What does it mean for QFT to be unitary?

I understand the statement that 'X QFT is unitary' is shorthand for saying 'the S-matrix of X QFT is unitary', cf. e.g. this Phys.SE post.

Is there some definition of unitarity that is stronger than this, which is a property of the theory and not just its S-matrix? The S-matrix is a very special time evolution operator which takes asymptotically free states at minus infinity in time to asymptotically free states at positive infinity. Is there a is there a definition of unitarity in terms of a finite time evolution operator which implies the S-matrix definition?

I have also heard people discussing whether a certain conformal field theory is unitary, which suggests to me that it could be a property of the theory itself. To my understanding, the S-matrix is defined in a CFT only by taking a limit in dimensional regularisation. Is it that when people say '$$d$$-dimensional CFT X is unitary', they mean 'the S-matrix of CFT X in dimensional regularisation where $$d' = d + \epsilon$$ for $$\epsilon >0$$ is unitary'?

This would seem strange to me, I believe that unitarity is a central property of a CFT, but the S-matrix is not always the most natural quantity to compute.

• I've never particularly liked the term 'unitary' as it is a mathematical term and not a physical one. It's also why I dislike the term Hilbert space. I'd say the physical concept is the conservation of probability and that is much more easily understandable - all the probabilities must add up to one: we're uncertain about what event will occur with what probability, but we are certain that something will occur. I also want to record my complaint that the name S-Matrix is, for novices, hard to understand, whereas Scattering Matrix is much easier to understand. Lets name things properly. Sep 30 '21 at 18:36
• Regarding conformal field theory see en.wikipedia.org/wiki/Conformal_field_theory#Unitarity
– Gold
Sep 30 '21 at 18:37
• Thanks @Gold, that clears up my CFT question. I wonder if the definition from Wikipedia is equivalent to the statement I came up with in dim reg?
– Joe
Sep 30 '21 at 18:49
• @MoziburUllah But the conservation of total probability does not equal to unitarity. You can have e.g. the Lindbladian evolution. Sure, it spoils the purity of the states but preserves the total probability. So "physically" you should add something extra, like the possibility to reconstruct the initial state from the final one, if you want to get the unitarity.
– OON
Sep 30 '21 at 19:31
• For generic QFTs, unitarity means that scalar products are positive definite, so e.g. $\langle\psi|\psi\rangle\ge0$, with equality only for the zero vector. Oct 3 '21 at 19:09

Maybe this is a boring answer but unitarity of a QFT means states at different times can be related to each other by $$$$\left | \Psi(t_2) \right > = U \left | \Psi(t_1) \right >$$$$ where $$U$$ is unitary. Since $$U$$ is known in terms of the Hamiltonian (which always exists when time translation is a symmetry), we can say that a QFT is unitary if and only if the Hamiltonian is self-adjoint. This definition implies unitarity at the level of the S-matrix since the S-matrix is essentially the special case for times of $$\pm \infty$$. It also implies the definition referenced above for CFTs since the dilation operator becomes a Hamiltonian in the radial quantization approach.

Perhaps a more interesting point is that the implication doesn't always go the other way. Starting off with CFTs, it is common to pretend that they are nothing but sets of correlation functions of local operators. Unitarity then becomes the statement that Wick rotated correlators are reflection positive. That is $$$$\left < \mathcal{O}_n(r_n^{-1}) \dots \mathcal{O}_1(r_1^{-1}) \mathcal{O}_1(r_1) \dots \mathcal{O}_n(r_n) \right > > 0. \quad\quad (1)$$$$ If we do not assume unitarity, then we do not know that local operators provide a basis for all states. I.e. it is conceivable for (1) could hold even when unitarity in the whole theory fails. Such a failure would be purely due to nonlocal operators which can distinguish theories whose local sectors are the same. An example, recently mentioned to me by Sylvain Ribault, is that unitarity in the Ising model can be spoiled if we extend it to include cluster connectivities.

Now this is not a rigorous statement but it seems to me that QFT people "defining" their theory using the S-matrix is morally on par with CFT people "defining" their theory using local operators. Part of this prejudice comes from AdS / CFT where a conformal correlator encodes a bulk obervable which becomes an S-matrix element in the flat space limit. The other part concerns the fact that Green's functions in a massive theory are fairly messy observables. A great set of notes by Clifford Cheung points out a large redundancy in the free scalar Lagrangian but a similar redundancy can be seen in the Green's functions as well. Problem 10.5 of Srednicki shows that if we take $$$$\phi \to \phi + \lambda \phi^2 \quad\quad (2)$$$$ in a free theory, the $$\phi\phi \to \phi\phi$$ amplitude is still the same as it was before (zero). However, the $$\left < \phi\phi\phi\phi \right >$$ correlator is not the same as it was before. It differs by a regular piece which the LSZ formula doesn't see. Based on this, it seems at the very least to be a secondary issue to consider the implications of unitarity for massive QFT observables with operators inserted at finite times. The reason why it's of primary importance for a CFT is that scale invariance provides a well defined principle for throwing out almost all field redefinitions of the form (2).

• This is not at all a boring answer ;-)
– M.Jo
Oct 1 '21 at 9:54
• Thanks for a great answer. I can't see where your equation (1) comes from, can you motivate that? What is $r^{-1}$, is it $r/|r|^2$? Also, I understand that you're saying that the standard definition that people use for unitarity in a CFT does not imply that the theory has unitary time evolution operator, is that correct? This seems quite counter intuitive to me, maybe I'll understand that better if I can understand your equation (1)
– Joe
Oct 1 '21 at 17:06
• Sorry. With full spatial dependence, local operators are $\mathcal{O}(r, \textbf{n})$ so $r$ and $r^{-1}$ are numbers. I'm just suppressing the unit vectors in (1). This standard definition implies that the CFT has a time evolution operator whose restriction to states given by $\mathcal{O}(0) \left | 0 \right >$ is unitary. It's just that there could be extra states. Oct 1 '21 at 19:30
• The origin of (1) is that if you evolve a state from $0$ to $t$, the conjugate state is evolved from $0$ to $-t$. Therefore the in radii are inverses of the out radii because $t = \log r$ is the radial quantization map. The paper arxiv.org/abs/1208.6449 has more information. Oct 1 '21 at 19:32