I did an exercise for my Quantum-Mechanics Lecture:
Let $\hbar$=2m=1. A particle in 1 dimension has $j(x)=2\ Im(\overline{\psi} (x) \ \psi'(x))$ and it's to show that there are superpositions $\psi (x) = a_1 e^{i k_1 x} + a_2 e^{i k_2 x}$, where $k_1, k_2 > 0$, of waves which propagate to the right at x=0 but j(0)<0.
You can show that by calculating j(0) which leads to a non positive semidefinite quadratic form in $a_1,\ a_2$.
(Remark: This superposition can not be normalized, but the exercise states that there are analogue waves which can.)
I have troubles understanding that. How can the wave (and therefore the probability of the particle to be at position x) propagate to the right when the current is negative? Maybe someone could explain me how to think about this?
Edit: The official solution of the exercise:
"With $\psi'=i(k_1 a_1 e^{i k_1 x} + k_2 a_2 e^{i k_2 x})$ is:
$\overline \psi(0) \psi'(0)=\sum_{i,j=1}^{2}i\ \overline{a}_i k_ja_j$ and
$j(0)=\sum_{i,j=1}^{2}(k_i + k_j) \overline{a}_i a_j$
This quadratic form in $a_1, a_2$ is not positive semi definite because the determinant is given by $-(k_1 – k_2)^2 < 0$"
Best Answer
To your question "How can the wave propagate to the right when the current is negative?" I will answer that your statement that "the wave propagates to the right" is not exactly correct: what you have to consider here is group velocity, not each individual phase velocity.
Since both plane waves propagate to the right with $k_1, k_2>0$, then you implicitly assume $\omega_1 \equiv \omega(k_1)>0$ and $\omega_2 \equiv \omega(k_2)>0$, but your problem gives no more information of the dispersion relation.
In order to have a better idea of what the flow of probability density is, you need to consider group velocity here given by $\dfrac{\Delta \omega}{\Delta k} = \dfrac{\omega_2-\omega_1}{k_2-k_1}$, which could indeed have any sign, depending on the dispersion relation.