What do physicists mean when they refer to a quantum field theory being unitary? Does this mean that all the symmetry groups of the theory act via unitary representations? I would appreciate if one could provide some references where the definition of a unitary QFT could be found. Especially in the case where there might not be a special direction singled out as "time".
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4$\begingroup$ No, that's a different story (unitary reps of symmetry groups). It means that all states $|n\rangle$ in the theory have positive norm, i.e. $\langle n|n \rangle > 0$, but the full answer is more involved and should involve a discussion of reflection positivity. $\endgroup$– VibertMay 24, 2013 at 19:45
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$\begingroup$ ...(and also the S-matrix and the Froissart bound, of course). $\endgroup$– VibertMay 24, 2013 at 20:30
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2$\begingroup$ In a quantum theory (quantum mechanics or quantum field theory), unitarity means conservation of probability (or conservation of information), that is, if a state $|\psi>$ evolves in a state $|\psi'>$, you will have $<\psi|\psi> = <\psi'|\psi'>$. This means that the operator which transforms $|\psi>$ into $|\psi'>$ must be unitary. Unitarity is mandatory for the probabilistic coherence of the quantum theory. $\endgroup$– TrimokMay 27, 2013 at 11:01
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To expand on @user26374's answer a little, the phrase "A QFT is unitary" comes from the requirement that the $S$-matrix is unitary, i.e. $S S^\dagger = S^\dagger S = 1$ which is equivalent to the statement that sum of probabilities is 1. Unitarity implies several serious constraints on how a QFT can be formulated. For example, unitarity implies the Froissart bound, $\sigma \leq s \log s$ ($\sigma$ is the total cross-section and $s$ is the center of mass energy). It also implies that the propagator for a field must go no faster than $\frac{1}{p^2}$ at large $p^2$.
Unitarity is discussed in Weinberg Vol. 1.
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$\begingroup$ Do you mean $\sigma(s)=O\left(s\,\log s\right)$? The LHS and RHS of that inequality don't have the same units also. $\endgroup$ May 7, 2019 at 23:02
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$\begingroup$ Thanks to the joyful SE system of erasing some users and their comments, we might never know what "@user26374" answer was, so this might not make a lot of sense to future infoarchaeologists. Please provide some further context if you are still around and remember any specifics $\endgroup$– lurscherJun 9, 2020 at 17:04
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$\begingroup$ In fact just by the fact of warning of this fact, after many years of being on this site, I just got my daily vote quotas limited $\endgroup$– lurscherJun 9, 2020 at 18:22
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$\begingroup$ I'd like to add that unitarity also implies a notion of no dissipation to an external (out of control/interest) environment. $\endgroup$ Sep 29, 2021 at 22:02
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1$\begingroup$ @xiaoshengyang - because we measure the $S$-matrix in a lab (actually, we measure cross sections and decay rates). Consequently, in order to verify whether a theory is actually unitary, we should attempt to recast it in terms of these measurable objects. $\endgroup$– PraharSep 9 at 5:54