How to Calculate Mean Momentum for a Particle in a 1D Infinite Well?

[n0_3sc]

Homework Statement

Particle in a 1D well of infinite depth, width 2a has wavefunction:
$$\psi = Asin(k_{n}x)$$
Its prepared in state:
$$\psi(x,t) = \frac{1}{\sqrt{2}} [\psi_{0}(x)+i\psi_{1}(x)]$$
I need to find the mean momentum as a function of time ie. prove this:
$$<\hat{p_{x}}(t)> = \frac{4\hbar}{3a} sin[\frac{3\hbar\pi^{2}}{8ma^{2}}t]$$

Homework Equations

I've worked out $$k_{n}$$ and the relevant normalisation constants...But I don't think its important to state them for my question since I need to know how to attempt this problem.

The Attempt at a Solution

I thought about doing it like this:
$$<\hat{p_{x}}>=\int\psi^*(x,t)\hat{p_{x}}\psi(x,t)\dx$$
where
$$\hat{p_{x}}=-i\hbar\frac{d}{dx}$$
...but this became far too tedious and I don't know where the 't' is meant to come in...
I also tried doing:
$$<\hat{p_{x}}>=m<\frac{d\hat{x}}{dt}>$$
(Ehrenfest's Theorem) but no luck...

Any guidance will help...

 

[Jhero]

Thank you for your question. It seems like you have already made some progress in finding the mean momentum, but have hit a roadblock. Let me provide some guidance on how to proceed.

Firstly, let's look at the given wavefunction. It is a superposition of two states, with one state being the real part and the other being the imaginary part. This means that the particle is in a state that is a combination of two different states with different momenta. So, the mean momentum will also be a combination of the momenta of these two states.

Now, let's look at the time dependence of the wavefunction. It is given as a function of time, which means that the state of the particle is changing with time. This means that the mean momentum will also change with time.

To find the mean momentum as a function of time, you will need to integrate over the entire wavefunction, taking into account the time dependence as well. This means that your integral should look something like this:

<\hat{p_{x}}(t)> = \int \psi^*(x,t) \hat{p_{x}}(t) \psi(x,t) dx

Notice that the momentum operator is also a function of time, and you will need to take this into account when evaluating the integral.

To simplify the integral, you can use the fact that the wavefunction is a superposition of two states. This means that you can split the integral into two parts, one for each state, and then combine the results at the end.

I hope this guidance helps you in solving the problem. If you have any further questions, please feel free to ask. Good luck!