# Probability current is zero for normalizable stationary state

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?

• Commented Apr 20, 2021 at 14:53
• @Jakob Thanks for the link. The answer there uses the non-degeneracy of the 1D SE, which is what I'm hoping to avoid. Commented Apr 20, 2021 at 15:16
• Ah okay, sorry, I misread the question. Commented Apr 20, 2021 at 15:17

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.
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.
• The definition of stationary state I am working with is states of the form $\psi(x) e^{-iEt/\hbar}$ - why can we assert that $\psi$ is real? I know we can choose $\psi$ to be real, but what if we don't? Commented Apr 20, 2021 at 22:42
• @acernine Because the "i" in the Schrodinger equation after separation of variables is on the time-dependent part, i.e. the one that gives you the phase factor solution. And since $|\psi|^2$ is what's important, any phase that you give $\psi$ that isn't the time-dependent phase factor is redundant and does not change the physics. So you choose it to be real. Commented Apr 20, 2021 at 23:28
• As I understand it for the physics to be identical we need the phase factor to be global, but a local phase factor could change observables, for example momentum. So I don't understand how (without invoking lack of degeneracy) we can say that there is a real wavefunction which is physically indistinguishable from $\psi$. Commented Apr 21, 2021 at 0:54