I was given the wavefunction $\psi(x)= C e^{-x/a}e^{ibx}, 0< x < \infty$ and asked to compute $J$, the probability current and $d\rho/dt$, the change in the probability density with respect to time. That is fairly straightforward, using $d\rho/dt = -dJ/dx$, however I'm a bit confused as to why $d\rho/dt$ is nonzero. If $\Psi(x,t) = \psi(x) e^{i\omega t}$, shouldn't it follow that $\rho$ is time independent?
Also, it seems that if the wavefunction did not have the $e^{ibx}$ factor, then $d\rho/dt= 0$, so is there an intuitive explanation as to why this happens?
Additional details: I wasn't given the Hamiltonian or told anything about eigenstates. The question simply gave me the above wavefunction and asked to find the following:
the probability density $\rho(x)$,
The probability current $J(x)$ and
$\partial\rho/\partial t$.
Here are the values I found for the above:
$|C|^2 = \frac{4}{a^3}\implies \rho(x) = \frac{4}{a^3}x^2e^{-2x/a}$
$\Psi(x,t) = \psi(x)e^{-i\omega t}\implies J(x) = \frac{i\hbar}{2m}(\Psi\frac{\partial\Psi^*}{\partial x} - \Psi^*\frac{\partial\Psi}{\partial x}) = \frac{4\hbar bx^2e^{-2x/a}}{ma^3}$
$\partial\rho/\partial t = -\partial J/\partial x = -\frac{8\hbar bx}{ma^3}e^{-2x/a}(1-\frac{x}{a})$