A Manipulation with the Dirac equation

[spaghetti3451]

I know that the Dirac equation is $i\gamma^{\mu}\partial_{\mu}\psi=m\psi$.

How do I use this to show that $(\partial_{\mu}\bar{\psi})\gamma^{\mu}=im\bar{\psi}$?

 

[vanhees71]

First of all you have (in the usual standard representations of the Dirac matrices)
$$\gamma^{\mu \dagger}=\gamma^0 \gamma^{\mu} \gamma^0,$$
the "pseudo-hermitecity relation" and the definition
$$\overline{\psi}=\psi^{\dagger} \gamma^0.$$
So now take the Dirac equation and first apply Hermitean conjugation:
$$-\mathrm{i} \partial_{\mu} \psi^{\dagger} \gamma^{\mu \dagger}=m\psi^{\dagger}.$$
Now use $(\gamma^0)^2=1$ to first write on the left-hand side
$$-\mathrm{i} \partial_{\mu} \overline{\psi} \gamma^0 \gamma^{\mu \dagger} = m \psi^{\dagger}.$$
Finally multiply this equation with $\gamma^0$ and use the pseudo-hermitecity relation of the Dirac matrices to finally get the claimed equation:
$$-\mathrm{i} \partial_{\mu} \overline{\psi} \gamma^{\mu}=m \overline{\psi}.$$

 

[dextercioby]

How do you get from Psi to Psi-bar in the absence of a symmetric Lagrangian? @vanhees71 this smells like a homework problem (incorrectly placed outside the HW section), so I wouldn't throw in the solution.

 

[vanhees71]

What has the Lagrangian to do with basic definitions of the $\gamma$ matrices and bispinors?

 

[dextercioby]

Well, the Dirac equation in line 1 is one of the two Euler-Lagrange equations for the symmetrized Lagrangian density. The other Euler-Lagrange equation is the one whose derivation he sought. Your solution is direct, I was trying to lead him there.