# States created by local unitaries in QFT

In quantum field theory, consider acting on the vacuum with a local unitary operator that belongs to the local operator algebra associated with a region. In such a way, can we obtain a state that is orthogonal to the vacuum? i.e., Can the one-point function of a local unitary be zero? If so, any examples?

According to the Reeh-Schlieder theorem (assuming quite standard hypotheses on your QFT in Minkowski spacetime) the subspace of vectors you consider is dense in the whole Hilbert space. Therefore every state orthogonal to the vacuum is the limit of a sequence $$U_n|vacuum\rangle$$ where the unitaries $$U_n$$ (e.g., Weyl operators) are localized in a (arbitrarily) given compact region.

The theorem refers to all bounded operators in a von Neumann algebra $$A(O)$$ localized in the relatively compact region $$O$$ of Minkowski spacetime. However it is easy to prove that unitary operators are sufficient. Indeed:

(a) complex combinations of selfadjoint operators in $$A(O)$$ have the same span as the wole algebra (that is because $$X= (X+X^*)/2 + i (X-X^*)/2i$$ and the two selfadjoint operators $$(X+X^*)/2$$ and $$(X-X^*)/2i$$ stays in $$A(O)$$ if $$X$$ does because $$A(O)$$ is an algebra),

(b) the span of selfadjoint operators $$X|vacuum\rangle$$ is the same as the one of the unitary operators spectrally constructed out of these selfadjoint operators (that is beacuase, from Stone's theorem, $$t^{-1}(e^{it X} |\psi\rangle - |\psi\rangle) \to iX|\psi\rangle$$ as $$t\to 0$$ for $$X=X^*$$ and $$e^{itX} \in A(O)$$ if $$X\in A(O)$$ because $$A(O)$$ is closed with respect to the strong operator topology and the unitary operator $$e^{itA}$$ is the strong limit of functions of $$X$$ [actually it holds also in the uniform operator topology as $$X$$ is bounded])

A subtle point is the notion of localized operator $$X\in A(O)$$. Essentailly it means that $$[X,Y]=0$$ if $$Y\in A(O')$$ with $$O$$ and $$O'$$ causally separated.

REMARK The R-S theorem says that $$X$$ localized in $$O$$ does not mean that $$X|vacuum\rangle$$ is not changed outside $$O$$! That is because $$X|vacuum\rangle$$ can be made arbitrarily close to every given vector of the Hilbert space through a suitable choice of $$X$$ in $$A(O)$$.

• I thought RS theorem says vectors generated by any locally supported operators are dense, but they are not restricted to by untiary. Commented Jun 27, 2023 at 9:52
• Yes but with unitaries you can reach everything you can reach with observables. I in fact mentioned the Weyl operators. Commented Jun 27, 2023 at 9:53
• Clearly, with only unitaries I cannot change the state outside the region, how is that dense? Commented Jun 27, 2023 at 9:55
• I do not understand what you are arguing. I completed my answer with details, Notice that $A(O)$ is a von-Neumann algebra... Commented Jun 27, 2023 at 10:08
• My point is that unitary is conceptually different from any bounded operators. Are you saying that unitary action localized in $O$ can change the observations for measurements outside $O$? Wouldn't that violate causality? Commented Jun 27, 2023 at 14:39

Yes. I can think of at least two (classes of) cases.

1. The simplest case is the two dimensional compact free scalar CFT. There you have vertex operators $$V_p(z) = \mathrm{e}^{\mathrm{i} p X(z)}$$. These are unitary. Of course it's a CFT, so one-point functions vanish (you can also show that in the path integral). These operators create states that are orthogonal to the vacuum. This is explained nicely in David Tong's string theory lecture notes (subsubsection 4.6.2). Of course this isn't restricted to the free scalar. Other CFTs have unitary vertex operators and you can run the same argument.

2. In morally any QFT, if you allow for topologically non-trivial regions. For example, if you take an ring-like region, $$R=\mathbb{S}^1\times\mathbb{R}^{d-2}$$ (or a compact version thereof), line operators wrapping the $$\mathbb{S}^1$$ can be constructed locally within the algebra. Thus they belong to the algebra of local operators on $$R$$. See for example arXiv:2205.03412 or arXiv:2008.11748. In a gauge theory, then, a Wilson loop is a local unitary in the above sense and can have a vanishing one-point function. You can convince yourself of that by playing with simple theories with Wilson lines. The simplest example would be Chern-Simons theory on flat space. Since flat space is topologically trivial all Wilson lines are contractible and therefore have vanishing one-point functions. It creates, thus, a state orthogonal to the vacuum.

• You mean the Wilson loop one pt function is zero or non-zero? Commented Sep 12, 2023 at 16:12
• @Shadumu sorry I had written non-zero, but I meant to write zero. Corrected now. Commented Sep 13, 2023 at 6:34
• Is it shown in the references that it can vanish? Commented Sep 13, 2023 at 17:05
• @Shadumu It is not shown. I updated the answer with an example where it vanishes. I also added another class of examples which is much simpler and cleaner. Commented Sep 13, 2023 at 18:23
• In your first example, what's the support of $X(z)$? Commented Sep 13, 2023 at 23:45