    [tex] V_{nk} \doteq b\sqrt{\hbar / (2m \omega)}\left( \begin{array}{ccccc} 0 & 1 & 0 & 0 & 0 \\1 & 0 & \sqrt{2}& 0 & 0 \\0 & \sqrt{2}& 0 & \sqrt{3} &0\\ 0 & 0 & \sqrt{3}&0&0 \end{array}
    [/tex]

    Then I choose n = 0, since ground state.

    [tex]\Delta _{0} \equiv E_0 - E^{(0)}_0 = \lambda V_{00} + \lambda^{2} \sum _{k\neq 0} \dfrac{\vert V_{0k}\vert^{2}}{E^{(0)}_0 - E^{(0)}_k} + . . . [/tex]

    I notice that [tex]V_{00} = 0[/tex] and [tex]V_{0k} [/tex]is zero for all k except 1; so that:

    [tex]V_{01} = b\sqrt{\hbar / (2m \omega)} [/tex]

    And

    [tex]E^{(0)}_0 - E^{(0)}_1} = \hbar \omega [/tex]

    So that
    [tex] \Delta _{0} = -b^2 / (2m \omega ^2) [/tex]

    I have no answer to this problem, does it look right to you?
    Thanx!
     
    Last edited: Sep 23, 2007
  2. jcsd
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