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Momentum Space Integral

  1. Aug 4, 2016 #1
    1. The problem statement, all variables and given/known data
    This integral has to do with the probability amplitude that a free particle at position x0 is found at x at some time t. With H = p2/(2m), this involves evaluating the integral
    1/(2π)3∫d3p e-i(p2/(2m))t eip(x-x0)
    The answer is
    (m/(2πit))3/2e(im(x-x0)2)/(2t)

    2. Relevant Equations

    H = p2/(2m)

    3. The attempt at a solution
    I am not sure how to work with d3p, since I don't know how to decompose it in terms of p besides dpxdpydpz. When I try to evaluate that integral Mathematica takes forever, so I'm not sure its the right approach. When I just use this instead and evaluate from -∞ to ∞ or 0 to ∞ I get e(im(x-x0)2)/(2t) times a factor that does not equal (m/(2πit))3/2 and with some combinations of erf functions which is a red flag. How do I evaluate this?
     
    Last edited: Aug 4, 2016
  2. jcsd
  3. Aug 4, 2016 #2

    TSny

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    What does this factor look like before you try to evaluate it. Is it an integral of some sort?
     
    Last edited: Aug 4, 2016
  4. Aug 14, 2016 #3
    The Gaussian integral is,
    ##\int_{-\infty}^\infty e^{-\alpha x^2}dx=\sqrt{\frac{ \pi} {\alpha}}##.
    You can rewrite the exponential in the integral as:
    ## e^{(\frac {-it} {2m})(p_z^2 + p_y^2 + p_x^2 + \frac {(x-x_0)(2m)} {t} p_x) } = e^{\frac {im(x-x_0)^2} {2t}} e^{(\frac {-it} {2m}) (p_z^2 + p_y^2 +( p_x - \frac {(x-x_0)m} {t} )^2)} ##
    The integral becomes:
    ##\frac {e^{\frac {im(x-x_0)^2} {2t}}} {(2\pi)^3} \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty dp_x dp_y dp_z e^{(\frac {-it} {2m}) (p_z^2 + p_y^2 +( p_x - \frac {(x-x_0)m} {t} )^2)} ##
    Making a change of variables in ## p_x## and using the Gaussian integral result for ## p_x, p_y, p_z## the answer follows.
     
  5. Sep 27, 2017 #4
    Is there any techinques to evalute integrals like this, with respect to spherical coordinate??
     
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