Does a non-stationary state for a single particle with a nondegenerate spectrum necessarily have fluctuations in probability density?
Your expansion is not quite the most general and should more generally read $$ \Psi(x,t)=\sum_m c_m e^{-i E_m t/\hbar} \psi_m(x) $$ so that \begin{align} \frac{d\rho}{dt}=0\quad\Rightarrow 0= \sum_{mn} c_m c_n^* (E_m-E_n) e^{-i(E_m-E_n)t/\hbar} \psi_n(x)^*\psi_m(x)\tag{1} \end{align} (up to a factor of $i\hbar$). For simplicity order the eigenvalues so that $E_0\le E_1\ldots $.
(1) must be valid for any $t$ which means that, for all those $m’,n’$ so that $\Delta E_{m’,n‘}=E_{m’}-E_{n‘}$ is the same you must have $$ \Delta E_{m’,n’}\sum_{m’,n’} c_{m’}c_{n’}^* \psi_{m’}(x)\psi_{n’}(x)^*=0\, . \tag{2} $$ Note that, if $c_{m’}c_{n’}^*\psi_{m’}(x)\psi_{n’}(x)^*$ appears in the sum (2), then its complex conjugate will not appear as then $\Delta E_{m’,n’}=-\Delta E_{n’,m’}$. In other words, each of the sums in (2) is of one of the following types:
- The first is over $m’>n’$ with $\Delta E_{m’,n’}> 0$,
- the next is over $m’<n’$ with $\Delta E_{m’,n’}<0$,
- the last is for $\Delta E_{m’,n’}=0$, which implies $m’=n’$ since the energies are assumed non-degenerate.
For the first case we have
$$
\Delta E_{m’n’}\sum_{m’>n’} c_{m’}c_{n’}^*\psi_{n’}(x)^*\psi_{m’}(x) = 0\, ,
\qquad
\Delta E_{m’n’}\ne 0\, . \tag{3}
$$
However, the set $\{\psi_{n’}(x)^*\psi_{m’}(x), m’>n’\}$ is a linearly independent basis set so there is no $c_{m’}c_{n’}^*$ to satisfy (2). Hence there is no solution unless you admit $\Delta E_{m’n’}=0$, a contradiction.
The second case is done in the same way. Only the third case leads to no contradiction, which shows the result.