# 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

## 1 Answer

In quantum field theory, the Lagrangian density is an operator, not a number. So it doesn't make sense to say it has to be real; "real" is a term that applies to numbers, not operators.

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.

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Thanks for clarifying :-) Could you explain what do you mean by operator here? I thought it is a function of fields and their derivatives. When you say operator, to my mind this means that it acts on something, but what does the Lagrangian density in QFT act on? – UserX Dec 11 '12 at 3:10
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
I see, that makes sense. Sorry if I'm asking stupid questions but I didn't even start studying this stuff formally yet. – UserX Dec 11 '12 at 3:27
Dear @David, I would mostly disagree that the Lagrangian isn't a number. The only meaningful quantum formalism that uses the actions and Lagrangians is the Feynman path-integral approach and in that approach, they are c-number-valued functions of classical observables that are integrated over. ... Also, L is often unreal if we deal with odd-order terms in the Euclidean space (like Chern-Simons etc). – Luboš Motl Dec 11 '12 at 8:29
@Luboš I wouldn't say the path integral approach is the only meaningful one... at least, not in the sense that everything needs to be done that way. Yes, all of QFT does trace back to path integrals in the end, but in practice when doing perturbative QFT we use operator-valued Lagrangians directly, and that seems to be the level that the question is asked at. – David Z Dec 11 '12 at 19:09