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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 an 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 $$\mathcal{H}=\mathcal{H}^{\dagger}\Rightarrow\pi_{\phi}\dot{\phi}-\mathcal{L}=\pi_{\phi}\dot{\phi}-\mathcal{L}^{\dagger}\Rightarrow\mathcal{L}=\mathcal{L}^{\dagger}.$$

Frederic Thomas
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user35305
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    I think you answered your question yourself. Hamiltonian, momentum and coordinate are all observables (and thus real) and Lagrangian is a combination of them, thus Lagrangian must be real. – Echows Feb 13 '17 at 12:53
  • Possible duplicates: http://physics.stackexchange.com/q/127797/2451 , http://physics.stackexchange.com/q/46528/2451 – Qmechanic Feb 13 '17 at 12:58

2 Answers2

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From Weinberg's QFT, Vol I., pages 300-301:

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.

AccidentalFourierTransform
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The Lagrangian must be Hermitian because the action by definition must be real/Hermitian, see also this question.

Community
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ACuriousMind
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  • I'd already had a look at the answers to that question and I still don't really understand why the action must be real/Hermitian? Is there any physical reasoning/intuition for why this must be the case? – user35305 Feb 13 '17 at 12:48
  • @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
  • Ah ok. The dimensional analysis argument has elucidated it for me. Thanks for your help! – user35305 Feb 13 '17 at 13:01