Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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".

share|improve this question
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. –  Vibert May 24 '13 at 19:45
...(and also the S-matrix and the Froissart bound, of course). –  Vibert May 24 '13 at 20:30
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. –  Trimok May 27 '13 at 11:01

2 Answers 2

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.

share|improve this answer

Analogically, the unitarity of the theory of field $\psi $ in the most cases means, that field $\psi$ must transforms by the irreducible unitary representations of the Poincare group with mass $m$, momentum $p^{\mu}$ and spin $s$ (or with chirality $\lambda $ and momentum $p^{\mu}$). The unitarity requirement, of course, leads from QM postulate that the density of propability for particle is $\psi^{+}\psi$.

But if we don't connect fields with particles, we can analyze the non-unitary representations for fields.

share|improve this answer
The finite-dimensional irreps of the Poincare group which you refer to in your first paragraph are not unitary. –  user1504 Oct 3 '13 at 18:41
@user1504. Do we talk about representation by generators $M_{\mu \nu}, P_{\mu}$ with Casimir operators $P_{\mu}P^{\mu}, W_{\mu}W^{\mu}$ (for massive case)? If yes, I think that you're wrong. –  PhysiXxx Oct 3 '13 at 19:01
@user1504 . Why did you assume that irreps are finite-dimensional? –  PhysiXxx Oct 3 '13 at 19:07
I assumed that because you said that the field transforms by unitary representations. I usually understand this phrase to refer to the irreps in which the classical fields take their values. If this is not what you mean, then your 2nd paragraph seems broken to me. –  user1504 Oct 3 '13 at 19:15
@user1504 . I wrote the first paragraph in sense that if some field $\psi$, determined in Minkowski space-time, an own transformation can be represented as $$ \psi ' = e^{ia^{\mu}\hat {P}_{\mu} + \frac{i}{2}\omega^{\mu \nu}\hat {M}_{\mu \nu}}\psi , $$ where $\hat {M}_{\mu \nu}, \hat {P}^{\nu}$ are unitary generators which satisfy Poincare algebra and $a^{\mu}, \omega^{\mu \nu}$ are the local coordinates of the group of representation, it realizes the unitary representation of the Poincare group with spin $s$, mass $m$ and impulse $p^{\mu}$, or with helicity $\lambda$ and impulse $p^{\mu}$. –  PhysiXxx Oct 3 '13 at 19:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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