# Solving for the momentum from eigenfunctions

When you have a solution to a time-dependent Schrodinger Equation, $$\Psi(x,t)=\exp\left({-\frac{i\hbar^2k_0^2t}{2m}}\right)\sin(k_0x), \tag{1}$$ and want to know the distribution of momentum at time t, you can decompose the sin function into $$\sin(k_0x)=\frac{\exp(ik_0x)-\exp(-ik_0x)}{2i}, \tag{2}$$ and say each is an eigenfunction of $$\hat{p}$$.

I guess the momenta are $$\frac{\exp(ik_0x)}{2i}$$ and $$\frac{-\exp(-ik_0x)}{2i}$$ with equal probability 1/2 because the coefficients squared $${(1/(2i))}^2$$ are equal.

I'm still not sure why the eigenvalue would be $$-k_0$$, $$k_0$$ and not $$-\hbar k_0$$, $$\hbar k_0$$.

EDIT: The momentum here means the distribution of momenta. Due some feedback, I decided to attach the original problem, citing Professor Irfan Siddiqi's exam from 2018:

A free particle of mass m moving in one dimension is known to be in the initial state

$$\psi(x,0)=\sin (k_0 x)$$

1. What is $$ψ(x,t)$$? [10 pts]
2. What value of momentum will a measurement yield at time t, and with what probabilities will these values occur? [10 pts]
• Thanks @Charlie for improving my post Sep 30, 2020 at 12:32
• the question is not well posed. What does knowing the momentum means? Do you mean the average momentum? Do you mean the distribution of momenta? Sep 30, 2020 at 12:35
• I updated my original post. Would this provide enough clarification? Sep 30, 2020 at 13:43

The wavefunction $$(1)$$ is not a momentum eigenstate, so does not have momentum eigenvalues. As mentioned in the comments, interesting questions you can instead ask of this wavefunction include "what is the average momentum of this state?" and "what is the distribution of momenta for this state?".
For those who might wonder, I'd like to attempt answering my own question. I realized the eigenvalues are indeed $$\hbar k_0, -\hbar k_0$$. Also, we need to normalize the eigenfunctions so that the coefficients become $$\pm \frac{1}{i}.$$