Probability current is zero for normalizable stationary state

Question

I'm asked to show that the probability current is zero for a normalizable stationary state of the Shrodinger Equation. So we have that $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$. Now using the conservation of probability we have $$0=\frac{\partial}{\partial t}|\psi|^2=\frac{\partial}{\partial t}|\Psi|^2=-\frac{\partial j}{\partial x}$$ so $j=j(t)$, but in the definition of $j$ all the $t$ dependence drops out to give $$j=\frac{-i\hbar}{2m}\left(\psi^*\psi'-\psi'^*\psi\right)$$ so $j=j(x)$ and so we must have $j=\text{const}$.

I want to say that for the state to be normalizable we must have $j\rightarrow 0$ as $|x|\rightarrow\infty$, and so $j=0$ everywhere. But this argument becomes complex as I have to rule out cases like $\sin(e^x)/x$ where $\psi\rightarrow 0$ but $\psi'\rightarrow\infty$. I know this argument is closely linked to showing that in 1D the SE has no degeneracy, but I am sure the exam question doesn't want me to use such a complex argument. I also don't think I can quote the lack of degeneracy.

Is there a simpler way to show this, or is it in fact equivalent to the non-degeneracy of the 1D SE?

Answer 1

I want to say that for the state to be normalizable we must have $j\rightarrow 0$ as $|x|\rightarrow\infty$, and so $j=0$ everywhere.

I don't know why should be the case.

Normalisable just means that $|\Psi|^2 \rightarrow 0$ "fast enough". The $j$ takes out the phase, which in $|\Psi|^2$ does not matter, so I don't think that $j \rightarrow 0$ is equivalent to normalisation.

Stationary states have zero probability current $j$ just by virtue of being in the form (real function)*(pure phase factor), so that the conjugates in the definition of $j$ do not give you a "net" term.