-
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
$$\mathcal{L}^\dagger =\psi^\dagger(i\gamma^0\gamma^\mu \gamma^0 \partial_\mu +m)\gamma^0\psi$$
$$ =\overline{\psi}(i\gamma^\mu \overleftarrow{\partial_\mu} +m)\psi \neq \mathcal{L}.$$ -
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.
-
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 ?
Best Answer
$$(\psi^\dagger \gamma^0 \psi)^* = \psi^\dagger \gamma^0 \psi$$ 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: $$ \begin{align} (\eta\xi)^*&=[(a+ib)(c+id)]^*\\ &=(ac-bd+ibc+iad)^*\\ &=ca-db-icb-ida\\ &=(c-id)(a-ib)=\xi^*\eta^* \end{align} $$