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QM expectation value relation <x^n>, <p^n>

  1. Apr 20, 2006 #1
    I need to calculate <x^n> and <p^n> for psi(x)=exp(-ax^2/2)
    for n even.
    For <x^n>:
    <x^n>=integral(exp(-ax^2)*x^n )dx from -inf to +inf
    then i use integration by parts to get an infinite series and i use a formula to find the finite sum of the series
    =[exp(-ax^2)*x^(n+1)/((n+1-2a*(n+1)^2)] with limits -infinity to +infinity
    i don't think this is sufficient as i believe the answer should result in a constant and when i use this to work out <x^2>, i can't get a finite answer. I am not sure of another way to work through this question.

    For <p^n> i have shown that for n odd, this vanishes, but for n even i am not sure what the answer is. I would appreciate any help or suggestions on where to start with this question, thanks.
     
  2. jcsd
  3. Apr 20, 2006 #2

    dextercioby

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    Here's a trick...What happens if n=odd integer for \langle \hat{x}^{n} \rangle ...?

    Daniel.

    P.S. I'm sure you did gaussian integrals in statistical physics seminar.
     
  4. Apr 20, 2006 #3

    nrqed

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    I don't see how you can get an *infinite* series!

    First, write down the result for the case n=0. Then, consider the case with n=2. after doing integrations by parts, you may write the result in terms of the n=0 expression.

    Now do it with n=4 which, after integrations by parts, may be written in terms of the result of n=2. At that point, the generalization should be obvious. The result is of the form "a" raised to a certain power times a factorial term.
     
  5. Apr 20, 2006 #4
    Thanks for your quick replies. I am now able to get the answer :smile:
    I should have done it that way from the start but for some reason that infinite series thing led me on a bit.
    Thanks for all the help.
     
  6. Apr 21, 2006 #5
    I just realised that i haven't been able to get the <p^n> part right where p=-i(h-bar)d/dx
    i get <p^2>=(h-bar)^2 *a/2*sqrt(pi/a)
    <p^4>=(-3/4)*(h-bar)^4 *a^2*sqrt(pi/a)
    <p^6>=(-165/8)*(h-bar)^6 *a^4*sqrt(pi/a)
    <p^8>=(4605/16)*(h-bar)^8 *a^6*sqrt(pi/a)
    And this doesn't give me a nice relationship for <p^n>
    i realise that p^n where n is odd =0.
    I think i must have done something wrong with the maths somewhere, but i can't find where. And i am 90% sure that the <p^2> is correct but i don't know about the others.
     
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