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Anharmonic Oscillator - Energy Shift Calculation Using 1st Order Perturbation

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]
    V(x) = \frac{1}{2}mw^{2}x^{2} + \lambdax^{4}

    [/tex]

    Using first-order perturbation theory to calculate the energy shift of:

    1. The ground state:

    [tex]

    \psi_{0}(x) = (2\pi\sigma)^{\frac{-1}{4}}\exp(\frac{-x^{2}}{4\sigma})

    [/tex]

    of the harmonic oscillator, where:

    [tex]

    \sigma = \frac{\hbar}{2m\omega}

    [/tex]

    2. Relevant equations

    Within the stated question.

    3. The attempt at a solution

    If [tex]\lambda = 0[/tex] then the potential would correspond to a harmonic oscillator of classical frequency [tex]\omega[/tex] with energy levels defined as:

    [tex]E_{0n} = (n+\frac{1}{2})\hbar\omega[/tex]

    And with the ground state energy eigenfunction as:

    [tex]U_{00} = (\frac{m\omega}{\pi\hbar})^{1/4}.exp(\frac{-m\omega}{2\hbar}x^{2})[/tex]

    The inclusion of the [tex]\lambdax^{4}[/tex] term is what results in the anharmonic problem.

    By substituting from the equation for the ground state energy eigenfunction above, into

    [tex]E_{1n} = H_{nn}^{'}[/tex]

    Get the result of:

    [tex]E_{10} = ((\frac{m\omega}{\pi\hbar})^{1/2})\int_{\infty}^{\infty}\lambda x^{4}[exp(\frac{-m\omega}{\hbar}x^{2})]dx[/tex]

    Therefore:

    [tex]E_{10} = \frac{3\hbar^{2}}{4m^{2}w^{2}}\lambda[/tex]

    Which I believe is the energy shift for the ground state.


    I do need to do the same for the first excited state, so with a slightly different expression for [tex]\psi_{1}(x)[/tex], but I think I could do that once I know how to do this.
     
  2. jcsd
  3. Feb 21, 2010 #2

    vela

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    What's your question?
     
  4. Feb 21, 2010 #3
    Yes E10 is the first order correction in ur perturbation equation.
    its the expectation value.
    similarly u can find out the higher order correction.
    and ur perturb potential is x^4 not 4.
     
  5. Feb 21, 2010 #4
    Sorry about the typo, the perturbation is indeed [tex]\lambda x^{4}[/tex].
     
  6. Feb 21, 2010 #5
    Ok, so I've now got that expression for the first order correction to the ground state energy shift.. but where does [tex]\psi_{o}(x)[/tex] factor into it?!
     
  7. Feb 21, 2010 #6
    $$E10=\int \psi_{0}^{*}(x)x^4\psi_{0}(x)$$
     
  8. Feb 21, 2010 #7
    [tex]E_{10}=\int \psi_{0}^{*}(x)x^4\psi_{0}(x)[/tex]

    Sorry, I just don't see how what I need to do next at the moment :S
     
  9. Feb 21, 2010 #8
    substitute the value of ground state wavefunction and integrate it.

    thanks for making my equation nice....
     
  10. Feb 21, 2010 #9
    [tex]\psi_{0}^{*}(x)\psi_{o}=

    ((2\pi\sigma)^{\frac{-1}{4}})^{2}(exp(\frac{-x^{2}}{4\sigma}))^{2} = (2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma})[/tex]

    [tex]x^{4}\psi_{0}^{*}(x)\psi_{o}= x^{4}(2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma})[/tex]

    [tex] \int x^{4}\psi_{0}^{*}(x)\psi_{o}= \int x^{4}(2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma}) = (2\pi\sigma)^{\frac{-1}{2}} \int x^{4}exp(\frac{-x^{2}}{2\sigma}) [/tex]

    [tex](2\pi\sigma)^{\frac{-1}{2}} \int x^{4}exp(\frac{-x^{2}}{2\sigma}) =


    \frac{(-2x)(2\pi\sigma)^{\frac{-1}{2}}}{2 \sigma}exp(\frac{-x^{2}}{2\sigma})

    [/tex]

    Hopefully this is along the right lines?
     
  11. Feb 21, 2010 #10

    vela

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    Oh, so your "attempt at a solution" above actually wasn't your attempt at a solution, but part of the problem statement.

    Everything up until you actually integrated looks fine. How did you come up with that result?
     
  12. Feb 21, 2010 #11
    Apologies for any confusion with my initial post on this thread.

    I did get that result, but that I believe is the purturbation?

    So this integration I'm going through should give me the energy shift then?
     
  13. Feb 21, 2010 #12

    vela

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    Yes, the integral you wrote down gives you the correction to the ground state energy, but you're not doing the actual integral correctly. What steps are you doing to try integrate that function? Your mistake is somewhere in those steps. (Also, don't forget the limits on the integral.)
     
  14. Feb 21, 2010 #13
    Sorry, I don't see where in those integration steps is incorrect? Is that what you're hinting at?
     
  15. Feb 21, 2010 #14

    vela

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    I'm referring to the last line you wrote. The first three lines have nothing to do with integration. They're just algebra. In the last line, you apparently tried to integrate the function, and something went wrong. (If you differentiate the RHS, you'll find you don't recover the integrand.)
     
  16. Feb 21, 2010 #15
    [tex] \int_{-\infty}^{\infty} exp \left( \frac{-x^{2}}{2\sigma} \right) = \left(\frac{-1}{x^{2}}\right)\left(-2x\right)exp \left( \frac{-x^{2}}{2\sigma} \right) = \left(\frac{2}{x}\right)exp\left(\frac{-x^{2}}{2\sigma}\right)

    [/tex]

    Therefore:

    [tex]\int_{-\infty}^{\infty} x^{4}exp(\frac{-x^{2}}{2\sigma}) = \left(\left(2x^{3}\right)exp\left(\frac{-x^{2}}{2\sigma}\right)\right) - \int_{-\infty}^{\infty}\left(\left(8x^{2}\right)exp\left(\frac{-x^{2}}{2\sigma}\right)\right)
    [/tex]

    Back on track?
     
    Last edited: Feb 21, 2010
  17. Feb 21, 2010 #16

    vela

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    Nope. I have no idea what you did in the first line. I'm kind of guessing you tried to use

    [tex]\int e^{ax}dx=\frac{1}{a}e^{ax}+c[/tex]

    which you can't do because the exponent in your integral is of the form [itex]ax^2[/itex] not just [itex]ax[/itex] and that makes all the difference in the world. Moreover, if that's what you did, I don't see where the factor of (-2x) comes from or what happened to the factor of [itex]2\sigma[/itex].

    You need to show or explain what steps you're taking rather than just posting your results. When all you write down is the integral and an incorrect answer, it's impossible to see where your calculations are going astray.
     
  18. Feb 21, 2010 #17
    I did try to use that rule, but that was incorrect method then. I don't know how else to compute the integral :|
     
  19. Feb 21, 2010 #18

    vela

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    Have you learned about the gamma function? If not, integrating by parts is probably the most appropriate way for you.
     
  20. Feb 21, 2010 #19
    No I havn't learned about that function, hence why I used integration by parts method.

    Just calculated the integral again:

    [tex] \int_{-\infty}^{\infty} x^{4}exp(\frac{-x^{2}}{2\sigma}) = 3\sqrt{2\pi}\sigma^{3/2} [/tex]
     
  21. Feb 21, 2010 #20

    vela

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    That's almost right. The numerical part of your answer is right.

    Whenever you do these complicated calculations, you should perform some sanity checks to see if you might have made a mistake. One of the quickest and most useful is to see if your answer has the right units. In this case, the LHS has units of length^5 (4 from the x^4 plus 1 from the dx) while the RHS has units of length^3 ([itex]\sigma[/itex] has units of length^2). So look for places in your calculations where you might have dropped factors of [itex]\sigma[/itex].
     
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