The conserved current for a field [tex]\phi[/tex] obeying the Klein-Gordon equation is (neglecting operator ordering) proportional to [tex]i\phi^{\dag}\partial_\mu \phi-i\phi\partial_\mu \phi^{\dag}[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

The conserved current for a four component field [tex]\psi[/tex] obeying the Dirac equation is [tex]\psi^{\dag}\gamma^0\gamma^\mu \psi[/tex], with no derivatives.

The Klein-Gordon conserved current cannot be interpreted as conservation of probability because the 0-component is not non-negative and so cannot be a pdf. On the other hand, it does make sense as a conservation of charge (or particles vs anti-particles).

The Dirac current does lend itself to an interpretation as conservation of probability.

So far, so good? I am simplifying for the sake of brevity, but let me know if I said anything wrong.

Now, what do we make of the fact that each of the individual components [tex]\psi_j[/tex], j=1,2,3,4 of the Dirac field separately obey the Klein-Gordon equation, and therefore also satisfy

[tex]\partial^\mu\{i\psi^{\dag}_j\partial_\mu \psi_j-i\psi_j\partial_\mu \psi^{\dag}_j\}=0[/tex] ?

That is, both currents are conserved--the Dirac current in terms of all four components, coupled, and the Klein-Gordon current, each component independently. I presume that the two conservations are not equivalent, since one involves derivatives and the other doesn't.

What is the physical significance of each of these conserved currents?

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# Dirac conserved current vs Klein-Gordon conserved current

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