# How to calculate the frequency of oscillation of superposition states [closed]

Been working on this question for a while and I'm not sure how to go about it. Could someone point me in the right direction, particularly for the frequency question.

The question is as follows:

A wavefunction is described by the superposition state $$\Phi(x,0)=\sqrt{\frac{1}{L}}\left[\sin\left(\frac{n\pi x}{L}\right)+\sin\left(\frac{(n+1)\pi x}{L}\right)\right]$$

Write down the wave function for any time $t>0$ and show that this wavefunction oscillates with frequency $\nu=\frac{(2n+1)\hbar}{4\pi m L^2}$.

Firstly, wavefunctions time dependence is just the wavefunction times $\mathrm e^{-iE_nt/\hbar}$ .Do I multiply each state in the superposition by their respective time dependence or just the whole sum? I am very confused...

Secondly, I haven't a clue on how to approach the frequency question. I'm not asking for a solution, just a push in the right direction.

$$\Phi(x,t) = \phi_n(x)e^{i E_n t/\hbar} + \phi_{n+1}(x)e^{i E_{n+1} t/\hbar}$$
here $\phi_n(x)$ is the first state and $\phi_{n+1}(x)$ is the second.
As for the frequency of this state, try to write it as $$\Phi(x,t) = e^{i \omega t} \left(\phi_n(x) + e^{-i2 \pi\nu t}\phi_{n+1}(x) \right)$$ where $\omega$ is some constant and $\nu$ should be $\nu=\frac{(2n+1)\hbar}{4\pi m L^2}$. Do you see that this state will have a density profile that oscillates with frequency $\nu$? If not, try computing the absolute square of $\Phi(x,t)$.
To compute the frequency $\nu$ you have to know the energy $E_n$ and $E_{n+1}$. From the form of the solutions is assume you are looking at a particle in a box. In this case the energy is $$E_n=\frac{\hbar^2k^2}{2m}=\frac{\hbar^2\pi^2n^2}{2mL^2}.$$ Note that the differnce in energy is $$E_{n+1}-E_{n}=\frac{\hbar^2\pi^2((n+1)^2-n^2)}{2mL^2}=\frac{\hbar^2\pi^2(2n-1)}{2mL^2}=\frac{2\pi^3}{\hbar}\nu$$ so the frequency that you seek is proportional to the energy difference of the two states.