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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

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1 Answer 1

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

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

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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
1  
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
1  
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

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