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.




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

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


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

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.

  • $\begingroup$ "$\bar{\psi}(i\gamma^{\mu} \overleftarrow{\partial}_{\mu}-m) = 0$", pls check the sign. $\endgroup$ – MadMax Mar 10 '20 at 15:00
  • $\begingroup$ Whoops, thank you $\endgroup$ – infinitezero Mar 10 '20 at 15:50
  • $\begingroup$ 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. $\endgroup$ – user400188 May 13 '20 at 7:43
  • 1
    $\begingroup$ I didn't do this anywhere? $\endgroup$ – infinitezero May 13 '20 at 8:10
  • 1
    $\begingroup$ Yes. Since $\gamma^\mu$ does not depend on $x$, it commutes with $\partial_\mu$. $\endgroup$ – infinitezero May 13 '20 at 9:06

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