Measurement in a General superposition state General superposition state (the $u_n (x)$ are the spatial part of the energy eigenfunctions):
$$\psi(x,t)=\sum_{n=1}^\infty c_n \psi_n(x,t) =\sum_{n=1}^\infty c_n u_n(x) \exp⁡\left(-\frac{iE_n t}{\hbar}\right)$$
The expansion coefficients $c_n$ are (complex) constants
If a measurement is made:

When a measurement of some observable $O$ occurs, the wave function
$\psi=\sum_nc_n \psi_n$ (where $\psi_n$ are the eigenfunctions with respect to $\hat{O}$
) is projected onto one of the various possible measurement outcomes,
with a probability of projection given by $|c_n |^2$.

However consider the example of a particle in a superposition state of the 1D infinite square well described by an equally weighted combination of the ground and the first state:
$$\psi(x,t)=\frac{1}{\sqrt{2}} \left[ u_1(x) \exp⁡\left(-\frac{iE_1 t}{\hbar}\right)+u_2(x) \exp⁡\left(-\frac{iE_2 t}{\hbar}\right) \right]$$
Probability density changes with time:      $$|\psi(x,t)|^2=\frac{1}{2} u_1^2 (x)+ \frac{1}{2} u_2^2 (x)+u_1 (x) u_2 (x)\cos\left(\frac{(E_2-E_1)t}{\hbar}\right)$$
How are both of these true? Since $|c_n|^2$ is constant but $|\psi(x,t)|^2$ is not and they represent the same things (I think??)
 A: They do not represent the same things.

*

*The $c_n$'s are the probability amplitude that the system will be in the energy eigenstate $n$ if you measure the energy at any point in time.

*The wavefunction $\psi(x,t)$, on the other hand, is the probability amplitude that the system will be at the position $x$ if you measure position at time $t$.

In  your case, the time-dependent energy eigenstates are
\begin{align}
\psi_n(x,t) \propto u_i~\exp(-iE_it/\hbar),
\end{align}
up to normalization. It's easy to see from your first equation that if these states are already normalized, then the coefficients must be $c_n = 1/\sqrt{2}$. If they are not, though, you can find out the coefficients by taking into account that the energy eigenstates must be orthonormal. Then
\begin{align}
\langle \psi_m, t\vert \psi, t\rangle
&= \sum_n c_n \langle \psi_m, t\vert \psi_n, t \rangle \\
&= \sum_n c_n \delta_{n,m} \\
&= c_m.
\end{align}
By expanding on position space:
\begin{align}
\langle \psi_m, t\vert \psi, t\rangle
&= \int \psi_m^*(x,t) \psi(x,t) dx \\
&= \int \psi_m^*(x,t) \frac{1}{\sqrt{2}} \left[u_1(x)e^{-iE_1t/\hbar} + u_2(x)e^{-iE_2t/\hbar}\right] \\
&= \int \psi_m^*(x,t) \frac{1}{\sqrt{2}} \left[\frac{\psi_1(x,t)}{N_1} + \frac{\psi_2(x,t)}{N_2}\right] \\
&= \frac{\delta_{m,1}}{\sqrt{2} ~ N_1}  + \frac{\delta_{m,2}}{\sqrt{2} ~ N_2},
\end{align}
where the $N_i$'s are the normalization factors of the $u_i$ distributions.
Finally,
\begin{align} 
\vert c_1 \vert^2
&= \frac{1}{2 ~ N_1^2} = \frac{1}{2 ~ \int \vert u_1(x)\vert^2 dx} \\
\vert c_2 \vert^2
&= \frac{1}{2 ~ N_2^2} = \frac{1}{2 ~ \int \vert u_2(x)\vert^2 dx}.
\end{align}
Remember: the probabilities depend on which observable you are measuring.
