How can I derive the Dirac equation from the Lagrangian density for the Dirac field?

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    $\begingroup$ Use the Euler-Lagarange equation for fields. $\endgroup$ – user7757 May 19 '13 at 8:24
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    $\begingroup$ This standard derivation is done in every QFT text. If you are getting stuck can you please elaborate about what exactly is giving you trouble? $\endgroup$ – Michael Brown May 19 '13 at 8:51
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    $\begingroup$ Yeah, it's like a 2-line derivation $\endgroup$ – user12345 May 19 '13 at 10:37

The Lagrangian density for a Dirac field is $$ \mathcal{L} = i\bar\psi\gamma^\mu\partial_\mu\psi -m \bar\psi\psi $$ The Euler-Lagrange equation reads $$ \frac{\partial\mathcal{L}}{\partial\psi} - \frac{\partial}{\partial x^\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\right] = 0 $$ We treat $\psi$ and $\bar\psi$ as independent dynamical variables. In fact, it is easier to consider the Euler-Lagrange for $\bar\psi$ $$ \frac{\partial\mathcal{L}}{\partial\bar\psi} - \frac{\partial}{\partial x^\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar\psi)}\right] = 0\\ \Rightarrow i\gamma^\mu\partial_\mu\psi -m\psi - \frac{\partial}{\partial x^\mu}[ 0] = 0\\ \Rightarrow i\gamma^\mu\partial_\mu\psi -m\psi=0 $$ The partial differentiation is trivial - remember that $\bar\psi$ and $\partial_\mu\bar\psi$ are treated as though independent. We recover the Dirac equation as expected. If we had instead chosen the Euler-Lagrange for $\psi$, we would have found the conjugate Dirac equation.

  • $\begingroup$ While treating $\psi$ and $\bar{psi}$ as distinct, are we also accepting that the Lagrangian is of a interacting theory? $\endgroup$ – zudumathics May 25 '17 at 5:17

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