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Homework Help: Harmonic Oscillator wave function

  1. Aug 24, 2007 #1
    Uncertainty - Harmonic Oscillator

    The Wave function for the ground state of a quantum harmonic oscillator is
    [tex]
    \psi=(\alpha/\pi)^{1/4}e^{-\alpha x^2/2}
    [/tex] where [tex] \alpha = \sqrt{ mk/ \hbar^2} [/tex].

    Compute [tex] \Delta x \Delta p [/tex]


    known:
    Heisenberg Uncertainty Principle:
    [tex] \Delta p \Delta x >= \hbar/2[/tex]

    In order to compute [tex] \Delta x \Delta p [/tex], what do I need to do? any integral?
     
    Last edited: Aug 24, 2007
  2. jcsd
  3. Aug 24, 2007 #2

    mgb_phys

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    Well done - but I think that is already known to science.
    Or did you have a further question about it?
     
  4. Aug 24, 2007 #3
    My question is how to find [tex] \Delta x \Delta p [/tex]
     
  5. Aug 24, 2007 #4

    Dick

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    You'll want to find the expectation values of x^2 and p^2 (since the expectation values of x and p are zero), and take their square roots. These are your deltas. So yes, you have to integrate to find <psi|x^2|psi> and <psi|p^2|psi>.
     
  6. Aug 24, 2007 #5
    Just to be pedantic, [tex]\Delta x^2 = \langle x^2 \rangle - \langle x \rangle^2[/tex]. Just that in this case, [tex]\langle x \rangle = 0[/tex].
     
  7. Aug 24, 2007 #6
    Thanks guys,
    but how the <x> and <p> are zero, could you please help me for integral part, what is the limits of integral in this case?
     
  8. Aug 24, 2007 #7

    Dick

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    The limits on the integral are +/- infinity. And both integrals have the general form of an integral of x*exp(-K*x^2). So the integrand is an odd function. It's integral is zero.
     
  9. Aug 24, 2007 #8
    ok, I find [tex]<x^2> =1/2\alpha[/tex] what should I do for <p^2>
    p=-i*hbar ?
     
  10. Aug 24, 2007 #9

    Dick

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    Uh, p=-i*hbar*d/dx. You need to apply that operator twice to psi since you are finding <psi|p^2|psi>. Your <x^2> looks good.
     
    Last edited: Aug 24, 2007
  11. Aug 24, 2007 #10
    I found [tex] \Delta x \Delta p=\hbar/\sqrt {2} [/tex]
    which means [tex] \Delta x \Delta p[/tex] is independent from the value of alpha, what do you think? somehow I think I should have gotten [tex] \Delta x \Delta p=\hbar/2 [/tex]
     
  12. Aug 24, 2007 #11

    Dick

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    You did really well, except yes, you should have gotten hbar/2. I did. Can you find the missing sqrt(2)? I can check intermediate results if you want to post them.
     
  13. Aug 24, 2007 #12
    Thanks for the reply, I think I found the missing sqrt(2)
     
  14. Aug 24, 2007 #13

    Dick

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    Well done.
     
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