Techniques for Evaluating Momentum Space Integrals with Spherical Coordinates?

[Nick Heller]

Homework Statement

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 e^ip(x-x₀)
The answer is
(m/(2πit))^3/2e^{(im(x-x₀)2)/(2t)}

2. Homework Equations
H = p²/(2m)

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 dp_xdp_ydp_z. 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?

 

[TSny]

I get e^{(im(x-x₀)2)/(2t)} times a factor that does not equal (m/(2πit))

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

 

[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.

 

[Moayd Shagaf]

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