Amplitude for a free particle

1. Feb 15, 2009

karlsson

I hope this is the correct place for my question. I posted it here, because it`s from Peskin & Schroeder:
"Consider the amplitude for a free particle to propagate from $$\mathbf{x}_{0}$$ to $$\mathbf{x}$$ :

$$U(t)=\left\langle \mathbf{x}\right|e^{-iHt}\left|\mathbf{x_{0}}\right\rangle$$

In nonrelativistic quantum mechanics we have E=p^2/2m, so

$$U(t)&=&\left\langle \mathbf{x}\right|e^{-i(\mathbf{p}^{2}/2m)t}\left|\mathbf{x}_{0}\right\rangle$$

$$=\int\frac{d^{3}p}{(2\pi)^{3}}\left\langle \mathbf{x}\right|e^{-i(\mathbf{p}^{2}/2m)t}\left|\mathbf{p}\right\rangle \left\langle \mathbf{p}\right.\left|\mathbf{x_{0}}\right\rangle$$

$$=\frac{1}{(2\pi)^{3}}\int d^{3}p\, e^{-i(\mathbf{p}^{2}/2m)t}\cdot e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_{0})}$$

$$=\left(\frac{m}{2\pi it}\right)^{3/2}\, e^{im(\mathbf{x}-\mathbf{x}_{0})/2t}}$$
."

I don't understand the last equation.
Why I can't use the fourier-transformation:

$$=\frac{1}{(2\pi)^{3}}\int d^{3}p\, e^{-i\mathbf{p}\cdot(\mathbf{x}_{0}-\mathbf{x})}\widetilde{f}(\mathbf{p})$$

$$=f(\mathbf{p})$$

$$=e^{-i(\mathbf{x}-\mathbf{x}_{0})^{2}/2m)t}$$

2. Feb 15, 2009

Marco_84

but is what you have don actually ;)

marco

3. Feb 16, 2009

karlsson

Sry, I don't understand. My solution is a different one as from Peskin & Schroeder. But why?

4. Feb 16, 2009

jensa

The last step is made by completing the square and evaluating the gaussian integral (although with complex coefficient). I suspect the expression should be ~exp[im(x-x_0)^2/2t] though...

The last step is precisely the evaluation of a fourier transform. In what you write next:

It looks like you have simply taken $$\widetilde{f}=f$$ and replaced the $\mathbf{p}$ argument with $\mathbf{x}-\mathbf{x}_0$.. If so then this is wrong. The fourier transform should be evaluated as in the last step of your first equation (i.e. completing the square as I wrote)

5. Feb 16, 2009

karlsson

You are right.

Thanks a lot.