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

[AlonsoMcLaren]

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

 

[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?

 

[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

 

[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?

 

[AlonsoMcLaren]

σK=(p/m)σp

what is p? does p= <p>?

 

[Matterwave]

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

 

[AlonsoMcLaren]

Then <p> should be zero..

 

[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