# Going from the Dirac Lagrangian to the adjoint Dirac equation

I am comfortable doing the following calculation, Derivation of the adjoint of Dirac equation, notably — going from the standard Dirac equation to the adjoint Dirac equation via using Dirac conjugation.

I am however not comfortable deriving the adjoint Dirac equation from the Euler Lagrange equation (for $$\psi$$, the EL eq for $$\overline{\psi}$$ leads to the standard Dirac equation) of the Dirac Lagrangian

$$\mathcal{L} =\overline{\psi}(i\gamma^\mu \partial_\mu -m)\psi$$.

My problem boils down to the following term,

$$\partial_\mu (i\overline{\psi}\gamma^\mu)$$

How can I get this to yield

$$-i\gamma^\mu \partial_\mu \overline{\psi}?$$

My only thought is to write out Dirac conjugate, write out the Einstein summation, and use the properties of the gamma matrices ($$\gamma^0\gamma^0=$$ the identity matrix, and $$\gamma^0\gamma^i=-\gamma^i\gamma^0$$), but it doesn't seem to yield anything fruitful.

Thoughts?

Cheers

The Euler-Lagrange-Equation is given by:

$$\frac{\partial\mathcal{L}}{\partial {\psi}} - \partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi)} = 0$$

Let us take both derivatives separately. We treat $$\psi$$ and $$\bar{\psi}$$ as independent fields. This gives

$$\frac{\partial}{\partial\psi}\bar{\psi}(i\gamma^\mu\partial_{\mu}-m)\psi = -m \bar{\psi}$$

and

$$\partial_{\mu}\frac{\partial}{\partial(\partial_{\mu}\psi)}\bar{\psi}(i\gamma^\mu\partial_{\mu}-m)\psi = i\partial_{\mu}\bar{\psi}\gamma^\mu$$

So by plugging this into the first equation we get:

$$-m\bar{\psi} - i\partial_{\mu}\bar{\psi}\gamma^\mu = 0$$

Now, to make this look nicer (and bring this in the usual form) we say, that the differential operator operates to the left:

$$\bar{\psi}(i\gamma^{\mu} \overleftarrow{\partial}_{\mu}+m) = 0$$

Your proposed solution $$i \partial_\mu \gamma^\mu \bar{\psi}$$ doesn't work dimensional wise. Since $$\bar{\psi}$$ is a row vector, it needs to be left of the $$\gamma^\mu$$, which is a $$4\times 4$$ square matrix.

• "$\bar{\psi}(i\gamma^{\mu} \overleftarrow{\partial}_{\mu}-m) = 0$", pls check the sign. – MadMax Mar 10 at 15:00
• Whoops, thank you – infinitezero Mar 10 at 15:50
• How is it that you pulled out $\bar\Psi$ from $\partial_\mu\bar\Psi\gamma^\mu$? If $\gamma^\mu$ is a matrix, it will not in general commute. – user400188 May 13 at 7:43
• I didn't do this anywhere? – infinitezero May 13 at 8:10
• Yes. Since $\gamma^\mu$ does not depend on $x$, it commutes with $\partial_\mu$. – infinitezero May 13 at 9:06