1. I'm wondering of the Dirac Lagrangian density $$\mathcal{L} =\overline{\psi}(-i\gamma^\mu \partial_\mu +m)\psi $$ is an hermitian operator, since upon complex conjugating one gets $$\mathcal{L}^\dagger =\psi^\dagger(i\gamma^0\gamma^\mu \gamma^0 \partial_\mu +m)\gamma^0\psi$$ $$ =\overline{\psi}(i\gamma^\mu \overleftarrow{\partial_\mu} +m)\psi \neq \mathcal{L}.$$

  2. And should a Lagrangian always be Hermitian ? I know that a Hermitian operator has real eigenvalues, which is desirable for a operator describing observables. But here the Lagrangian isn't really an observable since it is determined modulo an total derivative.

  3. I've found a related question on the site: were it is said that

    "the derivative $\partial_\mu$ in the Dirac Lagrangian is antihermitian" (Is the Lagrangian density in field theory real?)

can someone show me how to demonstrate this ?


3 Answers 3


$$(\psi^\dagger \gamma^0 \psi)^* = \psi^\dagger \gamma^0 \psi$$ because $\gamma^0$ is hermitian. Also,

$$ \begin{align} (\psi^\dagger i \gamma^0 \gamma^\mu \partial_\mu \psi)^* &= -i \partial_\mu\psi^\dagger \gamma^{\mu\dagger} \gamma^0 \psi\\ &= -i \partial_\mu\psi^\dagger (\gamma^0 \gamma^\mu \gamma^0)\gamma^0 \psi\\ &= -i \partial_\mu\psi^\dagger \gamma^0 \gamma^\mu \psi\\ &= i \psi^\dagger \gamma^0 \gamma^\mu \partial_\mu\psi + \mathrm{surface\,\,term}\\ \end{align} $$ For the second line I used $\gamma^{\mu\dagger} = \gamma^0 \gamma^\mu \gamma^0$ and for the last line I integrated by parts. I think your question hinges on this part, because the last "index" we sum over is the spacetime index $x^\mu$, i.e., integration. It is the same reason why the quantum mechanical momentum operator $p = i \tfrac{\partial}{\partial x}$ is hermitian.

Edit: Something I glossed over is that the spinors are also Grassmann numbers, so care has to be taken. In particular, this means that the components of the spinors satisfy

$$(\psi_i \phi_j)^* = \phi_j^* \psi_i^*$$

(more about that here). One already interchanges the objects when taking the Hermitian conjugate by the rules of matrix algebra, and there is a temptation to want to introduce a minus sign because they are Grassmann numbers, but this would be redundant. Borrowing from the linked math.se answer: $$ \begin{align} (\eta\xi)^*&=[(a+ib)(c+id)]^*\\ &=(ac-bd+ibc+iad)^*\\ &=ca-db-icb-ida\\ &=(c-id)(a-ib)=\xi^*\eta^* \end{align} $$

  • $\begingroup$ In the first paragraph you impose the hermiticity condition $\gamma^{\mu\dagger} = \gamma^0 \gamma^\mu \gamma^0$ doesn't that cost you relativistic invariance? $\endgroup$
    – R. Rankin
    Jun 26, 2018 at 8:46

In principle what matters is the action and not the Lagrangian density that, strictly speaking must not be considered as an observable, since it is defined up to boundary terms. Concerning the free Lagrangian you are considering, it is real up to a boundary term. You can also re-define it just by adding the Hermitean conjugate Lagrangian you wrote to the original one and taking one half the result. This new Lagrangian density is real and equivalent to the initial one.


The Lagrangian density is real (always). So by definition it is Hermitian.

  • 3
    $\begingroup$ This might be better if you provided a bit more exposition. $\endgroup$
    – Kyle Kanos
    Dec 25, 2018 at 0:11
  • 2
    $\begingroup$ This answer is false. The lagrangian density is not even always hermitian. However if it is not, you can add a term to it which does not affect the action (or the EOM) which makes it hermitian. This is not just by definition of what a Lagrangian density is, though. It has to be shown for each L individually. $\endgroup$ Oct 5, 2019 at 16:17

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