21
$\begingroup$

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".

$\endgroup$
3
  • 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. $\endgroup$
    – Vibert
    May 24, 2013 at 19:45
  • $\begingroup$ ...(and also the S-matrix and the Froissart bound, of course). $\endgroup$
    – Vibert
    May 24, 2013 at 20:30
  • 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. $\endgroup$
    – Trimok
    May 27, 2013 at 11:01

1 Answer 1

12
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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
  • $\begingroup$ 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 $\endgroup$
    – lurscher
    Jun 9, 2020 at 17:04
  • $\begingroup$ 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 $\endgroup$
    – lurscher
    Jun 9, 2020 at 18:22
  • $\begingroup$ 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
  • 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. $\endgroup$
    – Prahar
    Sep 9 at 5:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.