# Unitary quantum field theory

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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".

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Daniel
asked May 24, 2013
retagged Mar 23, 2014
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.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Vibert
...(and also the S-matrix and the Froissart bound, of course).

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Vibert
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.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Trimok

## 2 Answers

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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.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user Prahar
answered Jun 27, 2013 by (540 points)
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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.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
answered Oct 3, 2013 by (45 points)
The finite-dimensional irreps of the Poincare group which you refer to in your first paragraph are not unitary.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user user1504
@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.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
@user1504 . Why did you assume that irreps are finite-dimensional?

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
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.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user user1504
@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 $p^{\mu}$.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
This representation is infinite-dimensional. As for the second paragraph, I meaned that if we want, for example, to classificate the fields by their spin, we may not claim the unitarity of representation of the group, by which field is transformed, so we can use Lorentz group rep.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
Pardon, generators are hermite.

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
And the representation, of course, is unitary

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx
@user1504. So, why did you downvote?

This post imported from StackExchange Physics at 2014-03-23 12:07 (UCT), posted by SE-user PhysiXxx

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