# On the Equivalence of Two Deformation Schemes in Quantum Field Theory

Gandalf Lechner 111Institute for Theoretical Physics, University of Leipzig,
04104 Leipzig, Germany, email<EMAIL_ADDRESS>Jan Schlemmer
222Department of Physics, University of Vienna, 1090 Vienna, Austria, email:
<EMAIL_ADDRESS>GL and JS supported by FWF project P22929-N16
“Deformations of quantum field theories” Yoh Tanimoto 333Institute for
Theoretical Physics, University of Göttingen, 37077 Göttingen, Germany, email:
<EMAIL_ADDRESS>Supported by Deutscher
Akademischer Austauschdienst.

(September 11, 2012)

###### Abstract

Two recent deformation schemes for quantum field theories on two-dimensional
Minkowski space, making use of deformed field operators and Longo-Witten
endomorphisms, respectively, are shown to be equivalent.

Keywords: deformations of quantum field theories, two-dimensional models,
modular theory
MSC (2010): 81T05, 81T40

##  1 Deformations of QFTs by inner functions and their roots

In recent years, there has been a lot of interest in deformations of quantum
field theories [GL07, BS08, BLS11, DT11, LW11, LR12, MM11, Lec12, Tan12a,
BT12, Ala12, Muc12, Pla12] in the sense of specific procedures modifying
quantum field theoretic models on Minkowski space, mostly motivated by the
desire to construct new models in a non-perturbative manner. Various
constructions have been invented, relying on different methods such as smooth
group actions, non-commutative geometry, chiral conformal field theory,
boundary quantum field theory, and inverse scattering theory.

In many situations, it is possible to set up the deformation in such a way
that Poincaré covariance is completely preserved and locality partly. More
precisely, often the deformation introduces operators which are no longer
localized in arbitrarily small regions of spacetime, but rather in unbounded
regions like a Rindler wedge $W:=\\{x\in\mathbbm{R}^{d}\,:\,x_{1}>|x_{0}|\\}$.
In the operator-algebraic framework of quantum field theory [Haa96], such a
wedge-local Poincaré covariant model can be conveniently described by a so-
called Borchers triple $(\mathcal{M},U,\Omega)$ [Bor92, BLS11], consisting of
a von Neumann algebra $\mathcal{M}$ of operators localized in the wedge $W$, a
suitable representation $U$ of the translations, and an invariant (vacuum)
vector $\Omega$ (see Def. 1.2 below). Depending on the method at hand, the von
Neumann algebra $\mathcal{M}$ is generated by different objects, like deformed
field operators or twisted chiral observables.

It is the aim of this letter to show that some of the constructions on two-
dimensional Minkowski space are identical in the sense of unitary equivalence
of their associated Borchers triples. More precisely, we will show that the
deformations presented in [Tan12a], starting from a chiral field theory, are
equivalent to the deformations in terms of deformed field operators, presented
in [Lec12], for mass $m=0$ and dimension $d=1+1$ (Section 2). In the special
case of the so-called warped convolution deformation [BLS11], such an
equivalence was already observed in [Tan12a]. Here we prove that also for the
infinite family of deformations considered in [Lec12], one obtains the same
construction as in [Tan12a] in the chiral situation, where the deformation
amounts to a unitary equivalence transformation by a Longo-Witten endomorphism
[LW11] on each light ray. Furthermore, we will show that certain aspects of
the chiral construction carry over to the massive situation (Section 3).

The deformations we are interested in here take certain families of analytic
functions as input parameters, whose relations we will now clarify. We will
write $\mathbbm{H}\subset\mathbbm{C}$ for the open upper half plane,
$\mathrm{S}(0,\pi):=\\{\zeta\in\mathbbm{C}\,:\,0<\operatorname{Im}\zeta<\pi\\}$
for the strip, and $H^{\infty}(\mathbbm{H})$, $H^{\infty}(\mathrm{S}(0,\pi))$
for the Hardy spaces of bounded analytic functions on these domains. Recall
that for a function $f\in H^{\infty}(\mathbbm{H})$, the limit
$\lim_{\varepsilon\searrow 0}f(t+i\varepsilon)$ exists almost everywhere444By
“almost everywhere” (a.e.) and “almost all” we always refer to Lebesgue
measure on $\mathbbm{R}$. and defines a boundary value function in
$L^{\infty}(\mathbbm{R})$. The same holds for functions in
$H^{\infty}(\mathrm{S}(0,\pi))$ and their boundary values at $\mathbbm{R}$ and
$\mathbbm{R}+i\pi$.

######  Definition 1.1.

  1. i)

A symmetric inner function is a function $\varphi\in H^{\infty}(\mathbbm{H})$
whose boundary values on the real line satisfy
$\overline{\varphi(t)}=\varphi(t)^{-1}=\varphi(-t)$ for almost all
$t\in\mathbbm{R}$.

  2. ii)

A root of a symmetric inner function $\varphi$ is a function $R\in
L^{\infty}(\mathbbm{R})$ such that $\overline{R(t)}=R(t)^{-1}=R(-t)$ and
$R(t)^{2}=\varphi(t)$ for almost all $t\in\mathbbm{R}$. The family of all
roots of symmetric inner functions will be denoted $\mathcal{R}$.

  3. iii)

A scattering function is a function $S\in H^{\infty}(\mathrm{S}(0,\pi))$ whose
boundary values satisfy
$\overline{S(\theta)}=S(\theta)^{-1}=S(-\theta)=S(i\pi+\theta)$ for almost all
$\theta\in\mathbbm{R}$.

Symmetric inner functions provide the input into deformations making use of
Longo-Witten endomorphisms [LW11, Tan12a, LR12, BT12], whereas scattering
functions are used in inverse scattering approaches such as [Lec03, BLM11].
For convenience, the latter are usually defined with the additional
requirement of extending continuously to the closure of $\mathrm{S}(0,\pi)$.
However, going through the construction, say in [Lec03], one realizes that
this continuity assumption is not necessary. What is required is that the
boundary conditions on $S$ hold almost everywhere, the boundary values are
regular enough to define multiplication operators on $L^{2}(\mathbbm{R})$, and
for $f\in H^{\infty}(\mathrm{S}(0,\pi))$ with Schwartz boundary values,
$[0,\pi]\ni\lambda\mapsto\int_{\mathbbm{R}}d\theta\,f(\theta+i\lambda)S(\theta+i\lambda)$
is continuous. As this is the case for any $S\in
H^{\infty}(\mathrm{S}(0,\pi))$, one can just as well work with the more
general definition of scattering function given above.

We also note that scattering functions and symmetric inner functions are in
one to one correspondence by $S(\zeta):=\varphi(\sinh\zeta)$,
$\zeta\in\mathrm{S}(0,\pi)$. As $\sinh(i\pi+\zeta)=-\sinh\zeta=\sinh(-\zeta)$,
this identification produces the required properties of the boundary values in
Def. 1.1 iii). On the other hand, $\varphi(z):=S(\sinh^{-1}z)$,
$z\in\mathbbm{H}$, is well-defined and analytic because of the crossing
symmetry $S(i\pi+\theta)=S(-\theta)$ of $S$. This identification of the strip
and the half plane via $\sinh$ is the one encountered in massive theories
[GL07]. In massless theories, also the identification
$\exp:\mathrm{S}(0,\pi)\to\mathbbm{H}$ occurs [LW11], and under this
identification, scattering functions correspond to the subset of symmetric
inner functions with the additional symmetry
$\overline{\varphi(t)}=\varphi(t^{-1})$ , see (1.20) below.

Regarding Def. 1.1 ii), we note that each symmetric inner function has
infinitely many different roots, and the family of roots $\mathcal{R}$
contains all symmetric inner functions because Def. 1.1 i) is stable under
taking squares. These roots are the input into the deformations in [Lec12,
Ala12, Pla12], where under additional regularity assumptions, they are called
deformation functions. We note that these additional requirements are only
necessary when working on the tensor algebra of test functions [Lec12], but
not when working directly on a representation space such as in [Ala12]. In
particular, the roots will not be required to be analytic, and also the
condition $R(0)=1$ [Lec12], related to fixing a root of an inner function,
will not be assumed here.

In the following, we will be concerned with deformations of free field
theories of mass $m\geq 0$ on two-dimensional Minkowski space, and now set up
some standard notation for this. We will be working on the Bose Fock space

$\displaystyle\mathcal{H}:=\Gamma(\mathcal{H}_{1})\,,\qquad\mathcal{H}_{1}:=L^{2}(\mathbbm{R},\tfrac{dp}{\omega_{m}(p)})\,,\quad\omega_{m}(p):=(m^{2}+p^{2})^{1/2}\,.$

Its Fock vacuum will be denoted $\Omega$, and we have the usual representation
$\Gamma(U_{1})$ of the proper Poincaré group as the second quantization of

$\displaystyle[U_{1}(x,\lambda)\Psi_{1}](p)$
$\displaystyle:=e^{i(x_{0}\omega_{m}(p)-x_{1}p)}\,\Psi_{1}(\lambda
p)\,,\qquad[U_{1}(j)\Psi_{1}](p):=\overline{\Psi_{1}(p)}\,,$ (1.1)

where $x=(x_{0},x_{1})\in\mathbbm{R}^{2}$ is the translation,
$\lambda\in\mathbbm{R}$ denotes the boost rapidity parameter, $\lambda
p:=-\sinh\lambda\cdot\omega_{m}(p)+\cosh\lambda\cdot p$, and $j(x)=-x$ is the
space-time reflection. We will also write $U(x):=\Gamma(U_{1}(x,0))$ for the
translations.

From an operator-algebraic point of view, a wedge-local quantum field theory
is equivalent to a Borchers triple.

######  Definition 1.2.

A Borchers triple $(\mathcal{M},U,\Omega)$ on $\mathbbm{R}^{2}$ consists of a
von Neumann algebra $\mathcal{M}\subset\mathcal{B}(\mathcal{H})$, a strongly
continuous unitary positive energy representation $U$ of the translation group
$\mathbbm{R}^{2}$ on $\mathcal{H}$, and a $U$-invariant unit vector
$\Omega\in\mathcal{H}$ such that

  1. i)

$U(x)\mathcal{M}U(x)^{-1}\subset\mathcal{M}$ for any $x\in\overline{W}$,

  2. ii)

$\Omega$ is cyclic and separating for $\mathcal{M}$.

Two Borchers triples $(\mathcal{M},U,\Omega)$ and
$(\tilde{\mathcal{M}},\tilde{U},\tilde{\Omega})$ will be called equivalent,
written
$(\mathcal{M},U,\Omega)\cong(\tilde{\mathcal{M}},\tilde{U},\tilde{\Omega})$,
if there exists a unitary $V$ such that
$V\mathcal{M}V^{*}=\tilde{\mathcal{M}}$, $VU(x)V^{*}=\tilde{U}(x)$ for all
$x\in\mathbbm{R}^{2}$, and $V\Omega=\tilde{\Omega}$.

Recall that by a famous theorem of Borchers [Bor92], the representation $U$
can be extended to a (anti-) unitary representation $U_{\mathcal{M}}$ of the
proper Poincaré group $\mathcal{P}_{+}$ with the help of the modular data
$J_{\mathcal{M}},\Delta_{\mathcal{M}}$ of $(\mathcal{M},\Omega)$, by

$\displaystyle
U_{\mathcal{M}}(x,\lambda):=U(x)\Delta_{\mathcal{M}}^{-\frac{i\lambda}{2\pi}}\,,\qquad
U_{\mathcal{M}}(j):=J_{\mathcal{M}}\,.$ (1.2)

As is well known, a Borchers triple gives rise to a Poincaré-covariant net of
wedge algebras [Bor92], which can under further conditions be extended to a
net of double cone algebras [BL04, Lec08]. We will not discuss the extension
question here, but rather focus on the wedge-local aspects only. Note that the
net of wedge algebras generated from a Borchers triple
$(\mathcal{M},U,\Omega)$ will transform covariantly under a representation
$U_{\mathcal{M}}$ of the Poincaré group which depends on $\mathcal{M}$.
However, in the case of two equivalent Borchers triples
$(\mathcal{M},U,\Omega)\cong(\tilde{\mathcal{M}},\tilde{U},\tilde{\Omega})$,
modular theory tells us that the modular data of $(\mathcal{M},\Omega)$ and
$(\tilde{\mathcal{M}},\tilde{\Omega})$ are related by
$VJ_{\mathcal{M}}V^{*}=J_{\tilde{\mathcal{M}}}$,
$V\Delta_{\mathcal{M}}^{it}V^{*}=\Delta_{\tilde{\mathcal{M}}}^{it}$, i.e.
equivalence of Borchers triples implies equivalence of the associated wedge-
local nets including their representations $U_{\mathcal{M}}\cong
U_{\tilde{\mathcal{M}}}$ of the proper Lorentz group.

A particular example of a Borchers triple is provided by the model of a free
scalar quantum field: Let $a(\xi)$ and $a^{\dagger}(\xi):=a(\xi)^{*}$,
$\xi\in\mathcal{H}_{1}$, denote the standard CCR annihilation and creation
operators on $\mathcal{H}$, and for $f\in\mathscr{S}(\mathbbm{R}^{2})$, let

$\displaystyle\phi_{m}(f)$
$\displaystyle:=a^{\dagger}(f^{+})+a(\overline{f^{-}})\,,\qquad
f^{\pm}(p):=\tilde{f}(\pm\omega_{m}(p),\pm p)\,,$ (1.3)

denote the free Klein-Gordon field of mass $m\geq 0$ (with $f$ restricted to
directional derivatives of test functions in the case $m=0$ because of the
well-known infrared divergence in the measure $\frac{dp}{|p|}$). With the
wedge algebra

$\displaystyle\mathcal{M}_{m}:=\\{e^{i\phi_{m}(f)}\,:\,f\in\mathscr{S}_{\mathbbm{R}}(W)\\}^{\prime\prime}\,,$
(1.4)

the Fock translations $U$ (1.1) and the Fock vacuum $\Omega$, we then have a
Borchers triple $(\mathcal{M}_{m},U,\Omega)$. In this case, the modular data
of $(\mathcal{M}_{m},\Omega)$ reproduce the Poincaré representation (1.1),
i.e. $U_{\mathcal{M}_{m}}=\Gamma(U_{1})$. For convenience of notation, we will
write

$\displaystyle
J:=J_{\mathcal{M}_{m}}=\Gamma(U_{1}(j))\,,\qquad\Delta^{it}:=\Delta^{it}_{\mathcal{M}_{m}}=\Gamma(U_{1}(0,-2\pi
t))\,.$ (1.5)

Fixing the representation $U$ of the translations and the vector $\Omega$, the
algebra $\mathcal{M}_{m}$ is however by no means the only von Neumann algebra
completing $U,\Omega$ to a Borchers triple. In the following, we will
introduce for each $R\in\mathcal{R}$ two von Neumann algebras
$\mathcal{M}_{R,m}$, $\mathcal{N}_{R}$ with this property, obtained by
(generalizations of) the deformation procedures in [Lec12] and [Tan12a],
respectively. For $R=1$, both families reduce to the undeformed situation,
i.e. $\mathcal{M}_{1,m}=\mathcal{M}_{m}$, $\mathcal{N}_{1}=\mathcal{M}_{0}$.

To define the first set of deformed wedge algebras $\mathcal{M}_{R,m}$, we
introduce a unitary-valued function
$T_{R,m}:\mathbbm{R}\to\mathcal{U}(\mathcal{H})$ [Lec12]

$\displaystyle[T_{R,m}(p)\Psi]_{n}(p_{1},\ldots,p_{n})$
$\displaystyle:=\prod_{k=1}^{n}R_{m}(p,p_{k})\cdot\Psi_{n}(p_{1},\ldots,p_{n})\,.$
(1.6)

Here the function $R_{m}\in L^{\infty}(\mathbbm{R}^{2})$ is in the case of
positive mass defined as

$\displaystyle R_{m}(p,q)$
$\displaystyle:=R\big{(}\tfrac{1}{2}(\omega_{m}(q)p-\omega_{m}(p)q)\big{)}\,,\qquad
m>0\,,$ (1.7)

where the factor $\frac{1}{2}$ is a matter of convention. Taking the limit
$m\to 0$, one observes that the argument $\frac{1}{2}(|q|p-|p|q)$ of $R$
vanishes if $p$ and $q$ have the same sign. As the root $R$ is only defined up
to equivalence in $L^{\infty}(\mathbbm{R})$, its value at $0$ is not fixed. We
therefore define

$\displaystyle R_{0}(p,q):=\left\\{\begin{array}[]{ccl}R(-pq)&;&p>0,\,q<0\\\
R(+pq)&;&p<0,\,q>0\\\
1&;&p>0,\,q>0\;\;\text{or}\;\;p<0,\,q<0\end{array}\right.\;.$ (1.11)

Note that for any mass $m\geq 0$, we have for almost all $p,q\in\mathbbm{R}$

$\displaystyle R_{m}(q,p)=R_{m}(p,q)^{-1}\,,\qquad m\geq 0\,.$ (1.12)

The assignment $a_{R}(p):=a(p)T_{R,m}(p)$ defines an operator-valued
distribution for any $R\in\mathcal{R}$, which explicitly acts on a vector
$\Psi$ of finite particle number according to, $\xi\in\mathcal{H}_{1}$,

$\displaystyle[a_{R}(\xi)\Psi]_{n}(p_{1},\ldots,p_{n})=\sqrt{n+1}\int\frac{dq}{\omega_{m}(q)}\,\overline{\xi(q)}\,\prod_{k=1}^{n}R_{m}(q,p_{k})\Psi_{n+1}(q,p_{1},\ldots,p_{n})\,.$
(1.13)

Its adjoint is denoted $a^{\dagger}_{R}(\xi):=a_{R}(\overline{\xi})^{*}$, and
the corresponding deformed field operator is

$\displaystyle\phi_{R,m}(f):=a^{\dagger}_{R}(f^{+})+a_{R}(\overline{f^{-}})\,,\qquad
f\in\mathscr{S}(\mathbbm{R}^{2})\,.$ (1.14)

As $\phi_{R,m}(f)$ is essentially self-adjoint on the subspace of finite
particle number for real $f$, one can pass to the generated von Neumann
algebra

$\displaystyle\mathcal{M}_{R,m}:=\\{e^{i\phi_{R,m}(f)}\,:\,f\in\mathscr{S}_{\mathbbm{R}}(W)\\}^{\prime\prime}\,.$
(1.15)

######  Theorem 1.3.

Let $R\in\mathcal{R}$ and $m\geq 0$. Then $(\mathcal{M}_{R,m},U,\Omega)$ is a
Borchers triple with modular data $J_{\mathcal{M}_{R,m}}=J$ and
$\Delta^{it}_{\mathcal{M}_{R,m}}=\Delta^{it}$.

For $m>0$, this has been established in [Lec12], and for $m=0$, one can use
essentially the same proofs, so that we do not have to go into details here.
In fact, as for $m=0$ the mass shell decomposes into two half-rays which are
left invariant by the Lorentz boosts, one can in this case more generally
consider three roots $R,R_{1},R_{2}\in\mathcal{R}$, with the additional
requirement $\overline{R_{k}(t)}=R_{k}(t^{-1})$ for almost all
$t\in\mathbbm{R}$, $k=1,2$, and put

$\displaystyle R_{0}(p,q):=\left\\{\begin{array}[]{ccl}R(-pq)&;&p>0,\,q<0\\\
R(+pq)&;&p<0,\,q>0\\\ R_{1}(+\tfrac{p}{q})&;&p>0,\,q>0\\\
R_{2}(-\tfrac{p}{q})&;&p<0,\,q<0\end{array}\right.\;.$ (1.20)

Also with this more general definition of $R_{0}$, the algebra
$\mathcal{M}_{R,0}$ completes $U,\Omega$ to a Borchers triple. In the
terminology of [FS93], the functions $R_{1}$, $R_{2}$ govern the left-left and
right-right “scattering” of the model, whereas $R$ determines the left-right
(wave) scattering [Bis12, Buc77, DT11]. If $R=1$, the corresponding model is
chiral – this is in particular the case for the short distance scaling limits
of the models generated by the massive wedge algebras $\mathcal{M}_{R,m}$,
$m>0$. In this context one finds $R=1$,
$R_{1}(t)^{2}=R_{2}(t)^{2}=\varphi(t-t^{-1})$ with some symmetric inner
function $\varphi$ [BLM11]. For the purposes of this letter, we will however
restrict ourselves to the case $R_{1}=R_{2}=1$ (1.11), which corresponds to
the construction in [Tan12a].

To define the second set of deformed wedge algebras $\mathcal{N}_{R}$, one
works in the massless case $m=0$, and uses the chiral structure present in
this situation. Here the Fock space, the representation $U$, the invariant
vector $\Omega$, and the wedge algebra $\mathcal{M}_{0}$ split into two
(chiral) factors. With
$\mathcal{H}_{1}^{\pm}:=L^{2}(\mathbbm{R}_{\pm},\frac{dp}{|p|})$, we have
$\mathcal{H}_{1}=\mathcal{H}_{1}^{+}\oplus\mathcal{H}_{1}^{-}$ and

$\displaystyle\mathcal{H}$
$\displaystyle\cong\mathcal{H}^{+}\otimes\mathcal{H}^{-}\,,\qquad\mathcal{H}^{\pm}:=\Gamma(\mathcal{H}_{1}^{\pm})\,,$
(1.21) $\displaystyle\Omega$
$\displaystyle\cong\Omega_{+}\otimes\Omega_{-}\,,$ (1.22) $\displaystyle U(x)$
$\displaystyle\cong U_{+}(x_{-})\otimes U_{-}(x_{+})\,,\qquad
x_{\pm}:=x_{0}\pm x_{1}\,,$ (1.23) $\displaystyle\mathcal{M}_{0}$
$\displaystyle\cong\mathcal{M}_{0,+}\otimes\mathcal{M}_{0,-}\,,$ (1.24)

where $\Omega_{\pm}$ denotes the Fock vacuum in $\mathcal{H}^{\pm}$. The
canonical unitary $V:\mathcal{H}^{+}\otimes\mathcal{H}^{-}\to\mathcal{H}$
realizing the above isomorphisms is recalled in (2.4). Note that

$\displaystyle V^{*}\Gamma(U_{1}(x,\lambda))V$
$\displaystyle=\Gamma_{+}(U_{1,+}(x_{-},\lambda))\otimes\Gamma_{-}(U_{1,-}(x_{+},\lambda))\,,$
(1.25) $\displaystyle V^{*}\Gamma(U_{1}(j))V$
$\displaystyle=\Gamma_{+}(U_{1,+}(j))\otimes\Gamma_{-}(U_{1,-}(j))\,,$ (1.26)

where $\Gamma_{\pm}$ denotes second quantization on $\mathcal{H}^{\pm}$, with
$(U_{1,\pm}(x_{\mp},\lambda)\Psi_{1})(p)=e^{\pm
ipx_{\mp}}\Psi_{1}(e^{\mp\lambda}\cdot p)$ and
$(U_{1,\pm}(j)\Psi_{1})(p)=\overline{\Psi_{1}(p)}$. For the sake of a concise
notation, we will write

$\displaystyle J_{\otimes}$
$\displaystyle:=\Gamma_{+}(U_{1,+}(j))\otimes\Gamma_{-}(U_{1,-}(j))=V^{*}JV\,,$
(1.27) $\displaystyle\Delta_{\otimes}^{it}$
$\displaystyle:=\Gamma_{+}(U_{1,+}(0,-2\pi
t))\otimes\Gamma_{-}(U_{1,-}(0,-2\pi t))=V^{*}\Delta^{it}V\,.$ (1.28)

Given $R\in\mathcal{R}$, one introduces the unitary
$S_{R}\in\mathcal{U}(\mathcal{H}^{+}\otimes\mathcal{H}^{-})$ [Tan12a],

$\displaystyle\big{[}S_{R}\Psi\big{]}_{n,n^{\prime}}(p_{1},\ldots,p_{n},q_{1},\ldots,q_{n^{\prime}})=\prod_{\begin{subarray}{c}i=1\ldots
n\\\ j=1\ldots
n^{\prime}\end{subarray}}R_{0}(p_{i},q_{j})\cdot\Psi_{n,n^{\prime}}(p_{1},\ldots,p_{n},q_{1},\ldots,q_{n^{\prime}})\,,$
(1.29)

and defines the von Neumann algebra

$\displaystyle\mathcal{N}_{R}:=(\mathcal{M}_{0,+}\otimes 1)\vee
S_{R^{2}}(1\otimes\mathcal{M}_{0,-})S_{R^{2}}^{*}\,.$ (1.30)

######  Theorem 1.4.

Let $R\in\mathcal{R}$. Then $(\mathcal{N}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})$ is a Borchers triple with modular data
$J_{\mathcal{N}_{R}}=S_{R^{2}}J_{\otimes}$ and
$\Delta^{it}_{\mathcal{N}_{R}}=\Delta_{\otimes}^{it}$.

This theorem has been proven in [Tan12a]. From (1.30), it is clear that
$\mathcal{N}_{R}$ depends on $R$ only via the symmetric inner function
$R^{2}$.

Our results can now compactly be summarized as follows ($R\in\mathcal{R}$,
$m\geq 0$):

  * •

$(\mathcal{N}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})\cong(\mathcal{M}_{R,0},U,\Omega)$ (Theorem
2.4).

  * •

$(\mathcal{M}_{R_{1},m},U,\Omega)\cong(\mathcal{M}_{R_{2},m},U,\Omega)$ if and
only if $R_{1}^{2}=R_{2}^{2}$ (Proposition 3.2).

As $R^{2}$ is essentially the two-particle S-matrix of the model described by
the Borchers triple $(\mathcal{M}_{R,m},U,\Omega)$, the last result amounts to
a proof of uniqueness of the solution of the inverse scattering problem in the
setting of the deformations studied here. In case of continuous $R$, such an
effect was already observed in [Ala12]. For massless nets obeying a number of
natural conditions, uniqueness of the inverse scattering problem is known once
one fixes the asymptotic algebra [Tan12a]. Furthermore, explicit examples of
S-matrices not preserving the Fock space structure are known in the massless
case [BT12].

The convenient deformation formula (1.30) is a result of the chiral structure
present in the massless case, and has no direct analogue in the massive case.
In Section 3, we will discuss why the situation is more complex in the massive
case even though (formal) relations between deformed and undeformed creation
and annihilation operators still exist.

##  2 Equivalence of the two deformations in the massless case

The aim of this section is to demonstrate the equivalence of the mass zero
Borchers triples $(\mathcal{M}_{R,0},U,\Omega)$ (1.15) and
$(\mathcal{N}_{R},U_{+}\otimes U_{-},\Omega_{+}\otimes U_{-})$ (1.30) for
arbitrary roots $R\in\mathcal{R}$. From Theorem 1.3 and Theorem 1.4, we see
that the modular groups of these von Neumann algebras coincide with the one
parameter boost groups on their respective Hilbert spaces, but their modular
conjugations differ by $S_{R^{2}}$, i.e. we have
$J_{\mathcal{N}_{R}}=S_{R^{2}}V^{*}J_{\mathcal{M}_{R,0}}V$ with the canonical
unitary $V:\mathcal{H}^{+}\otimes\mathcal{H}^{-}\to\mathcal{H}$ (2.4). We will
therefore in a first step go over to an equivalent form of $\mathcal{N}_{R}$
which has modular conjugation $V^{*}J_{\mathcal{M}_{R,0}}V$, without the
factor $S_{R^{2}}$. In general, this can be accomplished by conjugating with a
root of the “S-matrix” $J_{\mathcal{N}_{R}}V^{*}J_{\mathcal{M}_{R,0}}V$
[Wol92], and in our present situation, this amounts to considering

$\displaystyle\hat{\mathcal{N}}_{R}:=S_{R}^{*}(\mathcal{M}_{0,+}\otimes
1)S_{R}\vee S_{R}(1\otimes\mathcal{M}_{0,-})S_{R}^{*}\,.$ (2.1)

######  Lemma 2.1.

Let $R\in\mathcal{R}$. Then $(\hat{\mathcal{N}}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})$ is a Borchers triple equivalent to
$(\mathcal{N}_{R},U_{+}\otimes U_{-},\Omega_{+}\otimes\Omega_{-})$, with
modular data

$\displaystyle
J_{\hat{\mathcal{N}}_{R}}=J_{\otimes}\,,\qquad\Delta^{it}_{\hat{\mathcal{N}}_{R}}=\Delta^{it}_{\otimes}.$
(2.2)

###### Proof.

As $S_{R}$ (1.29) satisfies $S_{R}^{2}=S_{R^{2}}$, we have the unitary
equivalence of algebras $\hat{\mathcal{N}}_{R}=S_{R}^{*}\mathcal{N}_{R}S_{R}$.
The unitary $S_{R}$ clearly commutes with all translations
$U_{+}(x_{-})\otimes U_{-}(x_{+})$ and leaves $\Omega_{+}\otimes\Omega_{-}$
invariant. Hence $(\hat{\mathcal{N}}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})\cong(\mathcal{N}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})$; this also shows that
$(\hat{\mathcal{N}}_{R},U_{+}\otimes U_{-},\Omega_{+}\otimes\Omega_{-})$ is a
Borchers triple.

Regarding the modular data of
$(\hat{\mathcal{N}}_{R},\Omega_{+}\otimes\Omega_{-})$, we first have by
modular theory

$\displaystyle
J_{\hat{\mathcal{N}}_{R}}=S_{R}^{*}J_{\mathcal{N}_{R}}S_{R}\,,\qquad\Delta_{\hat{\mathcal{N}}_{R}}^{it}=S_{R}^{*}\Delta_{\mathcal{N}_{R}}^{it}S_{R}=S_{R}^{*}\Delta_{\otimes}^{it}S_{R}\,.$
(2.3)

Taking into account the action of $\Gamma_{\pm}(U_{1,\pm}(0,\lambda))$, one
sees from (1.11) and (1.29) that $S_{R}$ commutes with the Lorentz boosts. As
these coincide with the modular unitaries $\Delta^{it}_{\otimes}$, the second
equation in (2.2) follows. To establish the claimed form of the modular
conjugation, we note
$J_{\otimes}S_{R}=S_{\overline{R}}J_{\otimes}=S_{R}^{*}J_{\otimes}$ and
compute

$\displaystyle J_{\hat{\mathcal{N}}_{R}}$
$\displaystyle=S_{R}^{*}J_{\mathcal{N}_{R}}S_{R}=S_{R}^{*}S_{R^{2}}J_{\otimes}S_{R}=S_{R}^{*}S_{R}^{2}S_{R}^{*}J_{\otimes}=J_{\otimes}\,.$

This completes the proof. ∎

The equivalence between the two deformed Borchers triples with wedge algebras
$\mathcal{M}_{R,0}$ and $\hat{\mathcal{N}}_{R}$ will now be established using
the creation and annihilation operators into which the fields generating
$\mathcal{M}_{0}$ can be decomposed. Corresponding to the splitting
$\mathcal{M}_{0}=\mathcal{M}_{0,+}\otimes\mathcal{M}_{0,-}$ we have creation
and annihilation operators $a_{\pm}$, $a_{\pm}^{\dagger}$ acting on
$\mathcal{H}^{\pm}$.

In the following, we will always suppress the canonical embeddings
$\iota^{n}_{\pm}:(\mathcal{H}^{\pm}_{1})^{\otimes
n}\to\mathcal{H}_{1}^{\otimes n}$,
$(\iota^{n}_{\pm}\Psi_{n}^{\pm})(p_{1},...,p_{n}):=\Psi_{n}^{\pm}(p_{1},...,p_{n})$
for $p_{1},...,p_{n}\in\mathbbm{R}_{\pm}$ and
$(\iota^{n}_{\pm}\Psi_{n}^{\pm})(p_{1},...,p_{n}):=0$ otherwise. With these
embeddings understood,
$\mathcal{H}_{1}=\mathcal{H}_{1}^{+}\oplus\mathcal{H}_{1}^{-}$, and hence
$\mathcal{H}\cong\mathcal{H}^{+}\otimes\mathcal{H}^{-}$. This isomorphism is
given explicitly by a unitary

$V:\mathcal{H}^{+}\otimes\mathcal{H}^{-}\to\mathcal{H}\,,$

which is uniquely determined by its action on the total set [Gui72] of
“exponential vectors”
$e^{\Psi_{1}}:=\sum_{n=0}^{\infty}\frac{1}{\sqrt{n!}}\Psi_{1}^{\otimes n}$ by

$V(e^{\Psi_{1}}\otimes
e^{\Phi_{1}}):=e^{\Psi_{1}\oplus\Phi_{1}}\,,\qquad\Psi_{1}\in\mathcal{H}_{1}^{+},\;\Phi_{1}\in\mathcal{H}_{1}^{-}\,.$
(2.4)

The definitions of $\mathcal{M}_{R,0}$ and $\mathcal{N}_{R}$ make use of the
realizations of $\mathcal{H}$ as
$\Gamma(\mathcal{H}_{1}^{+}\oplus\mathcal{H}_{1}^{-})$ and
$\Gamma(\mathcal{H}_{1}^{+})\otimes\Gamma(\mathcal{H}_{1}^{-})$, respectively.
We will work on
$\mathcal{H}=\Gamma(\mathcal{H}_{1}^{+}\oplus\mathcal{H}_{1}^{-})$, and first
compute an explicit expression of $S_{R}$ on this space.

######  Lemma 2.2.

Let $R\in\mathcal{R}$ and $\hat{S}_{R}:=VS_{R}V^{*}$. Then,
$\Psi\in\mathcal{H}$,

$\displaystyle\big{[}\hat{S}_{R}\Psi\big{]}_{n}(p_{1},\ldots,p_{n})=\prod_{i,j=1}^{n}R_{0}^{+}(p_{i},p_{j})\cdot\Psi_{n}(p_{1},\ldots,p_{n})\,,$
(2.5)

where

$R_{0}^{+}(p,q)=\begin{cases}R(-pq)&\ p>0,q<0\\\ 1&\text{ else
}\end{cases}\,.$

###### Proof.

As exponential vectors form a total set in $\mathcal{H}$, it is sufficient to
compute $\hat{S}_{R}$ on $e^{\Psi_{1}\oplus\Phi_{1}}$ to verify (2.5). The
action of $V$ from (2.4) on vectors
$\Xi=\sum_{n,m=0}^{\infty}\Xi_{n,m}\in\bigoplus_{n,m=0}^{\infty}((\mathcal{H}_{1}^{+})^{\otimes_{s}n}\otimes(\mathcal{H}_{1}^{-})^{\otimes_{s}m})=\mathcal{H}^{+}\otimes\mathcal{H}^{-}$
is explicitly given by

$\big{[}V\Xi]_{n}=\sum_{k=0}^{n}\binom{n}{k}^{1/2}\operatorname{Symm}_{n}\Xi_{k,n-k}\,,$
(2.6)

where for $f:\mathbbm{R}^{n}\to\mathbbm{C}$,

$[\operatorname{Symm}_{n}f](p_{1},\ldots,p_{n}):=\frac{1}{n!}\sum_{\pi\in\textfrak{S}_{n}}f(p_{\pi(1)},\ldots,p_{\pi(n)})$

denotes total symmetrization. Combining this with (1.29), we find

$\displaystyle[$
$\displaystyle\hat{S}_{R}(e^{\Psi_{1}\oplus\Phi_{1}})]_{n}(p_{1},\ldots,p_{n})=\sum_{k=0}^{n}\binom{n}{k}^{1/2}\operatorname{Symm}_{n}(S_{R}(e^{\Psi_{1}}\otimes
e^{\Phi_{1}})_{k,n-k})(p_{1},\ldots,p_{n})$
$\displaystyle=\sum_{k=0}^{n}\frac{\binom{n}{k}^{1/2}}{n!}\sum_{\pi\in\textfrak{S}_{n}}\prod_{\begin{subarray}{c}i=1\ldots
k\\\ j=k+1\ldots
n\end{subarray}}\\!\\!R_{0}^{+}(p_{\pi(i)},p_{\pi(j)})\frac{\Psi_{1}(p_{\pi(1)})\cdots\Psi_{1}(p_{\pi(k)})\cdot\Phi_{1}(p_{\pi(k+1)})\cdots\Phi_{1}(p_{\pi(n)})}{\sqrt{k!}\sqrt{(n-k)!}}\,.$

In the second line, $R_{0}$ was replaced by $R_{0}^{+}$, which does not change
the result since the factors of $\Psi_{1}$ and $\Phi_{1}$ (explicitly writing
out the embedding $\Psi_{1}\circ\iota_{+}$ and $\Phi_{1}\circ\iota_{-}$) are
equal to zero unless $p_{\pi(i)}>0$ and $p_{\pi(j)}<0$, and
$R_{0}(p,q)=R_{0}^{+}(p,q)$ for $p>0$, $q<0$.

Next we change the range of indices $i=1,...,k$, $j=k+1,...,n$ of the product
to $i=1,...,n$, $j=1,...,n$. This does not change the result because for the
indices $i,j$ which were not present before, we have
$R_{0}^{+}(p_{\pi(i)},p_{\pi(j)})=1$ on the support of the remaining factors.
After these manipulations,
$(p_{1},...,p_{n})\mapsto\prod_{i,j=1}^{n}R_{0}^{+}(p_{\pi(i)},p_{\pi(j)})$ is
a totally symmetric function independent of $\pi$ and $k$. Thus we get

$\displaystyle[\hat{S}_{R}$
$\displaystyle(e^{\Psi_{1}\oplus\Phi_{1}})]_{n}(p_{1},\ldots,p_{n})$
$\displaystyle=\prod_{i,j=1}^{n}R_{0}^{+}(p_{i},p_{j})\sum_{k=0}^{n}\binom{n}{k}^{1/2}\frac{1}{n!}\sum_{\pi\in\textfrak{S}_{n}}\frac{\Psi_{1}(p_{\pi(1)})\cdots\Psi_{1}(p_{\pi(k)})\cdot\Phi_{1}(p_{\pi(k+1)})\cdots\Phi_{1}(p_{\pi(n)})}{\sqrt{k!}\sqrt{(n-k)!}}$
$\displaystyle=\prod_{i,j=1}^{n}R_{0}^{+}(p_{i},p_{j})\left[V(e^{\Psi_{1}}\otimes
e^{\Phi_{1}})\right]_{n}(p_{1},\ldots,p_{n})$
$\displaystyle=\prod_{i,j=1}^{n}R_{0}^{+}(p_{i},p_{j})\left[e^{\Psi_{1}\oplus\Phi_{1}}\right]_{n}(p_{1},\ldots,p_{n})\,,$

and the proof is finished. ∎

After these preparations, we can now state the precise relation between the
generators appearing in the two types of deformations.

######  Proposition 2.3.

Let $R\in\mathcal{R}$ be a root of a symmetric inner function. Then,
$\psi_{\pm}\in\mathcal{H}_{1}^{\pm}$,

$\displaystyle a_{R}(\psi_{+})=VS_{R}^{*}(a_{+}(\psi_{+})\otimes
1)S_{R}V^{*}\,,\qquad a_{R}(\psi_{-})=VS_{R}(a_{+}(\psi_{-})\otimes
1)S_{R}^{*}V^{*}\,.$ (2.7)

###### Proof.

Let $\Psi\in\mathcal{H}$ be a vector of finite particle number. Using
$a\circ\iota_{+}=V(a_{+}\otimes 1)V^{*}$, $a\circ\iota_{-}=V(1\otimes
a_{-})V^{*}$ and the corresponding relations for $a^{\dagger}$ and
$a^{\dagger}_{\pm}$, we can equivalently show
$\hat{S}_{R}^{*}a(\psi_{+})\hat{S}_{R}=a_{R}(\psi_{+})$,
$\hat{S}_{R}a(\psi_{-})\hat{S}_{R}^{*}=a_{R}(\psi_{-})$. To this end, we
compute ($p_{1},...,p_{n}\in\mathbbm{R}$):

$\displaystyle[$
$\displaystyle\hat{S}_{R}a(\psi)\hat{S}_{R}^{*}\Psi]_{n}(p_{1},\ldots,p_{n})=\prod_{i,j=1}^{n}{R_{0}^{+}}(p_{i},p_{j})\cdot\sqrt{n+1}\int\frac{dq}{\lvert
q\rvert}\,\overline{\psi(q)}\,[\hat{S}_{R}^{*}\Psi]_{n+1}(q,p_{1},\ldots,p_{n})$
$\displaystyle=\sqrt{n+1}\prod_{i,j=1}^{n}\\!\\!R_{0}^{+}(p_{i},p_{j})\int\frac{dq}{\lvert
q\rvert}\overline{\psi(q)}\,\prod_{i^{\prime},j^{\prime}=1}^{n}\\!\\!\overline{{R_{0}^{+}}(p_{i^{\prime}},p_{j^{\prime}})}\cdot\prod_{k=1}^{n}\overline{{R_{0}^{+}}(q,p_{k}){R_{0}^{+}}(p_{k},q)}\,\Psi_{n+1}(q,p_{1},..,p_{n})$
$\displaystyle=\sqrt{n+1}\>\bigg{\\{}\int_{0}^{\infty}\frac{dq}{\lvert
q\rvert}\,\overline{\psi(q)}\,\prod_{k=1}^{n}\overline{{R_{0}^{+}}(q,p_{k})}\Psi_{n+1}(q,p_{1},\ldots,p_{n})$
(2.8)
$\displaystyle\qquad\qquad\qquad\qquad+\int_{-\infty}^{0}\frac{dq}{\lvert
q\rvert}\,\overline{\psi(q)}\,\prod_{k=1}^{n}\overline{{R_{0}^{+}}(p_{k},q)}\Psi_{n+1}(q,p_{1},\ldots,p_{n})\bigg{\\}}\,,$

where in the last equality $R_{0}^{+}(p,q)=1$ unless $p>0$ and $q<0$ was used.

On the other hand, using $R_{0}(p,q)=R_{0}^{+}(p,q)$ for $p>0,q<0$, and
$R_{0}(p,q)=\overline{R_{0}^{+}(q,p)}$ for $p<0,q>0$, we find

$\displaystyle[a_{R}(\psi)\Psi]_{n}(p_{1},\ldots,p_{n})$
$\displaystyle=\sqrt{n+1}\int\frac{dq}{\lvert
q\rvert}\,\overline{\psi(p)}\,\prod_{k=1}^{n}R_{0}(q,p_{k})\Psi_{n+1}(q,p_{1},\ldots,p_{n})$
$\displaystyle=\sqrt{n+1}\bigg{\\{}\int_{0}^{\infty}\frac{dq}{\lvert
q\rvert}\,\overline{\psi(q)}\,\prod_{k=1}^{n}R_{0}^{+}(q,p_{k})\Psi_{n+1}(q,p_{1},\ldots,p_{n})$
(2.9) $\displaystyle\qquad\qquad\qquad+\int_{-\infty}^{0}\frac{dq}{\lvert
q\rvert}\,\overline{\psi(q)}\,\prod_{k=1}^{n}\overline{R_{0}^{+}(p_{k},q)}\Psi_{n+1}(q,p_{1},\ldots,p_{n})\bigg{\\}}\,,$

from which we read off $\hat{S}_{R}a(\psi)\hat{S}_{R}^{*}=a_{R}(\psi)$ for
$\text{supp}\,\psi\subset\mathbbm{R}_{-}$. For
$\text{supp}\,\psi\subset\mathbbm{R}_{+}$, the remaining integrals in (2.8)
and (2.9) agree up to complex conjugation of $R_{0}^{+}$; this is compensated
by using $\hat{S}_{R}^{*}=\hat{S}_{\overline{R}}$ , i.e. in this case we have
$\hat{S}_{R}^{*}a(\psi)\hat{S}_{R}=a_{R}(\psi)$. ∎

To obtain the equivalence of Borchers triples, recall that the massless field
$\phi_{0}$ decomposes into chiral components $\phi_{0,\pm}$, each depending on
one light ray coordinate $x_{\mp}=x_{0}\mp x_{1}$ only, namely for $f$ which
is the derivative of a function in $\mathscr{S}(\mathbbm{R}^{2})$,

$\displaystyle\phi_{0}(f)$ $\displaystyle=V\big{(}\phi_{0,+}(f_{+})\otimes
1+1\otimes\phi_{0,-}(f_{-})\big{)}V^{*}\,,\qquad\phi_{0,\pm}(f_{\pm})=a^{\dagger}_{\pm}(\tilde{f}_{\pm}|_{\mathbbm{R}_{\pm}})+a_{\pm}(\widetilde{\overline{f_{\pm}}}|_{\mathbbm{R}_{\pm}})\,,$
$\displaystyle f_{\pm}(\mp x_{\mp})$
$\displaystyle=\frac{1}{2\sqrt{2\pi}}\int_{\mathbbm{R}}dx_{\pm}\,f\left(\tfrac{1}{2}(x_{+}+x_{-}),\tfrac{1}{2}(x_{+}-x_{-})\right)\,.$
(2.10)

The algebras in question are generated by these field operators (all of which
are essentially self-adjoint on their respective subspaces of finite particle
number) by

$\displaystyle\mathcal{M}_{0}$
$\displaystyle=\\{e^{i\phi_{0}(f)}\,:\,f\in\mathscr{S}_{\mathbbm{R}}(W)\\}^{\prime\prime}\,,$
(2.11) $\displaystyle\mathcal{M}_{0,\pm}$
$\displaystyle=\\{e^{i\phi_{0,\pm}(f_{\pm})}\,:\,f\in\mathscr{S}_{\mathbbm{R}}(W)\\}^{\prime\prime}=\\{e^{i\phi_{0,\pm}(g)}\,:\,g\in\mathscr{S}_{\mathbbm{R}}(\mathbbm{R}_{+})\\}^{\prime\prime}\,.$
(2.12)

We now come to the main result of this section.

######  Theorem 2.4.

Let $R\in\mathcal{R}$. Then
$(\mathcal{M}_{R,0},U,\Omega)\cong(\mathcal{N}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})$.

###### Proof.

The equivalence $(\mathcal{N}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})\cong(\hat{\mathcal{N}}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})$ was established already, and by
construction of $V$, we have $V(U_{+}(x_{-})\otimes U_{-}(x_{+}))V^{*}=U(x)$
and $V\Omega_{+}\otimes\Omega_{-}=\Omega$. Hence the claim follows once we
have shown $V\hat{\mathcal{N}}_{R}V^{*}=\mathcal{M}_{R,0}$.

By (2.1) and (2.12), $\hat{\mathcal{N}}_{R}$ is generated by the bounded
functions of the field operators $S_{R}^{*}(\phi_{0,+}(f_{+})\otimes 1)S_{R}$
and $S_{R}(1\otimes\phi_{0,-}(f_{-}))S_{R}^{*}$,
$f_{\pm}\in\mathscr{S}(\mathbbm{R}_{+})$. Conjugating with $V$, we have

$\displaystyle VS_{R}^{*}\big{(}\phi_{0,+}(f_{+})\otimes 1\big{)}S_{R}V^{*}$
$\displaystyle=\hat{S}_{R}^{*}\big{(}a^{\dagger}(\tilde{f}_{+}|_{\mathbbm{R}_{+}})+a(\widetilde{\overline{f_{+}}}|_{\mathbbm{R}_{+}})\big{)}\hat{S}_{R}$
$\displaystyle=a^{\dagger}_{R}(\tilde{f}_{+}|_{\mathbbm{R}_{+}})+a_{R}(\widetilde{\overline{f_{+}}}|_{\mathbbm{R}_{+}})\,,$

where in the second step, we have used Proposition 2.3, which also holds for
the creation operators by taking adjoints. Given $f_{+}$ which is the
derivative of a function in $\mathscr{S}(\mathbbm{R}_{+})$, we find $f$ which
is the derivative in $x_{+}$ direction of some function in $\mathscr{S}(W)$
such that $f_{+}$ is recovered from $f$ by (2.10) and $f_{-}=0$ (namely, one
can take the product of $f_{+}$ (which is a function of $x_{-}$) and a
function of $x_{+}$ with integral one). In this situation,
$f^{+}=\tilde{f}_{+}|_{\mathbbm{R}_{+}}$,
$\overline{f^{-}}=\widetilde{\overline{f_{+}}}|_{\mathbbm{R}_{+}}$, and thus

$\displaystyle VS_{R}^{*}\big{(}\phi_{0,+}(f_{+})\otimes
1\big{)}S_{R}V^{*}=\phi_{R,0}(f)\,.$

As all vectors of finite particle number are analytic for these field
operators, this equivalence also holds for their associated unitaries
$e^{i\phi_{R,0}(f)}$ and $e^{i\phi_{0,+}(f_{+})}$, and thus we have the
inclusion $VS_{R}^{*}(\mathcal{M}_{0,+}\otimes
1)S_{R}V^{*}\subset\mathcal{M}_{R,0}$.

Similarly, for the other light ray we obtain

$\displaystyle VS_{R}\big{(}1\otimes\phi_{0,-}(f_{-})\big{)}S_{R}^{*}V^{*}$
$\displaystyle=a^{\dagger}_{R}(\tilde{f}_{-}|_{\mathbbm{R}_{-}})+a_{R}(\widetilde{\overline{f_{-}}}|_{\mathbbm{R}_{-}})=\phi_{R,0}(f)$

for suitably chosen $f\in\mathscr{S}(W)$, and hence
$VS_{R}(1\otimes\mathcal{M}_{0,-})S_{R}^{*}V^{*}\subset\mathcal{M}_{R,0}$.
Thus $V\hat{\mathcal{N}}_{R}V^{*}\subset\mathcal{M}_{R,0}$. As $\Omega$ is
cyclic and separating for both $V\hat{\mathcal{N}}_{R}V^{*}$ and
$\mathcal{M}_{R,0}$, and their modular groups w.r.t. $\Omega$ coincide,
$\Delta_{V\hat{\mathcal{N}}_{R}V^{*}}^{it}=V\Delta_{\otimes}^{it}V^{*}=\Delta_{\mathcal{M}_{R,0}}^{it}$,
the equality of von Neumann algebras
$V\hat{\mathcal{N}}_{R}V^{*}=\mathcal{M}_{R,0}$ follows by Takesaki’s theorem
[Tak03] (see [Tan12b, Theorem A.1] for an explicit application). ∎

Recall that by construction, $\mathcal{N}_{R}$ (1.30) depends on $R$ only via
the symmetric inner function $R^{2}$, i.e.
$\mathcal{N}_{R_{1}}=\mathcal{N}_{R_{2}}$ if $R_{1}^{2}=R_{2}^{2}$. By the
equivalences

$\displaystyle(\mathcal{N}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})\cong(\hat{\mathcal{N}}_{R},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})\cong(\mathcal{M}_{R,0},U,\Omega)\,,$

this also implies
$(\mathcal{M}_{R_{1},0},U,\Omega)\cong(\mathcal{M}_{R_{2},0},U,\Omega)$ if
$R_{1}^{2}=R_{2}^{2}$.

##  3 Structure of massive deformations

The analysis in the previous section resulted in particular in two equivalence
properties of the massless deformed models: On the one hand, the two deformed
Borchers triples
$(\mathcal{M}_{R,0},U,\Omega)\cong(\mathcal{N}_{R,0},U_{+}\otimes
U_{-},\Omega_{+}\otimes\Omega_{-})$ depend only on the symmetric inner
function $\varphi=R^{2}$, i.e. choosing a different root of $\varphi$ results
in an equivalent model. On the other hand, the deformed and undeformed
(chiral) fields are unitarily equivalent. This equivalence however depends on
the light ray, and thus the triples $(\mathcal{M}_{R,0},U,\Omega)$ and
$(\mathcal{M}_{1,0},U,\Omega)$ are not equivalent for general roots
$R\in\mathcal{R}$ (the operator $S_{R^{2}}=S_{\varphi}$ appears as the
S-matrix, an invariant of Borchers triple [Tan12a, DT11]).

In this section, we show that the first property also holds in the massive
case, whereas the second one only holds in a weaker sense which is specified
below.

### Independence of the choice of root

We begin with a preparatory lemma.

######  Lemma 3.1.

Let $m\geq 0$ and $r\in\mathcal{R}$ be a root of the (trivial) symmetric inner
function $\varphi(t)=1$. Then the operator

$\displaystyle[Y_{r}\Psi]_{n}(p_{1},\ldots,p_{n}):=\prod_{1\leq i<j\leq
n}r_{m}(p_{i},p_{j})\cdot\Psi_{n}(p_{1},\ldots,p_{n})$ (3.1)

is a well-defined unitary on $\mathcal{H}$ which commutes with the
representation $U$, leaves $\Omega$ invariant, and satisfies

$\displaystyle Y_{r}\phi_{R,m}(f)Y_{r}^{*}\Psi=\phi_{r\cdot
R,m}(f)\Psi\,,\qquad
f\in\mathscr{S}(\mathbbm{R}^{2})\,,\;R\in\mathcal{R}\,,\;\Psi\in\mathcal{D}\,.$
(3.2)

###### Proof.

As $r$ is a root of $1$, it takes only the values $\pm 1$ and is in particular
real. Hence
$r_{m}(p_{j},p_{i})=\overline{r_{m}(p_{i},p_{j})}=r_{m}(p_{i},p_{j})$ is
symmetric and thus the product in (3.1) preserves the totally symmetric
subspace of $L^{2}(\mathbbm{R}^{n})$, and $Y_{r}$ defines a unitary on the
Bose Fock space $\mathcal{H}$.

It is clear that $Y_{r}$ commutes with translations and leaves $\Omega$
invariant. To establish (3.2), we first calculate for an annihilation operator
$a_{R}(\psi)$, $\psi\in\mathcal{H}_{1}$,

$\displaystyle[Y_{r}$ $\displaystyle
a_{R}(\psi)Y_{r}^{*}\Psi]_{n}(p_{1},\ldots,p_{n})$
$\displaystyle=\sqrt{n+1}\prod_{i<j}r_{m}(p_{i},p_{j})\int\frac{dq}{\omega_{m}(q)}\,\overline{\psi(q)}\,\prod_{k=1}^{n}R_{m}(q,p_{k})\cdot[Y_{r}^{*}\Psi]_{n+1}(q,p_{1},\ldots,p_{n})$
$\displaystyle=\sqrt{n+1}\prod_{i<j}|r_{m}(p_{i},p_{j})|^{2}\int\frac{dq}{\omega_{m}(q)}\,\overline{\psi(q)}\,\prod_{k=1}^{n}\Big{(}r_{m}(q,p_{k})R_{m}(q,p_{k})\Big{)}\Psi_{n+1}(q,p_{1},\ldots,p_{n})$
$\displaystyle=[a_{r\cdot R}(\psi)\Psi]_{n}(p_{1},\ldots,p_{n})\,.$

Thus $Y_{r}a_{R}(\psi)Y_{r}^{*}\Psi=a_{r\cdot R}(\psi)\Psi$, and by taking
adjoints, we also find
$Y_{r}a^{\dagger}_{R}(\psi)Y_{r}^{*}\Psi=a^{\dagger}_{r\cdot R}(\psi)\Psi$. As
$\phi_{R,m}(f)=a^{\dagger}_{R}(f^{+})+a_{R}(\overline{f^{-}})$, the claimed
equivalence (3.2) follows. ∎

With this lemma, it is now easy to show that the Borchers triple
$(\mathcal{M}_{R,m},U,\Omega)$ is independent of the choice of root up to
equivalence.

######  Proposition 3.2.

Let $R_{1},R_{2}\in\mathcal{R}$ be roots of the same symmetric inner function
$R_{1}^{2}=R_{2}^{2}$. Then
$(\mathcal{M}_{R_{1},m},U,\Omega)\cong(\mathcal{M}_{R_{2},m},U,\Omega)$,
$m\geq 0$.

###### Proof.

As $R_{1}^{2}=R_{2}^{2}$, the function $r(t):=R_{1}(t)R_{2}(t)^{-1}$ is a root
of $1$ as in Lemma 3.1, i.e. we have
$Y_{r}\phi_{R_{2},m}(f)Y_{r}^{*}=\phi_{R_{1},m}(f)$ (3.2) for any
$f\in\mathscr{S}(\mathbbm{R}^{2})$. But these field operators have the dense
subspace $\mathcal{D}$ of vectors of finite particle number as entire analytic
vectors [Lec12], and $Y_{r}\mathcal{D}=\mathcal{D}$. Hence the equivalence
(3.2) lifts to the unitaries $e^{i\phi_{R_{k},m}(f)}$, $k=1,2$,
$f\in\mathscr{S}_{\mathbbm{R}}(\mathbbm{R}^{2})$, and the von Neumann algebras
they generate, $Y_{r}\mathcal{M}_{R_{1},m}Y_{r}^{*}=\mathcal{M}_{R_{2},m}$.
Since $Y_{r}$ also commutes with $U$ and leaves $\Omega$ invariant, the
claimed equivalence of Borchers triples follows. ∎

As mentioned in Section 1, this result states that within the class of
Borchers triples considered here, the inverse scattering problem for the two-
particle S-matrix $R^{2}$ has a unique solution up to unitary equivalence. For
massless asymptotically complete nets, this uniqueness is known in the
stronger form that the wave S-matrix and the free net give an explicit formula
to construct the deformed Borchers triple [Tan12a]. It is an interesting open
problem to find its massive counterpart.

### Equivalence at fixed momentum

We now come to the discussion of equivalences between deformed and undeformed
field operators. In the massless case, this equivalence can be expressed as,
$R\in\mathcal{R}$,

$\displaystyle a_{R}(\xi)$
$\displaystyle=\begin{cases}\hat{S}_{R}^{*}a(\xi)\hat{S}_{R}&\text{supp}\,\xi\subset\mathbbm{R}_{+}\\\
\hat{S}_{R}a(\xi)\hat{S}_{R}^{*}&\text{supp}\,\xi\subset\mathbbm{R}_{-}\end{cases}\,,\qquad
m=0\,.$

For $m>0$, the Lorentz group acts transitively on the upper mass shell, so
that there is no invariant distinction between its left and right branch.
However, we still have an equivalence of the above form at sharp momentum.
Recall that for $p\in\mathbbm{R}$, the annihilator $a(p)$ is a well-defined
unbounded operator on the dense domain $\mathcal{D}_{0}\subset\mathcal{D}$ of
vectors $\Psi\in\mathcal{D}$ of finite particle number with continuous wave
functions $\Psi_{n}\in C(\mathbbm{R}^{n})$, $n\in\mathbbm{N}$.

To implement this equivalence, we define an operator-valued function
$\mathbbm{R}\ni p\mapsto\hat{S}_{R,m}(p)\in\mathcal{U}(\mathcal{H})$ by

$\displaystyle[\hat{S}_{R,m}(p)\Psi]_{n}(p_{1},\ldots,p_{n}):=\prod_{1\leq
i<j\leq n}R((p_{i}+p_{j})\wedge_{m}p)\cdot\Psi_{n}(p_{1},\ldots,p_{n})\,,$
(3.3)

where $p\wedge_{m}q:=\frac{1}{2}(\omega_{m}(q)p-\omega_{m}(p)q)$. Note that in
case the root $R$ is continuous, one has
$\hat{S}_{R,m}(p)\mathcal{D}_{0}=\mathcal{D}_{0}$.

Using $\overline{R(t)}=R(t)^{-1}=R(-t)$, the definition of $R_{m}$ (1.7), and
$(p+q)\wedge_{m}p=q\wedge_{m}p$, we then get

$\displaystyle[$
$\displaystyle\hat{S}_{R,m}(p)a(p)\hat{S}_{R,m}(p)^{*}\Psi]_{n}(p_{1},..,p_{n})$
$\displaystyle=\sqrt{n+1}\,\prod_{i<j}^{n}R((p_{i}+p_{j})\\!\wedge_{m}p)\cdot[\hat{S}_{R,m}(p)^{*}\Psi]_{n+1}(p,p_{1},..,p_{n})$
$\displaystyle=\sqrt{n+1}\,\prod_{i<j}^{n}\Big{(}R((p_{i}+p_{j})\wedge_{m}p)\overline{R((p_{i}+p_{j})\wedge_{m}p)}\Big{)}\cdot\prod_{k=1}^{n}\overline{R((p+p_{k})\wedge_{m}p)}\cdot\Psi_{n+1}(p,p_{1},...,p_{n})$
$\displaystyle=\sqrt{n+1}\,\prod_{k=1}^{n}\overline{R(p_{k}\wedge_{m}p)}\cdot\Psi_{n+1}(p,p_{1},...,p_{n})$
$\displaystyle=\sqrt{n+1}\,\prod_{k=1}^{n}R_{m}(p,p_{k})\cdot\Psi_{n+1}(p,p_{1},...,p_{n})$
$\displaystyle=[a_{R}(p)\Psi]_{n}(p_{1},...,p_{n})\,,$

where the last equality follows from comparison with (1.13). We thus have on
$\mathcal{D}_{0}$

$\displaystyle\hat{S}_{R,m}(p)a(p)\hat{S}_{R,m}(p)^{*}=a_{R,m}(p)\,.$ (3.4)

It should be noted that there is actually a big freedom in the choice of
$\hat{S}_{R,m}$ with this property, as it is only the adjoint action of
$\hat{S}_{R,m}(p)$ on $a(p)$ with the same momentum $p$ that matters in the
end. One manifestation of this freedom is the fact that $\hat{S}_{R,0}(p)$ for
$p>0$ does not agree with $\hat{S}_{R}^{*}$ (2.5), whereas their adjoint
action on $a(p)$ does. For $m>0$, another implementation of the equivalence is

$(\hat{S}_{R,m}(p)\Psi)_{n}(p_{1},\ldots p_{n})=\prod_{i<j}\overline{R({\rm
sgn}(\max(p_{j},p_{i})-p)|p_{i}\wedge_{m}p_{j}|)}\cdot\Psi_{n}(p_{1},\ldots,p_{n})\,,$

where the sign function sgn is defined with sgn$(0):=-1$. This can be checked
by a computation analogous to the previous one. Observe that if $p$ is
sufficiently large, this coincides with the root of the two-particle S-matrix
[Lec12], and for $p$ sufficiently small with its inverse. Hence this latter
implementation is analogous to the massless case, where the deformation is
given exactly by the S-matrix $\hat{S}_{R}$ and its adjoint.

By (formally) taking adjoints one gets the same relation between
$a^{\dagger}(p)$ and $a^{\dagger}_{R,m}(p)$. However, even when making this
adjoint rigorous (e.g. in the sense of quadratic forms) one cannot expect to
get an equivalence of the Fourier transform of the deformed and undeformed
field at sharp $p$. One has to keep in mind that the splitting into chiral
components is _not_ a splitting of the field according to momentum transfer
but related to a split of the one-particle Hilbert space into positive and
negative momentum parts. Thus creation and annihilation operators appear
either both with positive or both with negative momentum, so both are
transformed with $\hat{S}_{R,m}$ or $\hat{S}_{R,m}^{*}$. For the massive case
this mechanism is not available and therefore the relations between deformed
and undeformed creation and annihilation operators will not yield a
corresponding relation between the fields.

To conclude, the structure of the wedge algebra is deformed in a very
transparent manner in the chiral situation (1.30), but not for $m>0$, where
one has to rely on the use of generating fields. This observation is to some
extent in parallel with the simpler structure of the wave S-matrix in the
chiral case in comparison to the many particle S-matrix in the massive case,
and deserves further investigation.

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