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Perturbed in the harmonic oscillator

  1. Dec 21, 2012 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations

    289028545.png

    3. The attempt at a solution

    for part a I do not know how to write it in power series form ?

    for part b :
    I chose the perturbed H' is v(x)= (1+ε )K x^2 /2
    then I started integrate E_1 = ∫ H' ψ^2 dx

    the problem was , the result equals to ∞ !!

    shall I integrate in the interval [ 0,∞ ] !
     
    Last edited by a moderator: Dec 22, 2012
  2. jcsd
  3. Dec 21, 2012 #2

    pasmith

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    You are given [itex]E_n = (n + \frac12)\hbar\sqrt{k/m}[/itex]. [itex]k[/itex] is now [itex]k(1+\epsilon)[/itex], so [itex]E_n[/itex] is now
    [tex]
    E_n = \left(n + \frac12\right)\hbar\sqrt{\frac km(1 + \epsilon)} =
    \left(n + \frac12\right)\hbar\sqrt{\frac km}(1 + \epsilon)^{1/2}
    [/tex]
    Since [itex]|\epsilon| < 1[/itex], [itex](1 + \epsilon)^{1/2}[/itex] can be expanded in a binomial series.
     
  4. Dec 21, 2012 #3
    ok , I do it thanks

    any help for second part please
     
  5. Dec 21, 2012 #4

    pasmith

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    I can't help you with that unless you tell me how your textbook defines [itex]H'[/itex], [itex]E_n^1[/itex] and [itex]\psi_n^0[/itex]. I am prepared to guess that you have
    [tex]E_n = E_n^0 + \epsilon E_n^1 + O(\epsilon^2)\\
    \psi_n = \psi_n^0 +\epsilon\psi_n^1 + O(\epsilon^2)[/tex]
    but I'm not prepared to guess what [itex]H'[/itex] might be. However I suspect the method is to substitute the above into the Schrodinger equation to get
    [tex]
    (E_n^0 + \epsilon E_n^1)(\psi_n^0 + \epsilon \psi_n^1) = -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2} (\psi_n^0 + \epsilon\psi_n^1) + \frac{kx^2}{2}(1 + \epsilon)(\psi_n^0 +\epsilon\psi_n^1)
    [/tex]
    and then require that the coefficients of [itex]\epsilon^0[/itex] and [itex]\epsilon^1[/itex] should vanish. At some stage you may want to take an inner product with [itex]\langle\psi_n^0|[/itex], and recall that
    [tex]H_0 = -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2} + \frac{kx^2}{2}[/tex]
    is self-adjoint ([itex]\langle f | H_0 | g \rangle = \langle g | H_0 | f \rangle[/itex] for all [itex]f[/itex] and [itex]g[/itex]).
     
  6. Dec 22, 2012 #5

    vela

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    Your H' is wrong, or you're not expressing yourself clearly. H' is the difference between the complete Hamiltonian and the unperturbed Hamiltonian.

    Show us how you're getting that the integral diverges. The problem seems to lie in your evaluation of the integral.
     
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