# Hermiticity of the Lagrangian in QFT

Why must the Lagrangian (density) of a given quantum field theory (QFT) be Hermitian?

It's something that is mentioned, but not really explained (as far as I can tell) in Srednicki's QFT book, however, I was under the impression that the Lagrangian itself is not and observable (since it is not uniquely defined). Given this, why does one require it to be Hermitian? Is it simply because the Hamiltonian is an observable, and since it is related to the Lagrangian via a Legendre transform, it is required to be real?! e.g. for scalar QFT $$\mathcal{H}=\pi_{\phi}\dot{\phi}-\mathcal{L}$$ and so $$\mathcal{H}=\mathcal{H}^{\dagger}\Rightarrow\pi_{\phi}\dot{\phi}-\mathcal{L}=\pi_{\phi}\dot{\phi}-\mathcal{L}^{\dagger}\Rightarrow\mathcal{L}=\mathcal{L}^{\dagger}.$$

• @user35305 By definition. If nothing else, consider that the action has dimensions of $\text{energy}\cdot\text{time}$ and appears as a probability amplitude $\exp(-\mathrm{i}S/\hbar)$ in the path integral. What would you make of a complex-valued energy and a probability amplitude that isn't a complex unit vector? – ACuriousMind Feb 13 '17 at 12:52
In addition to being Lorentz-invariant, the action $I$ is required to be real. This is because we want just as many field equations as there are fields. By breaking up any complex field into their real and imaginary parts, we can always think of $I$ as being a functional only of a number of real fields, say $N$ of them. If $I$ were complex, with independent real and imaginary parts, then the real and imaginary parts of the conditions that $I$ be stationary (the Euler-Lagrange equations) would yield $2N$ field equations for $N$ fields, too many to be satisfied except in special cases. We will see in the next section that the reality if the action also ensures that the generators of various symmetry transformations are Hermitian operators.