an integral from nonrelativistic quantum mechanics

Question

I am trying to perform the following integral: $$\frac{1}{(2\pi)^3}\int d^3p\,e^{-i(\vec{p}^2/2m)t}\cdot e^{i\vec{p}\cdot\vec{r}}.$$ I tried like this: $$\frac{1}{(2\pi)^3}\int d^3p\,e^{-i(\vec{p}^2/2m)t}\cdot e^{i\vec{p}\cdot\vec{r}}\\ =\frac{1}{(2\pi)^3}\int dp\,d\theta\,d\varphi\, p^2\,\sin\theta\,e^{-itp^2/2m}\cdot e^{ipr\cos\theta}\\ =\frac{1}{(2\pi)^2}\int dp\,\int_{-1}^1 d\cos\theta\,p^2 e^{-itp^2/2m}\cdot e^{ipr\cos\theta}\\ =\frac{1}{(2\pi)^2}\int dp\,\,p^2 e^{-itp^2/2m} \frac{e^{ipr}-e^{-ipr}}{ipr}\\ =\frac{1}{(2\pi)^2}\int dp\,\,\frac{2p\sin pr}{r} e^{-itp^2/2m}.$$

But I can not finish it at the last step. Could anybody help?

Answer 1

It is easiest to solve this via completing the square. Observe that $$\frac{1}{(2\pi)^3} \int\!d^3p\,e^{-i(\vec{p}^2/2m)t} e^{i\vec{p}\cdot \vec{r}}= \frac{1}{(2\pi)^3} \int\!d^3p\,e^{-i(\vec{p} -m\vec{r}/t)^2 t/2m + i m \vec{r}^2/2t} = \frac{e^{i m \vec{r}^2/2t}}{(2\pi)^3} \int\!d^3 q\,e^{-i q^2 t/2m} $$ with $\vec{q} = \vec{p}-m \vec{r}/t$.

The last integral is a constant. I guess you know the Fresnel integral $$\int\!dx\,e^{-i t x^2/2m} = \sqrt{\frac{2\pi m}{i t}}.$$ The constant is the cube of this integral.

Answer 2

One has $$\int_{\mathbb{R}^{n}}\exp\Big[-\frac{1}{2}x^{T}Ax+J^{T}x\Big]d^{(n)}x=\sqrt{\frac{(2\pi)^{n}}{\det[A]}}\exp\Big[\frac{1}{2}J^{T}A^{-1}J\Big]$$ In your case $n=3$, $A=\frac{it}{m}I$, where $I$ is the identity, and $J=ir$. The determiant of $A$ is $\det[A]=(\frac{it}{m})^{3}=-i\frac{t^{3}}{m^{3}}$. The inverse of $A$ is $A^{-1}=-i\frac{m}{t}I$.