Measurement in a General superposition state

Question

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??)

Answer 1

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.