# Is the Lagrangian density in field theory real?

As the Lagrangian in classical mechanics corresponds to energy, it must be real. But is that the case in quantum field theory? I mean, it should still correspond to some sort of energy, but what about all the "$i$"s here and there, such as in the Dirac Lagrangian $i\bar{\psi}\gamma^\mu\partial_\mu \psi$ and the current density $J_\mu = ie[\dots]$ (see Griffiths for example)?

Another question is, how can it be hermitian, $\mathcal{L} = \mathcal{L}^\dagger$, when we have those "$i$"s? Wouldn't I get a minus sign if I complex-conjugated the interaction term and the Dirac field term? I'm really confused and hope someone can help

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Possible duplicates: physics.stackexchange.com/q/127797/2451 and links therein. – Qmechanic Jul 18 at 14:41

What does have to be true is that $\mathcal{L}$ has to have real expectation values in all physical states, and that in turn means it has to be hermitian (what mathematicians call self-adjoint). But hermiticity is not just a matter of being real. You can have other non-hermitian factors besides $i$. In particular, the derivative $\partial_\mu$ in the Dirac Lagrangian is antihermitian, and so the combination $i\partial_\mu$ as a whole is hermitian.
Fields are operators in QFT. And when you combine operators, you get another operator; that's why the Lagrangian, as a combination of fields and derivative operators, is an operator itself. The operators act on states of the universe, such as the vacuum state $\lvert 0\rangle$, or n-particle states made by applying the field operators to that state, e.g. $u\lvert 0\rangle$ is a state with one up quark. – David Z Dec 11 '12 at 3:19