Dirac equation for the conjugated field

1. Dec 12, 2009

phsopher

This is probably a stupid question, but when I apply the Euler-Lagrange equation to the Lagrangian density of the Dirac field I get for the conjugate field

$$\bar{\psi} (-i \partial_\mu \gamma^{\mu} -m) = 0$$ (derivative acts to the left).

But when I take a hermitian conjugate of the Dirac equation for the field I get an extra $$\gamma^0$$:

$$0 = \left[ (i \partial_\mu \gamma^{\mu} -m)\psi \right]^\dagger = \psi^\dagger (-i \partial_\mu (\gamma^{\mu})^\dagger -m) = \psi^\dagger (-i \partial_\mu \gamma^0 \gamma^{\mu} \gamma^0 -m) = \psi^\dagger \gamma^0(-i \partial_\mu \gamma^{\mu} \gamma^0 -m) = \bar{\psi} (-i \partial_\mu \gamma^{\mu} \gamma^0 -m)$$.

What am I missing?

2. Dec 12, 2009

dextercioby

Taking the normal hermitean conjugate of the <original> Dirac eqn will not give you the <conjugated> equation. You need an extra $\gamma_0$.

Actually, you took out the $\gamma_0$ from the paranthesis without it being there next to the <m>. That's wrong.

3. Dec 13, 2009

phsopher

Ah yes, of course. Then I can multiply with $$\gamma^0$$ from the right and get the same equation as from Euler-Lagrange. I'm such an idiot. Thanks a lot.