Quasi-free states on a class of algebras of multicomponent commutation
relations

Eugene Lytvynov
Department of Mathematics, Swansea University, Bay Campus, Swansea, SA1 8EN,
UK;
e-mail<EMAIL_ADDRESS>
Nedal Othman
Dubai Women’s College, Higher Colleges of Technology, Baghdad Street, Al
Nahda, Dubai, P.O. Box: 16062, The United Arab Emirates;
e-mail<EMAIL_ADDRESS>
Abstract

Multicomponent commutations relations (MCR) describe plektons, i.e.,
multicomponent quantum systems with a generalized statistics. In such systems,
exchange of quasiparticles is governed by a unitary matrix $Q(x_{1},x_{2})$
that depends on the position of quasiparticles. For such an exchange to be
possible, the matrix must satisfy several conditions, including the functional
Yang–Baxter equation. The aim of the paper is to give an appropriate
definition of a quasi-free state on an MCR algebra, and construct such states
on a class of MCR algebras. We observe a significant difference between the
classical setting for bosons and fermions and the setting of MCR algebras. We
show that the developed theory is applicable to systems that contain
quasiparticles of opposite type. An example of such a system is a two-
component system in which two quasiparticles, under exchange, change their
respective types to the opposite ones ($1\mapsto 2$, $2\mapsto 1$). Fusion of
quasiparticles means intuitively putting several quasiparticles in an
infinitely small box and identifying the statistical behaviour of the box. By
carrying out fusion of an odd number of particles from the two-component
system as described above, we obtain further examples of quantum systems to
which the developed theory is applicable.

Keywords: Anyon; Plekton; Fusion of quasiparticles; Quasi-free state

Mathematics Subject Classification (2020): Primary 81R10, 81V27; Secondary
47L90.

##  1 Introduction and preliminaries

Multicomponent commutations relations (MCR) describe plektons, i.e.,
multicomponent quantum systems with a generalized statistics. In such systems,
exchange of quasiparticles is governed by a unitary matrix that depends on the
position of quasiparticles. For such an exchange to be possible, the matrix
must satisfy several conditions, including the functional Yang–Baxter
equation.

The aim of the paper is to give an appropriate definition of a quasi-free
state on an MCR algebra and construct such states on a class of MCR algebras.
We observe a significant difference between the classical setting for bosons
and fermions and the setting of MCR algebras.

###  1.1 Multicomponent commutation relations

The first paper pointing out the possibility of a multicomponent quantum
system was the comment by Menikoff, Sharp and Goldin [10]. Such systems were
rigorously derived and studied by Liguori and Mintchev [13], see also Goldin
and Majid [11]. The paper [8] gave an overview of multicomponent quantum
systems with concrete examples when the number of components of a quantum
system is two. That paper actually treated a more general setting than the one
considered in [11, 13]. For other results related to multicomponent quantum
systems, see e.g. Bożejko, Speicher [6] and Jørgensen, Schmitt, Werner [12].

Let us briefly recall the definition of the MCR, for more detail see e.g. [8,
11, 13]. Let $X:=\mathbb{R}^{d}$ with $d\geq 2$ (the physically important case
being $d=2$) and let $V:=\mathbb{C}^{r}$, $r$ being the number of components
(types of quisiparticles) in the quantum system. Let $\\{e_{1},\dots,e_{r}\\}$
be the standard orthonormal basis in $V$ (the $i$th coordinate of $e_{i}$
being 1, the other coordinates being 0). Let $J$ be the complex conjugation in
$V$, i.e., the antilinear operator satisfying $Je_{i}=e_{i}$ for all $i$. Let
$\mathcal{L}(V^{\otimes 2})$ denote the space of all linear operators
(matrices) in $V^{\otimes 2}$. We fix a map
$Q:\mathbb{R}^{2}\to\mathcal{L}(V^{\otimes 2})$ such that, for each
$(y_{1},y_{2})\in\mathbb{R}^{2}$, $Q(y_{1},y_{2})$ is a unitary operator,
$Q^{*}(y_{1},y_{2})=Q(y_{2},y_{1})$, and the functional Yang–Baxter equation
is satisfied in $V^{\otimes 3}$:

$Q_{1}(y_{1},y_{2})Q_{2}(y_{1},y_{3})Q_{1}(y_{2},y_{3})=Q_{2}(y_{2},y_{3})Q_{1}(y_{1},y_{3})Q_{2}(y_{1},y_{2})$
(1)

for $(y_{1},y_{2},y_{3})\in\mathbb{R}^{3}$. In (1) and below, for a linear
operator $\mathcal{C}$ acting in $V^{\otimes 2}$ and $k\geq 3$, we denote by
$\mathcal{C}_{i}$ the linear operator in $V^{\otimes k}$ acting by
$\mathcal{C}$ on the $i$th and $(i+1)$th components of $V^{\otimes k}$. We
also define, with an abuse of notation, for
$x_{i}=(x_{i}^{1},\dots,x_{i}^{d})\in X$ ($i=1,2$),
$Q(x_{1},x_{2}):=Q(x_{1}^{1},x_{2}^{1})$.

Let $\mathcal{H}:=L^{2}(X;V)$ be the $L^{2}$-space (with respect to the
Lebesgue measure on $X$) of $V$-valued functions on $X$. We keep the notation
$J$ for the complex conjugation in $\mathcal{H}$. Let $A^{+}(f)$, $A^{-}(f)$
($f\in\mathcal{H}$) be linear operators acting in a separable Hilbert space
$\mathfrak{F}$, with a dense domain $\mathfrak{D}\subset\mathfrak{F}$. We
assume that the operators $A^{+}(f)$, $A^{-}(f)$ depend linearly on $f$,
$A^{+}(f)^{*}\restriction_{\mathfrak{D}}=A^{-}(Jf)$, and the operators
$A^{+}(f)$, $A^{-}(f)$ map $\mathfrak{D}$ into itself. We introduce operator-
valued distributions

$A^{+}(x)=(A_{1}^{+}(x),\dots,A^{+}_{r}(x)),\quad
A^{-}(x)=(A_{1}^{-}(x),\dots,A^{-}_{r}(x))\quad(x\in X)$ (2)

that satisfy, for all $f(x)=\sum_{k=1}^{r}f^{k}(x)e_{k}\in\mathcal{H}$,

$A^{\sharp}(f)=\sum_{k=1}^{r}\int_{X}f^{k}(x)A_{k}^{\sharp}(x)\,dx,\quad\sharp\in\\{+,-\\}.$
(3)

For $u,v\in V$, denote $\langle
u,v\rangle_{V}:=(u,Jv)_{V}=\sum_{k=1}^{r}u^{k}v^{k}$. Then we may write
formula (3) in the heuristic form $A^{\sharp}(f)=\int_{X}\langle
f(x),A^{\sharp}(x)\rangle_{V}\,dx$. Analogously, we write, for a product of
two operators,

$\displaystyle
A^{\sharp_{1}}(f_{1})A^{\sharp_{2}}(f_{2})=\int_{X^{2}}\big{\langle}f_{1}(x_{1})\otimes
f_{2}(x_{2}),A^{\sharp_{1}}(x_{1})\otimes
A^{\sharp_{2}}(x_{2})\big{\rangle}_{V^{\otimes
2}}\,dx_{1}\,dx_{2},\quad\sharp_{1},\sharp_{2}\in\\{+,-\\}.$

We say that the operators $A^{+}(f)$, $A^{-}(f)$ satisfy the $Q$-MCR if

$\displaystyle\int_{X^{2}}\big{\langle}f_{1}(x_{1})\otimes
f_{2}(x_{2}),A^{+}(x_{1})\otimes A^{+}(x_{2})\big{\rangle}_{V^{\otimes
2}}\,dx_{1}\,dx_{2}$
$\displaystyle\quad=\int_{X^{2}}\big{\langle}Q(x_{2},x_{1})f_{1}(x_{1})\otimes
f_{2}(x_{2}),A^{+}(x_{2})\otimes A^{+}(x_{1})\big{\rangle}_{V^{\otimes
2}}\,dx_{1}\,dx_{2},$ (4)
$\displaystyle\int_{X^{2}}\big{\langle}f_{1}(x_{1})\otimes
f_{2}(x_{2}),A^{-}(x_{1})\otimes A^{-}(x_{2})\big{\rangle}_{V^{\otimes
2}}\,dx_{1}\,dx_{2}$
$\displaystyle\quad=\int_{X^{2}}\big{\langle}\widehat{Q}(x_{2},x_{1})f_{1}(x_{1})\otimes
f_{2}(x_{2}),A^{-}(x_{2})\otimes A^{-}(x_{1})\big{\rangle}_{V^{\otimes
2}}\,dx_{1}\,dx_{2},$ (5)
$\displaystyle\int_{X^{2}}\big{\langle}f_{1}(x_{1})\otimes
f_{2}(x_{2}),A^{-}(x_{1})\otimes A^{+}(x_{2})\big{\rangle}_{V^{\otimes
2}}\,dx_{1}\,dx_{2}=\int_{X}\langle f_{1}(x),f_{2}(x)\rangle_{V}\,dx$
$\displaystyle\qquad+\int_{X^{2}}\big{\langle}\widetilde{Q}(x_{1},x_{2})f_{1}(x_{1})\otimes
f_{2}(x_{2}),A^{+}(x_{2})\otimes A^{-}(x_{1})\big{\rangle}_{V^{\otimes
2}}\,dx_{1}\,dx_{2}.$ (6)

In these formulas,
$\widehat{Q}(x_{1},x_{2}):=\mathbb{S}^{(2)}\,Q(x_{2},x_{1})\,\mathbb{S}^{(2)}$,
where the antilinear operator $\mathbb{S}^{(2)}$ in $V^{\otimes 2}$ is given
by $\mathbb{S}^{(2)}(u\otimes v):=(Jv)\otimes(Ju)$, and
$\widetilde{Q}(x_{1},x_{2})$ satisfies

$\big{\langle}\widetilde{Q}(x_{1},x_{2})e_{i}\otimes e_{j},e_{k}\otimes
e_{l}\big{\rangle}_{V^{\otimes 2}}=\big{\langle}Q(x_{1},x_{2})e_{k}\otimes
e_{i},e_{l}\otimes e_{j}\big{\rangle}_{V^{\otimes 2}}\quad\forall
i,j,k,l\in\\{1,\dots,r\\}.$ (7)

Note that the double integrals appearing on the right hand side of formulas
(4)–(6) are assumed to be well-defined. Below, in Section 4, we will recall
the Fock representation of the $Q$-MCR.

Note that the $Q$-MCR (4)–(6) can be written in the following shorthand form:

$\displaystyle A^{+}(x_{1})\otimes A^{+}(x_{2})$
$\displaystyle=Q(x_{2},x_{1})^{T}A^{+}(x_{2})\otimes A^{+}(x_{1}),$
$\displaystyle A^{-}(x_{1})\otimes A^{-}(x_{2})$
$\displaystyle=\widehat{Q}(x_{2},x_{1})^{T}A^{-}(x_{2})\otimes A^{-}(x_{1}),$
$\displaystyle A^{-}(x_{1})\otimes A^{+}(x_{2})$
$\displaystyle=\delta(x_{1}-x_{2})\operatorname{Tr}(\cdot)+\widetilde{Q}(x_{1},x_{2})^{T}A^{+}(x_{2})\otimes
A^{-}(x_{1}).$ (8)

Here, for $M\in\mathcal{L}(V^{\otimes 2})$, $M^{T}$ is the transposed of $M$;
for $v^{(2)}\in V^{\otimes 2}$,
$\operatorname{Tr}(v^{(2)}):=\sum_{k=1}^{r}(v^{(2)},e_{k}\otimes
e_{k})_{V^{\otimes 2}}$ is the trace111Note that, in the usual way,
$V^{\otimes 2}$ can be identified with $\mathcal{L}(V)$, and then
$\operatorname{Tr}(v^{(2)})$ becomes the trace of the linear operator
$v^{(2)}$. of $v^{(2)}$; and for $f_{1},f_{2}\in L^{2}(X;\mathbb{C})$,

$\int_{X^{2}}f_{1}(x_{1})f_{2}(x_{2})\delta(x_{1}-x_{2})dx_{1}\,dx_{2}:=\int_{X}f_{1}(x)f_{2}(x)dx.$

When $r=1$ (i.e., the quantum system has a single component), $Q(x_{1},x_{2})$
is just a complex-valued function satisfying $|Q(x_{1},x_{2})|=1$ and
$\overline{Q(x_{1},x_{2})}=Q(x_{2},x_{1})$ while equation (1) is trivially
satisfied. Furthermore, $\widehat{Q}(x_{1},x_{2})=Q(x_{1},x_{2})$ and
$\widetilde{Q}(x_{1},x_{2})=Q(x_{1},x_{2})$. Hence, formulas (8) become

$\displaystyle
A^{+}(x_{1})A^{+}(x_{2})=Q(x_{2},x_{1})A^{+}(x_{2})A^{+}(x_{1}),\quad
A^{-}(x_{1})A^{-}(x_{2})=Q(x_{2},x_{1})A^{-}(x_{2})A^{-}(x_{1}),$
$\displaystyle
A^{-}(x_{1})A^{+}(x_{2})=\delta(x_{1}-x_{2})+Q(x_{1},x_{2})A^{+}(x_{2})A^{-}(x_{1}).$
(9)

We will call formulas (9) the $Q$-anyon commutation relations ($Q$-ACR), see
e.g. [13, 5] and the references in [8]. For $Q(x_{1},x_{2})\equiv 1$, formulas
(9) become the canonical commutation relations (CCR), and for
$Q(x_{1},x_{2})\equiv-1$ they become the canonical anticommutation relations
(CAR).

###  1.2 Quasi-free states

In the theory of the CCR and CAR algebras, quasi-free states play a
fundamental role. We refer the reader to e.g. [1, 2], Chapter 5 in [7], or
Chapter 17 in [9].

Let $\mathbb{A}$ be the unital $*$-algebra generated by the operators
$A^{+}(f)$, $A^{-}(f)$ ($f\in\mathcal{H}$), satisfying either the CCR or CAR.
Define, for $f\in\mathcal{H}$, the field (or Segal-type) operators
$B(f)=A^{+}(f)+A^{-}(Jf)$. Since
$A^{+}(f)=\frac{1}{2}\big{(}B(f)-iB(if)\big{)}$ and
$A^{-}(f)=\frac{1}{2}\big{(}B(Jf)+iB(iJf)\big{)}$, the algebra $\mathbb{A}$ is
generated by $B(f)$ ($f\in\mathcal{H}$). Let $\tau$ be a state on
$\mathbb{A}$. Then $\tau$ is completely determined by the moments of $B(f)$
under $\tau$. Intuitively, the state $\tau$ being quasi-free means that the
moments of $B(f)$ can be calculated similarly to the case where $\tau$ is the
vacuum state on the Fock representation of the CCR or CAR, respectively. For
example, in the case of the CAR, this means that the odd moments are equal to
zero and

$\tau\big{(}B(f_{1})B(f_{2})\dotsm
B(f_{2n})\big{)}=\sum_{\xi}(-1)^{\operatorname{cross}(\xi)}\prod_{\begin{subarray}{c}\\{i,j\\}\in\xi\\\
i<j\end{subarray}}\tau\big{(}B(f_{i})B(f_{j})\big{)}.$ (10)

Here, the summation is over all partitions $\xi$ of the set
$\\{1,2,\dots,2n\\}$ into $n$ two-point sets and $\operatorname{cross}(\xi)$
denotes the number of crossings in $\xi$, i.e., the number of all choices of
$\\{i,j\\},\,\\{k,l\\}\in\xi$ with $i<k<j<l$.

A state $\tau$ is called gauge-invariant if, for any $q\in\mathbb{C}$,
$|q|=1$, $\tau$ remains invariant under the transformation $A^{+}(f)\mapsto
qA^{+}(f)$, $A^{-}(f)\mapsto\overline{q}A^{-}(f)$, equivalently

$\tau\big{(}B(qf_{1})\dotsm B(qf_{n})\big{)}=\tau\big{(}B(f_{1})\dotsm
B(f_{n})\big{)}\quad\forall n.$

Due to the commutation relations, any state $\tau$ is completely characterized
by the $n$-point functions,

$\mathbf{S}^{(m,n)}(f_{1},\dots,f_{m},g_{1},\dots,g_{n})=\tau\big{(}A^{+}(f_{1})\dotsm
A^{+}(f_{m})A^{-}(g_{1})\dotsm A^{-}(g_{n})\big{)}.$ (11)

As easily seen, a state $\tau$ is gauge-invariant if and only if
$\mathbf{S}^{(m,n)}=0$ if $m\neq n$. In fact, a state $\tau$ is gauge-
invariant quasi-free if and only if $\mathbf{S}^{(m,n)}=0$ if $m\neq n$ and

$\mathbf{S}^{(n,n)}(f_{n},\dots,f_{1},g_{1},\dots,g_{n})=\operatorname{per}\left[\mathbf{S}^{(1,1)}(f_{i},g_{j})\right]_{i,j=1,\dots,n}=\sum_{\pi\in
S_{n}}\prod_{i=1}^{n}\mathbf{S}^{(1,1)}(f_{i},g_{\pi(i)})$ (12)

for the CCR algebra, and

$\mathbf{S}^{(n,n)}(f_{n},\dots,f_{1},g_{1},\dots,g_{n})=\operatorname{det}\left[\mathbf{S}^{(1,1)}(f_{i},g_{j})\right]_{i,j=1,\dots,n}=\sum_{\pi\in
S_{n}}\text{sgn}(\pi)\prod_{i=1}^{n}\mathbf{S}^{(1,1)}(f_{i},g_{\pi(i)})$ (13)

for the CAR algebra.

Araki, Woods [3] and Araki, Wyss [4] presented an explicit construction of the
gauge-invariant quasi-free states on the CCR algebra and the CAR algebra,
respectively. This has been done by constructing a non-Fock representation of
the commutation relations in the symmetric, respectively antisymmetric Fock
space over $\mathcal{H}\oplus\mathcal{H}$ (doubling of the underlying space)
and applying the vacuum state to this representation.

In [14], gauge-invariant quasi-free states on the algebra of the anyon
commutation relations (ACR) were defined and constructed. The definition of
such a state was based on an extension of formulas (12), (13) to the ACR case.
In particular, a counterpart of the sign of the permutation $\pi$ in formula
(13) becomes the function

$Q_{\pi}(x_{1},\dots,x_{n})=\prod_{\begin{subarray}{c}1\leq i<j\leq n\\\
\pi(i)>\pi(j)\end{subarray}}Q(x_{i},x_{j}).$ (14)

Note that, for $Q(\cdot,\cdot)\equiv-1$, one indeed has
$Q_{\pi}(x_{1},\dots,x_{n})\equiv\text{sgn}(\pi)$. A gauge-invariant quasi-
free state on the ACR algebra was constructed as the vacuum state on a
representation of the ACR in the Fock space of $\mathbf{Q}$-symmetric
functions with the underlying space $\mathbf{X}:=X_{1}\sqcup X_{2}$. Here
$X_{1}$ and $X_{2}$ are two copies of the space $X$ and the function
$\mathbf{Q}$ is defined as follows: $\mathbf{Q}(x_{1},x_{2}):=Q(x_{1},x_{2})$
if both $x_{1}$ and $x_{1}$ are from the same part, $X_{i}$ ($i=1,2$), and
$\mathbf{Q}(x_{1},x_{2}):=Q(x_{2},x_{1})$ if $x_{1}$ and $x_{2}$ are from
different parts, $X_{i}$ and $X_{j}$, respectively, with $i\neq j$. In the
case of a CCR or CAR algebra this construction of gauge-invariant quasi-free
states reduces to the construction of Araki–Woods or Araki–Wyss, respectively.

###  1.3 A brief description of the results

In this paper, we consider a multicomponent system with a continuous operator-
valued function $Q(x_{1},x_{2})$. We define a $Q$-MCR algebra, which contains
multiple integrals as those appearing on the right-hand side of formulas
(4)–(6).

For the vacuum state defined on the Fock representation of the $Q$-MCR, we
calculate the moments of the field operators. Similarly to (10), this allows
us to come up with formulas which may potentially be required for a state on
the $Q$-MCR algebra to be quasi-free. However, it appears that even the gauge-
invariant quasi-free states on the $Q$-ACR algebra constructed in [14] do not
satisfy these conditions. The reason for this is the very definition of the
function $\mathbf{Q}(x_{1},x_{2})$. Indeed, when one evaluates the $n$-point
functions $\mathbf{S}^{(n,n)}$ for such a system, only the terms with all
points from $X_{1}$ do not vanish. But for such points, $x_{1},x_{2}\in
X_{1}$, we have $\mathbf{Q}(x_{1},x_{2})=Q(x_{1},x_{2})$. Hence, one comes up
with an extension of formulas (12), (13) which contains the function $Q_{\pi}$
given by (14) in place of the sign of the permutation $\pi$ in the case of the
CAR. However, when one evaluates $\tau\big{(}B(f_{1})\dotsm B(f_{2n})\big{)}$,
one has to deal with terms containing points from both parts $X_{1}$ and
$X_{2}$. Because of this, one is unable to come up with the function
$Q_{\pi}(x_{1},\dots,x_{n})$, unless $Q(x_{1},x_{2})=Q(x_{2},x_{1})$. But the
latter formula means that $Q(x_{1},x_{2})$ is either identically equal to $1$
(CCR), or identically equal to $-1$ (CAR).

Thus, instead of using the term a quasi-free state on the $Q$-MCR algebra, we
use the term a strongly quasi-free state on the $Q$-MCR algebra if a
counterpart of formula (10) holds for a state $\tau$ on the $Q$-MCR algebra.
As a result, the gauge-invariant quasi-free states constructed on the $Q$-ACR
algebra in [14] are not strongly quasi-free. If $\mathbf{S}^{(m,n)}=0$ for
$m\neq n$ and a counterpart of formulas (12), (13) holds for a state $\tau$ on
the $Q$-MCR algebra, then we call $\tau$ gauge-invariant quasi-free.

Our next aim is to construct gauge-invariant quasi-free states on the $Q$-MCR
algebra. One hopes that the construction from [14] of the gauge-invariant
quasi-free states on the $Q$-ACR algebra (which in turn includes the
Araki–Woods and Araki–Wyss constructions as special cases) can be extended to
the case of a multicomponent system. We show that this indeed can be achieved,
however under additional assumptions on the operator-valued function $Q$. It
should be stressed that these assumptions are essentially necessary for the
$Q$-MCR to hold for this construction.

Furthermore, if it additionally holds that
$\widetilde{Q}(x_{1},x_{2})=Q(x_{2},x_{1})$, then the corresponding gauge-
invariant quasi-free states are also strongly quasi-free.

We show that the developed theory is applicable to systems that contain
quasiparticles of opposite type. An example of such a system is a two-
component system in which two quasiparticles, under exchange, change their
respective types to the opposite ones ($1\mapsto 2$, $2\mapsto 1$). More
precisely, let $q_{1}$ and $q_{2}$ be complex-valued continuous functions on
$\mathbb{R}^{2}$ satisfying
$\overline{q_{i}(y_{1},y_{2})}=q_{i}(y_{2},y_{1})$, $|q_{i}(y_{1},y_{2})|=1$,
and let the permutation $\theta\in S_{2}$ be given by $\theta(1)=2$,
$\theta(2)=1$. Then the $Q$-MCR in this case are of the following form:

$\displaystyle A_{i}^{+}(x_{1})A_{i}^{+}(x_{2})$
$\displaystyle=q_{1}(y_{2},y_{1})A_{\theta(i)}^{+}(x_{2})A_{\theta(i)}^{+}(x_{1}),$
$\displaystyle A_{i}^{+}(x_{1})A_{j}^{+}(x_{2})$
$\displaystyle=q_{2}(y_{2},y_{1})A_{i}^{+}(x_{2})A_{j}^{+}(x_{1}),\quad i\neq
j,$ $\displaystyle A_{i}^{-}(x_{1})A_{i}^{-}(x_{2})$
$\displaystyle=q_{1}(y_{2},y_{1})A_{\theta(i)}^{-}(x_{2})A_{\theta(i)}^{-}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{j}^{-}(x_{2})$
$\displaystyle=q_{2}(y_{2},y_{1})A_{i}^{-}(x_{2})A_{j}^{-}(x_{1}),\quad i\neq
j,$ $\displaystyle A_{i}^{-}(x_{1})A_{i}^{+}(x_{2})$
$\displaystyle=\delta(x_{1}-x_{2})+q_{2}(y_{1},y_{2})A_{\theta(i)}^{+}(x_{2})A_{\theta(i)}^{-}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{j}^{+}(x_{2})$
$\displaystyle=q_{1}(y_{1},y_{2})A_{i}^{+}(y_{2})A_{j}^{-}(x_{1}),\quad i\neq
j,$ (15)

for $i,j\in\\{1,2\\}$. For the corresponding $Q$-MCR algebra $\mathbb{A}$, our
theory gives a class of gauge-invariant quasi-free states $\tau$ on
$\mathbb{A}$.

If we additionally assume that $q_{1}(y_{1},y_{2})=q_{2}(y_{2},y_{1})$, then
the condition $\widetilde{Q}(x_{1},x_{2})=Q(x_{2},x_{1})$ is satisfied. Hence,
in this case, each gauge-invariant quasi-free state $\tau$ constructed in the
paper is strongly quasi-free, a property that the gauge-invariant quasi-free
states on the ACR algebra do not possess.

Fusion of (non-abelian) anyons plays a central role in topological quantum
computation, see e.g. [15, Chapter 4] or [17, Section 12.1]. Intuitively,
fusion means putting several anyons in an infinitely small box and identifying
the statistical behaviour of the box. Let $k\geq 3$ be an odd number. We show
that fusion of $k$ quasiparticles that are described by the commutation
relations (15) leads to a new, nontrivial exchange function $Q(x_{1},x_{2})$
to which our theory of quasi-free states is applicable. Note that, in this
case, we have $V=(\mathbb{C}^{2})^{\otimes k}$, i.e., the quantum system has
$2^{k}$ components.

The paper is organized as follows. In Section 2, we define the $Q$-MCR algebra
and discuss some of its basic properties. In particular, we show that the
$Q$-MCR algebra allows Wick (normal) ordering. In Section 3, we briefly
discuss general states on the $Q$-MCR algebra. In Section 4, we construct the
Fock state on the $Q$-MCR algebra. The main result of this section, Theorem
4.2, gives an explicit formula for the moments of sums of creation and
annihilations operators with respect to the Fock state.

The main results of the paper are in Section 5 and 6. In Section 5, we use
Theorem 4.2 to define gauge-invariant quasi-free states and strongly quasi-
free states on the $Q$-MCR algebra. Under several assumptions on the exchange
(operator-valued) function $Q$, we explicitly construct such quasi-free states
(Theorem 5.7). In Section 6, we present examples of the function $Q$ that
satisfies the conditions of Theorem 5.7. In particular, Theorem 6.4 states the
form of a function $Q$ that satisfies the conditions of Theorem 5.7 and
describes a quantum system with quasiparticles of opposite type.

Finally, we remark that, in the case where the condition
$\widetilde{Q}(x_{1},x_{2})=Q(x_{2},x_{1})$ is satisfied and hence strongly
quasi-free states exist, one may hope to construct strongly quasi-free states
that are not gauge-invariant.

##  2 The $Q$-MCR algebra

Let $Q:X^{2}\to\mathcal{L}(V^{\otimes 2})$ be a continuous function as in
Subsection 1.1. Our aim is to define the $Q$-MCR algebra.

To simplify notation, we denote $Y:=\mathbb{R}$, $Z:=\mathbb{R}^{d-1}$, so
that $X=Y\times Z$. Respectively, for a given $x\in X$, we denote $y:=x^{1}\in
Y$ and $z:=(x^{2},\dots,x^{d})\in Z$, so that $x=(y,z)$. We denote
$\mathcal{G}:=L^{2}(Z;\mathbb{C})$. Then

$\mathcal{H}=L^{2}(X;V)=L^{2}(Y;V)\otimes\mathcal{G}.$

We denote by $C_{0}(Y^{n};V^{\otimes n})$ the vector space of all continuous
functions $\varphi^{(n)}:Y^{n}\to V^{\otimes n}$ with compact support. We
denote by $\mathfrak{F}^{(n)}$ the vector space spanned by functions
$f^{(n)}:X^{n}\to V^{\otimes n}$ of the form

$f^{(n)}(x_{1},\dots,x_{n})=\varphi^{(n)}(y_{1},\dots,y_{n})g_{1}(z_{1}),\dotsm
g_{n}(z_{n})$ (16)

with $\varphi^{(n)}\in C_{0}(Y^{n};V^{\otimes n})$ and
$g_{1},\dots,g_{n}\in\mathcal{G}$.

######  Remark 2.1.

Below we will use the trivial facts that, for $f^{(m)}\in\mathfrak{F}^{(m)}$,
$g^{(n)}\in\mathfrak{F}^{(n)}$ and a continuous function
$C:Y^{2}\to\mathcal{L}(V^{\otimes 2})$,

$f^{(m)}(x_{1},\dots,x_{m})\otimes
g^{(n)}(x_{m+1},\dots,x_{m+n})\in\mathfrak{F}^{(m+n)},$

and

$C_{i}(y_{i},y_{i+1})f^{(m)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{m})\in\mathfrak{F}^{(m)},\quad
i=1,\dots,m-1.$

Furthermore, for $u,v\in V$, $\langle
u,v\rangle_{V}=\operatorname{Tr}(u\otimes v)$.

We define the $Q$-MCR algebra $\mathbb{A}$ as follows. Consider the following
(formal for now) operator-valued integrals:

$\int_{X^{n}}\langle
f^{(n)}(x_{1},\dots,x_{n}),A^{\sharp_{1}}(x_{1})\otimes\dotsm\otimes
A^{\sharp_{n}}(x_{n})\rangle_{V^{\otimes n}}\,dx_{1}\dotsm dx_{n}$ (17)

with $f^{(n)}\in\mathfrak{F}^{(n)}$ and
$\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$ ($n\in\mathbb{N}$), the integral in
(17) depending on $f^{(n)}$ linearly. To shorten notation, we will denote the
integral in (17) by

$\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n})=\Phi(f^{(n)}(x_{1},\dots,x_{n});\sharp_{1},\dots,\sharp_{n}).$
(18)

As a vector space, $\mathbb{A}$ is spanned by the identity operator and
integrals of the form (17), equivalently (18). Then $\mathbb{A}$ is defined as
the unital $*$-algebra with the product

$\displaystyle\Phi(f^{(m)};\sharp_{1},\dots,\sharp_{m})\Phi(g^{(n)};\sharp_{m+1},\dots,\sharp_{m+n})$
$\displaystyle\quad=\Phi(f^{(m)}(x_{1},\dots,x_{m})\otimes
g^{(n)}(x_{m+1},\dots,x_{m+n});\sharp_{1},\dots,\sharp_{m+n})$ (19)

and the $*$-operation

$\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n})^{*}=\Phi(\mathbb{S}^{(n)}f^{(n)}(x_{n},\dots,x_{1});-\sharp_{n},\dots,-\sharp_{1}).$
(20)

Here, for $\sharp\in\\{+,-\\}$, $-\sharp$ denotes the opposite sign of
$\sharp$, and the antilinear operator $\mathbb{S}^{(n)}:V^{\otimes n}\to
V^{\otimes n}$ is defined by

$\mathbb{S}^{(n)}v_{1}\otimes\dotsm\otimes
v_{n}=(Jv_{n})\otimes\dotsm\otimes(Jv_{1}),\quad v_{1},\dots,v_{n}\in V.$

Furthermore, elements of $\mathbb{A}$ are subject to the commutation relations
(4)–(6). More precisely, if $i\in\\{1,\dots,n-1\\}$ and
$\sharp_{i}=\sharp_{i+1}=+$, then

$\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n})=\Phi\big{(}Q_{i}(x_{i},x_{i+1})f^{(n)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{n});\sharp_{1},\dots,\sharp_{n}\big{)},$
(21)

if $\sharp_{i}=\sharp_{i+1}=-$, then

$\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n})=\Phi\big{(}\widehat{Q}_{i}(x_{i},x_{i+1})f^{(n)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{n});\sharp_{1},\dots,\sharp_{n}\big{)},$
(22)

and if $\sharp_{i}=-,\,\sharp_{i+1}=+$, then

$\displaystyle\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n})$
$\displaystyle=\Phi(\widetilde{Q}_{i}(x_{i+1},x_{i})f^{(n)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{n});\sharp_{1},\dots,\sharp_{i+1},\sharp_{i},\dots,\sharp_{n})$
$\displaystyle\quad+\Phi(g^{(n-2)};\sharp_{1},\dots,\sharp_{i-1},\sharp_{i+2},\dots,\sharp_{n}),$
(23)

where

$g^{(n-2)}(x_{1},\dots,x_{n-2}):=\int_{X}\operatorname{Tr}_{i}f^{(n)}(x_{1},\dots,x_{i-1},x,x,x_{i},\dots,x_{n-2})\,dx\in\mathfrak{F}^{(n-2)}.$
(24)

Here $\operatorname{Tr}_{i}:V^{\otimes n}\to V^{\otimes(n-2)}$ acts by the
functional $\operatorname{Tr}:V^{\otimes 2}\to\mathbb{C}$ on the $i$th and
$(i+1)$th components of $V^{\otimes n}$.

For $m,n\geq 0$ with $m+n\geq 1$ and $f^{(m+n)}\in\mathfrak{F}^{(m+n)}$,
denote $W^{(m,n)}(f^{(m+n)}):=\Phi(f^{(m+n)};\sharp_{1},\dots,\sharp_{m+n})$
with $\sharp_{1}=\dots=\sharp_{m}=+$, $\sharp_{m+1}=\dots=\sharp_{m+n}=-$.
Thus, $W^{(m,n)}(f^{(m+n)})$ is a Wick-ordered element of the $Q$-MCR algebra
$\mathbb{A}$. Let also $\mathfrak{SF}^{(m,n)}$ consist of all elements
$f^{(m+n)}\in\mathfrak{F}^{(m+n)}$ that satisfy

$f^{(m+n)}(x_{1},\dots,x_{m+n})=Q_{i}(x_{i},x_{i+1})f^{(m+n)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{m+n})$
(25)

for all $i=1,\dots,m-1$, and

$\displaystyle
f^{(m+n)}(x_{1},\dots,x_{m+n})=\widehat{Q}_{i}(x_{i},x_{i+1})f^{(m+n)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{m+n})$

for all $i=m+1,\dots,m+n-1$. In words, $f^{(m+n)}$ is $Q$-symmetric in the
first $m$ variables and $\widehat{Q}$-symmetric in the last $n$ variables.

The following proposition states that $\mathbb{A}$ is a Wick algebra, i.e., it
allows Wick (normal) ordering.

######  Proposition 2.2.

Each element of the $Q$-MCR algebra $\mathbb{A}$ can be represented in the
form

$c\mathbf{1}+\sum_{m,n\geq 0,\ m+n\geq 1}W^{(m,n)}(f^{(m,n)})$ (26)

where $c\in\mathbb{C}$ and $f^{(m,n)}\in\mathfrak{SF}^{(m,n)}$. The sum in
(26) is finite.

###### Proof.

The commutation relation (23) implies that each element of $\mathbb{A}$ can be
represented in the form (26) with $c\in\mathbb{C}$ and
$f^{(m,n)}\in\mathfrak{F}^{(m+n)}$. So we only need to prove that the
functions $f^{(m,n)}$ can be chosen from $\mathfrak{SF}^{(m,n)}$.

For $i=1,\dots,m-1$, define
$U_{i}:\mathfrak{F}^{(m+n)}\to\mathfrak{F}^{(m+n)}$ by

$(U_{i}f^{(m+n)})(x_{1},\dots,x_{m+n}):=Q_{i}(x_{i},x_{i+1})f^{(m+n)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{m+n}).$
(27)

The functional Yang–Baxter equation (1) implies that, for $i=1,\dots,m-2$,
$U_{i}U_{i+1}U_{i}=U_{i+1}U_{i}U_{i+1}$ (the Yang–Baxter equation).
Furthermore, we obviously have $U_{i}U_{j}=U_{j}U_{i}$ if $|i-j|\geq 2$, and
$U_{i}^{2}=\mathbf{1}$. Therefore, we can construct a representation of the
symmetric group $S_{m}$ by setting $U_{\pi}:=U_{i_{1}}U_{i_{2}}\dotsm
U_{i_{k}}$ where $\pi\in S_{m}$ has an (arbitrary) representation
$\pi=\pi_{i_{1}}\pi_{i_{2}}\dotsm\pi_{i_{k}}$. Here, for $i=1,\dots,m-1$,
$\pi_{i}$ is the adjacent transposition of $i$ and $i+1$. Let
$P_{m}:=\frac{1}{m!}\sum_{\pi\in S_{m}}U_{\pi}$.

Next, for $i=m+1,\dots,m+n-1$, we define
$\widehat{U}_{i}:\mathfrak{F}^{(m+n)}\to\mathfrak{F}^{(m+n)}$ by

$(\widehat{U}_{i}f^{(m+n)})(x_{1},\dots,x_{m+n}):=\widehat{Q}_{i}(x_{i},x_{i+1})f^{(m+n)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{m+n}).$

It follows from (1) and the definition of $\widehat{Q}(\cdot,\cdot)$ that
$\widehat{Q}(\cdot,\cdot)$ also satisfies the functional Yang–Baxter equation.
Hence, we can similarly define $\widehat{P}_{n}:=\frac{1}{n!}\sum_{\pi\in
S_{n}}\widehat{U}_{\pi}$, where $S_{n}$ is interpreted as the group of
permutations of $m+1,\dots,m+n$.

It is easy to see that $P_{m}\widehat{P}_{n}$ maps $\mathfrak{F}^{(m+n)}$ onto
$\mathfrak{SF}^{(m,n)}$. Furthermore, by (21) and (22), for each
$f^{(m+n)}\in\mathfrak{F}^{(m+n)}$, we have

$W^{(m,n)}(f^{(m+n)})=W^{(m,n)}(P_{m}\widehat{P}_{n}f^{(m+n)}).\qquad\qed$

The following proposition shows that, under an additional assumption on the
operator-valued function $Q$, the algebra $\mathbb{A}$ allows anti-normal
ordering.

######  Proposition 2.3.

Assume that, for all $(x_{1},x_{2})\in X^{2}$, the operator
$\widetilde{Q}(x_{1},x_{2})\in\mathcal{L}(V^{\otimes 2})$ is invertible.
Furthermore, assume that there exists a constant $\varkappa\in\mathbb{R}$ such
that, for all $x\in X$ and $v^{(2)}\in V^{\otimes 2}$,

$\operatorname{Tr}\big{(}\widetilde{Q}(x,x)^{-1}v^{(2)}\big{)}=\varkappa\operatorname{Tr}v^{(2)},\quad
x\in X,\ v^{(2)}\in V^{\otimes 2}.$ (28)

Let $\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$ be such that, for some
$i\in\\{1,\dots,n-1\\}$, $\sharp_{i}=+$ and $\sharp_{i+1}=-$. Then, for all
$f^{(n)}\in\mathfrak{F}^{(n)}$,

$\displaystyle\Phi^{(n)}\left(f^{(n)};\sharp_{1},\dots,\sharp_{n}\right)$
$\displaystyle\quad=\Phi^{(n)}\left(\widetilde{Q}_{i}(x_{i},x_{i+1})^{-1}f^{(n)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{n});\sharp_{1},\dots,\sharp_{i+1},\sharp_{i},\dots,\sharp_{n}\right)$
$\displaystyle\qquad-\varkappa\Phi(g^{(n-2)};\sharp_{1},\dots,\sharp_{i-1},\sharp_{i+2},\dots,\sharp_{n}),$
(29)

where $g^{(n-2)}$ is given by (24).

###### Proof.

Applying formulas (23), (24) to

$\Phi^{(n)}\left(\widetilde{Q}_{i}(x_{i},x_{i+1})^{-1}f^{(n)}(x_{1},\dots,x_{i+1},x_{i},\dots,x_{n});\sharp_{1},\dots,\sharp_{i+1},\sharp_{i},\dots,\sharp_{n}\right)$

and using the assumption (28), we get (29). ∎

##  3 States on the $Q$-MCR algebra

To study states on the $Q$-MCR algebra $\mathbb{A}$, we need to make an
assumption on their continuity. To this end, we equip $C_{0}(Y^{n};V^{\otimes
n})$ with the Fréchet topology in which a sequence $(f_{k})_{k=1}^{\infty}$
converges if and only if there exists a compact set $K\subset Y^{n}$ such that
the support of each $f_{k}$ is a subset of $K$ and $(f_{k})_{k=1}^{\infty}$
converges uniformly on $K$. The dual space of $C_{0}(Y^{n};V^{\otimes n})$ is
the space $\mathcal{M}(Y^{n};V^{\otimes n})$ of $V^{\otimes n}$-valued Radon
measures on $Y^{n}$, i.e., $F^{(n)}$ is a continuous linear functional on
$C_{0}(Y^{n};V^{\otimes n})$ if and only if

$F^{(n)}(\varphi^{(n)})=\int_{Y^{n}}\langle\varphi^{(n)}(y_{1},\dots,y_{n}),m^{(n)}(dy_{1}\dots,dy_{n})\rangle_{V^{\otimes
n}},\quad\varphi^{(n)}\in C_{0}(Y^{n};V^{\otimes n}),$

where $m^{(n)}\in\mathcal{M}(Y^{n};V^{\otimes n})$, see e.g. [16, Chapter 7].
Note that $m^{(n)}$ is completely determined by its values on functions of the
form
$\varphi^{(n)}(y_{1},\dots,y_{n})=\varphi_{1}(y_{1})\otimes\dots\otimes\varphi_{n}(y_{n})$
with $\varphi_{1},\dots,\varphi_{n}\in C_{0}(Y;V)$.

Let $\tau:\mathbb{A}\to\mathbb{C}$ be a state on $\mathbb{A}$. For any
$\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$, we define a linear functional
$\tau^{(n)}_{\sharp_{1},\dots,\sharp_{n}}:C_{0}(Y^{n};V^{\otimes
n})\times\mathcal{G}^{n}\to\mathbb{C}$ by

$\displaystyle\tau^{(n)}_{\sharp_{1},\dots,\sharp_{n}}(\varphi^{(n)},g_{1},\dots,g_{n}):=\tau\left(\Phi^{(n)}(\varphi^{(n)}(y_{1},\dots,y_{n})g_{1}(z_{1})\dotsm
g_{n}(z_{n});\sharp_{1},\dots,\sharp_{n})\right).$ (30)

Obviously, these linear functionals uniquely identify the state $\tau$. We
assume that, for any fixed $g_{1},\dots,g_{n}\in\mathcal{G}$,
$\tau^{(n)}_{\sharp_{1},\dots,\sharp_{n}}(\cdot,g_{1},\dots,g_{n})$ is
continuous on $C_{0}(Y^{n};V^{\otimes n})$. Hence, there exists a $V^{\otimes
n}$-valued Radon measure
$m_{\sharp_{1},\dots,\sharp_{n}}^{(n)}[g_{1},\dots,g_{n}]\in\mathcal{M}(Y^{n};V^{\otimes
n})$ that can be identified with
$\tau^{(n)}_{\sharp_{1},\dots,\sharp_{n}}(\cdot,g_{1},\dots,g_{n})$.

Below, for $f\in\mathfrak{F}^{(1)}$ and $\sharp\in\\{+,-\\}$, we denote
$A^{\sharp}(f):=\Phi(f;\sharp)$, called creation, respectively annihilation
operators in the $Q$-MCR algebra $\mathbb{A}$, and

$B(f):=A^{+}(f)+A^{-}(Jf).$ (31)

######  Proposition 3.1.

Under the above assumption on the continuity of a state $\tau$ on a $Q$-MCR
algebra $\mathbb{A}$, the following statements hold.

(i) For $m,n\geq 0$, $m+n\geq 1$, we define

$\displaystyle\mathbf{S}^{(m,n)}(\varphi_{1}\otimes
g_{1},\dots,\varphi_{m}\otimes g_{m},\varphi_{m+1}\otimes
g_{m+1},\dots,\varphi_{m+n}\otimes g_{m+n})$
$\displaystyle\quad:=\tau\big{(}A^{+}(\varphi_{1}\otimes g_{1})\dotsm
A^{+}(\varphi_{m}\otimes g_{m})A^{-}(\varphi_{m+1}\otimes g_{m+1})\dotsm
A^{-}(\varphi_{m+n}\otimes g_{m+n})\big{)},$

where $\varphi_{i}\in C_{0}(Y;V)$, $g_{i}\in\mathcal{G}$, and
$(\varphi_{i}\otimes g_{i})(x):=\varphi_{i}(y)g_{i}(z)$ ($i=1,\dots,m+n$).
Then the functionals $\mathbf{S}^{(m,n)}$ uniquely identify the state $\tau$.

(ii) For $n\geq 1$, we define

$\mathbf{M}^{(n)}(\varphi_{1}\otimes g_{1},\dots,\varphi_{n}\otimes
g_{n}):=\tau\big{(}B(\varphi_{1}\otimes g_{1})\dotsm B(\varphi_{n}\otimes
g_{n})\big{)},$

where $\varphi_{i}\in C_{0}(Y;V_{\mathbb{R}})$, $g_{i}\in\mathcal{G}$
($i=1,\dots,n$). Here $V_{\mathbb{R}}:=\mathbb{R}^{m}\subset\mathbb{C}^{m}=V$.
Then the functionals $\mathbf{M}^{(n)}$ uniquely identify the state $\tau$.

###### Proof.

(i) By Proposition 2.2, the state $\tau$ is completely determined by the
functionals $\tau^{(m+n)}_{\sharp_{1},\dots,\sharp_{m+n}}$ with
$\sharp_{1}=\dots=\sharp_{m}=+$ and $\sharp_{m+1}=\dots=\sharp_{m+n}=-$, where
$m,n\geq 0$, $m+n\geq 1$. Statement (i) now follows from the assumed
continuity of the functionals $\tau^{(m+n)}_{\sharp_{1},\dots,\sharp_{m+n}}$.

(ii) First, we note that the result of part (i) remains true if all
$\varphi_{i}$’s are assumed to be from $C_{0}(Y;V_{\mathbb{R}})$. Next, for
each $\varphi\in C_{0}(Y;V_{\mathbb{R}})$ and $g\in\mathcal{G}$, we have

$A^{+}(\varphi\otimes g)=\frac{1}{2}\big{(}B(\varphi\otimes
g)-iB(\varphi\otimes ig)\big{)},\quad A^{-}(\varphi\otimes
g)=\frac{1}{2}\big{(}B(\varphi\otimes\bar{g})+iB(\varphi\otimes
i\bar{g})\big{)}.$

Hence, the functionals $\mathbf{M}^{(n)}$ uniquely identify the functionals
$\mathbf{S}^{(m,n)}$. ∎

##  4 The Fock state on the $Q$-MCR algebra

Let us briefly recall the Fock representation of the $Q$-MCR (4)–(6), for
detail see [13] or [8]. First, note that $\mathcal{H}^{\otimes
n}=L^{2}(X^{n};V^{\otimes n})$. Define a unitary operator
$U\in\mathcal{L}(\mathcal{H}^{\otimes 2})$ by
$(Uf^{(2)})(x_{1},x_{2}):=Q(x_{1},x_{2})f^{(2)}(x_{2},x_{1})$. For
$i=1,\dots,n-1$, let $U_{i}$ denote the operator $U$ acting on the $i$th and
$(i+1)$th components of $\mathcal{H}^{\otimes n}$, compare with (27).
Similarly to the proof of Proposition 2.2, define $P_{n}:=\sum_{\pi\in
S_{n}}U_{\pi}$. Then $P_{n}$ is an orthogonal projection in
$\mathcal{H}^{\otimes n}$. Denote by $\mathcal{H}^{\circledast n}$ the image
of $P_{n}$. This is the subspace of all functions from $\mathcal{H}^{\otimes
n}$ that are $Q$-symmetric, compare with (25). Let also
$\mathcal{H}^{\circledast 0}:=\mathbb{C}$. Define the $Q$-symmetric Fock space
$\mathcal{F}^{Q}(\mathcal{H}):=\bigoplus_{n=0}^{\infty}\mathcal{H}^{\circledast
n}n!$, i.e., $\mathcal{F}^{Q}(\mathcal{H})$ is the Hilbert space of all
infinite sequences $F=(f^{(n)})_{n=0}^{\infty}$ with
$f^{(n)}\in\mathcal{H}^{\circledast n}$ and
$\|F\|^{2}_{\mathcal{F}^{Q}(\mathcal{H})}=\sum_{n=0}^{\infty}\|f^{(n)}\|_{\mathcal{H}^{\circledast
n}}^{2}n!<\infty$. The vector $\Omega=(1,0,0,\dots)$ is called the vacuum. Let
$\mathcal{F}^{Q}_{\mathrm{fin}}(\mathcal{H})$ denote the (dense) subspace of
$\mathcal{F}^{Q}(\mathcal{H})$ consisting of all
$F=(f^{(n)})_{n=0}^{\infty}\in\mathcal{F}^{Q}(\mathcal{H})$ such that, for
some $N\in\mathbb{N}$ (depending on $F$), $f^{(n)}=0$ for all $n\geq N$.

Let $f\in\mathcal{H}$. We define a (standard) creation operator $a^{+}(f)$ as
the linear operator in $\mathcal{F}^{Q}_{\mathrm{fin}}(\mathcal{H})$ given by
$a^{+}(f)f^{(n)}:=P_{n+1}(f\otimes f^{(n)})$ for each
$f^{(n)}\in\mathcal{H}^{\circledast n}$. Next, we define a (standard)
annihilation operator $a^{-}(f)$ as the linear operator in
$\mathcal{F}^{Q}_{\mathrm{fin}}(\mathcal{H})$ given by

$(a^{-}(f)f^{(n)})(x_{1},\dots,x_{n-1})=n\int_{X}\langle
f^{(n)}(x,x_{1},\dots,x_{n-1}),f(x)\rangle_{V}\,dx.$ (32)

Here, for $u_{1},\dots,u_{n},v\in V$, we denote

$\langle u_{1}\otimes\dots\otimes u_{n},v\rangle_{V}:=\langle
u_{1},v\rangle_{V}\,u_{2}\otimes\dots\otimes u_{n}\,.$ (33)

Then $a^{-}(f)$ is the restriction to
$\mathcal{F}^{Q}_{\mathrm{fin}}(\mathcal{H})$ of $a^{+}(Jf)^{*}$.

Similarly to (2), (3) one defines operator-valued distributions $a^{+}(x)$,
$a^{-}(x)$. One shows that the operator-valued integrals on the right-hand
side of formulas (4)–(6), with $A^{+}(\cdot)$, $A^{-}(\cdot)$ being replaced
by $a^{+}(\cdot)$, $a^{-}(\cdot)$, are well-defined and formulas (4)–(6) hold.

Let us show that this construction can be easily extended to a representation
of the $Q$-MCA algebra. By formula (33) in [8], we have, for any
$f_{1},\dots,f_{m}\in\mathcal{H}$ and $f^{(k)}\in\mathcal{H}^{\circledast n}$,

$a^{+}(f_{1})a^{+}(f_{1})\dotsm a^{+}(f_{m})f^{(k)}=P_{m+k}(f_{1}\otimes
f_{2}\otimes\dotsm\otimes f_{m}\otimes f^{(k)}).$ (34)

Furthermore, by using (32) and (33), we have

$\displaystyle(a^{-}(f_{1})a^{-}(f_{2})\dotsm
a^{-}(f_{n})f^{(k)})(x_{1},\dots,x_{k-n})=(k)_{n}$
$\displaystyle\times\int_{X^{n}}(\mathbb{T}^{(2n)}\otimes\mathbf{1}_{k-n})(f_{1}\otimes\dots\otimes
f_{n})(x_{1}^{\prime},\dots,x_{n}^{\prime})\otimes
f^{(k)}(x_{n}^{\prime},x_{n-1}^{\prime},\dots,x_{1}^{\prime},x_{1},\dots,x_{k-n})$
$\displaystyle\quad\times dx_{1}^{\prime}\dotsm dx_{n}^{\prime}\,.$ (35)

Here, $(k)_{n}$ is the Pochhammer symbol,
$\mathbb{T}^{(2n)}:V^{\otimes(2n)}\to\mathbb{C}$ is the linear functional
given by

$\mathbb{T}^{(2n)}\,v_{1}\otimes\dots\otimes v_{2n}:=\prod_{i=1}^{n}\langle
v_{i},v_{2n-i+1}\rangle_{V},\quad v_{1},\dots,v_{2n}\in V,$ (36)

and $\mathbf{1}_{k-n}$ is the identity operator in $V^{\otimes(k-n)}$.

In view of formulas (34), (35), we define for
$f^{(m+n)}\in\mathfrak{F}^{(m+n)}$, a linear operator $W^{(m,n)}(f^{(m+n)})$
in $\mathcal{F}^{Q}_{\mathrm{fin}}(\mathcal{H})$ by

$\displaystyle\left(W^{(m,n)}\left(f^{(m+n)}\right)u^{(k)}\right)(x_{1},\dots,x_{m-n+k})=P_{m-n+k}\bigg{[}\int_{X^{n}}\left(\mathbf{1}_{m}\otimes\mathbb{T}^{(2n)}\otimes\mathbf{1}_{k-n}\right)$
$\displaystyle
f^{(m+n)}(x_{1},\dots,x_{m},x_{1}^{\prime},\dots,x_{n}^{\prime})\otimes
u^{(k)}(x_{n}^{\prime},\dots,x_{1}^{\prime},x_{m+1},\dots,x_{m-n+k})\,dx_{1}^{\prime}\dotsm
dx_{n}^{\prime}\bigg{]}$ (37)

for $u^{(k)}\in\mathcal{H}^{\circledast k}$. Due to the $Q$-MCR, the operators
$W^{(m,n)}(f^{m+n})$ uniquely identify operators
$\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n})$ with
$f^{(n)}\in\mathfrak{F}^{(n)}$ and $\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$.
Thus, we get a representation of the $Q$-MCR $*$-algebra $\mathbb{A}$.

Next, we define the vacuum state $\tau$ on $\mathbb{A}$ by
$\tau(a):=(a\Omega,\Omega)_{\mathcal{F}^{Q}(\mathcal{H})}$ for
$a\in\mathbb{A}$. Since, for each $f\in\mathfrak{F}^{(1)}$,
$a^{-}(f)\Omega=0$, $\tau$ is the Fock state on $\mathbb{A}$. Note also that
the state $\tau$ can be extended to the vector space of all linear operators
acting in $\mathcal{F}^{Q}_{\mathrm{fin}}(\mathcal{H})$.

Obviously, for the Fock state $\tau$, the corresponding functionals
$\tau^{(n)}_{\sharp_{1},\dots,\sharp_{n}}(\cdot,g_{1},\dots,g_{n})$ are
continuous on $C_{0}(Y^{n};V^{\otimes n})$. Hence, by Proposition 3.1, the
functionals $\mathbf{S}^{(m,n)}$ uniquely identify $\tau$, and so do the
functionals $\mathbf{M}^{(n)}$. It follows from the definition of $\tau$ that
all the functionals $\mathbf{S}^{(m,n)}$ are equal to zero. Our next aim is to
calculate the functionals $\mathbf{M}^{(n)}$. To this end, we will first prove
a slightly more general result. Before doing that, let us introduce required
notations.

For $n\in\mathbb{N}$, let $\mathcal{P}^{(2n)}$ denote the set of all
partitions of $\\{1,\dots,2n\\}$ into $n$ two-point sets. Fix a partition
$\xi\in\mathcal{P}^{(2n)}$. All the definitions below depend on the choice of
$\xi$, nevertheless for simplicity we will not show this dependence in some of
our notations.

Let

$\xi=\big{\\{}\\{i_{1},j_{1}\\},\dots,\\{i_{n},j_{n}\\}\big{\\}}$ (38)

with

$i_{1}<j_{1},\ i_{2}<j_{2},\dots,\ i_{n}<j_{n},\quad
i_{1}>i_{2}>\dotsm>i_{n}=1.$ (39)

Denote $I:=\\{i_{1},i_{2},\dots,i_{n}\\}$, $J:=\\{j_{1},j_{2},\dots,j_{n}\\}$.
For $k=1,2,\dots,n$, we define sets $J^{(k)}$ and $\mathbf{J}^{(k)}$ as
follows. Let

$J^{(1)}:=\\{j\in J\mid i_{1}<j\leq j_{1}\\},\quad\mathbf{J}^{(1)}:=\\{j\in
J\mid i_{1}<j\\},$

and for $k=2,\dots,n$,

$\displaystyle J^{(k)}$ $\displaystyle:=\big{\\{}j\in J\mid i_{k}<j\leq
j_{k},\ j\neq j_{1},\ j\neq j_{2},\dots,\ j\neq j_{k-1}\big{\\}},$
$\displaystyle\mathbf{J}^{(k)}$ $\displaystyle:=\left\\{j\in J\mid i_{k}<j,\
j\neq\min\mathbf{J}^{(1)},\ j\neq\min\mathbf{J}^{(2)},\dots,\
j\neq\min\mathbf{J}^{(k-1)}\right\\}.$ (40)

We write

$J^{(k)}=\big{\\{}j_{1}^{(k)},j_{2}^{(k)},\dots,j_{l_{k}}^{(k)}\big{\\}},\quad\mathbf{J}^{(k)}=\big{\\{}\mathbf{j}_{1}^{(k)},\mathbf{j}_{2}^{(k)},\dots,\mathbf{j}_{m_{k}}^{(k)}\big{\\}},$
(41)

with

$j_{1}^{(k)}<j_{2}^{(k)}<\dots<j_{l_{k}}^{(k)},\quad\mathbf{j}_{1}^{(k)}<\mathbf{j}_{2}^{(k)}<\dots<\mathbf{j}_{m_{k}}^{(k)}.$
(42)

Here $l_{k}$ and $m_{k}$ are the number of elements of the sets $J^{(k)}$ and
$\mathbf{J}^{(k)}$, respectively,

######  Remark 4.1.

Note that

$\displaystyle\xi^{\prime}=\big{\\{}\\{i_{1},\mathbf{j}_{1}^{(1)}\\},\\{i_{2},\mathbf{j}_{1}^{(2)}\\},\dots,\\{i_{n},\mathbf{j}_{1}^{(n)}\\}\big{\\}}$

belongs to $\mathcal{P}^{(2n)}$. Furthermore, since
$\mathbf{j}_{1}^{(k)}=\min\mathbf{J}^{(k)}$, the definition (40) implies that
$\xi^{\prime}$ is a non-crossing partition.

Recall that, for a linear operator $\mathcal{C}\in\mathcal{L}(V^{\otimes 2})$
and $i\in\\{1,2,\dots,m-1\\}$, we denote by $\mathcal{C}_{i}$ the linear
operator in $V^{\otimes m}$ acting as $\mathcal{C}$ on the $i$th and $(i+1)$th
components of the tensor product $V^{\otimes m}$. Now, similarly, for $1\leq
i<j\leq m$, we define a linear operator $\mathcal{C}[i,j]$ in $V^{\otimes m}$
that acts as the operator $\mathcal{C}$ on the $i$th and $j$th components of
$V^{\otimes m}$.

Let us now fix an arbitrary $(x_{1},\dots,x_{2n})\in X^{2n}$. For
$k=1,\dots,n$, we define a linear operator
$Q^{(k)}(\xi;x_{1},x_{2},\dots,x_{2n})$ in $V^{\otimes 2n}$ as follows. If
$J^{(k)}=\\{j_{k}\\}$, i.e., $l_{k}=1$, then
$Q^{(k)}(\xi;x_{1},x_{2},\dots,x_{2n})$ is the identity operator. If
$l_{k}>1$, then we set

$\displaystyle Q^{(k)}(\xi;x_{1},x_{2},\dots,x_{2n})$
$\displaystyle\quad:=Q(x_{i_{k}},x_{j_{1}^{(k)}})\big{[}\mathbf{j}_{1}^{(k)},\,\mathbf{j}_{2}^{(k)}\big{]}\,Q(x_{i_{k}},x_{j_{2}^{(k)}})\big{[}\mathbf{j}_{2}^{(k)},\,\mathbf{j}_{3}^{(k)}\big{]}\dotsm
Q(x_{i_{k}},x_{j_{{l_{k}}-1}^{(k)}})\big{[}\mathbf{j}_{{l_{k}}-1}^{(k)},\,\mathbf{j}_{l_{k}}^{(k)}\big{]}.$
(43)

Next we define a linear operator

$\displaystyle Q(\xi;x_{1},x_{2},\dots,x_{2n})$
$\displaystyle\quad:=Q^{(n)}(\xi;x_{1},x_{2},\dots,x_{2n})Q^{(n-1)}(\xi;x_{1},x_{2},\dots,x_{2n})\dotsm
Q^{(1)}(\xi;x_{1},x_{2},\dots,x_{2n}).$ (44)

Recall Remark 4.1. Similarly to (36), we define a linear functional
$\mathbb{T}^{(2n)}(\xi):V^{\otimes 2n}\to\mathbb{C}$ by

$\mathbb{T}^{(2n)}(\xi)v_{1}\otimes v_{2}\otimes\dotsm\otimes v_{2n}=\langle
v_{i_{1}},v_{\mathbf{j}_{1}^{(1)}}\rangle_{V}\langle
v_{i_{2}},v_{\mathbf{j}_{1}^{(2)}}\rangle_{V}\dotsm\langle
v_{i_{n}},v_{\mathbf{j}_{1}^{(n)}}\rangle_{V}.$ (45)

######  Theorem 4.2.

For $f\in\mathcal{H}$, let $a^{+}(f)$ and $a^{-}(f)$ denote the standard
creation and annihilation operators in the $Q$-symmetric Fock space
$\mathcal{F}^{Q}(\mathcal{H})$. Let
$\tau(\cdot)=(\cdot\,\Omega,\Omega)_{\mathcal{F}^{Q}(\mathcal{H})}$ denote the
Fock state. Let $n\in\mathbb{N}$ and
$f_{1},f^{\prime}_{1},\dots,f_{2n},f^{\prime}_{2n}\in\mathcal{H}$. Then

$\tau\big{(}(a^{+}(f_{1})+a^{-}(f^{\prime}_{1}))\dotsm(a^{+}(f_{2n-1})+a^{-}(f^{\prime}_{2n-1}))\big{)}=0$
(46)

and

$\displaystyle\tau\big{(}(a^{+}(f_{1})+a^{-}(f^{\prime}_{1}))\dotsm(a^{+}(f_{2n})+a^{-}(f^{\prime}_{2n}))\big{)}=\sum_{\xi\in\mathcal{P}^{(2n)}}\int_{X^{2n}}\bigotimes_{\begin{subarray}{c}\\{i,j\\}\in\xi\\\
i<j\end{subarray}}\sigma^{(2)}(dx_{i}\,dx_{j})$
$\displaystyle\qquad\times\mathbb{T}^{(2n)}(\xi)Q(\xi;x_{1},\dots,x_{2n})h_{1}(x_{1})\otimes
h_{2}(x_{2})\otimes\dotsm\otimes h_{2n}(x_{2n}).$ (47)

Here $\sigma^{(2)}$ is the measure on $X^{2}$ satisfying

$\int_{X^{2}}f^{(2)}(x_{1},x_{2})\sigma^{(2)}(dx_{1}\,dx_{2})=\int_{X}f^{(2)}(x,x)dx$
(48)

and $h_{k}(x)=f_{k}^{\prime}(x)$ if $k\in I$ and $h_{k}(x)=f_{k}(x)$ if $k\in
J$.

###### Proof.

Denote by
$\mathbb{F}(\mathcal{H}):=\bigoplus_{n=0}^{\infty}\mathcal{H}^{\otimes n}n!$
the full Fock space over $\mathcal{H}$, and by
$\mathbb{F}_{\mathrm{fin}}(\mathcal{H})$ the vector space of all sequences
$F=(f^{(n)})_{n=0}^{\infty}\in\mathbb{F}(\mathcal{H})$ such that, for some
$N\in\mathbb{N}$ (depending on $F$), $f^{(n)}=0$ for all $n\geq N$. For each
$f\in\mathcal{H}$, we define linear operators $\mathcal{A}^{+}(f)$ and
$\mathcal{A}^{-}(f)$ in $\mathbb{F}_{\mathrm{fin}}(\mathcal{H})$ by

$\displaystyle(\mathcal{A}^{+}(f)f^{(n)})(x_{1},\dots,x_{n+1}):=f(x_{1})\otimes
f^{(n)}(x_{2},\dots,x_{n+1}),$
$\displaystyle\left(\mathcal{A}^{-}(f)f^{(n)}\right)(x_{1},\dots,x_{n-1})=\int_{X}\Big{\langle}f(x),f^{(n)}(x,x_{1},\dots,x_{n-1})$
$\displaystyle\quad+\sum_{k=1}^{n-1}Q_{1}(x,x_{1})Q_{2}(x,x_{2})\dotsm
Q_{k}(x,x_{k})f^{(n)}(x_{1},\dots,x_{k},x,x_{k+1},\dots,x_{n-1})\Big{\rangle}_{V}\,dx$
$\displaystyle=\int_{X}\Big{[}\operatorname{Tr_{1}}\big{(}f(x)\otimes
f^{(n)}(x,x_{1},\dots,x_{n-1})\big{)}+\sum_{k=1}^{n-1}\operatorname{Tr_{1}}\big{(}Q_{2}(x,x_{1})Q_{3}(x,x_{2})\dotsm
Q_{k+1}(x,x_{k})$ $\displaystyle\quad f(x)\otimes
f^{(n)}(x_{1},\dots,x_{k},x,x_{k+1},\dots,x_{n-1})\big{)}\Big{]}dx,\quad
f^{(n)}\in\mathcal{H}^{\otimes n}.$ (49)

######  Lemma 4.3.

For any $f_{1},\dots,f_{n}\in\mathcal{H}$ and
$\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$, we have

$\tau\left(a^{\sharp_{1}}(f_{1})\dotsm
a^{\sharp_{n}}(f_{n})\right)=\left(\mathcal{A}^{\sharp_{1}}(f_{1})\dotsm\mathcal{A}^{\sharp_{n}}(f_{n})\Omega,\Omega\right)_{\mathbb{F}(\mathcal{H})}.$

###### Proof.

It is sufficient to prove that, for each $f^{(n)}\in\mathcal{H}^{\otimes n}$,

$a^{+}(f)P_{n}f^{(n)}=P_{n+1}\mathcal{A}^{+}(f)f^{(n)},\quad
a^{-}(f)P_{n}f^{(n)}=P_{n-1}\mathcal{A}^{-}(f)f^{(n)}.$ (50)

It easily follows from the definition of the operator $P_{n}$ that
$P_{n+1}(\mathbf{1}_{\mathcal{H}}\otimes P_{n})=P_{n+1}$, where
$\mathbf{1}_{\mathcal{H}}$ is the identity operator in $\mathcal{H}$. From
here the first formula in (50) follows.

Next, it follows from the proof of [6, Theorem 3.1] that

$P_{n}=\frac{1}{n}(\mathbf{1}_{\mathcal{H}}\otimes
P_{n-1})(\mathbf{1}+U_{1}+U_{1}U_{2}+\dots+U_{1}U_{2}\dotsm U_{n-1}).$ (51)

Denote by $a^{-}_{\mathrm{free}}(f)$ the linear operator in
$\mathbb{F}_{\mathrm{fin}}(\mathcal{H})$ given by

$(a^{-}_{\mathrm{free}}(f)f^{(n)})(x_{1},\dots,x_{n-1}):=n\int_{X}\langle
f(x),f^{(n)}(x,x_{1},\dots,x_{n-1})\rangle_{V}\,dx,\quad
f^{(n)}\in\mathcal{H}^{\otimes n}.$ (52)

By (32), (49), (51), and (52), we get, for each
$f^{(n)}\in\mathcal{H}^{\otimes n}$,

$\displaystyle a^{-}(f)P_{n}f^{(n)}$
$\displaystyle=a_{\mathrm{free}}^{-}(f)P_{n}f^{(n)}$
$\displaystyle=a_{\mathrm{free}}^{-}(f)\frac{1}{n}\,(\mathbf{1}_{\mathcal{H}}\otimes
P_{n-1})(\mathbf{1}+U_{1}+U_{1}U_{2}+\dots+U_{1}U_{2}\dotsm U_{n-1})f^{(n)}$
$\displaystyle=P_{n-1}\,\frac{1}{n}\,a_{\mathrm{free}}^{-}(f)(\mathbf{1}+U_{1}+U_{1}U_{2}+\dots+U_{1}U_{2}\dotsm
U_{n-1})f^{(n)}$ $\displaystyle=P_{n-1}\mathcal{A}^{-}(f)f^{(n)}.\qed$

Formula (46) trivially holds, so we only need to prove (47). By Lemma 4.3,

$\displaystyle\tau\big{(}(a^{+}(f_{1})+a^{-}(f^{\prime}_{1}))\dotsm(a^{+}(f_{2n})+a^{-}(f^{\prime}_{2n}))\big{)}$
$\displaystyle\quad=\big{(}(\mathcal{A}^{+}(f_{1})+\mathcal{A}^{-}(f^{\prime}_{1}))\dotsm(\mathcal{A}^{+}(f_{2n})+\mathcal{A}^{-}(f^{\prime}_{2n}))\Omega,\Omega\big{)}_{\mathbb{F}(\mathcal{H})}.$
(53)

For each $f\in\mathcal{H}$ and $k\in\mathbb{N}$, we define a linear operator
$\mathcal{A}^{-}(f,k)$ in $\mathbb{F}_{\mathrm{fin}}(\mathcal{H})$ as follows:
if $f^{(n)}\in\mathcal{H}^{\otimes n}$ and $n<k$, then
$\mathcal{A}^{-}(f,k)f^{(n)}=0$, and if $n\geq k$, then

$\displaystyle\left(\mathcal{A}^{-}(f,k)f^{(n)}\right)(x_{1},\dots,x_{n-1})=\int_{X}\operatorname{Tr}_{1}Q_{2}(x,x_{1})Q_{3}(x,x_{2})\dotsm
Q_{k}(x,x_{k-1})$ $\displaystyle\qquad\quad f(x)\otimes
f^{(n)}(x_{1},\dots,x_{k-1},x,x_{k},\dots,x_{n-1})\,dx.$ (54)

Here, for $k=1$, the operator $Q_{2}(x,x_{1})Q_{3}(x,x_{2})\dotsm
Q_{k}(x,x_{k-1})$ is understood as the identity operator. We may equivalently
write formula (54) as follows:

$\displaystyle\left(\mathcal{A}^{-}(f,k)f^{(n)}\right)(x_{1},\dots,x_{k-1},x_{k+1},\dots,x_{n})=\int_{X^{2}}\operatorname{Tr}_{1}Q_{2}(x,x_{1})Q_{3}(x,x_{2})\dotsm
Q_{k}(x,x_{k-1})$ $\displaystyle\qquad\quad f(x)\otimes
f^{(n)}(x_{1},\dots,x_{k-1},x_{k},x_{k+1},\dots,x_{n})\,\sigma^{(2)}(dx\,dx_{k}).$
(55)

We will say that the operator $\mathcal{A}^{-}(f,k)$ annihilates the $x_{k}$
variable.

By (49) and (54), $\mathcal{A}^{-}(f)=\sum_{k\geq 1}\mathcal{A}^{-}(f,k)$, or
equivalently, for each $f^{(n)}\in\mathcal{H}^{\otimes n}$,
$\mathcal{A}^{-}(f)f^{(n)}=\sum_{k=1}^{n}\mathcal{A}^{-}(f,k)f^{(n)}$. Hence,
we can express the right-hand side of (53) as a summation over all partitions
$\xi\in\mathcal{P}^{(2n)}$ so that each $\xi$ of the form (38), (39)
corresponds to the term in which the creation operators are at places
$j_{1},\dots,j_{n}$, the annihilation operators are at places
$i_{1},\dots,i_{n}$, and the annihilation operator at place $i_{k}$
annihilates the variable created by the creation operator at place $j_{k}$. By
(40)–(42), this gives

$\displaystyle\tau\big{(}(a^{+}(f_{1})+a^{-}(f^{\prime}_{1}))\dotsm(a^{+}(f_{2n})+a^{-}(f^{\prime}_{2n}))\big{)}$
$\displaystyle\quad=\sum_{\xi\in\mathcal{P}^{(2n)}}\mathcal{A}^{-}(h_{i_{n}},l_{n})\mathcal{A}^{+}(h_{i_{n}-1})\dotsm\mathcal{A}^{+}(h_{i_{n-1}+1})$
$\displaystyle\qquad\mathcal{A}^{-}(h_{i_{n-1}},l_{n-1})\mathcal{A}^{+}(h_{i_{n-1}+1})\dotsm\mathcal{A}^{+}(h_{i_{n-2}+1})\dotsm\mathcal{A}^{-}(h_{i_{1}},l_{1})\mathcal{A}(h_{i_{1}+1})\dotsm\mathcal{A}^{+}(h_{2n})\Omega.$
(56)

For a fixed $\xi\in\mathcal{P}^{(2n)}$, let us now calculate the value of the
expression in the sum appearing in (56). To this end, we introduce the
following notations. Assume $\mathbf{R}$ is a subset of $\\{1,\dots,2n\\}$ and
for some $k\in\\{1,\dots,2n\\}$ let
$\mathbf{S}=\\{k,k+1,\dots,2n\\}\setminus\mathbf{R}$. We write
$\mathbf{R}=\\{\mathbf{r}_{1},\mathbf{r}_{2},\dots,\mathbf{r}_{l}\\}$ and
$\mathbf{S}=\\{\mathbf{s}_{1},\mathbf{s}_{2},\dots,\mathbf{s}_{m}\\}$ with
$\mathbf{s}_{1}<\mathbf{s}_{2}<\dotsm<\mathbf{s}_{m}$. Then, for a function
$\psi^{(m)}:X^{m}\to\mathbb{C}$, we will use the notation

$\displaystyle\psi^{(m)}(x_{k},x_{k+1},\dots,x_{2n}\setminus
x_{\mathbf{r}_{1}},x_{\mathbf{r}_{2}},\dots,x_{\mathbf{r}_{l}}):=\psi^{(m)}(x_{\mathbf{s}_{1}},x_{\mathbf{s}_{2}},\dots,x_{\mathbf{s}_{m}}).$

Next let $\zeta_{1},\zeta_{2},\dots,\zeta_{2l}$ be different numbers from the
set $\\{k,k+1,\dots,2n\\}$. Let

$\\{k,k+1,\dots,2n\\}\setminus\\{\zeta_{1},\zeta_{2},\dots,\zeta_{2l}\\}=\\{\gamma_{1},\gamma_{2},\dots,\gamma_{2n-k+1-2l}\\}$

with $\gamma_{1}<\gamma_{2}<\dotsm<\gamma_{2n-k+1-2l}$. We define a linear
functional

$\mathbb{T}^{(2n-k+1)}\left(k;\,\zeta_{1},\zeta_{2}\mid\zeta_{3},\zeta_{4}\mid\dotsm\mid\zeta_{2l-1},\zeta_{2l}\right):V^{\otimes(2n-k+1)}\to
V^{\otimes(2n-k+1-2l)}$

by

$\displaystyle\mathbb{T}^{(2n-k+1)}\left(k;\,\zeta_{1},\zeta_{2}\mid\zeta_{3},\zeta_{4}\mid\dotsm\mid\zeta_{2l-1},\zeta_{2l}\right)v_{k}\otimes
v_{k+1}\otimes\dotsm\otimes v_{2n}$ $\displaystyle\quad=\langle
v_{\zeta_{1}},v_{\zeta_{2}}\rangle_{V}\langle
v_{\zeta_{3}},v_{\zeta_{4}}\rangle_{V}\dotsm\langle
v_{\zeta_{2l-1}},v_{\zeta_{2l}}\rangle_{V}\,v_{\gamma_{1}}\otimes
v_{\gamma_{2}}\otimes\dotsm\otimes v_{2n-k+1-2l}$

for $v_{k},\dots,v_{n}\in V$.

Recall that, for a linear operator $\mathcal{C}\in\mathcal{L}(V^{\otimes 2})$
and $1\leq i<j\leq m$, we defined a linear operator
$\mathcal{C}[i,j]\in\mathcal{L}(V^{\otimes m})$. Now, for $k\geq 1$ and $k\leq
i<j\leq k+m-1$, we define an operator
$\mathcal{C}[k;\,i,j]\in\mathcal{L}(V^{\otimes m})$ by

$\mathcal{C}[k;\,i,j]:=\mathcal{C}[i-k+1,j-k+1].$

This definition can be interpreted as follows. We enumerate the components of
$V^{\otimes m}$ as $k,k+1,\dots,k+m-1$ and the operator $\mathcal{C}[k;\,i,j]$
acts as the operator $\mathcal{C}$ on the variables $i$ and $j$. In
particular, $\mathcal{C}[1;i,j]=\mathcal{C}[i,j]$.

We have

$\displaystyle\left(\mathcal{A}^{+}(h_{{i_{1}}+1})\dotsm\mathcal{A}^{+}(h_{2n})\Omega\right)(x_{i_{1}+1},\dots,x_{2n})=h_{i_{1}+1}(x_{i_{1}+1})\otimes\dotsm\otimes
h_{2n}(x_{2n})$

and by (55),

$\displaystyle\left(\mathcal{A}^{-}(h_{i_{1}},l_{1})\mathcal{A}^{+}(h_{i_{1}+1})\mathcal{A}^{+}(h_{2n})\Omega\right)(x_{i_{1}},\dots,x_{i_{2n}}\setminus
x_{i_{1}},x_{j_{1}})$
$\displaystyle\quad=\int_{X^{2}}\sigma^{(2)}(dx_{i_{1}}\,dx_{j_{1}})\operatorname{Tr}_{1}Q_{2}(x_{i_{1}},x_{j_{1}^{(1)}})Q_{3}(x_{i_{1}},x_{j_{2}^{(1)}})$
$\displaystyle\qquad\dotsm
Q_{l_{1}}(x_{i_{1}},x_{j_{l_{1}-1}^{(1)}})h_{i_{1}}(x_{i_{1}})\otimes
h_{i_{2}}(x_{i_{2}})\otimes\dotsm\otimes h_{2n}(x_{2n})$
$\displaystyle\quad=\int_{X^{2}}\sigma^{(2)}(dx_{i_{1}}\,dx_{j_{1}})\,\mathbb{T}^{(2n-i_{1}+1)}(i_{1};\,i_{1},\mathbf{j}_{1}^{(1)})$
$\displaystyle\qquad
Q(x_{i_{1}},x_{j_{1}^{(1)}})[i_{1};\,\mathbf{j}_{1}^{(1)},\mathbf{j}_{2}^{(1)}]Q(x_{i_{1}},x_{j_{2}^{(1)}})[i_{1};\,\mathbf{j}_{2}^{(1)},\mathbf{j}_{3}^{(1)}]\dotsm
Q(x_{i_{1}},x_{j_{{l_{1}}-1}^{(1)}})[i_{1};\,\mathbf{j}_{l_{1}-1}^{(1)},\mathbf{j}_{l_{1}}^{(1)}]$
$\displaystyle\qquad h_{i_{1}}(x_{i_{1}})\otimes
h_{i_{2}}(x_{i_{2}})\otimes\dotsm\otimes h_{2n}(x_{2n}).$

Next, we have

$\displaystyle\left(\mathcal{A}^{+}(h_{i_{2}+1})\dotsm\mathcal{A}^{+}(h_{i_{1}-1})\mathcal{A}^{-}(h_{i_{1}},l_{1})\mathcal{A}^{+}(h_{{i_{1}}+1})\dotsm\mathcal{A}^{+}(h_{2n})\Omega\right)(x_{{i_{2}}+1},x_{{i_{2}}+2}\dots,x_{2n}\setminus
x_{i_{1}},x_{j_{1}})$
$\displaystyle\quad=\int_{X^{2}}\sigma^{(2)}(dx_{i_{1}}\,dx_{j_{1}})h_{{i_{2}}+1}(x_{{i_{2}}+1})\otimes\dotsm\otimes
h_{{i_{1}}-1}(x_{{i_{1}}-1})\otimes\mathbb{T}^{(2n-i_{1}+1)}(i_{1};\,i_{1},\mathbf{j}_{1}^{(1)})$
$\displaystyle\qquad
Q(x_{i_{1}},x_{j_{1}^{(1)}})[i_{1};\,\mathbf{j}_{1}^{(1)},\mathbf{j}_{2}^{(1)}]Q(x_{i_{1}},x_{j_{2}^{(1)}})[i_{1};\,\mathbf{j}_{2}^{(1)},\mathbf{j}_{3}^{(1)}]\dotsm
Q(x_{i_{1}},x_{j_{{l_{1}}-1}^{(1)}})[i_{1};\,\mathbf{j}_{l_{1}-1}^{(1)},\mathbf{j}_{l_{1}}^{(1)}]$
$\displaystyle\qquad h_{i_{1}}(x_{i_{1}})\otimes
g_{i_{2}}(x_{i_{2}})\otimes\dotsm\otimes g_{2n}(x_{2n})$
$\displaystyle\quad=\int_{X^{2}}\sigma^{(2)}(dx_{i_{1}}\,dx_{j_{1}})\mathbb{T}^{(2n-i_{2})}(i_{2}+1;\,i_{1},\mathbf{j}_{1}^{(1)})$
$\displaystyle\qquad
Q(x_{i_{1}},x_{j_{1}^{(1)}})[i_{2}+1;\,\mathbf{j}_{1}^{(1)},\mathbf{j}_{2}^{(1)}]Q(x_{i_{1}},x_{j_{2}^{(1)}})[i_{2}+1;\,\mathbf{j}_{2}^{(1)},\mathbf{j}_{3}^{(1)}]$
$\displaystyle\qquad\dotsm
Q(x_{i_{1}},x_{j_{l_{1}-1}^{(1)}})[i_{2}+1;\,\mathbf{j}_{l_{1}-1}^{(1)},\mathbf{j}_{l_{1}}^{(1)}]h_{i_{2}+1}(x_{i_{2}+1})\otimes
h_{i_{2}+2}(x_{i_{2}+2})\otimes\dotsm\otimes h_{2n}(x_{2n}).$

From here, using the commutativity of the functionals, respectively operators,
acting in different variables, we similarly get:

$\displaystyle\big{(}\mathcal{A}^{-}(h_{i_{2}},l_{2})\mathcal{A}^{+}(h_{{i_{2}}+1})\dotsm\mathcal{A}^{+}(h_{{i_{1}}-1})\mathcal{A}^{-}(h_{i_{1}},l_{1})\mathcal{A}^{+}(h_{{i_{1}}+1})$
$\displaystyle\quad\dotsm\mathcal{A}^{+}(h_{2n})\Omega\big{)}(x_{i_{2}},x_{{i_{2}}+1},\dots,x_{2n}\setminus
x_{i_{1}},x_{j_{1}},x_{i_{2}},x_{j_{2}})$
$\displaystyle\quad=\int_{X^{4}}\sigma^{(2)}(dx_{i_{2}}\,dx_{j_{2}})\sigma^{(2)}(dx_{i_{1}}\,dx_{j_{1}})\mathbb{T}^{(2n-i_{2}+1)}(i_{2};\,i_{1},\mathbf{j}_{1}^{(1)}\mid
i_{2},\mathbf{j}_{1}^{(2)})$ $\displaystyle\qquad
Q(x_{i_{2}},x_{j_{1}^{(2)}})[i_{2};\,\mathbf{j}_{1}^{(2)},\mathbf{j}_{2}^{(2)}]Q(x_{i_{2}},x_{j_{2}^{(2)}})[i_{2};\,\mathbf{j}_{2}^{(2)},\mathbf{j}_{3}^{(2)}]\dotsm
Q(x_{i_{2}},x_{j_{l_{2}-1}^{(2)}})[i_{2};\,\mathbf{j}_{l_{2}-1}^{(2)},\mathbf{j}_{l_{2}}^{(2)}]$
$\displaystyle\qquad
Q(x_{i_{1}},x_{j_{1}^{(1)}})[i_{2};\,\mathbf{j}_{1}^{(1)},\mathbf{j}_{2}^{(1)}]Q(x_{i_{1}},x_{j_{2}^{(1)}})[i_{2};\,\mathbf{j}_{2}^{(1)},\mathbf{j}_{3}^{(1)}]\dotsm
Q(x_{i_{1}},x_{j_{l_{1}-1}^{(1)}})[i_{2};\,\mathbf{j}_{l_{1}-1}^{(2)},\mathbf{j}_{l_{1}}^{(1)}]$
$\displaystyle\qquad h_{i_{2}}(x_{i_{2}})\otimes
h_{{i_{2}}+1}(x_{{i_{2}}+1})\otimes\dotsm\otimes h_{2n}(x_{2n}).$

Continuing by analogy, we get at the $k$th step:

$\displaystyle\big{(}\mathcal{A}^{-}(h_{i_{k}},l_{k})\mathcal{A}^{+}(h_{{i_{k}}+1})\dotsm\mathcal{A}^{+}(h_{{i_{k-1}}+1})\mathcal{A}^{-}(h_{i_{k-1}},l_{k-1})\mathcal{A}^{+}(h_{i_{k-1}+1})\dotsm
A^{+}(h_{{i_{k-2}}+1})$
$\displaystyle\qquad\dotsm\mathcal{A}^{-}(h_{i_{2}},l_{2})\mathcal{A}^{+}(h_{{i_{2}}+1})\dotsm\mathcal{A}^{+}(h_{{i_{1}}-1})\mathcal{A}^{-}(h_{i_{1}},l_{1})\mathcal{A}^{+}(h_{{i_{1}}+1})\dotsm\mathcal{A}^{+}(h_{2n})\Omega\big{)}$
$\displaystyle\qquad(x_{i_{k}},x_{i_{k}+1},\dots,x_{2n}\setminus
x_{i_{1}},x_{j_{1}},x_{i_{2}},x_{j_{2}},\dots,x_{i_{n}},x_{j_{n}})$
$\displaystyle\quad=\int_{X^{2k}}\sigma^{(2)}(dx_{i_{1}}dx_{j_{1}})\sigma^{(2)}(dx_{i_{2}}dx_{j_{2}})\dotsm\sigma^{(2)}(dx_{i_{k}}dx_{j_{k}})$
$\displaystyle\qquad\mathbb{T}^{(2n-i_{k}+1)}(i_{k};i_{1},\mathbf{j}_{1}^{(1)}\mid
i_{2},\mathbf{j}_{1}^{(2)}\mid\dotsm\mid i_{k},\mathbf{j}_{1}^{(k)})$
$\displaystyle\qquad
Q(x_{i_{k}},x_{j_{1}^{(k)}})[i_{k};\mathbf{j}_{1}^{(k)},\mathbf{j}_{2}^{(k)}]Q(x_{i_{k}},x_{j_{2}^{(k)}})[i_{k};\mathbf{j}_{2}^{(k)},\mathbf{j}_{3}^{(k)}]\dotsm
Q(x_{i_{k}},x_{j_{l_{k}-1}^{(k)}})[i_{k};\mathbf{j}_{l_{k}-1}^{(k)},\mathbf{j}_{l_{k}}^{(k)}]$
$\displaystyle\qquad
Q(x_{i_{k-1}},x_{j_{1}^{(k-1)}})[i_{k};\mathbf{j}_{1}^{(k-1)},\mathbf{j}_{2}^{(k-1)}]Q(x_{i_{k-1}},x_{j_{2}^{(k-1)}})[i_{k};\mathbf{j}_{2}^{(k-1)},\mathbf{j}_{3}^{(k-1)}]$
$\displaystyle\qquad\dotsm
Q(x_{i_{k-1}},x_{j_{l_{k-1}-1}^{(k-1)}})[i_{k};\mathbf{j}_{l_{k-1}-1}^{(k-1)},\mathbf{j}_{l_{k-1}}^{(k-1)}]$
$\displaystyle\qquad\dotsm
Q(x_{i_{1}},x_{j_{1}^{(1)}})[i_{k};\mathbf{j}_{1}^{(1)},\mathbf{j}_{2}^{(1)}]Q(x_{i_{1}},x_{j_{2}^{(1)}})[i_{k};\mathbf{j}_{2}^{(1)},\mathbf{j}_{3}^{(1)}]\dotsm
Q(x_{i_{1}},x_{j_{l_{1}-1}^{(1)}})[i_{k};\mathbf{j}_{l_{1}-1}^{(1)},\mathbf{j}_{l_{1}}^{(1)}]$
$\displaystyle\qquad h_{i_{k}}(x_{i_{k}})\otimes
h_{i_{k+1}}(x_{{i_{k}}+1})\otimes\dotsm\otimes h_{2n}(x_{2n}).$

Thus, after $n$ steps, we get, by using the definitions (43)–(45),

$\displaystyle\big{(}\mathcal{A}^{-}(h_{i_{n}},l_{n})\mathcal{A}^{+}(h_{i_{n}-1})\dotsm\mathcal{A}^{+}(h_{i_{n-1}+1})\mathcal{A}^{-}(h_{i_{n-1}},l_{n-1})\mathcal{A}^{+}(h_{i_{n-1}+1})\dotsm\mathcal{A}^{+}(h_{i_{n-2}+1})$
$\displaystyle\quad\dotsm\mathcal{A}^{-}(h_{i_{1}},l_{1})\mathcal{A}(h_{i_{1}+1})\dotsm\mathcal{A}^{+}(h_{2n})\Omega\big{)}(x_{1},x_{2},\dots,x_{n})$
(57)
$\displaystyle\quad=\int_{X^{2n}}\sigma^{(2)}(dx_{i_{1}}\,dx_{j_{1}})\sigma^{(2)}(dx_{i_{2}}\,dx_{j_{2}})\dotsm\sigma^{(2)}(dx_{i_{n}}\,dx_{j_{n}})$
$\displaystyle\qquad\times\mathbb{T}^{(2n)}(\xi)Q(\xi;x_{1},x_{2},\dots,x_{2n})h_{1}(x_{1})\otimes
h_{2}(x_{2})\otimes\dotsm\otimes h_{2n}(x_{2n}).$ (58)

Formulas (56) and (58) imply (47). ∎

######  Remark 4.4.

We can unify formulas (46) and (47) into a single formula

$\displaystyle\tau\big{(}(a^{+}(f_{1})+a^{-}(f_{1}^{\prime}))\dotsm(a^{+}(f_{n})+a^{-}(f_{n}^{\prime}))\big{)}=\sum_{\xi\in\mathcal{P}^{(n)}}\int_{X^{n}}\bigotimes_{\begin{subarray}{c}\\{i,j\\}\in\xi\\\
i<j\end{subarray}}\sigma^{(2)}(dx_{i}\,dx_{j})$
$\displaystyle\qquad\times\mathbb{T}^{(n)}(\xi)Q(\xi;x_{1},\dots,x_{n})h_{1}(x_{1})\otimes
h(x_{2})\otimes\dotsm\otimes h_{n}(x_{n})$ (59)

which holds for all $n\in\mathbb{N}$. Indeed, if $n$ is even, then (4.4)
becomes (47) and if $n$ is odd, then the set $\mathcal{P}^{(n)}$ is empty,
hence the right hand-side of (4.4) is equal to zero.

Recall that the operator $Q(\xi;x_{1},\dots,x_{n})$ was defined through
$\xi\in\mathcal{P}^{(n)}$ and operators $Q(x,x^{\prime})$. But, for $x=(y,z)$
and $x^{\prime}=(y^{\prime},z^{\prime})$, we have
$Q(x,x^{\prime})=Q(y,y^{\prime})$. Therefore, we can write the operator
$Q(\xi;x_{1},\dots,x_{n})$ as $Q(\xi;y_{1},\dots,y_{n})$.

Similarly to the definition (48) of the measure $\sigma^{(2)}$ on $X^{2}$, we
define the measure $\nu^{(2)}$ on $Y^{2}$.

In accordance with (31), we define $b(f):=a^{+}(f)+a^{-}(Jf)$. We can easily
see that Theorem 4.2 implies the following

######  Corollary 4.5.

Let $\tau$ be the Fock state. For any $g_{1},g_{2}\in\mathcal{G}$, we define a
complex-valued measure $\lambda^{(2)}[g_{1},g_{2}]$ on $Y^{2}$ by

$\lambda^{(2)}[g_{1},g_{2}](dy_{1}\,dy_{2})=(g_{2},g_{1})_{\mathcal{G}}\,\nu^{(2)}(dy_{1}\,dy_{2}).$
(60)

Then, for any $\varphi_{1},\dots,\varphi_{n}\in C_{0}(Y;V_{\mathbb{R}})$ and
$g_{1},\dots,g_{n}\in\mathcal{G}$, we have

$\displaystyle\tau\big{(}b(\varphi_{1}\otimes g_{1})\dotsm
b(\varphi_{n}\otimes g_{n})\big{)}$
$\displaystyle=\sum_{\xi\in\mathcal{P}^{(n)}}\int_{Y^{n}}\bigotimes_{\begin{subarray}{c}\\{i,j\\}\in\xi\\\
i<j\end{subarray}}\lambda^{(2)}[g_{i},g_{j}](dy_{i}\,dy_{j})$
$\displaystyle\quad\mathbb{T}^{(n)}(\xi)Q(\xi;y_{1},\dots,y_{n})\varphi_{1}(y_{1})\otimes\varphi_{2}(y_{2})\otimes\dotsm\otimes\varphi_{n}(y_{n}).$

Below we will also need a formula for

$\displaystyle\tau\left(a^{-}(f_{1})\cdots a^{-}(f_{n})a^{+}(f_{n+1})\cdots
a^{+}(f_{2n})\right)$ (61)

This is, of course, a special case of formula (47), which can be simplified in
the case of (61).

For each permutation $\pi\in S_{n}$, we define a partition
$\xi\in\mathcal{P}^{(2n)}$ as follows:

$\xi=\big{\\{}\\{1,n+\pi(n)\\},\\{2,n+\pi(n-1)\\},\dots,\\{n,n+\pi(1)\\}\big{\\}}.$

We denote by $\mathcal{S}_{n}$ the subset of $\mathcal{P}^{(2n)}$ consisting
of all such partitions. Equivalently, $\mathcal{S}_{n}$ consists of all
partitions $\xi\in\mathcal{P}^{(2n)}$ such that, for each $\\{i,j\\}\in\xi$
with $i<j$, we have $i\leq n$ and $j\geq n+1$. The following corollary is
immediate.

######  Corollary 4.6.

Let $\tau$ be the Fock state. Let $\varphi_{1},\dots,\varphi_{2n}\in
C_{0}(Y;V)$ and $g_{1},\dots,g_{2n}\in\mathcal{G}$. Let the complex-valued
measure $\lambda^{(2)}[g_{i},g_{j}]$ be defined by (60). We have

$\displaystyle\tau\big{(}a^{-}(\varphi_{1}\otimes g_{1})\cdots
a^{-}(\varphi_{n}\otimes g_{n})a^{+}(\varphi_{n+1}\otimes g_{n+1})\cdots
a^{+}(\varphi_{2n}\otimes g_{2n})\big{)}$
$\displaystyle\quad=\sum_{\xi\in\mathcal{S}_{n}}\int_{Y^{2n}}\bigotimes_{\begin{subarray}{c}\\{i,j\\}\in\xi\\\
i<j\end{subarray}}\lambda^{(2)}[g_{i},g_{j}](dy_{i}\,dy_{j})$
$\displaystyle\qquad\times\mathbb{T}^{(2n)}Q(\xi;y_{1},\dots,y_{2n})\varphi_{1}(y_{1})\otimes\varphi_{2}(y_{2})\otimes\cdots\otimes\varphi_{2n}(y_{2n}).$

Here $\mathbb{T}^{(2n)}:V^{\otimes 2n}\to\mathbb{C}$ is the linear functional
defined by (36) and the linear operator $Q(\xi;x_{1},\dots,x_{2n})$ is defined
by formula (44) in which

$\displaystyle Q^{(k)}(\xi;x_{1},\dots,x_{2n})=$ $\displaystyle
Q_{n+k}(x_{n-k+1},x_{j_{1}^{(k)}})Q_{n+k+1}(x_{n-k+1},x_{j_{2}^{(k)}})$
$\displaystyle\quad\cdots Q_{n+k+l_{k}-1}(x_{n-k+1},x_{j_{l_{k}-1}^{(k)}}),$

where, for
$\xi=\big{\\{}\\{1,j_{n}\\},\\{2,j_{n-1}\\},\dots,\\{n,j_{1}\\}\big{\\}}\in\mathcal{S}_{n}$,
we denote $J=\\{j_{1},j_{2},\dots,j_{n}\\}$ and

$\displaystyle J^{(k)}$ $\displaystyle=\big{\\{}j\in J\mid j\leq j_{k},\,j\neq
j_{1},\dots,j\neq j_{k-1}\big{\\}},$
$\displaystyle=\big{\\{}j_{1}^{(k)},j_{2}^{(k)},\dots,j_{l_{k}}^{(k)}\big{\\}},\quad
j_{1}^{(k)}<j_{2}^{(k)}<\dots<j_{l_{k}}^{(k)}.$

##  5 Quasi-free states on the $Q$-MCR algebra

Let $\tau$ be a state on the $Q$-MCR algebra $\mathbb{A}$. We assume that, for
any $\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$ and any
$g_{1},\dots,g_{n}\in\mathcal{G}$, the functional
$\tau^{(n)}_{\sharp_{1},\dots,\sharp_{n}}(\cdot,g_{1},\dots,g_{n})$ is
continuous on $C_{0}(Y^{n};V^{\otimes n})$. In view of Corollaries 4.5 and
4.6, we now give the following

######  Definition 5.1.

(i) Assume that, for any $g_{1},g_{2}\in\mathcal{G}$, there exists a complex-
valued measure $\lambda^{(2)}[g_{1},g_{2}]$ on $Y^{2}$ such that, for all
$\varphi_{1},\varphi_{2}\in C_{0}(Y;V_{\mathbb{R}})$,

$\displaystyle\tau\big{(}B(\varphi_{1}\otimes g_{1})B(\varphi_{2}\otimes
g_{2})\big{)}=\int_{Y^{2}}\langle\varphi_{1}(y_{1}),\varphi_{2}(y_{2})\rangle_{V}\,\lambda^{(2)}[g_{1},g_{2}](dy_{1}\,dy_{2}).$

We say that $\tau$ is a strongly quasi-free state, if all
$\varphi_{1},\dots,\varphi_{n}\in C_{0}(Y;V_{\mathbb{R}})$ and
$g_{1},\dots,g_{n}\in\mathcal{G}$, we have

$\displaystyle\tau\big{(}B(\varphi_{1}\otimes g_{1})\dotsm
B(\varphi_{n}\otimes
g_{n})\big{)}=\sum_{\xi\in\mathcal{P}^{(n)}}\int_{Y^{n}}\bigotimes_{\begin{subarray}{c}\\{i,j\\}\in\xi\\\
i<j\end{subarray}}\lambda^{(2)}[g_{i},g_{j}](dy_{i}\,dy_{j})$
$\displaystyle\qquad\times\mathbb{T}^{(n)}(\xi)Q(\xi;y_{1},\dots,y_{n})\varphi_{1}(y_{1})\otimes\varphi_{2}(y_{2})\otimes\dotsm\otimes\varphi_{n}(y_{n}).$
(62)

(ii) Assume that, for any $g_{1},g_{2}\in\mathcal{G}$, there exists a complex-
valued measure $\rho^{(2)}[g_{1},g_{2}]$ on $Y^{2}$ such that, for all
$\varphi_{1},\varphi_{2}\in C_{0}(Y;V)$,

$\displaystyle\tau\left(A^{+}(\varphi_{1}\otimes
g_{1})A^{-}(\varphi_{2}\otimes
g_{2})\right)=\int_{Y^{2}}\rho^{(2)}[g_{1},g_{2}](dy_{1}\,dy_{2})\langle\varphi_{1}(y_{1}),\varphi_{2}(y_{2})\rangle_{V}.$

We say that $\tau$ is a gauge-invariant quasi-free state if for all
$m,n\in\mathbb{N}$, $\varphi_{1},\dots,\varphi_{m+n}\in C_{0}(Y;V)$ and
$g_{1},\dots,g_{m+n}\in\mathcal{G}$, we have

$\displaystyle\tau\left(A^{+}(\varphi_{1}\otimes g_{1})\dotsm
A^{+}(\varphi_{m}\otimes g_{m})A^{-}(\varphi_{m+1}\otimes g_{m+1})\dotsm
A^{-}(\varphi_{m+n}\otimes g_{m+n})\right)$
$\displaystyle\quad=\delta_{m,n}\sum_{\xi\in\mathcal{S}_{n}}\int_{Y^{2n}}\bigotimes_{\begin{subarray}{c}\\{i,j\\}\in\xi\\\
i<j\end{subarray}}\rho^{(2)}[g_{i},g_{j}](dy_{i}\,dy_{j})$
$\displaystyle\qquad\times\mathbb{T}^{(2n)}Q(\xi;y_{1},\dots,y_{2n})\varphi_{1}(y_{1})\otimes\varphi_{2}(y_{2})\otimes\cdots\otimes\varphi_{2n}(y_{2n}).$
(63)

In formula (63), $\delta_{m,n}$ denotes the Kronecker delta.

######  Remark 5.2.

Both formulas (62) and (63) imply that the state $\tau$ is completely
determined by the operator-valued function $Q(y_{1},y_{2})$ and by the values

$\tau\big{(}B(\varphi_{1}\otimes g_{1})B(\varphi_{2}\otimes
g_{2})\big{)}\text{ and }\tau\left(A^{+}(\varphi_{1}\otimes
g_{1})A^{-}(\varphi_{2}\otimes g_{2})\right),$

respectively.

Corollary 4.5 implies the following

######  Proposition 5.3.

The Fock state on the $Q$-MCR algebra is strongly quasi-free and the
corresponding measure $\lambda^{(2)}[g_{1},g_{2}](dy_{1}\,dy_{2})$ is given by
formula (60). The Fock state is gauge-invariant quasi-free, with
$\rho^{(2)}[g_{1},g_{2}](dy_{1}\,dy_{2})=0$.

We will now construct a class of gauge-invariant quasi-free states and
strongly quasi-free states on the $Q$-MCR algebra. We will be able to do this
under additional assumptions on the operator-valued function $Q$.

######  Assumption 5.4.

We assume:

  * (i)

For all $y_{1},y_{2}\in Y$, $\widetilde{Q}(y_{1},y_{2})$ is unitary and
$\widetilde{Q}(y_{1},y_{2})^{*}=\widetilde{Q}(y_{2},y_{1})$.

  * (ii)

For all $y_{1},y_{2}\in Y$, we have
$\widetilde{\widetilde{Q}}(y_{1},y_{2})=Q(y_{1},y_{2})$.

  * (iii)

For all $y_{1},y_{2}\in Y$, we have $\widehat{Q}(y_{1},y_{2})=Q(y_{1},y_{2})$
and $\widehat{\widetilde{Q}}(y_{1},y_{2})=\widetilde{Q}(y_{1},y_{2})$.

  * (iv)

There exists a constant $\varkappa\in\mathbb{R}$ such that, for all $y\in Y$
and $v^{(2)}\in V^{\otimes 2}$, we have
$\operatorname{Tr}\big{(}\widetilde{Q}(y,y)v^{(2)}\big{)}=\varkappa\operatorname{Tr}v^{(2)}$.

  * (v)

Let $Y_{1}$ and $Y_{2}$ be two copies of $Y$ and let $\mathbf{Y}:=Y_{1}\sqcup
Y_{2}$. Define a (continuous) function
$\mathbf{Q}:\mathbf{Y}^{2}\to\mathcal{L}\left(V^{\otimes 2}\right)$ as
follows: ${\bf Q}(y_{1},y_{2}):=Q(y_{1},y_{2})$ if either $y_{1},y_{2}\in
Y_{1}$ or $y_{1},y_{2}\in Y_{2}$, and
$\mathbf{Q}(y_{1},y_{2}):=\widetilde{Q}(y_{2},y_{1})$ if either $y_{1}\in
Y_{1}$, $y_{2}\in Y_{2}$ or $y_{1}\in Y_{2}$, $y_{2}\in Y_{1}$. [Note that, by
(i) and (ii), for any $(y_{1},y_{2})\in\mathbf{Y}^{2}$,
$\mathbf{Q}(y_{1},y_{2})$ is a unitary operator and
$\mathbf{Q}(y_{1},y_{2})^{*}=\mathbf{Q}(y_{2},y_{1})$.] Then the function
$\mathbf{Q}:\mathbf{Y}^{2}\to\mathcal{L}\left(V^{\otimes 2}\right)$ satisfies
the functional Yang–Baxter equation point-wise, cf. (1).

Similarly to Assumption 5.4, (v), we define $\mathbf{X}:=X_{1}\sqcup X_{2}$
and let $\mathbf{Q}:\mathbf{X}^{2}\to\mathcal{L}(V^{\otimes 2})$ be defined by
$\mathbf{Q}(x_{1},x_{2}):=\mathbf{Q}(y_{1},y_{2})$, where
$x_{i}=(y_{i},z_{i})\in\mathbf{X}$, $i=1,2$. For

$\mathbf{f}\in L^{2}(\mathbf{X};V)=L^{2}(X_{1};V)\oplus
L^{2}(X_{2};V)=\mathcal{H}\oplus\mathcal{H},$

we construct the standard creation operator $\mathbf{a}^{+}(\mathbf{f})$ and
the standard annihilation operator $\mathbf{a}^{-}(\mathbf{f})$ in the
$\mathbf{Q}$-symmetric Fock space
$\mathcal{F}^{\mathbf{Q}}(L^{2}(\mathbf{X};V))$. Similarly to Section 4, we
then construct the $\mathbf{Q}$-MCR algebra, denoted by $\mathbf{A}$, and the
Fock state on $\mathbf{A}$, denoted by $\boldsymbol{\tau}$. In particular, we
think of $\mathbf{A}$ as the algebra spanned by operator-valued integrals

$\int_{{\bf X}^{n}}\langle\mathbf{f}^{(n)}(x_{1},\dots,x_{n}),{\bf
a}^{\sharp_{1}}(x_{1})\otimes\dotsm\otimes{\bf
a}^{\sharp_{n}}(x_{n})\rangle_{V^{\otimes n}}\,dx_{1}\dotsm dx_{n},$ (64)

where $\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$ and
$\mathbf{f}^{(n)}\in\boldsymbol{\mathfrak{F}}^{(n)}$, where the space
$\boldsymbol{\mathfrak{F}}^{(n)}$ is constructed similarly to
$\mathfrak{F}^{(n)}$ but by starting with $\mathbf{X}$ instead of $X$, compare
with (17)

Let $\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$, $i_{1},\dots,i_{n}\in\\{1,2\\}$
and let $f^{(n)}\in\mathfrak{F}^{(n)}$. We define

$\int_{X^{n}}\langle
f^{(n)}(x_{1},\dots,x_{n}),\mathbf{a}_{i_{1}}^{\sharp_{1}}(x_{1})\otimes\dotsm\otimes\mathbf{a}_{i_{n}}^{\sharp_{n}}(x_{n})\rangle_{V^{\otimes
n}}\,dx_{1}\dotsm dx_{n}$ (65)

to be equal to the operator-valued integral (64) in which

$\mathbf{f}^{(n)}(x_{1},\dots,x_{n})=\begin{cases}f^{(n)}(x_{1},\dots,x_{n}),&\text{if
}(x_{1},\dots,x_{n})\in X_{i_{1}}\times X_{i_{2}}\times\dotsm\times
X_{i_{n}},\\\ 0,&\text{otherwise}.\end{cases}$ (66)

In formula (66), we identified $(x_{1},\dots,x_{n})\in X_{i_{1}}\times
X_{i_{2}}\times\dotsm\times X_{i_{n}}$ with $(x_{1},\dots,x_{n})\in X^{n}$.
Thus, intuitively, for $i=1,2$ and $\sharp\in\\{+,-\\}$,
$\mathbf{a}^{\sharp}_{i}(\cdot)$ is the restriction of
$\mathbf{a}^{\sharp}(\cdot)$ to $X_{i}$.

As easily seen, the algebra $\mathbf{A}$ is spanned by the operator-valued
integrals of the form (65). Similarly to (18), we denote the integral in (65)
by $\boldsymbol{\Phi}(f^{(n)};i_{1},\sharp_{1},\dots,i_{n},\sharp_{n})$.

Formulas (21)–(24), Proposition 2.3 and Assumption 5.4 imply

######  Lemma 5.5.

Let $f^{(n)}\in\mathfrak{F}^{(n)}$, $i_{1},\dots,i_{n}\in\\{1,2\\}$,
$\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$, and $k\in\\{1,\dots,n-1\\}$. The
following commutation relations hold: if $\sharp_{k}=\sharp_{k+1}$ and
$i_{k}=i_{k+1}$,

$\boldsymbol{\Phi}(f^{(n)};i_{1},\sharp_{1},\dots,i_{n},\sharp_{n})=\boldsymbol{\Phi}(Q_{k}(x_{k},x_{k+1})f^{(n)}(x_{1},\dots,x_{k+1},x_{k},\dots,x_{n});i_{1},\sharp_{1},\dots,i_{n},\sharp_{n});$

if $\sharp_{k}=\sharp_{k+1}$ and $i_{k}\neq i_{k+1}$,

$\displaystyle\boldsymbol{\Phi}(f^{(n)};i_{1},\sharp_{1},\dots,i_{n},\sharp_{n})$
$\displaystyle\quad=\boldsymbol{\Phi}(\widetilde{Q}_{k}(x_{k+1},x_{k})f^{(n)}(x_{1},\dots,x_{k+1},x_{k},\dots,x_{n});i_{1},\sharp_{1},\dots,i_{k+1},\sharp_{k},i_{k},\sharp_{k+1},\dots,i_{n},\sharp_{n});$

if $\sharp_{k}=-$, $\sharp_{k+1}=+$ and $i_{k}=i_{k+1}$,

$\displaystyle\boldsymbol{\Phi}(f^{(n)};i_{1},\sharp_{1},\dots,i_{n},\sharp_{n})$
$\displaystyle\quad=\boldsymbol{\Phi}(\widetilde{Q}_{k}(x_{k+1},x_{k})f^{(n)}(x_{1},\dots,x_{k+1},x_{k},\dots,x_{n});i_{1},\sharp_{1},\dots,i_{k},\sharp_{k+1},i_{k+1},\sharp_{k},\dots,i_{n},\sharp_{n})$
$\displaystyle\qquad+\boldsymbol{\Phi}(g^{(n-2)};i_{1},\sharp_{1},\dots,i_{k-1},\sharp_{k-1},i_{k+2},\sharp_{k+2},\dots,i_{n},\sharp_{n}),$

where $g^{(n-2)}$ is given by (24); if $\sharp_{k}=-$, $\sharp_{k+1}=+$ and
$i_{k}\neq i_{k+1}$,

$\displaystyle\boldsymbol{\Phi}(f^{(n)};i_{1},\sharp_{1},\dots,i_{n},\sharp_{n})$
$\displaystyle\quad=\boldsymbol{\Phi}(Q_{k}(x_{k},x_{k+1})f^{(n)}(x_{1},\dots,x_{k+1},x_{k},\dots,x_{n});i_{1},\sharp_{1},\dots,i_{k+1},\sharp_{k+1},i_{k},\sharp_{k},\dots,i_{n},\sharp_{n});$

if $\sharp_{k}=+$, $\sharp_{k+1}=-$ and $i_{k}=i_{k+1}$,

$\displaystyle\boldsymbol{\Phi}(f^{(n)};i_{1},\sharp_{1},\dots,i_{n},\sharp_{n})$
$\displaystyle\quad=\boldsymbol{\Phi}(\widetilde{Q}_{k}(x_{k+1},x_{k})f^{(n)}(x_{1},\dots,x_{k+1},x_{k},\dots,x_{n});i_{1},\sharp_{1},\dots,i_{k},\sharp_{k+1},i_{k+1},\sharp_{k},\dots,i_{n},\sharp_{n})$
$\displaystyle\qquad-\varkappa\boldsymbol{\Phi}(g^{(n-2)};i_{1},\sharp_{1},\dots,i_{k-1},\sharp_{k-1},i_{k+2},\sharp_{k+2},\dots,i_{n},\sharp_{n}),$

where $g^{(n-2)}$ is given by (24); if $\sharp_{k}=+$, $\sharp_{k+1}=-$ and
$i_{k}\neq i_{k+1}$,

$\displaystyle\boldsymbol{\Phi}(f^{(n)};i_{1},\sharp_{1},\dots,i_{n},\sharp_{n})$
$\displaystyle\quad=\boldsymbol{\Phi}(Q_{k}(x_{k},x_{k+1})f^{(n)}(x_{1},\dots,x_{k+1},x_{k},\dots,x_{n});i_{1},\sharp_{1},\dots,i_{k+1},\sharp_{k+1},i_{k},\sharp_{k},\dots,i_{n},\sharp_{n}).$

Let us fix a bounded linear operator
$K\in\mathcal{L}\left(\mathcal{G}\right)$. In the case where $\varkappa\geq
0$, we assume that $K\geq 0$ and in the case where $\varkappa<0$, we assume
$0\leq K\leq-\frac{1}{\varkappa}$. Let $K_{1}:=\sqrt{K}$ and
$K_{2}:=\sqrt{1+\varkappa K}$. Furthermore, we assume that both operators
$K_{1}$ and $K_{2}$ are not equal to zero, equivalently $K\neq 0$ and
$K\neq-\frac{1}{\varkappa}$.

For a bounded linear operator $C\in\mathcal{L}(\mathcal{G})$, we denote
$C^{\prime}:=JCJ$, the complex conjugate of $C$. Here, just as above $J$ is
the antilinear operator of complex conjugation.

Let $f=\varphi\otimes g$ with $\varphi\in C_{0}(Y;V)$ and $g\in\mathcal{G}$.
We define

$\displaystyle A^{+}(f)=\int_{X}\langle\varphi(y)(K_{2}g)(z),{\bf
a}_{2}^{+}(x)\rangle_{V}\,dx+\int_{X}\langle\varphi(y)(K_{1}g)(z),{\bf
a}_{1}^{-}(x)\rangle_{V}\,dx,$ $\displaystyle
A^{-}(f)=\int_{X}\langle\varphi(y)(K_{2}^{\prime}g)(z),{\bf
a}_{2}^{-}(x)\rangle_{V}\,dx+\int_{X}\langle\varphi(y)(K_{1}^{\prime}g)(z),{\bf
a}_{1}^{+}(x)\rangle_{V}\,dx.$ (67)

We define the corresponding operator-valued distributions $A^{+}(x)$,
$A^{-}(x)$ through the formula

$A^{+}(f)=\int_{X}\langle f(x),A^{+}(x)\rangle_{V}\,dx,\quad
A^{-}(f)=\int_{X}\langle f(x),A^{-}(x)\rangle_{V}\,dx.$ (68)

######  Proposition 5.6.

We have $\left(A^{+}(f)\right)^{*}=A^{-}(Jf)$ and the operators $A^{+}(f)$,
$A^{-}(f)$ defined by (67) satisfy the $Q$-MCR, see (4)–(6).

###### Proof.

The proposition is a consequence of (67), Lemma 5.5, and the following
observation: for any $g_{1},g_{2}\in\mathcal{G}$,

$\displaystyle\int_{Z}(K_{2}^{\prime}g_{1})(z)(K_{2}g_{2})(z)dz-\varkappa\int_{Z}(K_{1}^{\prime}g_{1})(z)(K_{1}g_{2})(z)dz$
$\displaystyle\quad=(K_{2}g_{2},K_{2}Jg_{1})_{\mathcal{G}}-\varkappa(K_{1}g_{2},K_{1}Jg_{1})_{\mathcal{G}}$
$\displaystyle\quad=\big{(}(K_{2}^{2}-\varkappa
K_{1}^{2})g_{2},Jg_{1}\big{)}_{\mathcal{G}}=\int_{Z}g_{1}(z)g_{2}(z)dz.\qed$

Inspired by Proposition 5.6, we will now construct a $Q$-MCR algebra
$\mathbb{A}$. Formally, $\mathbb{A}$ is spanned by the operators

$\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n}):=\int_{X^{n}}\langle
f^{(n)}(x_{1},\dots,x_{n}),A^{\sharp_{1}}(x_{1})\otimes\dotsm\otimes
A^{\sharp_{n}}(x_{n})\rangle_{V^{\otimes n}}\,dx_{1}\dotsm dx_{n},$

where $f^{(n)}\in\mathfrak{F}^{(n)}$,
$\sharp_{1},\dots,\sharp_{n}\in\\{+,-\\}$. We now rigorously define
$\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n})$ as follows.

Denote $K(+,1):=K_{1}$, $K(+,2):=K_{2}$, $K(-,1):=K_{1}^{\prime}$,
$K(-,2):=K_{2}^{\prime}$, and denote $\gamma(+,1):=-$, $\gamma(+,2):=+$
$\gamma(-,1):=+$, $\gamma(-,2):=-$. Let $f^{(n)}\in\mathfrak{F}^{(n)}$ be of
the form (16). Then we define

$\displaystyle\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n}):=\sum_{i_{1},\dots,i_{n}\in\\{1,2\\}}\boldsymbol{\Phi}\big{(}\varphi^{(n)}(y_{1},\dots,y_{n})$
$\displaystyle\quad\times\left(K(\sharp_{1},i_{1})g_{1}\right)(z_{1})\dotsm\left(K(\sharp_{n},i_{n})g_{n}\right)(z_{n});i_{1},\gamma(\sharp_{1},i_{1}),\dots,i_{n},\gamma(\sharp_{n},i_{n})\big{)}.$
(69)

Next, we extend the definition of $\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n})$
by linearity to the case of an arbitrary $f^{(n)}\in\mathfrak{F}^{(n)}$.

By construction, each $\Phi(f^{(n)};\sharp_{1},\dots,\sharp_{n})$ belongs to
the $\mathbf{Q}$-MCR algebra $\mathbf{A}$. Let $\mathbb{A}$ denote the unital
$*$-algebra generated by these elements. Thus, $\mathbb{A}$ is a $*$-sub-
algebra of ${\bf A}$. We will denote by $\tau$ the restriction of the Fock
state $\boldsymbol{\tau}$ to $\mathbb{A}$.

######  Theorem 5.7.

(i) The $*$-algebra $\mathbb{A}$ constructed above is a $Q$-MCR algebra, i.e.,
it satisfies the conditions (19)–(24).

(ii) The state $\tau$ on $\mathbb{A}$ is gauge-invariant quasi-free and the
corresponding complex-valued measure $\rho^{(2)}[g_{1},g_{2}]$ on $Y^{2}$ is
given by

$\rho^{(2)}[g_{1},g_{2}](dy_{1}dy_{2})=\nu^{(2)}(dy_{1}dy_{2})\int_{Z}(Kg_{1})(z)g_{2}(z)\mu(dz).$
(70)

(iii) The state $\tau$ is strongly quasi-free if and only if

$\widetilde{Q}(y_{1},y_{2})=Q(y_{2},y_{1}).$ (71)

In the latter case, the corresponding complex-valued measure
$\lambda^{(2)}[g_{1},g_{2}]$ on $Y^{2}$ is given by

$\lambda^{(2)}\left[g_{1},g_{2}\right](dy_{1}\,dy_{2})=\nu^{(2)}(dy_{1}\,dy_{2})\big{(}(g_{2},g_{1})_{\mathcal{G}}+(Kg_{1},g_{2})_{\mathcal{G}}+\varkappa(g_{2},Kg_{1})_{\mathcal{G}}\big{)}.$
(72)

###### Proof.

(i) By using (69), one shows that $\mathbb{A}$ is a $Q$-MCR algebra similarly
to the proof of Proposition 5.6.

(ii) For any $(f_{1},f_{2})\in\mathcal{H}\oplus\mathcal{H}$, we denoted by
$\mathbf{a}^{\sharp}(f_{1},f_{2})=\mathbf{a}_{1}(f_{1})+\mathbf{a}_{2}(f_{2})$
($\sharp\in\\{+,-\\}$) the corresponding standard creation and annihilation
operators in the $\mathbf{Q}$-symmetric Fock space
$\mathcal{F}^{\mathbf{Q}}(\mathcal{H}\oplus\mathcal{H})$. Let
$f=\varphi\otimes g$ with $\varphi\in C_{0}(Y;V)$ and $g\in\mathcal{G}$. By
(67),

$\displaystyle A^{+}(f)$
$\displaystyle=\mathbf{a}_{2}^{+}(\varphi\otimes(K_{2}g))+\mathbf{a}_{1}^{-}(\varphi\otimes(K_{1}g)),$
$\displaystyle A^{-}(f)$
$\displaystyle=\mathbf{a}_{2}^{-}(\varphi\otimes(K_{2}^{\prime}g))+\mathbf{a}_{1}^{+}(\varphi\otimes(K_{1}^{\prime}g)).$

For $f_{1}=\varphi_{1}\otimes g_{1}$ and $f_{2}=\varphi_{2}\otimes g_{2}$, we
have

$\displaystyle\tau\big{(}A^{+}(f_{1})A^{-}(f_{2})\big{)}$
$\displaystyle=\tau\big{(}\mathbf{a}_{1}^{-}(\varphi_{1}\otimes(K_{1}g_{1}))\,\mathbf{a}_{1}^{+}(\varphi_{2}\otimes(K_{1}^{\prime}g_{2}))\big{)}$
$\displaystyle=\int_{Y}\langle\varphi_{1}(y),\varphi_{2}(y)\rangle_{V}\,dy\,(K_{1}g_{1},K_{1}Jg_{2})_{\mathcal{G}}$
$\displaystyle=\int_{Y^{2}}\langle\varphi_{1}(y_{1}),\varphi_{2}(y_{2})\rangle_{V}\,\rho^{(2)}[g_{1},g_{2}](dy_{1}\,dy_{2}),$
(73)

where $\rho^{(2)}[g_{1},g_{2}]$ is given by (70). Next, for any
$f_{i}=\varphi_{i}\otimes g_{i}$ ($i=1,\dots,m+n$), we have

$\displaystyle\tau\big{(}A^{+}(f_{1})\dotsm A^{+}(f_{m})A^{-}(f_{m+1})\dotsm
A^{-}(f_{m+n})\big{)}=\tau\big{(}\mathbf{a}_{1}^{-}(\varphi_{1}\otimes(K_{1}g_{1}))$
$\displaystyle\qquad\dotsm\mathbf{a}_{1}^{-}(\varphi_{m}\otimes(K_{1}g_{m}))\,\mathbf{a}_{1}^{+}(\varphi_{m+1}\otimes(K_{1}^{\prime}g_{m+1}))\dotsm\mathbf{a}_{1}^{+}(\varphi_{m+n}\otimes(K_{1}^{\prime}g_{m+n})\big{)}$
$\displaystyle\quad=\tau\big{(}a^{-}(\varphi_{1}\otimes(K_{1}g_{1}))\dotsm
a^{-}(\varphi_{m}\otimes(K_{1}g_{m}))\,a^{+}(\varphi_{m+1}\otimes(K_{1}^{\prime}g_{m+1}))$
$\displaystyle\qquad\dotsm
a^{+}(\varphi_{m+n}\otimes(K_{1}^{\prime}g_{m+n})\big{)}.$ (74)

In the last equality of (74), $a^{+}(\cdot)$ and $a^{-}(\cdot)$ denote the
standard creation and annihilation operators in the $Q$-symmetric Fock space
$\mathcal{F}^{Q}(\mathcal{H})$, and $\tau$ denotes the corresponding Fock
state. Now the statement (ii) of the theorem follows from (73), (74), and
Corollary 4.6.

(iii) For $f=\varphi\otimes g$ with $\varphi\in C_{0}(Y;V_{\mathbb{R}})$ and
$g\in\mathcal{G}$, the operator $B(f)=A^{+}(f)+A^{-}(Jf)$ is given by

$B(f)=\mathbf{a}^{+}(\varphi\otimes(JK_{1}g),\varphi\otimes(K_{2}g))+\mathbf{a}^{-}(\varphi\otimes(K_{1}g),\varphi\otimes(JK_{2}g)).$

Therefore, for such $f_{1}=\varphi_{1}\otimes g_{1}$ and
$f_{2}=\varphi_{1}\otimes g_{2}$,

$\displaystyle\tau\big{(}B(f_{1})B(f_{2})\big{)}$
$\displaystyle\quad=\int_{Y}\langle\varphi_{1}(y),\varphi_{2}(y)\rangle_{V}\,dy\,\bigg{(}\int_{Z}(K_{1}g_{1})(z)(JK_{1}g_{2})(z)dz+\int_{Z}(JK_{2}g_{1})(z)(K_{2}g_{2})(z)dz\bigg{)}$
$\displaystyle\quad=\int_{Y}\langle\varphi_{1}(y),\varphi_{2}(y)\rangle_{V},dy\,\big{(}(K_{1}g_{1},K_{1}g_{2})_{\mathcal{G}}+(K_{2}g_{2},K_{2}g_{1})_{\mathcal{H}}\big{)}$
$\displaystyle\quad=\int_{Y^{2}}\langle\varphi_{1}(y_{1}),\varphi_{2}(y_{2})\rangle_{V}\,\lambda^{(2)}[g_{1},g_{2}](dy_{1}\,dy_{2}),$
(75)

where $\lambda^{(2)}[g_{1},g_{2}]$ is given by (72).

Assume that condition (71) is not satisfied. Then, by a straightforward
calculation of $\tau\big{(}B(f_{1})B(f_{2})B(f_{3})B(f_{4})\big{)}$, we see
that formula (62) fails for the state $\tau$ when $n=4$. Hence, $\tau$ is not
strongly gauge-invariant.

On the other hand, if condition (71) is satisfied, then for all
$(y_{1},y_{2})\in\mathbf{Y}^{2}$ we have
$\mathbf{Q}(y_{1},y_{2})=Q(y_{1},y_{2})$, compare with Assumption 5.4 (v).
Then that the statement of the theorem about $\tau$ being strongly quasi-free
follows from (75) and a straightforward analog of Corollary 4.5 when $X$ is
replaced by $\mathbf{X}$. ∎

##  6 Examples

We will now consider several classes of examples of the operator-valued
function $Q$ to which our construction of quasi-free states is applicable.

###  6.1 Abelian anyons

Let $V=\mathbb{C}$, which implies that $Q:Y^{2}\to\mathbb{C}$ is a continuous
complex-valued function satisfying $|Q(y_{1},y_{2})|=1$ and
$\overline{Q(y_{1},y_{2})}=Q(y_{2},y_{1})$. Assumption 5.4 is now trivially
satisfied, with $\varkappa=Q(y,y)\in\\{-1,1\\}$ and
$\widetilde{Q}(y_{1},y_{2})=Q(y_{1},y_{2})$. Thus, for each appropriate choice
of a bounded linear operator $K$ in $\mathcal{G}$, we obtain the corresponding
gauge-invariant quasi-free state $\tau$ on the $*$-algebra $\mathbb{A}$ of the
$Q$-anyon commutation relations. This is essentially the construction from
[14]. Note that, in the latter paper, the function $Q$ was not assumed to be
continuous, which led to an arbitrary choice of the value of
$\varkappa=Q(y,y)\in\mathbb{R}$, and to a different definition of the
counterpart of the space $\mathfrak{F}^{(n)}$.

Unless $Q(y_{1},y_{2})$ is not identically equal to 1 (which is the case of
the CCR) or $-1$ (which is the case of the CAR), the function $Q$ does not
satisfy condition (71), hence the state $\tau$ is not strongly quasi-free.

###  6.2 Lifting of anyon commutation relations to multicomponent commutation
relations

Let $V=\mathbb{C}^{r}$ and, for each $i,j\in\\{1,\dots,r\\}$, we fix a
continuous complex-valued function $q(y_{1},y_{2},i,j)$ on $Y^{2}$ such that
$|q(y_{1},y_{2},i,j)|=1$ and

$\overline{q(y_{1},y_{2},i,j)}=q(y_{2},y_{1},j,i).$

We also assume that there exists a $\varkappa\in\\{-1,1\\}$ such that
$q(y,y,i,i)=\varkappa$ for all $y\in Y$ and $i\in\\{1,\dots,r\\}$.

For $(y_{1},y_{2})\in Y^{2}$, we define
$Q(y_{1},y_{2})\in\mathcal{L}\left(V^{\otimes 2}\right)$ by

$Q(y_{1},y_{2})\,e_{i}\otimes e_{j}:=q(y_{1},y_{2},i,j)\,e_{j}\otimes
e_{i},\quad i,j\in\\{1,\dots,r\\}.$ (76)

It is straightforward to see that $Q(y_{1},y_{2})$ is unitary,
$Q^{*}(y_{1},y_{2})=Q(y_{2},y_{1})$, and $Q(y_{1},y_{2})$ satisfies the
functional Yang–Baxter equation. Furthermore, $Q(y_{1},y_{2})$ satisfies
Assumption 5.4, with

$\widetilde{Q}(y_{1},y_{2})e_{i}\otimes e_{j}=q(y_{1},y_{2},j,i)e_{j}\otimes
e_{i}.$ (77)

In accordance with formula (8), for the corresponding operator-valued
distributions $A_{i}^{+}(x)$, $A_{i}^{-}(x)$ , as in (2) and (3), the
following commutation relations hold:

$\displaystyle A_{i}^{+}(x_{1})A_{j}^{+}(x_{2})$
$\displaystyle=q(y_{2},y_{1},i,j)A_{j}^{+}(x_{2})A_{i}^{+}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{j}^{-}(x_{2})$
$\displaystyle=q(y_{2},y_{1},i,j)A_{j}^{-}(x_{2})A_{i}^{-}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{j}^{+}(x_{2})$
$\displaystyle=\delta_{ij}\,\delta(y_{1}-y_{2})+q(y_{1},y_{2},j,i)A_{j}^{+}(x_{2})A_{i}^{-}(x_{1})$

for $i,j\in\\{1,\dots,r\\}$.

######  Remark 6.1.

We may think of $q(y_{1},y_{2},i,j)$ as a continuous complex-valued function
$q$ on $\big{(}Y\times\\{1,\dots,r\\}\big{)}^{2}$ which takes on values of
modulus 1 and satisfies, for all $v_{1},v_{2}\in Y\times\\{1,\dots,r\\}$,
$\overline{q(v_{1},v_{2})}=q(v_{2},v_{1})$. Hence, the multicomponent system
considered in this subsection can be thought of as a lifting of the anyon
commutation relations, for which the underlying space is
$X\times\\{1,\dots,r\\}$ and the exchange function $Q$ is given by
$q(y_{1},y_{2},j,i)$.

By Theorem 5.7 (iii) and (77), a gauge-invariant quasi-free state $\tau$ is
strongly quasi-free only in the trivial case where, for all $y_{1},y_{2}\in Y$
and $i,j\in\\{1,\dots,r\\}$, we have $Q(y_{1},y_{2},i,j)=q_{ij}$ with
$q_{ij}\in\\{-1,1\\}$, and $q_{11}=\dots=q_{rr}$.

######  Example 6.2 (Antiparticle).

Let us consider the following example. Let $r=3$, and let a function
$q(y_{1},y_{2},i,j)$ on $(Y\times\\{1,2,3\\})^{2}$ be defined as follows:
$q(y_{1},y_{2},1,1)=q(y_{1},y_{2},2,2)=q(y_{1},y_{2})$,
$q(y_{1},y_{2},1,2)=q(y_{1},y_{2},2,1)=q(y_{2},y_{1})$ and
$q(y_{1},y_{2},i,j)=1$ if at least one of the numbers $i$ and $j$ is 3. Here
$q:Y^{2}\to\mathbb{C}$ is a continuous function, $|q(y_{1},y_{2})|=1$, and
$q(y_{2},y_{1})=\overline{q(y_{1},y_{2})}$. Since $A_{3}^{+}(x_{1})$ commutes
with all $A_{i}(x_{2})$ ($i=1,2,3$), the particle of type 3 is a boson. On the
other hand, both particles of type 1 and 2 are (abelian) anyons:
$A_{i}^{+}(x_{1})A_{i}^{+}(x_{2})=q(y_{2},y_{1})A_{i}^{+}(x_{2})A_{i}^{+}(x_{1})$
($i=1,2$).

Recall the Introduction for an intuitive interpretation of fusion of
quasiparticles. The fusion of particles of type 1 and 2 can be understood as a
(heuristic) operator-valued distribution $B(x):=A^{+}_{1}(x)A^{+}_{2}(x)$. As
easily seen, $B(x_{1})$ also commutes with all $A_{i}(x_{2})$ ($i=1,2,3$), and
hence also with $B(x_{2})$. Hence, one can identify $B(x)$ with $A_{3}(x)$.
Thus the fusion of the particles of type 1 and $2$ gives a boson. This means
that particle 2 is the antiparticle of particle 1, compare with [15, Section
4.1.1].

######  Remark 6.3.

It follows from Example 6.2 that, for the construction of a gauge-invariant
quasi-free state on the algebra of the anyon commutation relations, one does
doubling of the underlying space by adding to the particle its antiparticle.
Indeed, in that case, the doubling of the space is equivalent to an addition
of a particle of type 2, and the resulting commutation relations are
$A_{i}^{+}(x_{1})A_{i}^{+}(x_{2})=Q(y_{2},y_{1})A_{i}^{+}(x_{2})A_{i}^{+}(x_{1})$
for $i=1,2$, and
$A_{i}^{+}(x_{1})A_{j}^{+}(x_{2})=Q(y_{1},y_{2})A_{j}^{+}(x_{2})A_{i}^{+}(x_{1})$
for $i\neq j$.

###  6.3 Systems with particles of opposite type

Let $V=\mathbb{C}^{r}$ ($r\geq 2$), and just as above, let
$\\{e_{1},\dots,e_{r}\\}$ be the standard orthonormal basis in $V$. Let
$\theta$ be a permutation from $S_{r}$ such that
$\theta^{2}=\operatorname{id}$, i.e., all cycles in $\theta$ are of length one
or two. If $\theta(i)=j$ and $i\neq j$, we may think of $i$ and $j$ as
opposite types of particles.

######  Theorem 6.4.

For each $i,j\in\\{1,\dots,r\\}$, let $q(\cdot,\cdot,i,j)$ be a continuous
complex-valued function on $Y^{2}$ that satisfies the following assumptions,
for all $y_{1},y_{2}\in Y$ and $i,j\in\\{1,\dots,r\\}$,

$\displaystyle|q(y_{1},y_{2},i,j)|=1,$ (78)
$\displaystyle\overline{q(y_{1},y_{2},i,j)}=q(y_{2},y_{1},i,j),$ (79)
$\displaystyle q(y_{1},y_{2},i,j)=q(y_{1},y_{2},j,i),$ (80) $\displaystyle
q(y_{1},y_{2},\theta(i),\theta(j))=q(y_{1},y_{2},i,j).$ (81)

We also assume that there exists a $\varkappa\in\\{-1,1\\}$ such that

$q(y,y,i,i)=\varkappa\quad\text{for all $y\in Y$ and $i\in\\{1,\dots,r\\}$}.$
(82)

For $(y_{1},y_{2})\in Y^{2}$, we define
$Q(y_{1},y_{2})\in\mathcal{L}\left(V^{\otimes 2}\right)$ by

$Q(y_{1},y_{2})\,e_{i}\otimes e_{j}:=q(y_{1},y_{2},i,j)\,e_{\theta(j)}\otimes
e_{\theta(i)},\quad i,j\in\\{1,\dots,r\\}.$ (83)

Then the following statements hold.

(i) $Q(y_{1},y_{2})$ is unitary, $Q^{*}(y_{1},y_{2})=Q(y_{2},y_{1})$, and the
operator-valued function $Q(y_{1},y_{2})$ satisfies the functional Yang–Baxter
equation.

(ii) $Q(y_{1},y_{2})$ satisfies Assumption 5.4, with

$\widetilde{Q}(y_{1},y_{2})\,e_{i}\otimes
e_{j}=q(y_{1},y_{2},\theta(i),j)\,e_{\theta(j)}\otimes
e_{\theta(i)}=q(y_{1},y_{2},i,\theta(j))\,e_{\theta(j)}\otimes e_{\theta(i)}.$
(84)

The corresponding $Q$-MCR are given by

$\displaystyle A_{i}^{+}(x_{1})A_{j}^{+}(x_{2})$
$\displaystyle=q(y_{2},y_{1},i,j)A_{\theta(j)}^{+}(x_{2})A_{\theta(i)}^{+}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{j}^{-}(x_{2})$
$\displaystyle=q(y_{2},y_{1},i,j)A_{\theta(j)}^{-}(x_{2})A_{\theta(i)}^{-}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{j}^{+}(x_{2})$
$\displaystyle=\delta_{i,\,j}\,\delta(x_{1}-x_{2})+q(y_{1},y_{2},\theta(i),j)A_{\theta(j)}^{+}(x_{2})A_{\theta(i)}^{-}(x_{1}).$
(85)

(iii) For each appropriate choice of the operator $K$, the corresponding
gauge-invariant quasi-free state $\tau$ is strongly quasi-free if and only if

$q(y_{1},y_{2},i,j)=q(y_{2},y_{1},\theta(i),j)\quad\text{for all
}i,j\in\\{1,\dots,r\\}.$ (86)

In the latter case, the commutation relation (85) becomes

$A_{i}^{-}(x_{1})A_{j}^{+}(x_{2})=\delta_{i,\,j}\,\delta(x_{1}-x_{2})+q(y_{2},y_{1},i,j)A_{\theta(j)}^{+}(x_{2})A_{\theta(i)}^{-}(x_{1}).$

###### Proof.

(i) The statement easily follows from (78)–(81).

(ii) Assumption 5.4 (i), (ii) and (iii) follow from (78)–(81) by
straightforward calculations. Formula (82) implies Assumption 5.4 (iv). To
show that Assumption 5.4 (v) is satisfied, we note that the operator
$\mathbf{Q}(y_{1},y_{2})$ acts as follows:

$\mathbf{Q}(y_{1},y_{2})e_{i}\otimes
e_{j}=\mathfrak{q}(y_{1},y_{2},i,j)e_{\theta(j)}\otimes e_{\theta(i)},$

where

$\mathfrak{q}(y_{1},y_{2},i,j)=\begin{cases}q(y_{1},y_{2},i,j),&\text{if
either $y_{1},y_{2}\in Y_{1}$ or $y_{1},y_{2}\in Y_{2}$},\\\
q(y_{1},y_{2},\theta(i),j)&\text{if either $y_{1}\in Y_{1}$, $y_{2}\in Y_{2}$
or $y_{1}\in Y_{2}$, $y_{2}\in Y_{1}$.}\end{cases}$ (87)

Formulas (78)–(81) and (87) imply that the functions
$\mathfrak{q}(y_{1},y_{2},i,j)$ satisfy the formulas (78)–(81) in which
$q(y_{1},y_{2},i,j)$ is replaced by $\mathfrak{q}(y_{1},y_{2},i,j)$, and
$y_{1},y_{2}\in\mathbf{Y}=Y_{1}\sqcup Y_{2}$. Now, the functional Yang–Baxter
equation for $\mathbf{Q}(y_{1},y_{2})$ can be checked completely analogously
to the proof of the functional Yang–Baxter equation for $Q(y_{1},y_{2})$.

The corresponding commutation relations hold in accordance with formula (8).

(iii) The statement follows immediately from Theorem 5.7 (iii). ∎

We will now consider some special cases of the operator-valued function
$Q(y_{1},y_{2})$ as in Theorem 6.4.

####  6.3.1 Two-component systems

Let $V=\mathbb{C}^{2}$. Let $\theta\in S_{2}$ be given by $\theta(1)=2$,
$\theta(2)=1$. We fix two continuous functions $q_{i}:Y^{2}\to\mathbb{C}$
($i=1,2$) such that, for all $(y_{1},y_{2})\in Y^{2}$,
$|q_{i}(y_{1},y_{2})|=1$ and $\overline{q_{i}(y_{1},y_{2})}=q(y_{2},y_{1})$.
We define $Q(y_{1},y_{2})\in\mathcal{L}(V^{\otimes 2})$ by

$\displaystyle Q(y_{1},y_{2})e_{i}\otimes e_{i}$
$\displaystyle=q_{1}(y_{1},y_{2})e_{\theta(i)}\otimes e_{\theta(i)},$
$\displaystyle Q(y_{1},y_{2})e_{i}\otimes e_{j}$
$\displaystyle=q_{2}(y_{1},y_{2})e_{i}\otimes e_{j},\quad i\neq j.$ (88)

Thus, formula (83) holds with $q(y_{1},y_{2},i,i)=q_{1}(y_{1},y_{2})$ and
$q(y_{1},y_{2},i,j)=q_{2}(y_{1},y_{2})$ if $i\not=j$. The assumptions
(78)–(82) are obviously satisfied. The corresponding $Q$-MCR are then given by
(15).

Each corresponding gauge-invariant quasi-free state $\tau$ is strongly quasi-
free if and only if $q_{1}(y_{1},y_{2})=q_{2}(y_{2},y_{1})=:q(y_{1},y_{2})$
for all $(y_{1},y_{2})\in Y^{2}$. In the latter case, the $Q$-MCR are given by

$\displaystyle A_{i}^{+}(x_{1})A_{i}^{+}(x_{2})$
$\displaystyle=q(y_{2},y_{1})A_{\theta(i)}^{+}(x_{2})A_{\theta(i)}^{+}(x_{1}),$
$\displaystyle A_{i}^{+}(x_{1})A_{j}^{+}(x_{2})$
$\displaystyle=q(y_{1},y_{2})A_{i}^{+}(x_{2})A_{j}^{+}(x_{1}),\quad i\neq j,$
$\displaystyle A_{i}^{-}(x_{1})A_{i}^{-}(x_{2})$
$\displaystyle=q(y_{2},y_{1})A_{\theta(i)}^{-}(x_{2})A_{\theta(i)}^{-}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{j}^{-}(x_{2})$
$\displaystyle=q(y_{1},y_{2})A_{i}^{-}(x_{2})A_{j}(x_{1}),\quad i\neq j,$
$\displaystyle A_{i}^{-}(x_{1})A_{i}^{+}(x_{2})$
$\displaystyle=\delta(x_{1}-x_{2})+q(y_{2},y_{1})A_{\theta(i)}^{+}(x_{2})A_{\theta(i)}^{-}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{j}^{+}(x_{2})$
$\displaystyle=q(y_{1},y_{2})A_{i}^{+}(x_{2})A_{j}^{-}(x_{1}),\quad i\neq j,$

for $i,j\in\\{1,2\\}$.

####  6.3.2 Three-component systems

We will now construct a three-component system by adding an abelian anyon to
the two-component from § 6.3.1.

Let $V=\mathbb{C}^{3}$, and let $\theta\in S_{3}$ be given by $\theta(1)=2$,
$\theta(2)=1$, $\theta(3)=3$. We fix four continuous functions
$q_{i}:Y^{2}\to\mathbb{C}$ ($i=1,2,3,4$) such that, for all $(y_{1},y_{2})\in
Y^{2}$, $|q_{i}(y_{1},y_{2})|=1$ and
$\overline{q_{i}(y_{1},y_{2})}=q(y_{2},y_{1})$. We define
$Q(y_{1},y_{2})\in\mathcal{L}(V^{\otimes 2})$ by (88) for $i,j\in\\{1,2\\}$,
and

$\displaystyle Q(y_{1},y_{2})e_{3}\otimes e_{3}$
$\displaystyle=q_{3}(y_{1},y_{2})e_{3}\otimes e_{3},$ $\displaystyle
Q(y_{1},y_{2})e_{i}\otimes e_{3}$
$\displaystyle=q_{4}(y_{1},y_{2})e_{3}\otimes e_{\theta(i)},$ $\displaystyle
Q(y_{1},y_{2})e_{3}\otimes e_{i}$
$\displaystyle=q_{4}(y_{1},y_{2})e_{\theta(i)}\otimes e_{3},\quad i=1,2.$

We also assume that $q_{1}(y,y)=q_{3}(y,y)=\varkappa$. The assumptions
(78)–(82) are obviously satisfied. The corresponding $Q$-MCR are given by (15)
for $i,j\in\\{1,2\\}$. The commutation relations for the operators
$A^{+}_{3}(x)$, $A^{-}_{3}(x)$ are the usual abelian anyon commutation
relations governed by the function $q_{3}$. Finally, for $i\in\\{1,2\\}$,

$\displaystyle A_{i}^{+}(x_{1})A_{3}^{+}(x_{2})$
$\displaystyle=q_{4}(y_{2},y_{1})A_{3}^{+}(x_{2})A_{\theta(i)}^{+}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{3}^{-}(x_{2})$
$\displaystyle=q_{4}(y_{2},y_{1})A_{3}^{-}(x_{2})A_{\theta(i)}^{-}(x_{1}),$
$\displaystyle A_{i}^{-}(x_{1})A_{3}^{+}(x_{2})$
$\displaystyle=q_{4}(y_{1},y_{2})A_{3}^{+}(x_{2})A_{\theta(i)}^{-}(x_{1}),$
$\displaystyle A_{3}^{-}(x_{1})A_{i}^{+}(x_{2})$
$\displaystyle=q_{4}(y_{1},y_{2})A_{\theta(i)}^{+}(x_{2})A_{3}^{-}(x_{1}).$

Each corresponding gauge-invariant quasi-free state $\tau$ is strongly quasi-
free if and only if $q_{1}(y_{1},y_{2})=q_{2}(y_{2},y_{1})=:q(y_{1},y_{2})$
for all $(y_{1},y_{2})\in Y^{2}$, and both functions $q_{3}$ and $q_{4}$ are
real-valued. In particular, this implies that the particles of type 3 must be
either bosons or fermions.

###  6.4 Fusion of quasiparticles

Let $k\geq 3$ be odd. We will now discuss fusion of $k$ quasiparticles
described by the commutation relations (15).

Recall that, for the quasiparticles discussed in § 6.3.1,

$A^{+}_{i}(x_{1})A^{+}_{j}(x_{2})=q(y_{2},y_{1},i,j)A^{+}_{\theta(j)}(x_{2})A^{+}_{\theta(i)}(x_{1}),\quad
i,j\in\\{1,2\\},$ (89)

where $\theta(1)=2$, $\theta(2)=1$ and

$q(y_{1},y_{2},i,j)=\begin{cases}q_{1}(y_{1},y_{2}),&\text{if }i=j,\\\
q_{2}(y_{1},y_{2}),&\text{if }i\not=j.\end{cases}$ (90)

We start with a heuristic calculation of the corresponding exchange function
$Q(y_{1},y_{2})$. To describe $k$ quasiparticles at point $x$, we define, for
$\mathbf{i}=(i_{1},i_{2},\dots,i_{k})\in\\{1,2\\}^{k}$,

$A^{+}_{\mathbf{i}}(x):=A^{+}_{i_{1}}(x)A^{+}_{i_{2}}(x)\dotsm
A^{+}_{i_{k}}(x).$ (91)

(Note that the right hand side of (91) cannot be rigorously understood as an
operator-valued distribution.) Since $k$ is an odd number, the commutation
relation (89) easily implies that, for any $x_{1},x_{2}\in X$ and
$\mathbf{i},\mathbf{j}\in\\{1,2\\}^{k}$,

$A^{+}_{\mathbf{i}}(x_{1})A^{+}_{\mathbf{j}}(x_{2})=\mathbf{q}(y_{2},y_{1},\mathbf{i},\mathbf{j})A^{+}_{\theta(\mathbf{j})}(x_{2})A^{+}_{\theta(\mathbf{i})}(x_{1}).$
(92)

Here, we denote
$\theta(\mathbf{i}):=(\theta(i_{1}),\theta(i_{2}),\dots,\theta(i_{k}))$, and

$\mathbf{q}(y_{1},y_{2},\mathbf{i},\mathbf{j}):=\prod_{l,m=1}^{k}q\big{(}y_{1},y_{2},\theta^{m-1}(i_{l}),\theta^{l-1}(j_{m})\big{)}.$
(93)

In view of the heuristic formula (92), we now proceed with a rigorous
construction. Let $\\{e_{1},e_{2}\\}$ be the standard orthonormal basis in
$\mathbb{C}^{2}$, let $V=(\mathbb{C}^{2})^{\otimes k}$, and we choose in $V$
the orthonormal basis
$\left\\{e_{\mathbf{i}}\mid\mathbf{i}\in\\{1,2\\}^{k}\right\\}$, where
$e_{\mathbf{i}}:=e_{i_{1}}\otimes e_{i_{2}}\otimes\dots\otimes e_{i_{k}}$. We
define $Q:Y^{2}\to\mathcal{L}(V^{\otimes 2})$ by

$Q(y_{1},y_{2})e_{\mathbf{i}}\otimes
e_{\mathbf{j}}=\mathbf{q}(y_{1},y_{2},\mathbf{i},\mathbf{j})e_{\theta(\mathbf{j})}\otimes
e_{\theta(\mathbf{i})},$ (94)

where $\mathbf{q}(y_{1},y_{2},\mathbf{i},\mathbf{j})$ is defined by (93).

######  Proposition 6.5.

Assume $q_{1}(y,y)=q_{2}(y,y)=\varkappa\in\\{-1,1\\}$ for $y\in Y$. Then the
functions $\mathbf{q}(y_{1},y_{2},\mathbf{i},\mathbf{j})$ satisfy the
conditions of Theorem 6.4. For each appropriate choice of the operator $K$,
the corresponding gauge-invariant quasi-free state $\tau$ is strongly quasi-
free if and only if $q_{1}(y_{1},y_{2})=q_{2}(y_{2},y_{1})=:q(y_{1},y_{2})$
for all $(y_{1},y_{2})\in Y^{2}$.

###### Proof.

The statement easily follows from the fact that the function
$q(y_{1},y_{2},i,j)$ satisfies the conditions of Theorem 6.4 and the
definition (93). ∎

######  Remark 6.6.

Let $Q:Y^{2}\to\mathcal{L}(V^{\otimes 2})$ be an arbitrary continuous function
such that $Q(y_{1},y_{2})$ is a unitary operator in $V^{\otimes 2}$,
$Q^{*}(y_{1},y_{2})=Q(y_{2},y_{1})$, and $Q(y_{1},y_{2})$ satisfies the
functional Yang–Baxter equation. Let $k\geq 2$ and $W:=V^{\otimes k}$. Let
$Q_{k}:Y^{2}\to\mathcal{L}(W^{\otimes 2})$ be a function derived from fusion
of $k$ quasiparticles with the exchange function $Q$. Then one can show that
the continuous function $Q_{k}$ is also such that $Q_{k}(y_{1},y_{2})$ is a
unitary operator in $W^{\otimes 2}$, $Q^{*}(y_{1},y_{2})=Q(y_{2},y_{1})$, and
$Q(y_{1},y_{2})$ satisfies the functional Yang–Baxter equation. Note that, in
the case where the operator $Q$ is as in § 6.3.1, we proved a stronger result:
the operator-valued function $Q_{k}$ also satisfies Assumption 5.4, and if $Q$
satisfies (71), then so does $Q_{k}$.

#### Acknowledgements

This paper was mostly written when N.O. was a PhD student in the Department of
Mathematics of Swansea University. N.O. is grateful to the Department for
their constant support during his studies.

## References

  * [1] H. Araki, On quasifree states of CAR and Bogoliubov automorphisms, Publ. Res. Inst. Math. Sci. 6 (1970/71) 385–442.
  * [2] H. Araki and M. Shiraishi, On quasifree states of the canonical commutation relations. I, II, Publ. Res. Inst. Math. Sci. 7 (1971/72) 105–120 & 121–152.
  * [3] H. Araki and E. Woods, Representations of the C.C.R. for a nonrelativistic infinite free Bose gas, J. Math. Phys. 4 (1963) 637–662.
  * [4] H. Araki and W. Wyss, Representations of canonical anticommutation relations, Helv. Phys. Acta 37 (1964) 136–159.
  * [5] M. Bożejko, E. Lytvynov and J. Wysoczański, Noncommutative Lévy processes for generalized (particularly anyon) statistics, Comm. Math. Phys. 313 (2012) 535–569.
  * [6] M. Bożejko and R. Speicher, Completely positive maps on Coxeter groups, deformed commutation relations, and operator space, Math. Ann. 300 (1994) 97–120.
  * [7] O. Bratteli and D. W. Robinson, Operator Algebras and Quantum-Statistical Mechanics. II. Equilibrium States. Models in Quantum-Statistical Mechanics (Springer-Verlag, New York-Berlin, 1981).
  * [8] A. Daletskii, A. Kalyuzhny, E. Lytvynov and D. Proskurin, Fock representations of multicomponent (particularly non-Abelian anyon) commutation relations, Rev. Math. Phys. 32 (2020) 2030004, 36 pp.
  * [9] J. Dereziński and C. Gérard, Mathematics of Quantization and Quantum Fields (Cambridge University Press, Cambridge, 2013).
  * [10] G. A. Goldin, R. Menikoff and D. H. Sharp, Comments on: “General theory for quantum statistics in two dimensions” by Y. S. Wu, Phys. Rev. Lett. 54 (1985) 603.
  * [11] G. A. Goldin and S. Majid, On the Fock space for nonrelativistic anyon fields and braided tensor products, J. Math. Phys. 45 (2004) 3770–3787.
  * [12] P. E. T. Jørgensen, L. M. Schmittn and R. F. Werner, Positive representations of general commutation relations allowing Wick ordering, J. Funct. Anal. 134 (1995) 33–99.
  * [13] A. Liguori and M. Mintchev, Fock representations of quantum fields with generalized statistics, Comm. Math. Phys. 169 (1995) 635–652.
  * [14] E. Lytvynov, Gauge-invariant quasi-free states on the algebra of the anyon commutation relations, Comm. Math. Phys. 351 (2017) 653–687.
  * [15] J. K. Pachos, Introduction to Topological Quantum Computation (Cambridge Univ. Press, Cambridge, 2012).
  * [16] M. Simonnet, Measures and Probabilities (Springer-Verlag, New York, 1996).
  * [17] T. D. Stanescu, Introduction to Topological Quantum Matter & Quantum Computation (CRC Press, Boca Raton, 2017).