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QM: expectation value of a harmonic oscillator

  1. Feb 16, 2009 #1
    first post! but for bad reasons lol

    Im trying to find <x> and <p> for the nth stationary state of the harmonic potential: V(x)=(1/2)mw^2x^2

    i solved for x: x=sqrt(h/2mw)((a+)+(a-))
    so <x> integral of si x ((a+)+(a-)) x si.
    therefor the integral of si(n+1) x si + si(n-1) x si.
    si(n+1) x si as far as I know is always 0 so this would mean <x>=0?
    this same convention would be used for <p>.

    sorry for the sloppy work but I'm pretty sure <x>, and <p> shouldnt = 0

    please help, and thanks for the posts!
  2. jcsd
  3. Feb 16, 2009 #2


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    Your result is correct! They are both zero. You can see it directly if you think in terms of raising and lowering operator since, for any eigenstate |n>, we have

    [tex] \langle n| a | n\rangle = \langle n | a^\dagger | n \rangle = 0 [/tex]
  4. Feb 16, 2009 #3
    Alright thanks a lot! I have absolutely no idea how to find <T>. If you can give me a pointer in the right direction that would be greatly appreciated. Thank you very much for all the help.
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