Knowing that the free Dirac Lagrangian is :

$$\tag{1} \mathcal{L}= \bar{\psi} (i \gamma^\mu \partial_\mu -m ) \psi$$

and that the Euler-Lagrange equation is:

$$\tag{2} \frac{\partial \mathcal{L}}{\partial \psi}= \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\right)$$

I am trying to obtain the standard form of the Dirac equation $(3)$ without differentiating by $\bar{\psi}$:

$$\tag{3} (i\gamma^\mu \partial_\mu - m) \psi =0$$

I've been told that the relations $(\gamma^0)^2=1$ and $\gamma^{\dagger \mu}= \gamma^0 \gamma^\mu \gamma^0$ might be useful (I don't see how)

My take on it:

After expanding $(1)$: $$\tag{4} \frac{\partial \mathcal{L}}{\partial \psi}=- \bar{\psi}m$$


$$\tag{5} \frac{\partial \mathcal{L}}{\partial ( \partial_\mu \psi)}=\bar{\psi} i \gamma^\mu$$


$$\tag{6} -\bar{\psi}m - \partial_\mu (\bar{\psi}i \gamma^\mu)=0$$

this is where I become lost:

I think I can use the first term on $(6)$ and rewrite it as $ -\bar{\psi}m= m \bar{\psi}$, but how do I proceed from here?

I think I must find a way to place on $\bar{\psi}$ on the RHS of both terms in $(6)$, proceed by multiplying everything by $\psi$ twice, once to cancel the $\bar{\psi}$ and twice to terminate in the standard form shown in $(3)$. How do I do this? Is there another way?

I have seen other questions and links on the website but these don't quite do it as I intend and tend to differentiate by $\bar{\psi}$ which isn't what I want. Some of the links visited are: Going from the Dirac Lagrangian to the adjoint Dirac equation , Derivation of Dirac equation using the Lagrangian density for Dirac field .

  • 1
    $\begingroup$ Hint: Derive the Euler-Lagrange-Equation of $\bar\psi$. $\endgroup$ – Toffomat Jun 2 '20 at 11:00
  • $\begingroup$ That has already been done (in one of the links above it is done that way) but I am trying to do it by differentiating via $\psi$ and not $\bar{\psi}$. Thank you for your help $\endgroup$ – user7077252 Jun 2 '20 at 11:02
  • $\begingroup$ Why did you consider $\psi$ and $\bar{\psi}$ as independent variables? $\endgroup$ – Nikita Jun 30 '20 at 10:27

You start with $$\bar{\psi}(i\gamma^\mu\overleftarrow{\partial}_\mu+m) = 0$$ if you take the hermitian conjugate of all

$$\begin{align}\left(\bar{\psi}(i\gamma^\mu\overleftarrow{\partial}_\mu+m)\right)^\dagger &= \left(\psi^\dagger\gamma^0(i\gamma^\mu\overleftarrow{\partial}_\mu+m)\right)^\dagger= (i\gamma^\mu\overleftarrow{\partial}_\mu+m)^\dagger\gamma^{0\dagger}\psi\\ &= (-i\gamma^{\mu\dagger}\partial_\mu+m)\gamma^0\psi = (-i\gamma^0\gamma^\mu\gamma^0\partial_\mu+m)\gamma^0\psi\\ &=(-i\gamma^0\gamma^\mu\gamma^0\gamma^0\partial_\mu+m\gamma^0)\psi =(-i\gamma^0\gamma^\mu\partial_\mu+\gamma^0m)\psi\\ &=\gamma^0(-i\gamma^\mu\partial_\mu+m)\psi= 0 \end{align}$$ Now multiply from the left with a $\gamma^0$ and use the property $\gamma^0\gamma^0=1$ $$(-i\gamma^\mu\partial_\mu+m)\psi=0$$ which is exactily $$(i\gamma^\mu\partial_\mu-m)\psi = 0$$

  • $\begingroup$ I understand that in the first line you can move the $\gamma^0$ outside of the differentiation because it is just a number, but how can you just move the $\psi^\dagger$ by multiplying it by $\psi$? I thought we couldn't do it, because it is being differentiated. $\endgroup$ – user7077252 Jun 2 '20 at 13:27
  • $\begingroup$ Nobody said that we're multiplying by $\psi$ since by doing so you won't achieve anything at all. $\psi$ is a spinor, a function, it makes no sense multiply something by it. In the first line we only used the fact that transposition swaps the terms in a product $(AB)^T = B^T A^T$ $\endgroup$ – Davide Morgante Jun 2 '20 at 14:48

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