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Math Help - expected value of x in Gaussian wave packet

  1. #1
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    expected value of x in Gaussian wave packet

    This problem is killing me.

    \Psi(x) = Ae^{-\frac{(x-x_0)^2}{2K^2}}

    I normalized the wave packet

    \int^\infty_{-\infty} \Psi\Psi*\,dx = 1 where * is the complex conjugate

    using a table integral:

    \int^{\infty}_{-\infty}{e^{-\frac{(x-b)^2}{c^2}}}\,dx=c\sqrt{\pi}

    A^2 = \frac{1}{K\sqrt{\pi}}

    Now I need the expected value of x

    <x>=\int^\infty_{-\infty} x\Psi\Psi*\,dx
    =A^2\int^\infty_{-\infty} xe^{-\frac{(x-x_0)^2}{K^2}}\,dx

    which I expect to be x_0

    but I can't get it to work out. I tried substitution with u = the exponent but looking back over it I don't think it will work right because I can't seem to get rid of all the references to x when I substitute.

    Any pointers?
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  2. #2
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    Quote Originally Posted by stevedave View Post
    I need the expected value of x

    <x>=\int^\infty_{-\infty} x\Psi\Psi*\,dx
    =A^2\int^\infty_{-\infty} xe^{-\frac{(x-x_0)^2}{K^2}}\,dx

    which I expect to be x_0

    but I can't get it to work out.
    Substitute y = x-x_0. Then \int^\infty_{-\infty}\!\!\! xe^{-\frac{(x-x_0)^2}{K^2}}\,dx = \int_{-\infty}^\infty\!\!\!(y+x_0)e^{-y^2/K^2}dy = \int_{-\infty}^\infty\!\!\! ye^{-y^2/K^2}dy\ +\  x_0\int_{-\infty}^\infty\!\!\! e^{-y^2/K^2}dy. In the first of those two integrals the integrand is an odd function, so the integral will be 0. The second integral gives the answer that you want.
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  3. #3
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    Thanks, that works out nice. I think my problem was not changing the limits of integration when I tried to substitute for the entire exponent. The square relationship would have square the limits making them both positive and giving the same answer.
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