# Intuitive explanation of probability density and probability current

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

• 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. – Jakob Feb 1 at 17:22
• @Jakob Sorry I should've been more clear. $a$ and $b$ are just constants, and I want the probability density of $\Psi(x,t)$. – Luis Fe Feb 1 at 17:29
• You could also provide your calculations that show how you arrive that the derivative of $\rho$ is non-zero. – Jakob Feb 1 at 17:31
• Is this for a free particle? If not what is the Hamiltonian? – By Symmetry Feb 1 at 17:52
• ...but Ψ is not a stationary state, no? – Cosmas Zachos Feb 1 at 17:53

## 1 Answer

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).