Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I don't know how the following can be proved. Let $U(t)$ be the following integral over $\mathbb{R}^3$:

$$ U(t,\vec{x}-\vec{x}_0)=\int\frac{d^3p}{(2\pi)^3}e^{-i(p^2/2m)t}e^{i\vec{p}\cdot(\vec{x}-\vec{x}_0)}$$

where $p^2=p_{x}^2+p_{y}^2+p_{z}^2$ and $\vec{x}, \vec{x}_0$ arbritary vectors. Define $\vec{u}\equiv\vec{x}-\vec{x}_0$ and $u^2\equiv\vec{u}\cdot\vec{u}$ Then

$$ U(t,\vec{u})=\left(\frac{m}{2\pi it}\right)^{3/2}e^{imu^2/2t}$$

My attempt

$$ U(t,\vec{u})=\int\frac{dp_{x}dp_{y}dp_{z}}{(2\pi)^3} \exp\left(-i\left(\frac{p_x^2+p_y^2+p_z^2}{2m}\right)t\right)\exp\left( i(p_xu_x+p_yu_y+p_zu_z)\right)\\=\int\frac{dp_x}{2\pi}\exp\left(-i\frac{p_x^2}{2m}t\right)\exp\left( ip_xu_x\right)\int\frac{dp_y}{2\pi}\exp\left(-i\frac{p_y^2}{2m}t\right)\exp\left( ip_yu_y\right)\int\frac{dp_z}{2\pi}\exp\left(-i\frac{p_z^2}{2m}t\right)\exp\left( ip_zu_z\right)=f(t,u_{x})f(t,u_{y})f(t,u_{z})$$

So I only need to do the integral once for a general $u$. I try to write as a Fourier transform $\mathcal{F}[f](u)$

$$f(t,u)=\int\frac{dp}{2\pi}\exp(ipu)\exp\left( -i\frac{p^2t}{2m}\right)= \mathcal{F}\left( \exp\left( i\frac{p^2t}{2m}\right)\right)(u) \propto\delta\left(u-\frac{p^2t}{2m}\right)$$

I'm not sure about the last $\propto$ .. but I can't get the exponential that I should

Any hint will be appreciated thanks

share|cite|improve this question
up vote 2 down vote accepted

Hint: your evaluation of the integral over a chirped (quadratic) phase function is incorrect. Such an integral produces a result surprisingly like a gaussian:

$$\int_{-\infty}^{\infty} \: dx e^{-i x^2} = \sqrt{\frac{\pi}{i}} = \sqrt{\pi} e^{-i \pi/4}$$

You can prove this using Cauchy's integral theorem. Here is an outline of a proof of this.

To apply this to your problem, consider

$$\begin{align}\int_{-\infty}^{\infty} dx\: e^{-i a x^2} e^{i b x} &= \int_{-\infty}^{\infty} \: dx e^{-i a (x^2 - b/a x)} \\ &= \underbrace{\int_{-\infty}^{\infty} dx\: e^{-i a (x^2 - b/a x + b^2/(4 a^2))} e^{i b^2/(4 a)}}_{\text{completing the square}}\\ &= e^{i b^2/(4 a)}\int_{-\infty}^{\infty} dx\: e^{-i a (x-b/(2 a))^2}\\ &= e^{i b^2/(4 a)} \sqrt{\frac{\pi}{i a}}\end{align}$$

share|cite|improve this answer
Hi Ron thanks for the hint. Maybe should I write the exponential as $\exp[i(pu-ap^2)]=\exp(iw^2)$ but then If I try to do the integral with respect to $w$, $dw$ avoids the integral to have the suggested form $e^{-iw^2}$. I can't really see the full implications of your hint, sorry. – Jorge Mar 15 '13 at 15:49
Complete the square - everything has an $i$ in front in the exponential. Evaluating that integral is analogous to evaluating the integral over a gaussian. – Ron Gordon Mar 15 '13 at 15:51
Thank you for all your edits and time in this question, in particular the last one. – Jorge Mar 15 '13 at 16:03
My pleasure. If I had a quarter for every time I used this integral in my science life, I would have retired years ago. – Ron Gordon Mar 15 '13 at 16:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.