# Probability current vs. direction of wave function

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 $-(k_1 - k_2)^2 < 0$"

• When I have solved mentioned problem I have got the probability current density whose direction is oriented along the wave vectors. I used Euler's formulas. Also, don't forget to take imaginary part only. – freude Oct 26 '13 at 16:30
• The task explicitly states that the situation is like I described. Furthermore I got the solution of this exercise: See edit. – Prook Oct 26 '13 at 17:02
• I just recognized that there was also an error in the formula of j(x). That might have caused your result. I'm sorry for that. Just corrected it. – Prook Oct 26 '13 at 17:11

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.

I happened to stumble upon this old question, but I think it is still worth to give an answer for future reference.

Dominique Geoffrey is right in pointing out the fact that a dispersion relation is needed. However I think that this problem is instructive even in the most simple case of a material particle evolving freely, i. e. with a dispersion relation $\omega(k) = \frac{\hbar k^2}{2m}$ (I consider the nonrelativistic limit of course).

In this scenario it is completely possible to have a wavepacket with only positive momenta which give rise to a negative current for some time intervals. This counterintuitive phenomenon is well known and it has been studied for quite a long time in quantum mechanics (mainly in connection to the problem of determining the time of arrival distribution of a massive particle described by quantum mechanics).

In the literature this phenomenon is called "backflow effect": see http://arxiv.org/abs/1301.4893 for an introduction to the subject and the references therein if you are interested.