- #1

FaraDazed

- 347

- 2

## Homework Statement

Prove following integral of the probability current is true for all wavefunctions [itex]\Psi (x,t)[/itex] and comment on what this means phsically

[tex]

\int^{\infty}_{-\infty} j(x,t) dx = \frac{<p>}{m}

[/tex]

It also says that a hint is to integrate by parts, and that can assume that [itex]|\Psi (x,t)|[/itex] tends to 0 as x tends to infinity.

## Homework Equations

[tex]

j=\frac{-i \hbar}{2m}(\Psi^* \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^*}{\partial x}) \\

<p> = \int^{\infty}_{-\infty} \Psi^* \hat{p} \Psi dx

[/tex]

## The Attempt at a Solution

I am a bit lost on this one, I can't see where or how I can use integration by parts.

What I have done so far is rewriting the equaton for the probability current in terms of the momentum operator like..

[tex]

j=\frac{-i \hbar}{2m}(\Psi^* \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^*}{\partial x})\\

j = \frac{1}{2m}(\Psi^* \hat{p} \Psi - \Psi \hat{p} \Psi^* )\\

[/tex]

So that the integral of that over all space would then equal

[tex]

\int^{\infty}_{-\infty} j(x,t) dx = \frac{<p>}{2m} - \int^{\infty}_{-\infty} \Psi \hat{p} \Psi^* dx \\

[/tex]

or

[tex]

\int^{\infty}_{-\infty} j(x,t) dx = \frac{<p>}{2m} + \frac{i \hbar}{2m} \int^{\infty}_{-\infty} \Psi \frac{ \partial \Psi^*}{\partial x} dx

[/tex]

I am unsure of what changing the order of the conjugate does for the expectation value, if anything. This is probably not even the correct way to approach the problem.

In terms of what it means physically, obviously the expectation of momentum over mass is just the average velocity (I assume), but other than that I don't know what else I could say.

Any help/advice on this at all would be greatly appreciated! Thanks.