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??)

  • $\begingroup$ The relative phase of the contributing energy eigenstates matters for the time evolution. The simple magnitude squares of the coefficients doesn't contain that information. An object that has the equivalent information would be either a vector with all complex coefficients or a density matrix. You could for example not decide whether the oscillating term in your example is a sine or a cosine(or anything in between determined by the relative phase of the coefficients), if you were only given the coefficients squared. $\endgroup$
    – Hans Wurst
    Jun 28, 2021 at 19:01
  • $\begingroup$ Also note that the $c_n$ in the quote can be different from the expansion coefficients with respect to the energy eigenstates. The time independent probability interpretation of the coefficients squared is only applicable for the expansion coefficients with respect to the eigenbasis of the measured operator. $\endgroup$
    – Hans Wurst
    Jun 28, 2021 at 19:42

1 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.

  • $\begingroup$ The quoted bit in my question is copied directly from my notes, so i am assuming that is 100% correct. Since our conditions is for any observable and not just energy and since position is an observable, why would it not be true for position. $\endgroup$
    – Jack Jack
    Jun 28, 2021 at 21:21
  • $\begingroup$ The probability $\vert c_n \vert^2$ is not constant in time for any observable, only for observables that commute with the Hamiltonian. For a square well the position operator $\hat{x}$ does not commute with the Hamiltonian $\hat{H} = \hat{p}^2/2m + V(\hat{x})$, so the probability coefficients $\vert\psi(x,t)\vert^2$ are not constant in time. I suppose that in the expression $ψ=∑_nc_n ψ_n$ the lecturer meant is that these wavefunctions and coefficients should be taken at the time of measurement $t_0$. $\endgroup$ Jun 28, 2021 at 21:34

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