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**Gordon decomposition an "right" current**

I've got a question about the Dirac equation and the Gordon decomposition of the dirac current. The Gordon decomposition is

[tex]\bar \psi \gamma^\mu \psi = - \frac{1}{m}\Im (\bar \psi \partial^\mu \psi) - \frac{1}{2m}\partial_\mu ( \bar \psi \sigma^{\mu\nu} \psi) [/tex]

where h=c=1. As is well known the divergence of the left hand side vanishes if we simply use the dirac equation and its adjoint. But the divergence of the second term on the right hand side (the so called spin current) vanishes as well because it is antisymmetric in [tex] (\mu,\nu)[/tex]. As a consequence the first current on the rhs (the so called convective) must also be conserved.

So in total we have three conserved currents, namely

[tex]j^\mu = \bar \psi \gamma^\mu \psi [/tex]

[tex]j^\mu_{conv} = - \frac{1}{m}\Im (\bar \psi \partial^\mu \psi) [/tex]

[tex]j^\mu_{spin} = -\frac{1}{2m} \partial_\mu (\bar \psi \sigma^{\mu\nu} \psi) [/tex]

Firstly: that seems a bit much to me. Especially considering the fact that charge conservation results from gauge invariance I would expect that there is exactly one conserved current as there is exactly one gauge generator for electromagnetism.

Secondly: how do I know that it is really the first (i.e. dirac) current which enters the Maxwell equations as its sources ? Of course we could look at the well known Lagrangian of Dirac+Maxwell and say "oh it works, why bother". But couldn't we assume that it is actually the convective current or the spin current or even an unspecified linear combination of the three which acts as the source of the electromagnetic field ?