Homework Help: Momentum Space Integral

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1. Aug 4, 2016

Nick Heller

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)
(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. Aug 4, 2016

TSny

What does this factor look like before you try to evaluate it. Is it an integral of some sort?

Last edited: Aug 4, 2016
3. Aug 14, 2016

Fred Wright

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.

4. Sep 27, 2017

Moayd Shagaf

Is there any techinques to evalute integrals like this, with respect to spherical coordinate??