Momentum information from position space wave function

[jmd_dk]

I am trying to develop a graphical, interactive simulation of a wave function in position space, given an arbitrary potential. It works great, but as a final touch, I would like the user to be able to collapse the wave function, either by measuring its position or its momentum.

My problem is then, how do I calculate the momentum space wave function, given the position space wave function? I know that a simple Fourier transformation should do the job, but when I calculate it I always end up with some even momentum space wave function.

Example: Make a 1D Gaussian wave packet in position space, traveling with speed v. Its momentum space wave function will also be Gaussian, centered on the origin. But it should be centered on v! How do you capture the information of movement from the position wave function alone, at a single instant?

Thank you.

 

[vanhees71]

The Gaussian wave packet centered around p_0 is (\hbar=1)
\tilde{\psi}(p)=N \exp \left (-\frac{(p-p_0)^2}{4 \Delta p^2} \right),
The normalization factor should be chosen such that
\int_{\mathbb{R}} \mathrm{d} p \tilde{\psi}(p)=1 \; \Rightarrow \; N=\frac{1}{(2 \pi \Delta p^2)^{1/4}}.
Further \Delta p is the standard deviation of p.

The same state in position space is
\psi(x)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} \mathrm{d} p \tilde{\psi}(p)=\left (\frac{2 \Delta p^2}{\pi} \right)^{1/4} \exp \left (-\Delta p^2 x^2+\mathrm{i} p_0 x \right).
Note that from this we read off
2 \Delta x^2=\frac{1}{2 \Delta p^2} \; \Rightarrow \; \Delta x=\frac{1}{2 \Delta p},
which implies that
\Delta x \Delta p=\frac{1}{2}.
This shows that the Gaussian wave packet has the minimal product of \Delta x \Delta p according to the Heisenberg-Robertson uncertainty relation. It's a coherent state, and one can show that the Gaussian wave packets are the only ones with that minimum-uncertainty property.