Covariance of the Continuity Equation

As mentioned previously, we still need to show the conserved current is a 4-vector. We have

$$j^0 = c\rho = c\psi^\dagger\psi = c\psi^\dagger\gamma^0\gamma^0\psi,$$ (5.158)

$$j^k = c\psi^\dagger\alpha^k\psi = c\psi^\dagger\gamma^0\gamma^0\alpha^k\psi = c\psi^\dagger\gamma^0\gamma^k\psi,$$ (5.159)

Therefore

$$j^\mu(x) = c\psi^\dagger(x)\gamma^0\gamma^\mu\psi(x) .$$ (5.160)

Under a Lorentz transformation we have

$${j^\prime}^\mu(x^\prime) = c\psi^\dagger(x) S^\dagger\gamma^0\gamma^\mu S\psi(x)$$

$$= c\psi^\dagger(x)\gamma^0 S^{-1}\gamma^\mu S\psi(x)$$

$$= c\psi^\dagger(x)\gamma^0 a^\mu_{\ \nu}\gamma^\nu\psi(x)$$

$$= a^\mu_{\ \nu} j^\nu(x).$$ (5.161)

Thus $j^\mu(x)$ is a Lorentz 4-vector.