I tried the following integration

$$\frac{1}{(2 \pi \hbar)^3} \int_{-\infty}^{\infty} dp^3 exp[i\frac{\vec{p}\vec{r}}{\hbar}-i\frac{ip^2t}{2m\hbar}]$$

with this code but failed to have the right answer.

With[{r = {x, y, z}, p = {px, py, pz}}, 
 Integrate[1/(2 Pi h)^3 Exp[I (p.r) - I (p.p) t/(2 m h)], p \in Ball[]]]

Could anyone give me a hint? Any help would be appreciated!

  • 1
    $\begingroup$ What answer did you get? What did you expect? What is the syntax you are attempting with the use of \in? Do you also see that you have not properly closed up the Exp? $\endgroup$ Jan 26, 2022 at 4:42
  • $\begingroup$ Thanks for your reply. They are typos when I edited in the Stacks (sorry about that), and \in means the mathematical symbol \in (I don't know how to type it in the code block). I can run the code in my Mathematica, but it just didn't give me a correct answer. $\endgroup$
    – QubitTy
    Jan 26, 2022 at 5:26
  • 1
    $\begingroup$ Do you mean Sphere[] (the shell) or Ball[] (the region within the sphere) $\endgroup$
    – Bob Hanlon
    Jan 26, 2022 at 5:27
  • $\begingroup$ The answer should be : $$(\frac{m}{2\pi i \hbar t})^{3/2} \exp [i \frac{m}{2\hbar t} r^2], t\geq 0$$ $\endgroup$
    – QubitTy
    Jan 26, 2022 at 5:28
  • $\begingroup$ @BobHanlon Thank you. I should use Ball. But the result is still not right. It simply doesn't give me a explicit answer. $\endgroup$
    – QubitTy
    Jan 26, 2022 at 5:44

1 Answer 1


Try this 1D integration

Assuming[Element[r, Reals] && t > 0 && m > 0 && h > 0,
  1/(2 Pi h) Exp[I (p *r) - (p*p) t/(2 m h)], {p, -Infinity, 

$$\frac{e^{-\frac{h m r^2}{2 t}}}{\sqrt{2 \pi } \sqrt{\frac{h t}{m}}}$$

The 3D integral factorizes into 3 such 1D integrals.

  • $\begingroup$ Thanks a lot! Sorry for the late acknowledgement. $\endgroup$
    – QubitTy
    Feb 13, 2022 at 16:22

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