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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:

  1. the probability density $\rho(x)$,

  2. The probability current $J(x)$ and

  3. $\partial\rho/\partial t$.

Here are the values I found for the above:

  1. $|C|^2 = \frac{4}{a^3}\implies \rho(x) = \frac{4}{a^3}x^2e^{-2x/a}$

  2. $\Psi(x,t) = \psi(x)e^{-i\omega t}\implies J(x) = \frac{i\hbar}{2m}(\Psi\frac{\partial\Psi^*}{\partial x} - \Psi^*\frac{\partial\Psi}{\partial x}) = \frac{4\hbar bx^2e^{-2x/a}}{ma^3}$

  3. $\partial\rho/\partial t = -\partial J/\partial x = -\frac{8\hbar bx}{ma^3}e^{-2x/a}(1-\frac{x}{a})$

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  • $\begingroup$ Do $a$ and $b$ have any meaning? Do you want to calculate the probability density of $\Psi(x,t)$ or of $\psi(x)$? Anyway you are right that if you consider a stationary state, the associated probability density is time-independent. $\endgroup$ – Jakob Feb 1 at 17:22
  • $\begingroup$ @Jakob Sorry I should've been more clear. $a$ and $b$ are just constants, and I want the probability density of $\Psi(x,t)$. $\endgroup$ – Luis Fe Feb 1 at 17:29
  • $\begingroup$ You could also provide your calculations that show how you arrive that the derivative of $\rho$ is non-zero. $\endgroup$ – Jakob Feb 1 at 17:31
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    $\begingroup$ Is this for a free particle? If not what is the Hamiltonian? $\endgroup$ – By Symmetry Feb 1 at 17:52
  • $\begingroup$ ...but Ψ is not a stationary state, no? $\endgroup$ – Cosmas Zachos Feb 1 at 17:53
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You should think about $\psi(x)$ as an initial condition for which the Schrodinger equation will define a time dependence, hence a possibly time dependent $\rho$, whose time dependence is determined by the continuity equation $\partial_x J + \partial_t \rho = 0$. It doesn't matter what the Hamiltonian is for the continuity equation to hold (as long as it's Hermitian).

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