4

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?

Qmechanic
  • 201,751
acernine
  • 248

1 Answers1

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.

SuperCiocia
  • 24,596
  • 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? – acernine Apr 20 '21 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. – SuperCiocia Apr 20 '21 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 $\psi$. – acernine Apr 21 '21 at 00:54