# Is the conserved current probability or charge in Klein-Gordon and Dirac equations?

The conserved currents in KG and Dirac are:

K-G : $j^{\mu}=i(\phi^*\partial^{\mu}\phi-\phi\partial^{\mu}\phi^*)$

Dirac: $j^{\mu}=\bar{\psi}\gamma^{\mu}\psi$

$j^0$ is positive definite in Dirac's but not on KG's. One simple way to see this is noticing that, in KG, the conserved current can be written us $j^{\mu}=2p^{\mu}|N|^2$ where $N$ is a normalization factor. When energy is negative, so is $j^0$.

This problem is solved thinking of $j^{\mu}$ as a charge current and not probability. This makes sense because solutions with positive and negative energy (particles and antiparticles) have opposite charge densities.

How does this interpretation apply to the Dirac current? it is positive definite so it can't change sign for negative energy solutions...

• in QFT $j^\mu$ is an operator, and therefore it doesnt make sense to write $j^\mu=2p^\mu|N|^2$. You have to express it in terms of ladder operators $a_{\boldsymbol p},a_{\boldsymbol p}^\dagger$. – AccidentalFourierTransform Feb 12 '17 at 21:37
• Okey, that makes sense. I'm trying to understand classical fields first though. – P. C. Spaniel Feb 12 '17 at 21:41
• The wikipedia article on the Gordon decomposition has a discussion of the Dirac current as $p^\mu \rho$ where $\rho$ is the charge density. But as AFT said in the first comment, $\rho$ can be negative as well as positive – mike stone Feb 12 '17 at 22:02