Bound state operators and wedge-locality in integrable quantum field theories

We consider scalar two-dimensional quantum field theories with the factorizing S-matrix which has poles in the physical strip. In our previous work, we introduced the bound state operators and constructed candidate operators for observables in wedges. Under some additional assumptions on the S-matrix, we show that, in order to obtain their strong commutativity, it is enough to prove the essential self-adjointness of the sum of the left and right bound state operators. This essential self-adjointness is shown up to the two-particle component.


Introduction
The goal of this series of papers [6,7,9,35] is to construct more interacting two-dimensional integrable quantum field theories in the Haag-Kastler framework. The novelty of the recent progresses is that one first constructs observables localized in infinitely-extended wedge regions, then obtains compactly localized observables in an indirect way. In the preceding paper [6], we constructed candidates for wedge-local observables, however, they have subtle domain properties and their strong commutativity remained open. In this paper, we provide a possible way to settle this question, which is a crucial step towards construction of Haag-Kastler nets.
The main strategy was explained in [6] and we only briefly summarize it. See also [21] for an overview of the program. The operator-algebraic framework for quantum field theory (QFT), or a Haag-Kastler net [16], concerns the collection of algebras of local observables. As in any mathematical framework for relativistic QFT, it is a very difficult problem to construct interacting examples of Haag-Kastler net. For certain integrable models with prescribed S-matrix, Schroer proposed [31] to construct certain wedge-local observables, then obtain compactly localized observables as the intersection of two wedge-algebras. Some constructions of wedge-local algebras have been obtained for a class of S-matrices [18,22,34], as well as operator-algebraic constructions or deformations [1,2,5,12,20,33,34]. The final step to show the existence of local observables has been completed in certain cases, by showing modular nuclearity [4,19], or by split property [34].
In [18], Lechner constructed wedge-local fields for scalar factorizing S-matrices without poles in the physical strip. We attempted to extend this construction to the cases with poles in [6], however, certain spectral properties remained unsolved. More precisely, our candidates for arXiv:1602.04696v2 [math-ph] 19 Oct 2016 wedge-local fields were a slight modification of the fields of [18], but contained a very singular term, the bound state operator, which has been introduced exactly in order to treat the poles. The strong commutativity of these candidates remained unclear. We studied the one-particle component of these operators in detail in [35] and found convenient self-adjoint extensions. In the present work, we propose a strategy to prove the strong commutativity between the candidate operators.
Self-adjointness of unbounded operators is in general a hard problem. There are several criteria (e.g., see [26]), but we saw in [6,35] that most of them do not apply to our situation: the reason is that the domain of self-adjointness depends very much on the observables and one cannot find either a single domain or a single dominating operator. We try to solve this problem by a front attack, namely, we should find an appropriate common domain for any pair of wedge-observables and show their strong commutativity. Thanks to the Driessler-Fröhlich theorem, the problem of strong commutativity is reduced to the (essential) self-adjointness of the sum of the left and right bound state operators. We show this self-adjointness up to the two-particle component. It is worth mentioning that, because of all these technicalities, the candidate observable φ is no longer an operator-valued distribution.
This paper is organized as follows. In Section 2 we recall our approach to integrable QFT with bound states. Especially, we extend the domain of the bound state operators introduced in [6] using the convenient extension found in [35]. Then in Section 3 it is shown that the strong commutativity of wedge-observable candidates follows from the essential self-adjointness of the sum bound state operator. We do this by combining the Driessler-Fröhlich theorem. With this choice of domain, the covariance of the candidate operators also follows, which allows one to proceed to Borchers triples. In Section 4, we show the essential self-adjointness of the one and two particle components of the sum bound state operator. We also present some observations on n-particle components which demonstrates how the poles and zeros in the S-matrix affect the analytic properties of the observables.

Wedge-local nets and Borchers triples
Let us summarize the construction of [6]. Our goal is to construct interacting Haag-Kastler nets in two dimensions. As an intermediate step, we need to construct nets of observables localized in wedges, or wedge-local nets (of von Neumann algebras) for short. This notion is equivalent to that of Borchers triples, which we will use in this work. We refer to [6,Section 2.1] for an overview of the construction program through Borchers triples.
A (two-dimensional) Borchers triple (M, U, Ω) consists of a von Neumann algebra M, a unitary representation of the proper orthochronous Poincaré group P ↑ + and a vector Ω which satisfy the following conditions (see [6,Section 2

.1]):
• Ω is cyclic and separating for M; • the restriction of U to the translation subgroup R 2 has the joint spectrum included in V + = {(a 0 , a 1 ) : a 0 ≥ |a 1 |}; • M is covariant with respect to U , namely, Ad U (g)M ⊂ M for g ∈ P ↑ + which preserves the standard right-wedge W R .
In [6], we constructed for test functions f , g supported in W L , W R respectively, unbounded operators φ(f ), φ (g) which commute weakly on a dense domain. We also showed that, if φ(f ) and φ (g) commute strongly, then Ω is cyclic and separating for M := e i φ (g) : supp g (strong commutativity is important for the separating property). Furthermore, if we can show the covariance of φ with respect to U , then the covariance of the Borchers triple follows and the construction is complete. We will investigate strict locality, namely whether this Borchers triple defines a Haag-Kastler net, in a separate paper [8] (actually, for this purpose the definition of φ must be slightly extended as we will do in Section 2.4). Therefore, our most urgent problem is the strong commutativity.
We have φ(f ) = φ(f ) + χ(f ), where φ(f ) is very well under control, while χ(f ) is a new sort of operator and we investigated its one-particle component χ 1 (f ) in detail. We review the construction of these operators below.

Factorizing S-matrix with poles
As an input, we fix a two-particle scattering function S, which is at first a meromorphic function on R + i(0, π), the region we call the physical strip, and later extended to C which satisfies the following properties: (S1) Unitarity. S(θ) −1 = S(θ) for θ ∈ R.
We showed in [6] that (S6) follows from (S1)-(S5). It turns out crucial in the consideration of domains of our operators. We also classified all functions which satisfy these conditions [6, Appendix A]. We show that (S8) follows from (S1)-(S5) in Appendix A.2.
(S7) and (S9) are new requirements. The property (S7) had not appeared either in the literature on form factor program or in our previous works [6,7]. Yet, it is necessary in order to apply our methods, for some properties of the sum bound state operators. The last assumption (S9) will be necessary also in the proof of modular nuclearity (cf. [19]), and we use it here already for wedge-observables, in Lemma E.5. With the finiteness of zeros in the physical strip, the finiteness of S κ is equivalent to the absence of the singular part in S (cf. [23, Appendix A]). From the condition (S6), there must be at least one zero on the interval i(0, π 3 ), hence κ < π 3 , as we saw in [6].
We check in Appendix A that there are concrete examples of S which satisfy all these requirements. In the following, we mark explicitly the points where (S7) is needed.

The wedge-local observables
Some materials here are parallel to those of [18]. Assume that our model has only a single species of particle with mass m > 0. We fix a two-particle scattering function S, which satisfies the properties explained in Section 2.2.
Our Hilbert space is as follows. The one-particle space H 1 is simply L 2 (R) with the Lebesgue measure. We consider n-particle space H ⊗n 1 . An element of H ⊗n 1 can be considered as a function in L 2 (R n ). On each of these spaces, there is the representation of the symmetric group S n : where τ k is the transposition of k and k + 1. This indeed extends to a unitary representation of S n . We say that Ψ n ∈ H ⊗n 1 is S-symmetric if D n (σ)Ψ n = Ψ n for σ ∈ S n . This notion is equivalent to that for each k Ψ n (θ 1 , . . . , θ n ) = S(θ k+1 − θ k )Ψ n (θ 1 , . . . , θ k+1 , θ k , . . . , θ n ).
We denote the orthogonal projection onto the S-symmetric functions by P n and the space of S-symmetric functions by H n = P n H ⊗n 1 . The physical Fock space is the direct sum where H 0 = C. Let us also introduce an auxiliary Hilbert space, the unsymmetrized Fock space, which is H Σ = n H ⊗n 1 . By putting P S = n P n , we have H = P S H Σ . For ψ ∈ H 1 , the unsymmetrized creation and annihilation operators are defined on the algebraic direct sum of H ⊗n 1 by By our convention a † (ψ) is linear in ψ, while a(ψ) is antilinear. The S-symmetrized creation and annihilation operators are z † (ψ) = P S a † (ψ)P S , z(ψ) = P S a(ψ)P S . The one-particle CPT operator J 1 is given by (J 1 ψ)(θ) = ψ(θ), and the n-particle component J n is (J n Ψ n )(θ 1 , . . . , θ n ) = Ψ n (θ n , . . . , θ 1 ).
It is easy to see that J n commutes with P n , hence its preserves the subspace H n . The full CPT operator is J = n J n , where J 0 is the complex conjugation on H 0 = C.
For a test function f in R 2 , we define As f is compactly supported, its Fourier transform decays fast, and so do f ± , therefore, they belong to H 1 . On the level of test functions, we define the CPT transformation by f j (x) = f (−x). If S had no pole, then Lechner took the field φ(f ) = z † (f + ) + z(J 1 f − ) and the reflected field φ (g) = z † (g + ) + z (J 1 g + ), where z † (ψ) = Jz † (J 1 ψ)J and z (ψ) = Jz(J 1 ψ)J and proved that they were wedge-local [18]: namely, if supp f ⊂ W L and supp g ⊂ W R , then φ(f ) and φ (g) strongly commute.
In other words, a function ξ ∈ H 2 (S a,b ) is an analytic function on S a,b whose L 2 -norms on horizontal lines ξ( · + iλ) are bounded uniformly for λ ∈ (a, b). It is well-known that for λ → a, b the limits ξ( · + iλ) exist in the L 2 -sense [32, Corollary III.2.10]. As 0 ∈ [a, b], we can idenfity H 1 = L 2 (R) as a subspace of H 2 (S a,b ) by considering the value ξ(θ), θ ∈ R. We denote the L 2 -norm by ξ , while the Hardy space norm is ξ 2 := sup Similarly, H ∞ (S a,b ) is the space of bounded analytic functions on S a,b and we define the norm The operators z † (ξ) and z(ξ) are defined for vectors ξ ∈ H 1 , hence with the understanding that H 1 = L 2 (R) ⊃ H 2 (S 0,π ) by taking the boundary value at Im ζ = 0, z † and z are also defined on H 2 (S 0,π ). The reality condition f (a) = f (a) on a test function f translates to an element ξ ∈ H 2 (S 0,π ) as ξ(ζ + iπ) = ξ(ζ). It is straightforward to see that the proof of wedge-locality for S without pole [18] works for such ξ's: the Cauchy theorem works for H 2 -functions (see, e.g., [6,Lemma B.2]).

The bound state operator
We studied the one-particle component χ 1 (ξ) in detail in [35]: actually, we studied the domain H 2 (S −π,0 ) rather than H 2 S − π 3 ,0 , but the adaptation is straightforward. The important lessons obtained there are that χ 1 (ξ), when defined on H 2 S − π 3 ,0 , does not always have a nice selfadjoint extension and even if it does, its domain varies as ξ varies. This is due to the fact that the deficiency subspaces of χ 1 (ξ) is closely related with the zeros and the decay rate of ξ, and χ 1 (ξ) does not have any self-adjoint extension if ξ has an odd number of zeros and does not decay sufficiently fast as Re ζ → ±∞. And even if ξ has an even number of zeros, it is not obvious which extension is the suited one for our purpose of finding quantum observables. Therefore, we restrict ourselves to the certain subclass of functions.

Towards strong commutativity
In this Section we show that, in order to construct the Borchers triple associated with S with poles, it is enough to prove that χ(ξ) + χ (η) is essentially self-adjoint.

A general criterion
This Section is technically independent from QFT. For unbounded operators, strong commutativity is a hard problem. It may fail even for a two self-adjoint operators which commute on a common invariant core [25,Section 10], [26, Section VIII.5, Example 1]. We also exhibit more examples of weakly-commuting but not strongly-commuting operators in Appendix B. A sufficient condition which is used very widely in the context of QFT (e.g., [2,10,15]) is the Driessler-Fröhlich theorem [11]. The theorem says roughly that, if there is a positive self-adjoint operator T ("Hamiltonian") which can nicely bound the weakly commuting symmetric operators A, B, and their commutators with T , then A and B are actually self-adjoint and commute strongly (see Appendix C for the precise statements).
The trouble in our situation is that the physical Hamiltonian is not strong enough to estimate our candidates which contain the bound state operator χ(ξ): the Hamiltonian is the multiplication operator, while χ(ξ) contains an analytic continuation. We have no idea for any other operator which dominates χ(ξ) "nicely", and anyway there cannot be a single operator which bounds all χ(ξ)'s with different ξ's, as χ 1 (ξ)'s have already different domains of self-adjointness. In such a case, the question of strong commutativity is considerably difficult (cf. [10]).
Yet, the Driessler-Fröhlich theorem can be used to reduce the problem of strong commutativity to self-adjointness of a certain positive operator.
(2) for all ϕ, ψ ∈ D, Then A and B are essentially self-adjoint on any core of T and they strongly commute.
Proof . As we have Aψ, Bϕ = Bψ, Aϕ by assumption and T = A + B + T 0 by definition, it follows that for all ϕ, ψ ∈ D Hence we can apply Theorem C.1, by taking T as the positive self-adjoint operator.
The point of this proposition is that, instead of using a single "Hamiltonian" for all cases, we take the operator T = A+B +T 0 which depends on A and B. Of course, as weak commutativity does not imply strong commutativity, the essential self-adjointness hypothesis of T and the estimate of A, B by T are crucial.
We have the following case in mind: A = φ(ξ), B = φ (η), T 0 = c(N + 1), where N is the number operator on the S-symmetric Fock space and c ∈ R + . Furthermore, we will see in Section 3.3 that T = A + B + T 0 is a Kato-Rellich perturbation of the operator χ(ξ) + χ (η) + c(N + 1). This last operator preserves each n-particle subspace, and the question of (essential) self-adjointness is reduced to each n.
This self-adjointness is far from obvious. By a slight modification, the self-adjointness of a similar operator easily fails (see Appendix D).

Proof of weak commutativity on the intersection domain
We show that φ(ξ) and φ (η) commute weakly on Dom( φ(ξ)) ∩ Dom( φ (g)). Thanks to our choice of the extension (Section 2.4), this will be a straightforward modification of the proof of [6], with the help of some techniques in complex analysis of several variables in Appendix E.
The commutator [χ(ξ), z (η)]. We can compute this commutator as operators as before [6]. Although we have an extended domain for χ(ξ), the expression (2.2) is valid as we explained before. By computing the operators term by term, we get where in the last equation we used also the symmetry property η(θ) = η(θ − iπ) and the Ssymmetry of Ψ n . Let us look at the integrand without the S-factor and shift the first variable θ by πi , We consider the function by S-symmetry of Ψ. Furthermore, again from the S-symmetry of Ψ n and from the property (S6), it has zeros at any point where two of the variables coincide: θ = θ j . By applying Lemma E.7, we obtain that still belongs to the same domain. Now we can remove extra ξ 0 's with variables θ j , which does not affect the domain of ∆ 1 ⊗ 1 ⊗ · · · ⊗ 1 and get that belongs to Dom ∆ 1 6 1 ⊗1⊗· · ·⊗1 . Therefore, we can apply the Cauchy theorem (on Hilbert-space valued functions) and The commutator [z(ξ), χ (η)]. This can be computed in a similar way as the previous one: which coincides with the result of the commutator [χ(f ), z (J 1 g − )] up to a sign, therefore they cancel each other.
The commutators [z † (ξ), χ (η)] and [χ(ξ), z † (η)]. One can show that these two commutators cancel each other by taking the adjoints and repeat the computations as in the commutators before.
The commutator [φ(ξ), φ (η)]. This term has been essentially computed in [18]. The only difference is that ξ does not come from the Fourier transform of a test function f supported in W L , but the properties needed in the computations are that ξ ∈ H 2 (S 0,π ) (which allows one to apply the Cauchy theorem) and the symmetry property ξ(θ) = ξ(θ + iπ) (which assures that the terms appearing in the commutator is the boundary values of the same analytic function), and the corresponding properties of η. Here is no effect of the domain and the result is essentially taken from [18, before Proposition 2]: The commutator [χ(ξ), χ (η)]. We have the expressions (2.2) and those for χ n (η). This allows one to expand the commutator into the sum of n 2 terms as in [6] by a straightforward computation. Of them, there are 2n(n − 1) terms which come from the actions of χ(ξ), χ (η) on different variables in the sense of (2.2), and cancel each other. For that, we only need that χ n (ξ) is the projection onto P n H ⊗n of the operator χ 1 (ξ) ⊗ 1 ⊗ · · · ⊗ 1 acting on the first component (and the corresponding property of χ n (η) and the S-symmetry of the vectors). The extended domains do not affect this.
The remaining terms are those where χ 1 (ξ) and χ 1 (η) act on the same variable in the sense of (2.2): where ρ k , ρ n−k+1 are the cyclic permutations and in the second equality we used that S has no pole except πi 3 , 2πi 3 and applied the (Hilbert space-valued) Cauchy theorem: the pole at πi 3 makes no problem by the fact that Ψ n ∈ Dom(χ n (ξ)), Φ n ∈ Dom(χ n (η)). Now, the question is the pole of S at 2πi 3 . Let us forget at the moment the S-factors and look at the further shifted integrand: Let us observe that . We use the same trick as before: consider the function which belongs to From this expression, it is clear that Ψ has a zero at θ l − θ k = 0, k < l. A parallel argument holds for l < k.
By applying Lemma E.7, we obtain that still belongs to the same domain. We remove extra ξ 0 's with variables θ l and get that Therefore, we can apply the Cauchy theorem and By recalling (S6) which implies R = i|R|, this cancels the first contribution from the commutator The remaining terms in χ(ξ)Φ, χ (η)Ψ are, by an analogous argument as above, which is equal to the second contribution from the commutator Φ, [φ(f ), φ (g)]Ψ up to the sign, therefore, they cancel each other.
and therefore, Proof . We may assume that Ψ ∈ D n (ξ, η). We use the following expressions: By proceeding as in Section 3.2, we see that χ(ξ)Ψ, χ (η)Ψ consists of n(n − 1) times and F k is the corresponding cyclic permutation of the variables. We have which is bounded by 1 by (S8). Similarly, , which is again bounded by 1 and F n−k+1 is the corresponding cyclic permutation. We define c(ξ, η) = χ 1 (ξ) from which the first statement follows directly. The second statement is easy because which extends to Ψ = (Ψ n ) ∈ D(ξ, η) by replacing n by the number operator N , because the operators appearing here preserve H n . Therefore, by dropping the positive terms at each n in the right-hand side, we conclude: It is important that the estimate does not grow too fast with n, thanks to (S8).

Construction of Borchers triple
We now proceed to von Neumann algebras, always assuming that χ n (ξ) + χ n (η) is essentially self-adjoint on D n (ξ, η). Among the properties of Borchers triples, important are the separating property of the vacuum Ω and the endomorphic action of the Poincaré group. The former is an immediate consequence of Corollary 3.4 and the latter follows from the canonical correspondence from ξ to χ(ξ).
For the next lemma, we need to recall the Beurling factorization [ . In [35], we exploited this decomposition in order to construct the analytic function η 0 (see Section 2.4).
From here on, we switch to the objects with , because they are by convention those which generate the von Neumann algebra.
Proof . First note that, if η ∈ H 2 (S −π,0 ) ∩ H ∞ (S −π,0 ), then it follows from (U 1 (a, λ) Next, let us remember how and η − is an outer function on S − 2π where η out is the outer part of η.
We define Then, (M, U, Ω) is a Borchers triple.
Proof . Ω is cyclic for M by the argument of [6, Section 3, Reeh-Schlieder property] and the density of the linear span of V in H 1 . It is separating because Ω is cyclic for e i φ(ξ) : ξ ∈ V , which commutes with M by Corollary 3.4. Finally, for a ∈ W R , Ad U (a, λ)M ⊂ M follows immediately from Lemma 3.5 and the assumption that V is invariant under U 1 (a, λ) for a ∈ W L , hence J 1 V is invariant under U 1 (a, λ), a ∈ W R .
Note that, if the assumption of strong commutativity is dropped, the totality of V is easy, as (cf. [6]). We only have to find sufficiently many ξ's and η's which satisfy strong commutativity.
Observations on closure Lemma 4.1. Let X, Y be closed operators and assume that there is c ∈ R such that c from which it follows that {XΨ m } and {Y Ψ m } are both convergent, and therefore, Ψ ∈ Dom(X) ∩ Dom(Y ), namely, X + Y is closed.
From this it follows that X(Ψ n − Ψ m ) → 0 and Y (Ψ n − Ψ m ) → 0, namely XΨ n and Y Ψ n are separately convergent. As X and Y are closable, we obtain XΨ n → 0 and Y Ψ n → 0 and Φ = 0, namely X + Y is closable.
To show X + Y ⊂ X + Y , we only have to repeat the argument with Ψ n → Ψ to obtain Ψ ∈ Dom(X) ∩ Dom(Y ) = Dom(X + Y ).
from which it follows that {XΨ m } and {Y Ψ m } are both convergent, and therefore, Ψ ∈ Dom(X) ∩ Dom(Y ), namely, X + Y + c is closed. This is equivalent to that X + Y is closed.
A key observation is that χ 1 (ξ) can be written in the form X * X where X = ∆ 1 12 3 ) . It is clear that X is densely defined, as ξ 0 ∈ H ∞ (S π 3 , 2π 3 ), and closed as M ξ 0 (· + 2πi 3 ) is bounded. Furthermore, it holds that X * = M * Some ideas in the following lemma are due to Henning Bostelmann.
Lemma 4.4. Let X, Y be closed, densely defined operators such that • Y * X and X * Y are bounded, • X + Y and X * + Y * are densely defined, Proof . By definition, Dom(X + Y ) = Dom(X) ∩ Dom(Y ). Under the assumption, X + Y is closed. Indeed, we have XΨ, Y Ψ = Ψ, X * Y Ψ which is bounded by X * Y · Ψ 2 and we can apply Lemma 4.1.
Next, note that Dom(X + Y ) = Dom(X) ∩ Dom(Y ) is a core of X. Indeed, any Ψ ∈ Dom(X) can be decomposed as Ψ X ⊕ Ψ Y by assumption. On the other hand, Dom(Y ) is dense and admits the decomposition Dom(Y ) = (Dom(Y ) ∩ H X ) ⊕ (Dom(Y ) ∩ H Y ). As Dom(X + Y ) is dense by assumption, Dom(X + Y ) ∩ H X is dense in H X . But X is bounded on H X , thus Dom(X) includes H X . Namely, Dom(X) = H X ⊕(Dom(X)∩H Y ). Similarly, we have Dom Next we see that Dom((X +Y ) * ) = Dom(X * )∩Dom(Y * ). For each vector Ψ ∈ Dom(X +Y ) = Dom(X) ∩ Dom(Y ), we take the decomposition Ψ = Ψ X ⊕ Ψ Y which is given by the assumption. By the definition of the adjoint operator, a vector Φ is in Dom((X + Y ) * ) if and only if the map Ψ → Φ, (X + Y )Ψ is continuous for Ψ ∈ Dom(X + Y ) (namely, there is c such that | Φ, (X + Y )Ψ | ≤ c Ψ ). Then we have the following chain of equivalences: We saw that Dom(X + Y ) is a core both for X and Y , thus, the continuity of Φ → Φ, XΨ for Ψ ∈ Dom(X + Y ) extends to Ψ ∈ Dom(X), which implies that Φ ∈ Dom(X * ). Analogously, we obtain Φ ∈ Dom(Y * ). This implies that (X + Y ) * = X * + Y * .
It is well-known [26, Theorem X.25] that, because X + Y is closed, on the domain Dom((X + Y ) * (X +Y )) = {Ψ ∈ Dom(X +Y ) : (X +Y )Ψ ∈ Dom((X +Y ) * ))} the operator (X +Y ) * (X +Y ) is self-adjoint. From the above observations, this domain is actually where in the last line we used the assumptions that X Dom(X) ⊂ Dom(Y * ) and Y Dom(Y ) ⊂ Dom(X * ). Now we have (X + Y ) * (X + Y ) = X * X + X * Y + Y * X + Y * Y and this is different from X * X + Y * Y only by a bounded operator.
Proof . This is a straightforward consequence of this Lemma 4.4.
Indeed, we only check the existence of the decomposition H 1 = H X ⊕ H Y . We consider log ∆ and its spectral decomposition: we take H X as the spectral subspace of log ∆ corresponding to (−∞, 0], while H Y corresponds to (0, ∞). Now the required properties are shown as follows.
Let us denote the spectral projections onto these subspaces by P X and P Y . Let us take an element Ψ ∈ Dom(X) = Dom ∆ 1 12 M ξ 0 (· + 2πi 3 ) . This means, by definition, ξ 0 (θ + 2πi 3 )Ψ(θ) is in H 2 − π 6 , 0 . In addition, Ψ(θ) alone is L 2 . While it holds that Ψ = P X Ψ ⊕ P Y Ψ, P X Ψ is obviously in the domain of ∆ 1 12 , hence also in the domain of X = ∆ 1 12 M ξ 0 (· + 2πi 3 ) . This means that P Y Ψ = Ψ − P X Ψ is also in the domain of X. The restriction of X on H X is bounded, since there it holds that XΨ X = M ξ 0 (· + πi 2 ) ∆ 1 12 , where ξ 0 (θ + πi 2 ) is bounded. Similarly, the decomposition H X ⊕ H Y is compatible with Dom(Y ) and Y is bounded on H Y .

Two-particle components
In this Section, under (S7), we show that χ 2 (ξ) + χ 2 (η) = P 2 (χ 1 (ξ) ⊗ 1 + 1 ⊗ χ 1 (η))P 2 is essentially self-adjoint. It should be stressed that this expression does not involve Friedrichs extension. Simply the restriction to P 2 H ⊗2 1 is already essentially self-adjoint. This is important, because we have to be able to compute the weak commutator. In the course we also show that χ 2 (ξ) is essentially self-adjoint. It is also closed under (S7).
We provide different proofs. The first one is longer and tricky, but one may hope that with some more ideas the cases n ≥ 3 could be treated. The other two works only for χ 2 (ξ), but are relatively simple, and give the hints of why the general case requires a better understanding of the poles and zeros of S. It should be noted that operators of the form ∆ 1 M * f may in general fail to be self-adjoint (see Appendix D). We have to properly use the properties of S.

Fubini's theorem
We first show that χ 2 (ξ) is self-adjoint. For this purpose, let us note that the unitary operator M ξ 0 ⊗ M ξ 0 obviously commutes with M * S , hence also with D 2 (τ 1 ) and P 2 . As we have the question is now reduced to the self-adjointness of P 2 ∆ 1 6 1 ⊗ 1 P 2 .
We start with some observations on L 2 -functions.
Coming back to our operators, let us summarize the situation.

Proof . It is well-known that
Let us consider the joint spectral decomposition with respect to log ∆ 1 ⊗ 1 and 1 ⊗ log ∆ 1 .
Any vector in H 1 ⊗ H 1 can be considered as a two-variable L 2 -function on the spectral space of log ∆ 1 . Now, note that M S(· + πi 6 ) F is a bounded operator. If we put Φ n = M S(· + πi 6 ) F Ψ n , then Ψ n and Φ n are convergent to Ψ and Φ in the L 2 -sense, respectively.
). It is straightforward to observe that AB and BA are bounded. Our goal is the essential self-adjointness of P 2 ( In order to apply Lemma 4.2, it is enough to check that c Ψ 2 ≤ Re P 2 (A ⊗ 1)Ψ, P 2 (1 ⊗ B)Ψ . This holds indeed: and by noting that A ⊗ B is positive, while (A ⊗ 1)M S (B ⊗ 1) = M S(· + πi 6 ) (AB ⊗ 1) is bounded. We show that, if (A ⊗ 1 + 1 ⊗ B)Ψ ∈ Dom(P 2 (A ⊗ 1)), then it is in Dom(A ⊗ 1). This follows the idea of Proposition 4.7 (see the third and fourth paragraphs). Namely, we consider the following formal expression withΨ = (A ⊗ 1 + 1 ⊗ B)Ψ, This is formal, yet the whole expression has the meaning as an L 1 integral over the spectral space with respect to log A ⊗ 1 and 1 ⊗ log A: it is L 1 because has the meaning as an L 2 -function. AsΨ = (A⊗1+1⊗B)Ψ, this inner product can be formally decomposed as integral. Yet the whole expression is L 1 by the defining property of Ψ (that (A ⊗ 1 + 1 ⊗ B)Ψ ∈ Dom(P 2 (A ⊗ 1))) and the second term is positive (actually in this case it must be positive infinite as well), thus the last term must be negative infinite again in the sense of Lebesgue integral: a mixed infinity is impossible, because then the whole integral would have a measurable set on which it is positive infinite, which contradicts the assumed L 1 -property. On the spectral space of log A, we can write A as the multiplication operator by e t . The crossing term in question (A ⊗ 1)Ψ, (1 ⊗ A)M S(· + πi 6 ) FΨ is the following form: whereΦ = M S(· + πi 6 ) FΨ. When the real part of an integral is negative infinite, one can apply the Fubini theorem. We insert the spectral projection Q N of (A ⊗ A) corresponding to e N −1 , e N , which amounts to consider a strip N − 1 < t 1 + t 2 ≤ N . This is equivalent to the change of variable and integrating first with respect to t 2 − t 1 , which is legitimate by Fubini's theorem. On this strip, the integral is L 1 , becauseΨ andΦ are L 2 . Now recall thatΨ = (A ⊗ 1 + 1 ⊗ B)Ψ, therefore, the above integral consists of four contributions. Even though the expression above is formal, note that Q N commutes with A ⊗ 1 and 1 ⊗ A, while AB, BA are bounded. Therefore, if we take two of the contributions which implies that the only remaining term must have the real part which is negative infinite in the sense of Lebesgue integral. However, we have computed this and it is positive under (S7). Hence the whole integral is bounded below, even after removing Q N . This contradicts the assumption that the whole integral was negative infinite, hence we obtain that each term of the formal expression is finite, namely, Furthermore, we note that A ⊗ 1 and (A ⊗ 1 + 1 ⊗ B) commute strongly. By considering the joint spectral decomposition, it is immediate to see that To summarize, we proved Similarly, we obtain By the above consideration, we obtain the equality which is self-adjoint, because of the first expression.
We show that P 2 (A 2 ⊗ 1 + 1 ⊗ B 2 )P 2 is essentially self-adjoint, using the commutator Theorem C.1, by taking P 2 A 2 ⊗ 1 + 2(A ⊗ B) + 1 ⊗ B 2 + c1 P 2 as the reference operator with some c > 0. Let us check the required estimates In order to show P 2 ( is enough to have a lower bound on the following term: where the first term is positive, while we have x is a bounded operator. This is bounded by x · (A ⊗ 1)Ψ 2 = x · (A 2 ⊗ 1)Ψ, Ψ , therefore, we take c > 2 x and c 1 = 1. We can estimate analogously the term containing 1 ⊗ B 2 .
A bound of the weak commutator reduces to the weak commutator [P 2 (A ⊗ B)P 2 , P 2 A 2 ⊗ 1 + 1 ⊗ B 2 P 2 ], and its bound by follows from the estimate above of the term P 2 A 2 ⊗ 1 + 1 ⊗ B 2 Ψ, P 2 (A ⊗ B)Ψ , up to the constant c above, by noting that A ⊗ B commutes with A 2 ⊗ 1 and 1 ⊗ B 2 .

A shorter proof for χ 2 (ξ)
We here present a proof of the essential self-adjointness of χ 2 (ξ). Although it is not of our direct interest as we need the self-adjointness of χ 2 (ξ) + χ 2 (η), we believe it clarifies the reason why our method does not work for n ≥ 3.
Proof . We consider the operator and Dom(Y * ) = Dom A − 1 2 ⊗ A 1 2 . One can easily check the assumptions of Lemma 4.4: as for the decomposition of the Hilbert space, we change the variables t + := θ 2 + θ 1 , t − := θ 2 − θ 1 . Then the problem reduces to one variable t − and one can take the same decomposition as Proposition 4.5, by noting that S is bounded in R + i 0, π 6 . Finally, both A ⊗ A −1 and M S A −1 ⊗ A M * S commute with A ⊗ A strongly. Therefore, the bounded operator Q N (A ⊗ A) commute with both of them, where Q N is the spectral projection of A ⊗ A onto e N −1 , e N and the product is S is an extension of the direct sum of these components, therefore, it is essentially self-adjoint.
It appears difficult to apply the same idea to P 2 A 2 ⊗ 1 + 1 ⊗ B 2 P 2 , as A and B do not commute.
If n ≥ 3, A ⊗ A ⊗ A cannot reduce the problem to one variable. We know no other convenient operator.

Konrady's trick
We give a less simpler proof which however gives an insight of why the problem is complicated, cf. Section 4.3. For it, we need an additional assumption on S.
We indicate some supporting evidence in Appendix A.3 that there should exist examples of S which satisfy this condition. We do not pursue this condition further, as our main result (Proposition 4.9) has been proved without it. The general idea used here is called Konrady's trick [17], [26, Section X.2].
Lemma 4.11. Let X and Y be symmetric operators on Dom(X) and Dom(Y ), respectively. Assume that X + Y is self-adjoint on Dom(X + Y ) = Dom(X) ∩ Dom(Y ) and it holds that Re XΨ, Y Ψ ≥ 0 for Ψ ∈ Dom(X + Y ). Then X and Y are essentially self-adjoint on Dom(X + Y ).
Proof . Compute: now one can use Wüst's theorem [26, Theorem X.14] to see that X = (X + Y ) − Y and Y = (X + Y ) − X are essentially self-adjoint on Dom(X + Y ).
Proof . First, ∆ 1 12 1 is self-adjoint by the spectral theorem for a constant c > 0. We choose c > |S(θ + πi 6 )|, θ ∈ R. Then, we claim that ∆ Next, we show that ∆ Here, the left-hand side is self-adjoint, and the right-hand side is symmetric, hence it must be an equality and the right-hand side is self-adjoint. The second and the third terms are bounded, therefore, they have no effect on domains and ∆ 1 6 As N is arbitrary, we obtain the claim. 1 ⊗ 1 ⊗ · · · ⊗ 1 D n (ρ k ) * on P n H n . To prove it by induction, we may assume that χ n−1 (ξ), hence χ n−1 (ξ) ⊗ 1 is closed. To apply Lemma 4.1, we only have to check that the following is positive:
Proof . We follow the idea of Proposition 4.12. We set S k (θ θ θ) := j<k S(θ k − θ j ) as before. There because S k is uniformly continuous in a small neighborhood of R n as we assume (S9) and S k (θ θ θ)S k (θ θ θ) = 1.
The method of this proof is invalid for = 1 6 , or any fixed . Indeed, as k increases, Re S k (θ θ θ + π 6 )S k (−θ θ θ) may take negative values and the Konrady's trick fails. In order to have the right domain of self-adjointness of P n ∆ 1 6 1 ⊗ 1 · · · ⊗ 1 P n , we may have to correctly incorporate the zeros and poles of S to the domain (cf. [35]).

Outlook
In order to complete the construction of Borchers triples, we have to prove that χ n (ξ) + χ n (η) is (essentially) self-adjoint. We also hope to clarify the role played by the assumption (S7), and to drop it if possible. With the same assumption, one should be able to prove the Bisognano-Wichmann property and modular nuclearity [4,19] for an inclusion with a minimal distance [1], hence it should be possible to construct Haag-Kastler nets with minimal size, which will be published elsewhere [8].
On the other hand, if the strong commutativity of our candidates fails, then the construction of the Haag-Kastler nets for a given S-matrix with poles will be really a hard problem: if the net exists at all, the polarization-free generators are canonically constructed [3], while we checked that our φ(f )'s are formally compatible with the form factor program [6, Section 4.1]. If our candidates are not the right polarization-free generators, the formal computations must fail, which means that the right polarization-free generators should have even subtler domains. A related problem is whether the S-matrix is a complete invariant for asymptotically complete nets. This is in general open even for the simplest case where S = 1, if the temperateness of the polarization-free generators is not assumed [24].

A Properties of the S-matrix
Here we show the existence of two-particle S-matrix which satisfies some additional properties. Recall that the most general S-matrix satisfying (S1)-(S6) and (S9) takes the form 5 where − π 6 < ε < π 6 and S Blaschke (θ) is a finite Blaschke product which satisfies the conditions of [6, Appendix A] 6 .
We use freely sinh α · sinh β = cosh(α+β)−cosh(α−β) 2 , in particular, For S, we have B Weakly but not strongly commuting operators Let A, B be self-adjoint operators with domains Dom(A), Dom(B) with the dense intersection Dom(A) ∩ Dom(B). It is well-known that, even if A and B weakly commute, namely it holds that AΨ 1 , BΨ 2 = BΨ 1 , AΨ 2 for Ψ k ∈ Dom(A)∩Dom(B), it does not imply that the spectral projections of A and B commute. A classical (counter)example is due to Nelson [25, Section 10]: one takes a Riemann surface with a nontrivial topology and considers it as a real manifold and the L 2 -space on it. If one chooses two different translations, their generators on the L 2 -space commute weakly (actually as operators on a common core), as the derivations in these directions commute. However, the two translations do not commute, as the Riemann surface has a nontrivial topology.
The example above shows clearly why two operators weakly commute but not strongly: weak commutation is a local property, while strong commutation is global. The question of strong commutativity is not just technical and, if it fails, there is often a good reason for it. See [29], [30,Example 5.5,Exercise 6.16 ] for some other examples. We present here two more families of counterexamples.

B.1 From canonical commutation relation
This example is essentially due to Faddeev and Volkov [14]. Consider a CCR pair X = M id , where id(t) = t is the identity function and P = −i d dt on L 2 (R). Note that (e s 1 X ξ)(t) = e s 1 t ξ(t), (e s 2 P ξ)(t) = ξ(t − is 2 ) on suitable domains. If we take s 2 = 2π s 1 , then (e s 1 X e s 2 P ξ)(t) = (e s 2 P e s 1 X ξ)(t) = e s 1 t ξ(t − 2π s 2 ). It is not difficult to find a common dense domain of e s 1 X and e s 2 P , and they weakly commute. Yet they do not strongly commute. Indeed we know that {e s 1 X , e s 2 P } = B(L 2 (R)), while strong commutativity would imply {e s 1 X , e s 2 P } to be abelian, which is not true.

B.2 From bound state operators
Take two Blaschke products f 1 , f 2 on R+i(−π, 0) with the symmetry condition f j (θ−πi) = f j (θ). They extend meromorphically to C and satisfy f j (θ − 2πi) = f j (θ). Form . They are manifestly self-adjoint and, by a similar consideration as [35, Section 5.2], one can conclude that their domains are determined by the zeros and poles of f j . Especially one can see that the domains Dom(A 1 A 2 ) and Dom(A 2 A 1 ) are dense: indeed, there are analytic functions with specified zeros at the poles of f 1 , f 2 in R + i(−4π, 0). On such a function ξ it is clear that A 1 A 2 ξ = A 2 A 1 ξ, because on this domain it holds that M f j ∆ξ = ∆M f j ξ, as f j are 2πi-periodic. Yet, A 1 and A 2 do not strongly commute. Indeed, the intersection of their domains is not a core for any of them if f 1 and f 2 have different zeros or poles, which also follows from the ideas of [35, Section 3].

C The Driessler-Fröhlich theorem
First let us recall the commutator theorem of Glimm-Jaffe-Nelson. The theorem roughly says the following: if T is a positive self-adjoint operator, A is a symmetric operator and if A and [T, A] can be estimated by T , then A is essentially self-adjoint. Yet, depending on how to make such estimates, there are certain variations in the hypothesis of the theorem.
In one of such variations, one estimates A and [T, A] as bilinear forms, defined on Dom T as an operator, namely Aψ ≤ c T ψ for ψ ∈ Dom(T ), or for a core of T . For our situation, this latter version has a better chance to apply (see [26,Theorem X.37]): Theorem C.1 (the commutator theorem). Let T be a self-adjoint operator with T ≥ 1. Suppose that A is a symmetric operator defined on a core D of T so that (1) Aψ ≤ c 1 T ψ for all ψ ∈ D, Then, A is essentially self-adjoint on any core of T . Although the essential idea is the same as the original [11], we here present the adapted proof for the sake of clarity. Note that the domain of the weak commutativity is also modified: the original proof requires only the weak commutativity on a core of T 2 , while here we need it on a core of T . By the same reason as explained in [26, remark after Theorem X.37], the results might fail by a tiny change in assumptions, therefore, one should not underestimate the importance of these careful examinations. (2) for all ϕ, ψ ∈ D, for all ϕ, ψ ∈ D, Aψ, Bϕ = Bψ, Aϕ for all ψ, ϕ ∈ D.
Then A, B are essentially self-adjoint on any core of T and they strongly commute.
Proof . Note that all the assumptions hold on Dom(T ). Indeed, for any pair ψ, ϕ ∈ Dom(T ), there are {ψ n }, {ϕ n } ⊂ D such that ψ n → ψ, T ψ n → T ψ, therefore, T 1 2 ψ n → T 1 2 ψ and ϕ n → ϕ, T ϕ n → T ϕ, therefore, T 1 2 ϕ n → T 1 2 ϕ. From this, A and B can be naturally extended to Dom(T ) and all the statements and the weak commutativity hold there. Now, the claim that A and B are essentially self-adjoint of any core of T follows from the assumptions (1), (2) and Theorem C.1.

D Nontrivial extensions of sum operators
We show here that an operator which looks close to χ 2 (ξ) fails to be self-adjoint.
Proposition D.1. Let f be a bounded analytic function in R + i(− π 3 , 0), |f (θ)| = 1 and a ∈ R. Then a∆ 1 M f . We saw that this is not always self-adjoint, depending on f and a.

E Observations on S-symmetric functions
For a Hilbert space K, let L 2 (R, K) denote the space of K-valued L 2 -functions, and for an open subset U of R d , H 2 (R d + iU, K) be the Hardy space of K-valued functions Ψ which have an analytic continuation in R d + iU such that Ψ( · + iλ λ λ), λ λ λ ∈ U, belongs to L 2 (R d ) and their L 2 -norms are uniformly bounded with respect to λ.
Lemma E.2. Let {Ψ n } be a sequence in L 2 (R, K) and assume that there is an ψ ∈ L 2 (R) which satisfies Ψ n (θ) ≤ ψ(θ) (almost everywhere). If Ψ n is almost everywhere pointwise convergent to Ψ, then Ψ is again L 2 and Ψ n → Ψ in the L 2 -sense.
In the latter expression, the first factorΨ(k)e ikζ (e αk + e −αk ) is L 2 by assumption, and the second factor is an L 2 -function in k. Therefore, the integral gives a vector in K for any ζ in the above interval by Lemma E.1. This shows that the former expression is L 1 and it does not depend on α. Let us denote it by Ψ(ζ). This notation coincides with Ψ when ζ is real.
By assumption (S9), S has only finitely many zeros in the physical strip whose distance for the real line is larger than 0 < κ (< π 3 ). Let {α j } be the poles of S, − π 3 < Im α j < −κ (which correspond to the zeros in the physical strip by (S1) and (S2)).
This region is represented by the large triangle in Fig. 1. This can be considered as an L 2 -version of the Malgrange-Zerner theorem (cf. [13]).
Proof . By looking at the variable ζ k and ζ l , we obtain the analyticity in the large triangle C in Fig. 3. By changing the variable to (ζ k − ζ l , ζ k + ζ l ), and considering the small rectangle R , we can apply Lemma E.4 to conclude that for > 0, f l (θ k − θ l )Ψ n (θ 1 , . . . , θ n ) has a continuation to an element in H 2 (R 2 + iR , L 2 (R n−2 )), as Ψ n has a zero at θ k − θ l = 0 by assumption. Furthermore, as f l has no other pole, it continues further to R 2 + iC . By K-valued three line theorem (cf. [27,Theorem 12.9]), we obtain that the H 2 -norm as an element of H 2 (R 2 + iC , L 2 (R n−2 )) is determined at the corners of the triangle C which have finite distance from the line θ k − θ l = 0. As S is bounded at these corners and the edges, we obtain an H 2 -bound which is uniform in . Finally, as < is arbitrary, actually we obtain that f l (θ k − θ l )Ψ n (θ 1 , . . . , θ n ) ∈ H 2 (R 2 + iC, L 2 (R n−2 )). This is equivalent to the vector f l (θ k − θ l )Ψ n (θ 1 , . . . , θ n ) being in the same domain, as the analyticity in the other variables are not affected.
Repeating this n − 1 times, we obtain the claim.