A question on QED In QED it's given a current of the particle described by wave function $\psi$
$$j^\mu=\bar{\psi}\gamma^\mu\psi.$$
If we substitute positive- and negative-energy solutions we get that
$$j_{+}^0=j^0_-,$$
but
$$\vec{\jmath}_+=-\vec{\jmath}_-.$$
Is there a problem with that? That means that positrons would create the same electric potential as an electron, but theirs magnetic fields would be negative to each other. That seems odd, since positrons have charge negative to electron's charge.
 A: The negative-energy solutions are not, on their own, representative of positrons.  Instead, a positron is the absence of a negative-energy electron.  So the negative-energy electron has a positive charge, because the hole where a negative-energy electron could be has positive charge. This explains why the negative-energy electron generates a charge density $j^{0}$ with the usual sign.  In fact, it also generates a three-current $\vec{\jmath}$ that that is the same as for a positive-energy electron, provided that they have the same three-velocities $\vec{v}$.
However, there is an extra complication with the negative-energy solutions—that the momentum $\vec{p}$ and the velocity $\vec{v}$ point in opposite directions!  This is what is responsible for the sign change of $\vec{\jmath}$ noted in the question; if you compare the positive- and negative-energy electron modes with the same momentum $\vec{p}$, they are actually moving in opposite directions.  The reason for the negative sign in the velocity in the negative-energy case just comes from the definition of the group velocity $\vec{v}_{g}=\vec{\nabla}_{\vec{p}}E$, which points in the opposite of the expected direction when $E=-\sqrt{m^{2}+p^{2}}$ is negative.
