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Following Ramond, 1.5 Field Theory, it is mentioned that the classical Lagrangian density in (workable for HEP) QFT theories has to be Real, otherwise total probability is not conserved. Can someone give an example or elucidate more? How it is related to the non-unitarity of the S-matrix? A non real action implies a non unitarity of the S-matrix?

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marked as duplicate by CR Drost, Sebastian Riese, ACuriousMind, Qmechanic Feb 21 '16 at 1:59

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The S-matrix is defined by $S_{\beta\alpha}=\langle\beta,{\rm out}|\alpha,{\rm in}\rangle$ where $|\alpha,{\rm in}\rangle$ is the Heisenberg picture state such that if the state is measured at $t\rightarrow -\infty$ the particle content denoted by the label $\alpha$ will be observed. Similarly for the state $|\beta,{\rm out}\rangle$ at $t\rightarrow +\infty$. These in and out states are eigenstates of the full Hamiltonian; i.e free part + interaction.

Unitarity follows from this without having to mention the structure of the Hamiltonian.

You can then define an operator $S$ whose matrix elements between free particle eigenstates are S-matrix elements.

$S_{\beta\alpha}=\langle\beta|S|\alpha\rangle$ where $|\alpha,\beta\rangle$ are eigenstates of the free Hamiltonian $H_0$.

You can then go on to show (Weinberg Vol.1, Ch.3) that the S operator can be written as

$S=T{\rm exp}\left(-i\displaystyle\int^{+\infty}_{-\infty}H_I(t)dt\right)$

wherer $T$ is the time-ordered symbol. But this entire construction relies on the Hamiltonian, the full one and the free one, being Hermitian.

Also, since $H_I(t)=e^{iH_0t}H_{int}e^{-iH_0t}$ where $H_{int}$ is the interaction part of the Hamiltonian, we can see that we must have both $H_0$ and $H_{int}$ Hermitian for $S$ to be unitary (but this is neither here nor there since you need a Hermitian Hamiltonian to be able to get this formula for the S-matrix in the first place.)

Since a Hermitian Hamiltonian implies a Hermitian Lagrangian which implies a real classical Lagrangian we see that a real Action is necessary for unitarity of the S-matrix.

EDIT: Actually a non-Hermitian Hamiltonian would be really bad for another reason - it would result in a non unitary time translation operator.

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