Editing Dirac equation (section)

=== Conserved current ===
A [[conserved current]] of the theory is
<math display="block">J^\mu = \bar{\psi}\gamma^\mu\psi.</math>
{{math proof | title = Proof of conservation from Dirac equation | proof =
Adding the Dirac and adjoint Dirac equations gives
<math display="block">i((\partial_\mu\bar\psi)\gamma^\mu\psi+\bar\psi\gamma^\mu \partial_\mu\psi) = 0</math>
so by Leibniz rule,
<math display="block">i\partial_\mu(\bar\psi\gamma^\mu\psi) = 0</math>
}}

Another approach to derive this expression is by variational methods, applying [[Noether's theorem]] for the global <math>\text{U}(1)</math> symmetry to derive the conserved current <math>J^\mu.</math>

{{math proof | title = Proof of conservation from Noether's theorem | proof =
Recall the Lagrangian is
<math display="block">\mathcal{L} = \bar\psi(i\gamma^\mu \partial_\mu - m)\psi.</math>
Under a <math>U(1)</math> symmetry that sends 
<math display="block">\begin{align}
\psi &\mapsto e^{i\alpha}\psi, \\
\bar\psi &\mapsto e^{-i\alpha}\bar\psi,
\end{align}</math>
we find the Lagrangian is invariant.

Now considering the variation parameter <math>\alpha</math> to be infinitesimal, we work at first order in <math>\alpha</math> and ignore <math>\mathcal{O}{\alpha^2}</math> terms. From the previous discussion we immediately see the explicit variation in the Lagrangian due to <math>\alpha</math> is vanishing, that is under the variation,
<math display="block">\mathcal{L}\mapsto \mathcal{L} + \delta\mathcal{L},</math>
where <math>\delta\mathcal{L} = 0</math>.

As part of Noether's theorem, we find the implicit variation in the Lagrangian due to variation of fields. If the equation of motion for <math>\psi, \bar\psi</math> are satisfied, then
{{NumBlk||<math display="block">\delta\mathcal{L} = \partial_\mu\left(\frac{\partial \mathcal{L}}{\partial (\partial_\mu \psi)}\delta\psi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \bar\psi)}\delta\bar\psi\right) </math>|{{EquationRef|<nowiki>*</nowiki>}}}}
This immediately simplifies as there are no partial derivatives of <math>\bar\psi</math> in the Lagrangian. <math>\delta\psi</math> is the infinitesimal variation
<math display="block">\delta\psi(x) = i\alpha\psi(x).</math>
We evaluate
<math display="block">\frac{\partial \mathcal{L}}{\partial (\partial_\mu \psi)} = i\bar\psi\gamma^\mu.</math>
The equation ({{EquationNote|*}}) finally is
<math display="block">0 = -\alpha\partial_\mu(\bar\psi\gamma^\mu\psi)</math>
}}