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Expectation value of p^2

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    I am trying to find the variance of p for a wave function [tex]\Psi[/tex](x,0)=A(a^2-x^2)

    I'm confused about how to set up the integral.

    it should be something like [tex]-i^2h^2\int_{-a}^a A(a^2-x^2) (\frac{\partial\Psi}{\partial x})^2 dx[/tex]

    I'm confused about the partial derivative squared. My technique was to set up the integral like this:

    [tex]hA^3\int_{-a}^a (a^2-x^2) (-2x^2)^2 dx[/tex] but I'm pretty sure my answer is way wrong. I don't know how to deal with the partial derivative. please help.
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 15, 2009 #2

    kuruman

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    For the variance, you need to find <p> and <p2>, then σ2=<p2>-(<p>)2. With

    [tex]\psi(x)=A(a^{2}-x^{2})[/tex]

    You should use

    [tex]<p^{2}>=A^{2}\int(a^{2}-x^{2})(-\hbar^{2})\frac{d^{2}}{dx^{2}}(a^{2}-x^{2}) dx[/tex]

    and

    [tex]<p>=A^{2}\int(a^{2}-x^{2})(-i\hbar)\frac{d}{dx}(a^{2}-x^{2}) dx[/tex]

    Take the second or first derivatives under the integral sign, before you integrate.
     
    Last edited: Sep 15, 2009
  4. Sep 15, 2009 #3
    Thank you. I already found p, but my question is particularly about the part of the integrand that contains the partial derivative squared.

    [tex]\frac{d^{2}}{dx^{2}}(a^{2}-x^{2}) dx[/tex]

    I don't know what to do with that.
     
  5. Sep 15, 2009 #4
    Is it just [tex](-2x)^2[\tex]
     
  6. Sep 15, 2009 #5

    gabbagabbahey

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    Forget about the dx on the end of that expression until you are ready to integrate.

    Just take the second derivative of [itex]a^2-x^2[/itex] with respect to [itex]x[/itex]....that's exactly what [itex]\frac{d^{2}}{dx^{2}}(a^{2}-x^{2})[/itex] means.
     
  7. Sep 15, 2009 #6
    So the second derivative is just taking the derivative twice, right? Which means the answer is 2?
     
  8. Sep 15, 2009 #7

    gabbagabbahey

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    Yes, [itex]\frac{d^2}{dx^2}f(x)[/itex] is just another way of writing [itex]f''(x)[/itex].
     
  9. Sep 15, 2009 #8

    kuruman

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    - 2.
     
  10. Sep 15, 2009 #9
    This can't be right. The integral is between -a and a, and this would set <p^2> to 0. But <p> is 0, and that would make the variance 0, which would violate the uncertainty principle.
     
  11. Sep 15, 2009 #10

    gabbagabbahey

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    No,

    [tex]\langle p^2\rangle=2\hbar^2A^2\int_{-a}^{a} (a^2-x^2)dx\neq 0[/tex]
     
  12. Sep 15, 2009 #11
    No, it's not. I got the right answer. Thanks for all your help!
     
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