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$$\mathcal{L}=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi,$$ taking Euler-Lagrange equation on $\bar\psi$ gives the more familiar Dirac equation $$(i\gamma^\mu\partial_\mu-m)\psi=0$$ and its adjoint version $$\bar\psi(i\gamma^\mu\stackrel{\leftarrow}{\partial_\mu}+m)=0 .$$

Taking E-L equation on $\psi$ however gives $$\bar\psi(i\gamma^\mu\partial_\mu-m)=\partial_\mu(\bar\psi i\gamma^\mu)$$ which clearly constrains $\bar\psi$ differently. What is going on here?

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  • $\begingroup$ The r.h.s. of your third equation should not be there. I would advice you to check your derivation. $\endgroup$
    – Christophe
    Mar 14, 2023 at 6:51
  • $\begingroup$ For some reason you kept the $\partial_\mu$-term on the l.h.s. of the last equation, which is supposed to be $\frac{\partial\mathcal{L}}{\partial\psi}$. If you drop it, you're fine. $\endgroup$
    – go_science
    Mar 14, 2023 at 17:56

1 Answer 1

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(I haven’t checked the equations of motion you have written, I am assuming they are right)

In quantum field theory fields $(\phi)$are operators but positions $x$ are not. So $\partial_\mu$ is also not an operator. That means conjugate of $ {\partial_\mu\psi}$ is just $ {\partial_\mu\bar\psi}$

So the adjoint of EOM with respect to $\psi$ is actually $$(i\gamma^\mu\partial_\mu+m) \bar\psi =0 $$

Now use product rule on EL equation on $\psi$ to see that they are the same.

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