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The standard Dirac action $$ S = \int d^4 x \bar \psi (i \gamma^\mu \partial_\mu - m) \psi $$ is invariant under Lorentz transformation.

In David Tong's lecture note, eq (4.96) lists that the corresponding Noether current is $$ (J^\mu)^{\rho \sigma} = x^\rho T^{\mu \sigma} - x^\sigma T^{\mu \rho} - i \bar \psi\gamma^\mu S^{\rho \sigma} \psi \ , $$ where $T$ is the conserved stress tensor and $S^{ab} \propto \gamma^a\gamma^b - \gamma^b \gamma^a$.

The first two terms clear conserve. However, I fail to show that the second term is conserved: $$ \begin{align} \partial_\mu (\bar \psi i \gamma^\mu S^{\rho \sigma}\psi) = & \ (\partial_\mu\bar \psi i \gamma^\mu S^{\rho \sigma}\psi) + (\bar \psi i \gamma^\mu S^{\rho \sigma}\partial_\mu\psi)\\ = & \ - m (\bar \psi S^{\rho \sigma}\psi) + (\bar \psi i \gamma^\mu S^{\rho \sigma}\partial_\mu\psi)\ . \end{align} $$ I'm not sure how to deal with the second term, since $\gamma^\mu$ sits away from $\partial_\mu$, and $\gamma$ does not simply commute with $S^{\rho \sigma}$.

I wonder if there is some magical trick I should do or if the current is incorrect?

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    $\begingroup$ In the spinor theoy only the total angular momentum $J=L+S$ is conserved. $\endgroup$ May 20 at 8:46
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    $\begingroup$ The first two terms are conserved only if $T^{\mu\nu} = T^{\nu\mu}$ which is not true here (assuming $T$ is the canonical stress tensor, not the Belinfante stress tensor). $\endgroup$
    – Prahar
    May 20 at 13:24
  • $\begingroup$ @Prahar you are right, I mistaken the said $T$ with the symmetric one. $\endgroup$
    – Lelouch
    May 21 at 7:19

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