[Physics] Probability current density confusion

Question

As we all know, the probability current density in quantum mechanics is defined as: $$\textbf{J}=\dfrac{\hbar}{2mi}(\Psi^* \nabla \Psi-\Psi \nabla \Psi^*)$$ For simplicity let us work in one dimension and let us suppose a wave function $\Psi= A\ \text{cos}\ {kx}$. Applying the above definition and thus using $$J=\dfrac{\hbar}{2mi}\Big(\Psi^* \dfrac{\partial \Psi}{\partial x}-\Psi \dfrac{\partial \Psi^*}{\partial x}\Big)\quad\quad \text{we get:}\quad\quad J=0$$
Using the equation of continuity this means that:
$$\dfrac{\partial \rho}{\partial t}=0,$$ which after solving gives us: $\rho=f(x)$. Thus the probability density at any point is independent of time.
Now, this result will follow even if we take $\Psi= A\ \text{cos}\ {(kx-\omega t)}$. But here we can clearly see that the probability density i.e.
$$|\Psi|^2=|A|^2\ \text{cos}^2\ {(kx-\omega t)}$$ is time dependent. Is it $A$ which carries the time dependence and is responsible for this apparent discrepancy?

Answer 1

A solution of the free one-dimensional Schroedinger equation:

$$ i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}\,\,\,\quad \text{(1)} $$

is:

$$\psi = A e^{i(kx -\omega t)} \quad\quad\quad \text{(2)} $$

where $\omega$ fulfills the condition $\hbar \omega = \frac{(\hbar k)^2}{2m}$.

If tentatively one tries to construct a $\cos$-solution one would write

$$\psi = \frac{A}{2} e^{i(kx -\omega t)} + \frac{A}{2} e^{-i(kx -\omega t)} = A \cos (kx -\omega t)$$

Upon checking if $$\psi = A e^{-i(kx -\omega t)}$$ solves the Schroedinger equation one would only find a solution only if the following condition is fulfilled:

$$E = \hbar \omega = -\frac{(\hbar k)^2}{2m}$$

However, negative energy solutions are not allowed in the non-relativistic theory, therefore this solution has to be discarded, consequently also the $\cos$-solution has also to be discarded. This can, of course, be directly checked by inserting the $\cos (kx-\omega t)$ in the free Schroedinger-equation (1); it is not a solution. So one cannot expect it to fulfill the continuity equation.

So the only reasonable solutions in this context are either (2) or

$$\psi(x) = \cos(kx)\quad\quad\quad \text{(3)} $$

for the free time-independent Schroedinger equation

$$ \frac{\partial^2 \psi}{\partial x^2} +\frac{2m}{\hbar^2}E =0$$

with the condition $\frac{(\hbar k)^2}{2m} =E$.

Both solutions (2) and (3) fulfill the continuity equation, even if in the case of (3) it turns out to be quite uninteresting.

Solution (3) can of course be upgraded to a time-dependent solution by choosing

$$\psi(x,t) = e^{-i\omega t} \cos(kx)$$

Of course appropriate superpositions of either (2) or (3) would also be solutions, but using the right sign of $i$ in case of time-dependent solutions.

EDIT In case of the time-dependent solution (2) the probability current $J$ is non-zero, but its gradient is zero, therefore even if $\dot{\rho}=0$

$$ \dot{\rho} + \nabla J =0$$

is fulfilled.