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A gaussian wavefunction of the harmonic oscillator

  1. Oct 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m in the harmonic oscillator potential V(x) = (mω2x2)/2 is described at time
    t = 0 by the wavefunction

    χ(x, t = 0) = 1/[(2πσ2)1/4] exp[-x2/(4σ2)]

    What is <E> at time t?

    2. Relevant equations


    3. The attempt at a solution
    <T>+<V>= <E>

    I've found the expectation value of V, which is the easy part. just finding x^2 expectation value is enough.
    for <T> the formula is pretty straight forward BUT the integral is getting out of hand.
    How can I just simplify it?
    There is a hint "Hint: Use generating function to calculate overlap integrals" but no matter what I try to do it doesn't look like the integral formula given by the hint.

    ey2-(s-y)2= sigma [Hn(y)/n!] sn

    Basically it should look like a hermite polynomials. but how =(
     
  2. jcsd
  3. Oct 13, 2014 #2

    Simon Bridge

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    After a few lines of <T> you end up with having to evaluate a gaussian integral right?
    How would you normally do that?
     
  4. Oct 13, 2014 #3

    vela

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    Please show us what you've done so far.
     
  5. Oct 13, 2014 #4
    I've found <V>= S Y*(x,t) (mw2x2/2) Y(x,t) which was easy to integrate

    for the energy part in the wave function I wrote down e-i[(p2/2m)+V(x)]

    <T>= (-h2/2m) S Y* (a2Y /ax2Y)

    I am trying to find this. I know that the wave function is the zeroth state's wave function. So <E> should be hbar.w/2

    (sorry for my notation)
     
  6. Oct 13, 2014 #5

    vela

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  7. Oct 13, 2014 #6

    vela

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    How is ##\sigma## in the wave function defined? Is it an arbitrary constant or does it have a specific value in this problem?
     
  8. Oct 13, 2014 #7
    it is (mw/pi hbar)= 1/σ2
    but I don't think I can use that, you need to find the answer w/o knowing σ. so that I will be able to show <E>= hbar w/2

    if I use σ as it is, calculating that integral becomes a painful activity.
     
  9. Oct 13, 2014 #8

    vela

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    I don't think you calculated ##\langle V \rangle## correctly. You may want to rethink your approach to this problem.

    The idea behind using the generating function is that
    $$\int e^{-2xt-t^2}f(x)\,dx = \int \left[\sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}\right]f(x)\,dx
    = \sum_{n=0}^\infty \frac{t^n}{n!}\left[\int H_n(x) f(x)\,dx\right].$$ We evaluate the integral on the left, expand the result in powers of ##t##, and identify it with the expression on the right.

    For example, consider the integral
    $$\int e^{-2xt-t^2}x^2 e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2} (1+2t^2),$$ which you can verify relatively easily. This result tells us that
    \begin{eqnarray*}
    \int H_0(x)x^2 e^{-x^2}\,dx &= \frac{\sqrt{\pi}}{2}, \\
    \int H_2(x)x^2 e^{-x^2}\,dx &= \frac{\sqrt{\pi}}{2} 2\times 2! = 2\sqrt{\pi},
    \end{eqnarray*} and all of the other integral vanish.
     
  10. Oct 13, 2014 #9
    <YlHlY> this gives the expectation value of energy right?

    then <V> gives hbarw/4 which is the answer for the expectation value of the potential energy.
    I am 100% sure about that part.
    well I'll see about the next part. thank anyway.
     
  11. Oct 14, 2014 #10

    vela

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    If you don't know the value of ##\sigma##, you can't say that ##\psi(x,0)## is an energy eigenstate, so how did you deal with the time dependence when trying to calculate ##\langle \psi(x,t) \lvert V \rvert \psi(x,t) \rangle##?
     
  12. Oct 22, 2014 #11
    for harmonic oscillator's wavefunctions.
    just like stated in the question.
    time doesn't have a role. after calculating the Cn's you see the <E(t)> is equal to <E(t=0)>
    after doing that it's simple. I've found the rest as well. Thanks for the replies.
     
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