What's the kinetic energy uncertainty for gaussian wave packet?

1. Oct 5, 2011

AlonsoMcLaren

What's the kinetic energy uncertainty for gaussian wave packet ψ(x)=((α/pi)^(1/4))exp(-αx^2/2)?

2. Oct 5, 2011

Matterwave

Can you compute the uncertainty in p? If you know the uncertainty in p and you know that K=p^2/2m can you compute the uncertainty in K?

3. Oct 5, 2011

AlonsoMcLaren

I guess you can get the EXPECTATION VALUE of K, not the uncertainty in K, by squaring the uncertainty in p and dividing it by 2m

4. Oct 5, 2011

Matterwave

$$K=\frac{p^2}{2m}$$

$$<K>=<\frac{p^2}{2m}>=\frac{<p^2>}{2m}$$

$$\sigma_K=\frac{\partial K}{\partial p}\sigma_p$$

Does that look reasonable?

5. Oct 5, 2011

AlonsoMcLaren

σK=(p/m)σp

what is p? does p= <p>?

6. Oct 5, 2011

Matterwave

Yes, as usual, you put the expectation values in there.

7. Oct 5, 2011

AlonsoMcLaren

Then <p> should be zero..

8. Oct 5, 2011

Matterwave

Ah, well I think there may be something subtle happening here that prevents my formula from working. In the mean time then, you can compute this by brute force:

$$\sigma_K=\sqrt{\langle \psi | K^2|\psi\rangle-(\langle \psi | K|\psi\rangle)^2}=\frac{1}{m}\sqrt{\langle \psi |\frac{p^4}{4}|\psi\rangle-(\langle \psi | \frac{p^2}{2}|\psi\rangle)^2}$$

This involves the 4th moment of the Gaussian though...which may be kind of hard...maybe someone more enlightened can fill us in then...XD