Prove that the following integral of the probability current is true

[FaraDazed]

Homework Statement

Prove following integral of the probability current is true for all wavefunctions $\Psi (x,t)$ and comment on what this means phsically
$$\int^{\infty}_{-\infty} j(x,t) dx = \frac{<p>}{m}$$

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

Homework Equations

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

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..
$$j=\frac{-i \hbar}{2m}(\Psi^* \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^*}{\partial x})\\ j = \frac{1}{2m}(\Psi^* \hat{p} \Psi - \Psi \hat{p} \Psi^* )\\$$
So that the integral of that over all space would then equal
$$\int^{\infty}_{-\infty} j(x,t) dx = \frac{<p>}{2m} - \int^{\infty}_{-\infty} \Psi \hat{p} \Psi^* dx \\$$
or
$$\int^{\infty}_{-\infty} j(x,t) dx = \frac{<p>}{2m} + \frac{i \hbar}{2m} \int^{\infty}_{-\infty} \Psi \frac{ \partial \Psi^*}{\partial x} dx$$

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.

 

[mfb]

You can swap the part that gets differentiated with partial integration. You'll get an additional term that can be shown to vanish using the hint in the problem statement.

Using some nice tools from complex analysis looks quicker, but partial integration doesn't need additional mathematics.

 

[FaraDazed]

You can swap the part that gets differentiated with partial integration. You'll get an additional term that can be shown to vanish using the hint in the problem statement.

Using some nice tools from complex analysis looks quicker, but partial integration doesn't need additional mathematics.

ok thanks, I have tried the below, but still getting stuck, I can't see how that one term vanishes, as it is not a part of the integral,

$$\int^{\infty}_{-\infty} j(x,t) \> dx = \frac{<p>}{2m} + \frac{i \hbar}{2m} \int^{\infty}_{-\infty} \Psi \frac{ \partial \Psi^*}{\partial x} dx \\$$
And then with the integrand on the right, using parts, letting $dv=\frac{ \partial \Psi^*}{\partial x}$ and letting $u = \Psi$, so then that makes $v = \Psi^*$ and $du = \frac{ \partial \Psi}{\partial x}$
So then using the by parts formula that changes to...
$$\int^{\infty}_{- \infty} j(x,t) \> dx = \frac{<p>}{2m} + \frac{i \hbar}{2m} (uv - \int v \> du) \\ \int^{\infty}_{- \infty} j(x,t) \> dx = \frac{<p>}{2m} + \frac{i \hbar}{2m} (\Psi \Psi^* - \int \Psi^* \frac{ \partial \Psi}{\partial x} ) \\ \int^{\infty}_{- \infty} j(x,t) \> dx = \frac{<p>}{2m} + \frac{i \hbar}{2m} (|\Psi|^2 - \int \Psi^* \frac{ \partial \Psi}{\partial x}) \\$$

 

[blue_leaf77]

The last two terms in your last expression result from the partial integration, therefore they must be evaluated at the given integration limits. Especially for the last term, doesn't it look familiar?

 

[FaraDazed]

The last two terms in your last expression result from the partial integration, therefore they must be evaluated at the given integration limits. Especially for the last term, doesn't it look familiar?

Ah right, of course. With the last term, as there is that factor of $i \hbar$ then it is the expectation value of the momentum right? How does that work though, if the $uv$ is evaluated in the liimts, then the v.last expression has to be as well? I have not had all that much experience in integration by parts.

I.e. the last bit would be

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

 

[blue_leaf77]

How does that work though, if the uvuv is evaluated in the liimts, then the v.last expression has to be as well?

For the last term since this is an integral, the integration limits will translate to how they are named - they will become the upper and lower limits of the integral.

With the last term, as there is that factor of iℏi \hbar then it is the expectation value of the momentum right?

So how will that combine with the first term?

 

[FaraDazed]

For the last term since this is an integral, the integration limits will translate to how they are named - they will become the upper and lower limits of the integral.

So how will that combine with the first term?

Ah ok, that makes sense now, and can see how the answer comes about. Thank you for you help!

Essentially just $\frac{<p>}{2m}+ \frac{<p>}{2m} = \frac{<p>}{m}$.

About what it means physically though, I assume the <p>/m is just the time derivative of <x>, i.e. the velocity. But other than that, what could I say?

 

[mfb]

It is a connection between the expectation values for momentum and j. What does j represent?

 

[FaraDazed]

It is a connection between the expectation values for momentum and j. What does j represent?

well j represents the flow of probability, I know that is shows that the particle might be in motion even if the probability density has no specific time dependence.

 

[mfb]

You are comparing an expectation value for momentum with an integral over the spatial flow of probability, and they are equal.

 

[FaraDazed]

You are comparing an expectation value for momentum with an integral over the spatial flow of probability, and they are equal.

Yeah, I suppose I could also say that it shows that if the particle is not moving, the probability current will be zero (as expected).