Probability Current identically zero if and only if statement

Question

I have just started a course on quantum theory and have been stuck on this problem.

Suppose $\Psi(x,t)$ satisfies the one-dimensional time-dependent Schrödinger equation with real potential $V(x)$. That is,

$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi$

With $\rho(x,t) = |\Psi(x,t)|^2$ and $j(x,t) = \frac{i\hbar}{2m}(\Psi\frac{\partial \bar{\Psi}}{\partial x} - \bar{\Psi}\frac{\partial \Psi}{\partial x} )$ where $\bar{\Psi}$ denotes the complex conjugate of $\Psi$, we have the continuity equation, that $\frac{\partial \rho}{\partial t} + \frac{\partial j}{\partial x} = 0$

I need to show that j vanishes identically if and only if there exists a nowhere zero function $\lambda(t)$ such that $\lambda(t)\Psi(x,t)$ takes only real values. The if direction is simple, but I do not know how to make progress on the only if direction.

Answer 1

The probability current $j$ vanishes if and only if the phase $\theta$ of $\Psi$ is constant w.r.t. $x,$ i.e. only depends on $t.$ Take $\lambda=e^{-i\theta}.$

Answer 2

t is irrelevant for the current, effectively a constant, so we drop it from the discussion. Consider the polar representation $$ \Psi(x)=r(x)e^{i\theta (x)}, $$ for real r and θ.

Thus, $$\Psi\bar\Psi_x-\Psi_x\bar\Psi= -2ir^2 \theta_x.$$

So, for zero current, θ is a constant function, except at the zeros of r. Take the phase of λ to cancel the phase of θ when it's constant, so $\lambda \Psi$ is real; or 0, which is real.

You saw how the reverse holds: $$\bar \lambda \lambda (\Psi\bar\Psi_x-\Psi_x\bar\Psi)=0 ~~\implies ~~ \Psi\bar\Psi_x-\Psi_x\bar\Psi= 0.$$