# Quantum $U$-channels on $S$-spaces

Priyabrata Bag  Department of Mathematics, School of Science, Gandhi Institute
of Technology and Management (Deemed-to-be University), Doddaballapur Taluk,
Bengaluru, Karnataka 561203, INDIA<EMAIL_ADDRESS><EMAIL_ADDRESS>,
Azad Rohilla* Centre for Mathematical and Financial Computing, Department of
Mathematics, The LNM Institute of Information Technology, Rupa ki Nangal,
Post-Sumel, Via-Jamdoli Jaipur-302031, (Rajasthan) INDIA
<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Harsh Trivedi  Centre for
Mathematical and Financial Computing, Department of Mathematics, The LNM
Institute of Information Technology, Rupa ki Nangal, Post-Sumel, Via-Jamdoli
Jaipur-302031, (Rajasthan) INDIA<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

If the symmetry, (an operator $J$ satisfying $J=J^{*}=J^{-1}$) which defines
the Krein space, is replaced by a (not necessarily self-adjoint) unitary, then
we have the notion of an $S$-space which was introduced by Szafraniec. In this
paper, we consider $S$-spaces and study the structure of completely
$U$-positive maps between the algebras of bounded linear operators. We first
give a Stinespring-type representation for a completely $U$-positive map. On
the other hand, we introduce Choi $U$-matrix of a linear map and establish the
equivalence of the Kraus $U$-decompositions and Choi $U$-matrices. Then we
study properties of nilpotent completely $U$-positive maps. We develop the
$U$-PPT criterion for separability of quantum $U$-states and discuss the
entanglement breaking condition of quantum $U$-channels and explore $U$-PPT
squared conjecture. Finally, we give concrete examples of completely
$U$-positive maps and examples of $3\otimes 3$ quantum $U$-states which are
$U$-entangled and $U$-separable.

###### Key words and phrases:

Completely positive maps, Krein space, Quantum channels, Choi decomposition,
Stinespring decomposition, $S$-space.

###### 2010 Mathematics Subject Classification:

46E22, 46L05, 46L08, 47B50, 81T05.

*corresponding author

##  1\. Introduction

The Gelfand-Naimark-Segal (GNS) construction for a given state on a
$C^{*}$-algebra provides us a representation of the $C^{*}$-algebra on a
Hilbert space and a generating vector. A linear map $\tau$ from a
$C^{*}$-algebra $\mathcal{B}$ to a $C^{*}$-algebra $\mathcal{C}$ is said to be
completely positive (CP) if
$\sum_{i,j=1}^{n}c_{j}^{*}\tau(b_{j}^{*}b_{i})c_{i}\geq 0$ whenever
$b_{1},b_{2},\ldots,b_{n}\in\mathcal{B}$;
$c_{1},c_{2},\ldots,c_{n}\in\mathcal{C}$ and $n\in\mathbb{N}$. Stinespring’s
theorem (cf. [18, Theorem 1]), which characterizes operator-valued completely
positive maps, is a generalization of the GNS construction. Choi decomposition
(cf. [6]) for completely positive maps is a pioneering work in Matrix
Analysis.

Dirac [10] and Pauli [14] were among the pioneers to explore the quantum field
theory using Krein spaces, defined below. For our study, we require the
following important definitions:

######  Definition 1.1.

Assume $(\mathcal{K},\langle\cdot,\cdot\rangle)$ to be a Hilbert space and $J$
to be a symmetry, that is, $J=J^{*}=J^{-1}$. Define a map
$[\cdot,\cdot]:\mathcal{K}\times\mathcal{K}\to\mathbb{C}$ by

$\displaystyle[x,y]_{J}:=\langle Jx,y\rangle~{}\mbox{for
all}~{}x,y\in\mathcal{K}.$ (1.1)

The tuple $(\mathcal{K},J)$ is called a Krein space (cf. [3]).

######  Definition 1.2.

For each $V\in B(\mathcal{K}),$ there exists an operator
$V^{\natural}:=JV^{*}J\in B(\mathcal{K})$ such that

$\displaystyle[Vx,y]_{J}$ $\displaystyle=\langle JVx,y\rangle=\langle
x,V^{*}Jy\rangle=\langle x,J^{*}JV^{*}Jy\rangle$ $\displaystyle=\langle
Jx,JV^{*}Jy\rangle=\langle Jx,V^{\natural}y\rangle=[x,V^{\natural}y]_{J}.$

The operator $V^{\natural}$ is called the $J$-adjoint of $V.$

In the definition of the Krein space, if we replace the symmetry $J$ by a (not
necessarily self-adjoint) unitary $U$, then we arrive at the following
generalized notion due to Szafraniec [19]:

######  Definition 1.3.

Let $(\mathcal{H},\langle\cdot,\cdot\rangle)$ be a Hilbert space and let $U$
be a unitary on $\mathcal{H}$, that is, $U^{*}=U^{-1}$. Then we can define a
sesquilinear form by

$\displaystyle[x,y]_{U}:=\langle x,Uy\rangle~{}\mbox{for
all}~{}x,y\in\mathcal{H}.$ (1.2)

In this case, we call $(\mathcal{H},U)$ as an $S$-space.

The following definition is given by Phillipp, Szafraniec and Trunk, see [15,
Definition 3.1]:

######  Definition 1.4.

For each $V\in B(\mathcal{H}),$ there exists an operator
$V^{\\#}:=UV^{*}U^{*}\in B(\mathcal{H})$ such that

$\displaystyle[x,Vy]_{U}$ $\displaystyle=\langle x,UVy\rangle=\langle
V^{*}U^{*}x,y\rangle=\langle U^{*}UV^{*}U^{*}x,y\rangle$
$\displaystyle=\langle UV^{*}U^{*}x,Uy\rangle=[V^{\\#}x,y]_{U}.$

The operator $V^{\\#}$ is called the $U$-adjoint of $V.$

Phillipp, Szafraniec and Trunk [15] investigated invariant subspaces of self-
adjoint operators in Krein spaces by using results obtained through a detailed
analysis of S-spaces. Recently, in [16], Felipe-Sosa and Felipe introduced and
analyzed the notions of state and quantum channel on spaces equipped with an
indefinite metric in terms of a symmetry $J$. This study was further taken up
by Heo, in [11], where equivalence of Choi $J$-matrices and Kraus
$J$-decompositions was obtained and applications to $J$-PPT criterion and
$J$-PPT squared conjuncture were discussed. The notion of completely
$U$-positive maps was studied by Dey and Trivedi in [8, 9]. Motivated by these
inspiring works, in this paper, we develop structure theory of quantum
$U$-channels and its applications to the entanglement breaking.

The plan of the paper is as follows: In Section $\ref{sec1},$ we give
Stinespring-type representation for a completely $U$-positive map. In Section
$\ref{sec3},$ Choi $U$-matrix is introduced and the equivalence of Kraus
$U$-decompositions and Choi $U$-matrices is established. In Section
$\ref{new},$ some properties of nilpotent $U$-CP maps are discussed. In
Sections 5 and $\ref{sec5},$ we develop $U$-PPT criterion for separability of
quantum $U$-states and discuss the entanglement breaking condition of quantum
$U$-channels and explore $U$-PPT squared conjecture. Finally, in Section
$\ref{sec6},$ we give concrete examples of completely $U$-positive maps and
examples of $3\otimes 3$ quantum $U$-states which are $U$-entangled and
$U$-separable.

###  1.1. Background and notations

Let $(\mathcal{H},U)$ be an $S$-space. Then, $\mathcal{H}^{n}$ is the direct
sum of $n$-copies of the Hilbert space $\mathcal{H},$ and we denote by
$(\mathcal{H}^{n},U^{n})$ the $S$-space with the indefinite inner-product

$\displaystyle[{\bf h},{\bf k}]_{U^{n}}=\left<{\bf h},U^{n}{\bf
k}\right>=\sum^{n}_{j=1}\left<h_{j},Uk_{j}\right>=\sum^{n}_{j=1}[h_{j},k_{j}]_{U}$
(1.3)

where $U^{n}=\mbox{diag}(U,U,\ldots,U)\in M_{n}(B(\mathcal{H}))$ and ${\bf
h}=(h_{1},\ldots,h_{n}),~{}{\bf k}=(k_{1},\ldots,k_{n})\in\mathcal{H}^{n}.$

######  Definition 1.5.

Let $(\mathcal{H},U)$ be an $S$-space with the indefinite inner-product
$[\cdot,\cdot]_{U}.$ We denote by $B(\mathcal{H})^{U+}$ the set of all
$U$-positive linear operator $V$ on $\mathcal{H},$ that is,

$0\leq[Vh,h]_{U}:=\left<Vh,Uh\right>=\left<U^{*}Vh,h\right>,\mbox{ for all
}~{}h\in\mathcal{H}.$

Hence $V$ is $U$-positive if and only if $U^{*}V$ is positive with respect to
the usual inner product $\langle\cdot,\cdot\rangle.$

######  Definition 1.6.

Let $(\mathcal{H}_{i},U_{i})~{}(i=1,2)$ be an $S$-space with the indefinite
inner-product $[\cdot,\cdot]_{U_{i}}.$ Let $\phi:B(\mathcal{H}_{1})\to
B(\mathcal{H}_{2})$ be a linear map. Then $\phi$ is called
$(U_{1},U_{2})$-Hermitian if
$\phi(U_{1}V^{*}U^{*}_{1})=U_{2}\phi(V^{*})U^{*}_{2}$ for $V\in
B(\mathcal{H}_{1}).$ We say that a $(U_{1},U_{2})$-Hermitian linear map $\phi$
is

  1. (1)

$(U_{1},U_{2})$-positive if $\phi(B(\mathcal{H}_{1})^{U+})\subset
B(\mathcal{H}_{2})^{U+},$ that is, if $V\in(B(\mathcal{H}_{1}))^{U+}$ (or $V$
is $U_{1}$-positive), then $\phi(V)$ is $U_{2}$-positive. In simple words, if
$U_{1}^{*}V$ is positive with respect to the usual inner product
$\langle\cdot,\cdot\rangle_{\mathcal{H}_{1}},$ then $U_{2}^{*}\phi(V)$ is
positive with respect to the usual inner product
$\langle\cdot,\cdot\rangle_{\mathcal{H}_{2}}.$

  2. (2)

completely $(U_{1},U_{2})$-positive or $(U_{1},U_{2})$-CP if for each
$l\in\mathbb{N}$ the $l$-fold amplification
$\phi^{l}:I_{l}\otimes\phi:M_{l}(\mathbb{C})\otimes B(\mathcal{H}_{1})\to
M_{l}(\mathbb{C})\otimes B(\mathcal{H}_{2})$ defined by

$\phi^{l}([V_{ij}])=[\phi(V_{ij})],\>\>\>\>\text{for}\>\>\>\>[V_{ij}]\in
M_{l}(B(\mathcal{H}_{1}))$

satisfies

$\phi^{l}(M_{l}(B(\mathcal{H}_{1}))^{U+})\subset
M_{l}(B(\mathcal{H}_{2}))^{U+},$

that is, if $V=[V_{ij}]_{i,j}\in M_{l}(B(\mathcal{H}_{1}))^{U+}$ (i.e., $V$ is
$U_{1}^{l}$-positive), then $\phi^{l}(V)$ is $U_{2}^{l}$-positive. Here
$M_{l}(B(\mathcal{H}_{i}))^{U+}=B(\mathcal{H}^{l}_{i})^{U+}$ is the set of all
$U^{l}_{i}$-positive linear operators on $S$-spaces
$(\mathcal{H}^{l}_{i},U^{l}_{i}),$ and $U^{l}_{i}=\mbox{diag}(U,U,\ldots,U)$
$\in M_{l}(B(\mathcal{H}_{i}))$ for $i=1,2.$

  3. (3)

$U$-positive (and completely $U$-positive ($U$-CP) )if
$\mathcal{H}_{1}=\mathcal{H}_{2}=\mathcal{H}$ and $U_{1}=U_{2}=U$ and it is
($U_{1},U_{2})$-positive (and $(U_{1},U_{2})$-CP, respectively).

##  2\. Completely $U$-positive and completely $U$-co-positive maps

Our main objective in this section is to obtain Stinespring-type theorem for
completely $U$-positive maps. Let $(\mathcal{H}_{i},U_{i})~{}(i=1,2)$ be an
$S$-space with the indefinite inner product $[\cdot,\cdot]_{U_{i}}.$ Suppose
$\phi:B(\mathcal{H}_{1})\to B(\mathcal{H}_{2})$ is a linear map. Define a
linear map $\psi$ from $B(\mathcal{H}_{1})$ to $B(\mathcal{H}_{2})$ by
$\psi(X):=U_{2}\phi(U_{1}^{*}X)$ where $X\in B(\mathcal{H}_{1}).$ For any
$l\in\mathbb{N}$ and $V=[V_{ij}]\in M_{l}(B(\mathcal{H}_{1})),$ we obtain

$\displaystyle\psi^{l}(V)$
$\displaystyle=[\psi(V_{ij})]_{i,j}=[U_{2}\phi(U_{1}^{*}V_{ij})]_{i,j}=\begin{pmatrix}U_{2}\phi(U_{1}^{*}V_{11})&\cdots&U_{2}\phi(U_{1}^{*}V_{1l})\\\
\vdots&\ddots&\vdots\\\
U_{2}\phi(U_{1}^{*}V_{l1})&\cdots&U_{2}\phi(U_{1}^{*}V_{ll})\end{pmatrix}$
$\displaystyle=\begin{pmatrix}U_{2}&&0\\\ &\ddots&\\\
0&&U_{2}\end{pmatrix}\begin{pmatrix}\phi(U_{1}^{*}V_{11})&\cdots&\phi(U_{1}^{*}V_{ll})\\\
\vdots&\ddots&\vdots\\\
\phi(U_{1}^{*}V_{l1})&\cdots&\phi(U_{1}^{*}V_{ll})\end{pmatrix}=U_{2}^{l}\phi^{l}(U_{1}^{l^{*}}V).$

Similarly, we can easily show that $\phi^{l}(V)=U^{l^{*}}_{2}\psi(U_{1}^{l}V)$
where $\phi(V_{ij})=U^{*}_{2}\psi(U_{1}V_{ij}).$

The following result is a generalization of [16, Theorem 20] and [11,
Proposition 2.2] in the setting of $S$-spaces:

######  Proposition 2.1.

Let $(\mathcal{H}_{i},U_{i})~{}(i=1,2)$ be an $S$-space with the indefinite
inner product $[\cdot,\cdot]_{U_{i}}.$ Suppose $\phi:B(\mathcal{H}_{1})\to
B(\mathcal{H}_{2})$ is a linear map, then $\phi$ is CP if and only if the
corresponding linear map $\psi$ from $B(\mathcal{H}_{1})$ to
$B(\mathcal{H}_{2})$ defined by $\psi(X):=U_{2}\phi(U_{1}^{*}X)$ is
$(U_{1},U_{2})$-CP, where $X\in B(\mathcal{H}_{1}).$

###### Proof.

Let $\phi$ be a linear map from $B(\mathcal{H}_{1})$ to $B(\mathcal{H}_{2}).$
First assume that $\phi$ is CP. We have to prove that $\psi$ is
$(U_{1},U_{2})$-CP. For this purpose, let $V=[V_{ij}]\in
M_{l}(B(\mathcal{H}_{1}))^{U+},$ that is, $U_{1}^{l^{*}}V\in
M_{l}(B(\mathcal{H}_{1}))$ is positive, that is,

$0\leq[V{\bf{h}},{\bf{h}}]_{U_{1}^{l}}=\langle
V{\bf{h}},U_{1}^{l}{\bf{h}}\rangle=\langle
U_{1}^{l^{*}}V{\bf{h}},{\bf{h}}\rangle,$

where ${\bf{h}}\in\mathcal{H_{1}}^{l}.$ Consider

$\displaystyle[\psi^{l}(V){\bf{h^{\prime}}},{\bf{h^{\prime}}}]_{U_{2}^{l}}$
$\displaystyle=\langle\psi^{l}(V){\bf{h^{\prime}}},U_{2}^{l}{\bf{h^{\prime}}}\rangle=\langle
U_{2}^{l^{*}}\psi^{l}(V){\bf{h^{\prime}}},{\bf{h^{\prime}}}\rangle$
$\displaystyle=\langle
U^{l}_{2}\phi^{l}(U_{1}^{l^{*}}V){\bf{h^{\prime}}},U_{2}^{l}{\bf{h^{\prime}}}\rangle=\langle\phi^{l}(U_{1}^{l^{*}}V){\bf{h^{\prime}}},{\bf{h^{\prime}}}\rangle\geq
0,$

where ${\bf{h^{\prime}}}\in\mathcal{H_{2}}^{l}.$ Therefore $\langle
U_{2}^{l^{*}}\psi^{l}(V){\bf{h^{\prime}}},{\bf{h^{\prime}}}\rangle\geq 0,$
that is, $U_{2}^{l^{*}}\psi^{l}(V)$ is positive. This proves that $\psi(V)$ is
$U_{2}$-positive. Thus $\psi$ is $(U_{1},U_{2})$-CP.

Conversely, suppose that $\psi$ is $(U_{1},U_{2})$-CP. Since
$\psi(\cdot)=U_{2}\phi(U_{1}^{*}\cdot),$ we get
$\phi(U_{1}^{*}\cdot)=U^{*}_{2}\psi(\cdot).$ Therefore
$\phi(\cdot)=U^{*}_{2}\psi(U_{1}\cdot).$ Let $V=[V_{ij}]\in
M_{l}(B(\mathcal{H}_{1}))^{+},$ then we have to show that
$\phi^{l}(V)=[\phi({V_{ij}})]\in M_{l}(B(\mathcal{H}_{2}))^{+}.$ Now

$0\leq\langle V{\bf{h}},{\bf{h}}\rangle=\langle
U_{1}^{l}V{\bf{h}},U_{1}^{l}{\bf{h}}\rangle=[U_{1}^{l}V{\bf{h}},{\bf{h}}]_{U_{1}^{l}},$

where ${\bf{h}}\in\mathcal{H_{1}}^{l},$ it means, $U_{1}^{l}V\in
M_{l}(B(\mathcal{H}_{1}))^{U+}.$ Therefore

$\displaystyle\langle\phi^{l}(V){\bf{h^{\prime}}},{\bf{h^{\prime}}}\rangle$
$\displaystyle=\langle
U^{l^{*}}_{2}\psi(U_{1}^{l}V){\bf{h^{\prime}}},{\bf{h^{\prime}}}\rangle=\langle\psi(U_{1}^{l}V){\bf{h^{\prime}}},U^{l}_{2}{\bf{h^{\prime}}}\rangle$
$\displaystyle=[\psi(U_{1}^{l}V){\bf{h^{\prime}}},{\bf{h^{\prime}}}]_{U_{2}^{l}}\geq
0,$

where ${\bf{h^{\prime}}}\in\mathcal{H_{2}}^{l}$ and the last inequality
follows from the fact that $U_{1}^{l}V\in M_{l}(B(\mathcal{H}_{1}))^{U+}$ and
hence $\psi$ is $(U_{1},U_{2})$-CP.

∎

######  Theorem 2.2.

Let $(\mathcal{H}_{i},U_{i})~{}(i=1,2)$ be an $S$-space. Assume that a linear
map $\psi$ from $B(\mathcal{H}_{1})$ to $B(\mathcal{H}_{2})$ defined by
$\psi(V):=U_{2}\phi(U_{1}^{*}V)$ for all $V\in B(\mathcal{H}_{1})$ is
$(U_{1},U_{2})$-CP. Then there exist an $S$-space $(\mathcal{H},U),$ a
$*$-representation $\pi$ of $B(\mathcal{H}_{1})$ on the Hilbert space
$\mathcal{H}$ and a bounded linear operator $R:\mathcal{H}_{2}\to\mathcal{H}$
such that

$\psi(V)=R^{\\#}\pi(V)R$

where $U=\pi(U_{1}),$ and $R^{\\#}:=U_{2}R^{*}U^{*}.$ Moreover, if
$\psi(U_{1})=U_{2},$ then $R^{*}R=I_{\mathcal{H}_{2}}.$

###### Proof.

Suppose a linear map $\psi$ is $(U_{1},U_{2})$-CP. Then with the help of
Proposition 2.1, we get that $\phi$ defined by $\phi(V)=U^{*}_{2}\psi(U_{1}V)$
is CP. Then using Stinespring’s theorem [18, Theorem 1], there exist a Hilbert
space $\mathcal{H},$ a representation (a unital $*$-homomorphism ) $\pi$ of
$B(\mathcal{H}_{1})$ on the Hilbert space $\mathcal{H}$ and a bounded linear
operator $R:\mathcal{H}_{2}\to\mathcal{H},$ such that $\phi(V)=R^{*}\pi(V)R$
for every $V\in B(\mathcal{H}_{1}).$

Let $U=\pi(U_{1})\in B(\mathcal{H}),$ where $U$ is a fundamental unitary, that
is, $U^{*}=U^{-1},$ so that $(\mathcal{H},U)$ becomes an $S$-space. Define
$R^{\\#}:=U_{2}R^{*}U^{*},$ then

$\displaystyle\psi(V)=U_{2}\phi(U_{1}^{*}V)=U_{2}R^{*}\pi(U_{1}^{*}V)R=U_{2}R^{*}U^{*}\pi(V)R=R^{\\#}\pi(V)R.$

Furthermore, if $\psi(U_{1})=U_{2},$ then

$\displaystyle
U_{2}=\psi(U_{1})=U_{2}\phi(U_{1}^{*}U_{1})=U_{2}R^{*}\pi(U_{1}^{*}U_{1})R=U_{2}R^{*}R,$

hence $R^{*}R=I_{\mathcal{H}_{2}}.$ ∎

######  Theorem 2.3.

Suppose $\phi:B(\mathcal{H}_{1})\to B(\mathcal{H}_{2})$ is a linear map. If
$\phi$ satisfies the following conditions for all $V\in B(\mathcal{H}_{1}):$

$\displaystyle\phi(U_{1}^{*}V)=U_{2}^{*}\phi(V)\quad\mbox{and}\quad\phi(U_{1}V)=U_{2}\phi(V),$

then $\phi$ is a CP map if and only if $\phi$ is $(U_{1},U_{2})$-CP.

###### Proof.

First assume $\phi$ to be a CP map. Let $V=[V_{ij}]\in
M_{l}(B(\mathcal{H}_{1}))^{U+}.$ Observe that

$\displaystyle\phi^{l}(U_{1}^{l^{*}}V)$
$\displaystyle=[\phi(U_{1}^{{*}}V_{ij})]_{i,j}=\begin{pmatrix}\phi(U_{1}^{*}V_{11})&\cdots&\phi(U_{1}^{*}V_{1l})\\\
\vdots&\ddots&\vdots\\\
\phi(U_{1}^{*}V_{l1})&\cdots&\phi(U_{1}^{*}V_{ll})\end{pmatrix}$
$\displaystyle=\begin{pmatrix}U_{2}^{*}&&0\\\ &\ddots&\\\
0&&U_{2}^{*}\end{pmatrix}\begin{pmatrix}\phi(V_{11})&\cdots&\phi(V_{ll})\\\
\vdots&\ddots&\vdots\\\
\phi(V_{l1})&\cdots&\phi(V_{ll})\end{pmatrix}=U_{2}^{l^{*}}\phi^{l}(V).$

Similarly, we obtain $\phi^{l}(U_{1}^{l}V)=U_{2}^{l}\phi^{l}(V).$ Now consider

$\displaystyle[\phi^{l}(V){\bf{h^{\prime}}},{\bf{h^{\prime}}}]_{U_{2}^{l}}$
$\displaystyle=\langle\phi^{l}(V){\bf{h^{\prime}}},U_{2}^{l}{\bf{h^{\prime}}}\rangle=\langle
U_{2}^{l^{*}}\phi^{l}(V){\bf{h^{\prime}}},{\bf{h^{\prime}}}\rangle$
$\displaystyle=\langle\phi^{l}(U_{1}^{l^{*}}V){\bf{h^{\prime}}},{\bf{h^{\prime}}}\rangle\geq
0,$

where ${\bf{h^{\prime}}}\in\mathcal{H}_{2}^{l}.$ Therefore $\langle
U_{2}^{l^{*}}\phi^{l}(V){\bf{h^{\prime}}},{\bf{h^{\prime}}}\rangle\geq 0,$
that is, $U_{2}^{l^{*}}\phi^{l}(V)$ is positive with respect to the usual
inner product $\langle\cdot,\cdot\rangle.$ This proves that $\phi(V)$ is
$U_{2}$-positive. Thus $\phi$ is $(U_{1},U_{2})$-CP.

Conversely, suppose that $\phi$ is $(U_{1},U_{2})$-CP. Let $V=[V_{ij}]\in
M_{l}(B(\mathcal{H}_{1}))^{+}.$ Then we have to show that
$\phi^{l}(V)=[\phi({V_{ij}})]\in M_{l}(B(\mathcal{H}_{2}))^{+}.$ Since

$0\leq\langle V{\bf{h}},{\bf{h}}\rangle=\langle
U_{1}^{l}V{\bf{h}},U_{1}^{l}{\bf{h}}\rangle=[U_{1}^{l}V{\bf{h}},{\bf{h}}]_{U_{1}^{l}},$

where ${\bf{h}}\in\mathcal{H_{1}}^{l},$ it means $U_{1}^{l}V\in
M_{l}(B(\mathcal{H}_{1}))^{U+}.$ Then

$\displaystyle\langle\phi^{l}(V){\bf{h^{\prime}}},{\bf{h^{\prime}}}\rangle$
$\displaystyle=\langle
U^{l}_{2}\phi(V){\bf{h^{\prime}}},U^{l}_{2}{\bf{h^{\prime}}}\rangle=\langle\phi(U_{1}^{l}V){\bf{h^{\prime}}},U^{l}_{2}{\bf{h^{\prime}}}\rangle$
$\displaystyle=[\phi(U_{1}^{l}V){\bf{h^{\prime}}},{\bf{h^{\prime}}}]_{U_{2}^{l}}\geq
0,$

where ${\bf{h^{\prime}}}\in\mathcal{H_{2}}^{l}$ and the last inequality
follows from the fact that $U_{1}^{l}V\in M_{l}(B(\mathcal{H}_{1}))^{U+}$ and
$\phi$ is $(U_{1},U_{2})$-CP. ∎

######  Remark 2.4.

In particular, if $\mathcal{H}_{1}=\mathcal{H}_{2}=\mathcal{H}$ and
$U_{1}=U_{2}=U,$ and if a linear map $\phi:B(\mathcal{H})\to B(\mathcal{H})$
satisfies $\phi(U^{*}V)=U^{*}\phi(V)\mbox{ and }\phi(UV)=U\phi(V)$ for all
$V\in B(\mathcal{H}_{1}),$ then $\phi$ is CP if and only if $\phi$ is $U$-CP.

######  Definition 2.5.

Let $(\mathcal{H}_{i},U_{i})~{}(i=1,2)$ be an $S$-space. Assume that $\psi$ is
a linear map from $B(\mathcal{H}_{1})$ to $B(\mathcal{H}_{2}).$ Then

  1. (1)

for each $l\in\mathbb{N},$ $\psi$ is $l$-$(U_{1},U_{2})$-co-positive if
$\tau_{l}\otimes\psi:M_{l}(\mathbb{C})\otimes B(\mathcal{H}_{1})\to
M_{l}(\mathbb{C})\otimes B(\mathcal{H}_{2})$ is $(I_{l}\otimes
U_{1},I_{l}\otimes U_{2})$-positive where $\tau_{l}$ is the transpose map on
$M_{l}(\mathbb{C}).$

  2. (2)

$\psi$ is completely $(U_{1},U_{2})$-co-positive if it is
$l$-$(U_{1},U_{2})$-co-positive for each $l\in\mathbb{N}.$

  3. (3)

$\psi$ is $(U_{1},U_{2})$-positive partial transpose ($(U_{1},U_{2})$-PPT) if
it is $(U_{1},U_{2})$-CP and completely $(U_{1},U_{2})$-co-positive.

  4. (4)

In particular, if $\mathcal{H}_{1}=\mathcal{H}_{2}=\mathcal{H}$ and
$U_{1}=U_{2}=U,$ then we simply call it completely $U$-co-positive (and
$U$-positive partial transpose ($U$-PPT)) if it is completely
$(U_{1},U_{2})$-co-positive (and $(U_{1},U_{2})$-positive partial transpose,
respectively).

######  Proposition 2.6.

Let $(\mathcal{H}_{i},U_{i})~{}(i=1,2)$ be an $S$-space. Suppose
$\phi:B(\mathcal{H}_{1})\to B(\mathcal{H}_{2})$ is a linear map, then $\phi$
is completely co-positive if and only if the corresponding linear map $\psi$
from $B(\mathcal{H}_{1})$ to $B(\mathcal{H}_{2})$ defined by
$\psi(X):=U_{2}\phi(U_{1}^{*}X)$ is completely $(U_{1},U_{2})$-co-positive,
where $X\in B(\mathcal{H}_{1}).$

###### Proof.

Let $V=[V_{ij}]\in M_{l}(\mathbb{C})\otimes B(\mathcal{H}_{1})$ be such that
$(I_{l}\otimes U_{1}^{*})V\geq 0.$ Then

$\displaystyle(\tau_{l}\otimes\psi)(V)$
$\displaystyle=\begin{pmatrix}\psi(V_{11})&\cdots&\psi(V_{l1})\\\
\vdots&\ddots&\vdots\\\
\psi(V_{1l})&\cdots&\psi(V_{ll})\end{pmatrix}=\begin{pmatrix}U_{2}\phi(U_{1}^{*}V_{11})&\cdots&U_{2}\phi(U_{1}^{*}V_{l1})\\\
\vdots&\ddots&\vdots\\\
U_{2}\phi(U_{1}^{*}V_{1l})&\cdots&U_{2}\phi(U_{1}^{*}V_{ll})\end{pmatrix}$
$\displaystyle=\begin{pmatrix}U_{2}&&0\\\ &\ddots&\\\
0&&U_{2}\end{pmatrix}\begin{pmatrix}\phi(U_{1}^{*}V_{11})&\cdots&\phi(U_{1}^{*}V_{l1})\\\
\vdots&\ddots&\vdots\\\
\phi(U_{1}^{*}V_{1l})&\cdots&\phi(U_{1}^{*}V_{ll})\end{pmatrix}$
$\displaystyle=(I_{l}\otimes U_{2})(\tau_{l}\otimes\phi)(I_{l}\otimes
U_{1}^{*})V.$

Hence $(I_{l}\otimes U_{2}^{*})(\tau_{l}\otimes\psi)(V)$ is positive as $\phi$
is completely co-positive map.

Conversely, for any $V=[V_{ij}]\in M_{l}(B(\mathcal{H}_{1}),$ we have

$0\leq\langle V{\bf{h}},{\bf{h}}\rangle=\langle
U_{1}^{l}V{\bf{h}},U_{1}^{l}{\bf{h}}\rangle=[U_{1}^{l}V{\bf{h}},{\bf{h}}]_{U_{1}^{l}},$

where ${\bf{h}}\in\mathcal{H_{1}}^{l},$ it means $U_{1}^{l}V\in
M_{l}(B(\mathcal{H}_{1}))^{U+}.$ We obtain

$\displaystyle(\tau_{l}\otimes\phi)(V)$
$\displaystyle=\begin{pmatrix}\phi(V_{11})&\cdots&\phi(V_{l1})\\\
\vdots&\ddots&\vdots\\\
\phi(V_{1l})&\cdots&\phi(V_{ll})\end{pmatrix}=\begin{pmatrix}U_{2}^{*}\psi(U_{1}V_{11})&\cdots&U_{2}^{*}\psi(U_{1}V_{l1})\\\
\vdots&\ddots&\vdots\\\
U_{2}^{*}\psi(U_{1}V_{1l})&\cdots&U_{2}^{*}\psi(U_{1}V_{ll})\end{pmatrix}$
$\displaystyle=\begin{pmatrix}U_{2}^{*}&&0\\\ &\ddots&\\\
0&&U_{2}^{*}\end{pmatrix}\begin{pmatrix}\psi(U_{1}V_{11})&\cdots&\psi(U_{1}V_{l1})\\\
\vdots&\ddots&\vdots\\\
\psi(U_{1}V_{1l})&\cdots&\psi(U_{1}V_{ll})\end{pmatrix}$
$\displaystyle=U_{2}^{l^{*}}(\tau_{l}\otimes\psi)(U_{1}^{l}V).$

Therefore
$(\tau_{l}\otimes\phi)(V)=U_{2}^{l^{*}}(\tau_{l}\otimes\psi)(U_{1}^{l}V).$
Since $U_{1}^{l}V\in M_{l}(B(\mathcal{H}_{1}))^{U+}$ and $\psi$ is completely
$(U_{1},U_{2})$-co-positive, $\phi$ is co-positive.

∎

##  3\. Kraus $U$-decomposition and Choi $U$-matrix

In this section, we derive Kraus $U$-decomposition and Choi $U$-matrix and
establish their relation with the completely $U$-positive maps. Let
$M_{m}(\mathbb{C})$ denote the set of all $m\times m$-complex matrices. Kraus
proved that $\phi:M_{m}(\mathbb{C})\to M_{n}(\mathbb{C})$ is a CP map if and
only if

$\phi(V)=\sum_{i=1}^{l}R_{i}^{*}VR_{i},$ (3.1)

where $V=[V_{ij}]_{i,j}\in M_{m}(\mathbb{C})$ and for each $i,$ $R_{i}\in
M_{m,n}(\mathbb{C}).$ The expression in above equation is called a Kraus
decomposition.

Denote $M_{A}:=M_{m}(\mathbb{C})$ and $M_{B}:=M_{n}(\mathbb{C}).$ Let $U_{A}$
and $U_{B}$ be the fundamental unitaries in $M_{A}$ and $M_{B},$ respectively.
Define a linear map $\psi:M_{A}\to M_{B}$ by

$\psi(V):=\sum_{i=1}^{l}R_{i}^{\\#_{A,B}}VR_{i},$ (3.2)

where $R_{i}^{\\#_{A,B}}=U_{B}R^{*}_{i}U_{A}^{*}.$ Then $\psi$ is
$(U_{A},U_{B})$-CP. Indeed, for any $k\in\mathbb{N},$ take a
$U_{A}^{k^{*}}$-positive matrix $V=[V_{ij}]\in M_{k}(M_{A})^{U+}.$ Since
$V=[V_{ij}]\in M_{k}(M_{A})^{U+},$ $U_{A}^{k^{*}}V\in M_{k}(M_{A})^{+},$ that
is,

$\displaystyle U_{A}^{k^{*}}V$ $\displaystyle=\begin{pmatrix}U_{A}^{*}&&0\\\
&\ddots&\\\ 0&&U_{A}^{*}\end{pmatrix}\begin{pmatrix}V_{11}&\cdots&V_{1k}\\\
\vdots&\ddots&\vdots\\\ V_{k1}&\cdots&V_{kk}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}U_{A}^{*}V_{11}&\cdots&U_{A}^{*}V_{1k}\\\
\vdots&\ddots&\vdots\\\ U_{A}^{*}V_{k1}&\cdots&U_{A}^{*}V_{kk}\end{pmatrix}\in
M_{k}(M_{A})^{+}.$

Consider

$\displaystyle\psi^{k}(V)$
$\displaystyle=\psi^{k}\begin{pmatrix}V_{11}&\cdots&V_{1k}\\\
\vdots&\ddots&\vdots\\\
V_{k1}&\cdots&V_{kk}\end{pmatrix}=\begin{pmatrix}\psi(V_{11})&\cdots&\psi(V_{1k})\\\
\vdots&\ddots&\vdots\\\ \psi(V_{k1})&\cdots&\psi(V_{kk})\end{pmatrix}$
$\displaystyle=\sum_{i=1}^{l}\begin{pmatrix}R_{i}^{\\#_{A,B}}V_{11}R_{i}&\cdots&R_{i}^{\\#_{A,B}}V_{1k}R_{i}\\\
\vdots&\ddots&\vdots\\\
R_{i}^{\\#_{A,B}}V_{k1}R_{i}&\cdots&R_{i}^{\\#_{A,B}}V_{kk}R_{i}\end{pmatrix}$
$\displaystyle=\sum_{i=1}^{l}\begin{pmatrix}U_{B}R^{*}_{i}U_{A}^{*}V_{11}R_{i}&\cdots&U_{B}R^{*}_{i}U_{A}^{*}V_{1k}R_{i}\\\
\vdots&\ddots&\vdots\\\
U_{B}R^{*}_{i}U_{A}^{*}V_{k1}R_{i}&\cdots&U_{B}R^{*}_{i}U_{A}^{*}V_{kk}R_{i}\end{pmatrix}$
$\displaystyle=\sum_{i=1}^{l}\begin{pmatrix}U_{B}&&0\\\ &\ddots&\\\
0&&U_{B}\end{pmatrix}\begin{pmatrix}R_{i}^{*}&&0\\\ &\ddots&\\\
0&&R_{i}^{*}\end{pmatrix}U_{A}^{k^{*}}V\begin{pmatrix}R_{i}&&0\\\ &\ddots&\\\
0&&R_{i}\end{pmatrix}$
$\displaystyle=U_{B}^{k}\sum_{i=1}^{l}\begin{pmatrix}R_{i}^{*}&&0\\\
&\ddots&\\\ 0&&R_{i}^{*}\end{pmatrix}U_{A}^{k^{*}}V\begin{pmatrix}R_{i}&&0\\\
&\ddots&\\\ 0&&R_{i}\end{pmatrix},$

and since $U_{A}^{k^{*}}V\in M_{k}(M_{A})^{+},$ by using the Kraus
decomposition

$\sum_{i=1}^{l}R_{i}^{*^{k}}U_{A}^{k^{*}}VR_{i}^{k}\in M_{k}(M_{A})^{+},$

we obtain $U_{B}^{k^{*}}\psi^{k}(V)\geq 0.$ Hence $\psi^{k}(V)$ is a
$U_{B}$-positive matrix, that is, $\psi$ is $(U_{A},U_{B})$-CP map.

######  Theorem 3.1.

Let $U_{A}$ and $U_{B}$ be the fundamental unitaries in $M_{A}$ and $M_{B},$
respectively. A linear map $\psi:M_{A}\to M_{B}$ is a $(U_{A},U_{B})$-CP map
if and only if it has a decomposition of the form (3.2).

###### Proof.

Assume that $\psi$ is a $(U_{A},U_{B})$-CP map. Since a linear map
$\phi:M_{A}\to M_{B}$ defined by $\phi(V)=U^{*}_{B}\psi(U_{A}V)$ is CP, $\phi$
has a Kraus decomposition, that is,

$\phi(V)=\sum_{i=1}^{l}R_{i}^{*}VR_{i},$

where $V\in M_{m}(\mathbb{C})$ and for each $i,$ $R_{i}\in
M_{m,n}(\mathbb{C}).$ Thus we have

$\displaystyle\psi(V)=U_{B}\phi(U_{A}^{*}V)=U_{B}\sum_{i=1}^{l}R_{i}^{*}U_{A}^{*}VR_{i}=\sum_{i=1}^{l}U_{B}R_{i}^{*}U_{A}^{*}VR_{i}=\sum_{i=1}^{l}R_{i}^{\\#}VR_{i}.$

Therefore $\psi$ is a $(U_{A},U_{B})$-CP map if and only if $\psi$ has the
expression $\psi(V)=\sum_{i=1}^{l}R_{i}^{\\#}VR_{i},$ we call $\psi$ has a
Kraus $U$-decomposition in this case. ∎

Suppose $\\{e_{ij}\>\>|\>\>1\leq i,j\leq m\\}$ are the matrix units of
$M_{m}(\mathbb{C}).$ We observe that $D=[U_{A}e_{ij}]_{1\leq i,j\leq m}$ is
$I_{m}\otimes U_{A}$-positive. Indeed,

$\displaystyle(I_{m}\otimes U_{A}^{*})D$
$\displaystyle=\begin{pmatrix}U_{A}^{*}&&0\\\ &\ddots&\\\
0&&U_{A}^{*}\end{pmatrix}\begin{pmatrix}U_{A}e_{11}&\cdots&U_{A}e_{1m}\\\
\vdots&\ddots&\vdots\\\ U_{A}e_{m1}&\cdots&U_{A}e_{mm}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}e_{11}&\cdots&e_{1m}\\\ \vdots&\ddots&\vdots\\\
e_{m1}&\cdots&e_{mm}\end{pmatrix}\in M_{m^{2}}^{+}(\mathbb{C}).$

It implies from the above proposition that $[\psi(U_{A}e_{ij})]_{1\leq i,j\leq
m}$ is $I_{m}\otimes U_{B}$-positive.

######  Theorem 3.2.

Let $\psi:M_{A}\to M_{B}$ be a linear map. Then $\psi$ is $(U_{A},U_{B})$-CP
if and only if $[U_{B}^{*}\psi(U_{A}e_{ij})]_{1\leq i,j\leq m}$ is positive.

###### Proof.

The proof directly follows from [6, Theorem 2]. ∎

Let $\phi:M_{m}(\mathbb{C})\to M_{n}(\mathbb{C})$ be a linear map. Choi [6]
defined $C_{\phi}=\sum_{i,j=1}^{m}e_{ij}\otimes\phi(e_{ij}),$ called the Choi
matrix, and proved that it is positive if and only if $\phi$ is a CP map.

######  Definition 3.3.

Let $\psi:M_{m}(\mathbb{C})\to M_{n}(\mathbb{C})$ be a linear map. We define
$C_{\psi}^{U}:=\sum_{i,j=1}^{m}e_{ij}\otimes\psi(U_{A}e_{ij}).$ The matrix
$C_{\psi}^{U}$ is called the Choi $U$-matrix.

######  Theorem 3.4.

Let $U_{A}$ and $U_{B}$ be the fundamental unitaries in $M_{A}$ and $M_{B},$
respectively, where $M_{A}=M_{m}(\mathbb{C})$ and $M_{B}=M_{n}(\mathbb{C}).$
Then a linear map $\psi:M_{A}\to M_{B}$ is a $(U_{A},U_{B})$-CP map if and
only if $C_{\psi}^{U}$ is $I_{A}\otimes U_{B}$-positive in $M_{A}\otimes
M_{B}.$

###### Proof.

Let $\phi:M_{A}\to M_{B}$ be the linear map defined by
$\phi(V):=U^{*}_{B}\psi(U_{A}V)$ where $V\in M_{A}.$ Then by Proposition 2.1,
$\phi$ is CP if and only if $\psi$ is a $(U_{A},U_{B})$-CP map. It is known
from [6] that $\phi$ is CP if and only if $C_{\phi}$ is positive semi-
definite. Since, for any ${\bf{h}},{\bf{h^{\prime}}}\in\mathbb{C}^{mn},$ we
have

$\displaystyle[C_{\psi}^{U}{\bf{h}},{\bf{h^{\prime}}}]_{U_{B}^{m}}$
$\displaystyle=\langle
C_{\psi}^{U}{\bf{h}},{U_{B}^{m}}{\bf{h^{\prime}}}\rangle=\langle{U_{B}^{m^{*}}}C_{\psi}^{U}{\bf{h}},{\bf{h^{\prime}}}\rangle$
$\displaystyle=\langle\begin{pmatrix}U_{B}^{*}\psi(U_{A}e_{11})&\cdots&U_{B}^{*}\psi(U_{A}e_{1m})\\\
\vdots&\ddots&\vdots\\\
U_{B}^{*}\psi(U_{A}e_{m1})&\cdots&U_{B}^{*}\psi(U_{A}e_{mm})\end{pmatrix}{\bf{h}},{\bf{h^{\prime}}}\rangle$
$\displaystyle=\langle\begin{pmatrix}\phi(e_{11})&\cdots&\phi(e_{1m})\\\
\vdots&\ddots&\vdots\\\
\phi(e_{1m})&\cdots&\phi(e_{mm})\end{pmatrix}{\bf{h}},{\bf{h^{\prime}}}\rangle$
$\displaystyle=\langle C_{\phi}{\bf{h}},{\bf{h^{\prime}}}\rangle,$

that is, $C_{\phi}$ is positive if and only if $C_{\psi}^{U}$ is $I_{A}\otimes
U_{B}$-positive in $M_{A}\otimes M_{B},$ which completes the proof. ∎

##  4\. Nilpotent $U$-CP maps

Nilpotent CP maps were studied by Bhat and Mallick in [2]. Let ${\mathcal{H}}$
be a finite dimensional Hilbert space and $\phi:B(\mathcal{H})\to
B(\mathcal{H})$ be a CP map. Suppose $\phi$ is a nilpotent map of order $p,$
that is, $\phi^{p}=0$ and $\phi^{p-1}\neq 0.$ Define ${\mathcal{H}}_{1}:={\rm
ker}\,(\phi(U))$ and ${\mathcal{H}}_{k}:={\rm ker}\,(\phi^{k}(U))\ominus{\rm
ker}\,(\phi^{k-1}(U)),$ where $2\leq k\leq p.$ Then
$\cap_{k=1}^{p}{\mathcal{H}}_{k}=\emptyset$ and
$\mathcal{H}={\mathcal{H}}_{1}\oplus{\mathcal{H}}_{2}\oplus\cdots\oplus{\mathcal{H}}_{p}.$
Let $b_{i}:={\rm dim}\,({\mathcal{H}}_{i})$ for $1\leq i\leq p.$ Then
$(b_{1},b_{2},\ldots,b_{p})$ is called the CP nilpotent type of $\phi.$ In
this section, we introduce $U$-CP nilpotent type of $U$-CP maps.

######  Proposition 4.1.

Let ${\mathcal{H}}$ be a finite dimensional Hilbert space and
$(\mathcal{H},U)$ be an $S$-space with the indefinite inner product
$[\cdot,\cdot]_{U}.$ Suppose $\phi:B(\mathcal{H})\to B(\mathcal{H})$ is a CP
map, then the corresponding linear map $\psi$ from $B(\mathcal{H})$ to
$B(\mathcal{H})$ defined by $\psi(X):=U\phi(U^{*}X)$ is $U$-CP, with the Kraus
$U$-decomposition $\psi(X)=\sum_{i=1}^{l}R_{i}^{\\#}XR_{i},$ where $X\in
B(\mathcal{H})$ and $R_{i}^{\\#}=UR_{i}^{*}U^{*}$ for each $1\leq i\leq l.$
Then

  1. (1)

${\rm ker}\,(\psi(U))=\cap_{i=1}^{l}{\rm ker}\,(UR_{i}),$

  2. (2)

For $U$-positive $X,$ $\psi(X)=0$ if and only if
$\text{ran}(X)\subseteq\cap_{i=1}^{l}{\rm ker}\,(R_{i}^{*}U^{*}),$

  3. (3)

$\\{h\in{\mathcal{H}}\>\>|\>\>\psi(|Uh\rangle\langle
h|)=0\\}=\cap_{i=1}^{l}{\rm ker}\,(R_{i}^{*}U^{*}),$

  4. (4)

$\text{ran}(\psi(U))=\overline{span}\\{UR_{i}^{*}h\>\>|\>\>h\in{\mathcal{H}},\>\>1\leq
i\leq l\\}.$

###### Proof.

$(1)$ Consider

$\displaystyle{\rm ker}\,(\psi(U))$
$\displaystyle=\\{h\in{\mathcal{H}}\>\>|\>\ \psi(U)h=0\\}$
$\displaystyle=\\{h\in{\mathcal{H}}\>\>|\>\
\sum_{i=1}^{l}R_{i}^{\\#}UR_{i}h=0\\}$
$\displaystyle=\\{h\in{\mathcal{H}}\>\>|\>\
\sum_{i=1}^{l}[R_{i}^{\\#}UR_{i}h,h]_{U}=0\\}$
$\displaystyle=\\{h\in{\mathcal{H}}\>\>|\>\
\sum_{i=1}^{l}[UR_{i}h,R_{i}h]_{U}=0\\}$
$\displaystyle=\\{h\in{\mathcal{H}}\>\>|\>\ \sum_{i=1}^{l}\langle
UR_{i}h,UR_{i}h\rangle=0\\}$ $\displaystyle=\\{h\in{\mathcal{H}}\>\>|\>\
\sum_{i=1}^{l}\|UR_{i}h\|^{2}=0\\}$
$\displaystyle=\\{h\in{\mathcal{H}}\>\>|\>\ UR_{i}h=0,\>\>\text{for
each}\>\>1\leq i\leq l\\}$ $\displaystyle=\bigcap_{i=1}^{l}{\rm
ker}\,(UR_{i}).$

$(2)$ Suppose $\psi(X)=U\phi(U^{*}X)=0$ where $X$ is $U$-positive. It follows
that $\phi(U^{*}X)=0,$ and since $\phi$ is a CP map, using the Kraus
decomposition, we obtain $\sum_{i=1}^{l}R_{i}^{*}U^{*}XR_{i}=0.$ As $X$ is
$U$-positive ($U^{*}X$ is positive), we get $R_{i}^{*}U^{*}XR_{i}=0$ for each
$i.$ Note that $R_{i}^{*}(U^{*}X)^{\frac{1}{2}}=0.$ It implies that
$R_{i}^{*}U^{*}X=0.$ Let $h_{1}\in{\rm ran}(X),$ then there exists
$h_{2}\in\mathcal{H}$ such that $X(h_{2})=h_{1}.$ Now by applying
$R_{i}^{*}U^{*}$ on both the sides, we get $R_{i}^{*}U^{*}{h}_{1}=0$ for each
$i.$ Hence ${\rm ran}(X)\subseteq\cap_{i=1}^{l}{\rm ker}\,(R_{i}^{*}U^{*}).$

Conversely, let ${\rm ran}(X)\subseteq\cap_{i=1}^{l}{\rm
ker}\,(R_{i}^{*}U^{*}),$ then
$\psi(X)=\sum_{i=1}^{l}R_{i}^{\\#}XR_{i}=\sum_{i=1}^{l}UR_{i}^{*}U^{*}XR_{i}=0.$

$(3)$ One can easily see that $|Uh\rangle\langle h|$ is $U$-positive. Indeed,
$U^{*}|Uh\rangle\langle h|=|h\rangle\langle h|\geq 0.$ Also, we have
$\psi(|Uh\rangle\langle h|)=0,$ and ${\rm ran}(|Uh\rangle\langle
h|)=\mathbb{C}h,$ therefore it directly follows from $(2)$ that
$\\{h\in{\mathcal{H}}\>\>|\>\>\psi(|Uh\rangle\langle
h|)=0\\}=\cap_{i=1}^{l}{\rm ker}\,(R_{i}^{*}U^{*}).$

$(4)$ Let $h_{1}\in{\rm ran}(\psi(U))={\rm
ran}(\sum_{i=1}^{l}R_{i}^{\\#}UR_{i})={\rm
ran}(\sum_{i=1}^{l}UR_{i}^{*}R_{i}).$ Then
$\sum_{i=1}^{l}UR_{i}^{*}R_{i}h_{2}=h_{1}$ for some $h_{2}\in{\mathcal{H}}.$
Therefore
$h_{1}\in\overline{span}\\{UR_{i}^{*}h\>\>|\>\>h\in{\mathcal{H}},\>\>1\leq
i\leq l\\}.$ Hence ${\rm
ran}(\psi(U))\subseteq\overline{span}\\{UR_{i}^{*}h\>\>|\>\>h\in{\mathcal{H}},\>\>1\leq
i\leq l\\}.$

Conversely, let
$h\in\overline{span}\\{UR_{i}^{*}h\>\>|\>\>h\in{\mathcal{H}},\>\>1\leq i\leq
l\\}.$ Then $h=\sum_{i=1}^{l}\alpha_{i}UR_{i}^{*}h_{i}$ where
$\alpha_{i}\in\mathbb{C},$ $h_{i}\in{\mathcal{H}}.$ We have to show that
$h\in{\rm ran}(\psi(U))={\rm ran}(U\sum_{i=1}^{l}R_{i}^{*}R_{i}).$ It is
equivalent to show that $h\in{\rm
ker}\,(\sum_{i=1}^{l}R_{i}^{*}R_{i}U^{*})^{\perp},$ that is, $\langle
h,h^{\prime}\rangle_{\mathcal{H}}=0$ for all $h^{\prime}\in{\rm
ker}\,(\sum_{i=1}^{l}R_{i}^{*}R_{i}U^{*}).$

Consider $h^{\prime}\in$ ${\rm ker}\,(\sum_{i=1}^{l}R_{i}^{*}R_{i}U^{*}),$
then we have

$0=\sum_{i=1}^{l}[R_{i}^{*}R_{i}U^{*}h^{\prime},h^{\prime}]_{U^{*}}=\sum_{i=1}^{l}\langle
R_{i}^{*}R_{i}U^{*}h^{\prime},U^{*}h^{\prime}\rangle.$

It follows that $R_{i}U^{*}h^{\prime}=0$ for each $i.$ Observe that

$\displaystyle\langle h,h^{\prime}\rangle$
$\displaystyle=\sum_{i=1}^{l}\alpha_{i}\langle
UR_{i}^{*}h_{i},h^{\prime}\rangle=\sum_{i=1}^{l}\alpha_{i}\langle
h_{i},R_{i}U^{*}h^{\prime}\rangle=0,$

which proves that
$\text{ran}(\psi(U))=\overline{span}\\{UR_{i}^{*}h\>\>|\>\>h\in{\mathcal{H}},\>\>1\leq
i\leq l\\}.$ ∎

######  Proposition 4.2.

Let $(\mathcal{H},U)$ be an $S$-space with the indefinite inner product
$[\cdot,\cdot]_{U}.$ Suppose $\phi:B(\mathcal{H})\to B(\mathcal{H})$ is a CP
map, then the corresponding linear map $\psi$ from $B(\mathcal{H})$ to
$B(\mathcal{H})$ defined by $\psi(X):=U\phi(U^{*}X)$ is $U$-CP, with the Kraus
$U$-decomposition $\psi(X)=\sum_{i=1}^{l}R_{i}^{\\#}XR_{i},$ where $X\in
B(\mathcal{H})$ and $R_{i}^{\\#}=UR_{i}^{*}U^{*}$ for each $1\leq i\leq l.$
Then the followings are equivalent:

  1. (1)

$\psi^{p}(X)=0$ for all $X\in B(\mathcal{H});$

  2. (2)

$R_{i_{1}}R_{i_{2}}\cdots R_{i_{p}}=0$ for all $i_{1},i_{2},\ldots,i_{p}.$

###### Proof.

$(1)\implies(2):$ Let us assume for each $X\in B(\mathcal{H}),$ we have

$\displaystyle
0=\psi^{p}(X)=\sum_{i_{1},i_{2},\ldots,i_{p}=1}^{l}R_{i_{p},\ldots,i_{1}}^{\\#}XR_{i_{1}}R_{i_{2}}\cdots
R_{i_{p}},$

where $R_{i_{p},\ldots,i_{1}}^{\\#}=UR_{i_{p}}^{*}R_{i_{p-1}}^{*}\cdots
R_{i_{1}}^{*}U^{*}.$ Therefore

$0=\psi^{p}(I)=\sum_{i_{1},i_{2},\ldots,i_{p}=1}^{l}R_{i_{p},\ldots,i_{1}}^{\\#}R_{i_{1}}R_{i_{2}}\cdots
R_{i_{p}}.$

Now observe that

$\displaystyle\\{h\in{\mathcal{H}}\>\>|\>\
\sum_{i_{1},i_{2},\ldots,i_{p}=1}^{l}R_{i_{p},\ldots,i_{1}}^{\\#}R_{i_{1}}R_{i_{2}}\cdots
R_{i_{p}}h=0\\}$ $\displaystyle=$ $\displaystyle\\{h\in{\mathcal{H}}\>\>|\>\
\sum_{i_{1},i_{2},\ldots,i_{p}=1}^{l}[R_{i_{p},\ldots,i_{1}}^{\\#}R_{i_{1}}R_{i_{2}}\cdots
R_{i_{p}}h,h]_{U}=0\\}$ $\displaystyle=$
$\displaystyle\\{h\in{\mathcal{H}}\>\>|\>\
\sum_{i_{1},i_{2},\ldots,i_{p}=1}^{l}[R_{i_{1}}R_{i_{2}}\cdots
R_{i_{p}}h,R_{i_{1}}R_{i_{2}}\cdots R_{i_{p}}h]_{U}=0\\},$

which concludes the desired equality $(2).$

$(2)\implies(1):$ Trivial. ∎

Suppose $\psi$ is a $U$-CP map from $B(\mathcal{H})$ to $B(\mathcal{H})$
defined by $\psi(X)=U\phi(U^{*}X).$ Let $\psi$ be a nilpotent map of order
$p.$ Define ${\mathcal{K}}_{1}:={\rm ker}\,(\psi(U))$ and
${\mathcal{K}}_{k}:={\rm ker}\,(\psi^{k}(U))\ominus{\rm
ker}\,(\psi^{k-1}(U)),$ where $2\leq k\leq p.$ Then
$\cap_{k=1}^{p}{\mathcal{K}}_{k}=\emptyset$ and
$\mathcal{H}={\mathcal{K}}_{1}\oplus{\mathcal{K}}_{2}\oplus\cdots\oplus{\mathcal{K}}_{p}.$

######  Definition 4.3.

Let $c_{i}:={\rm dim}\,({\mathcal{K}}_{i})$ for $1\leq i\leq p.$ Then
$(c_{1},c_{2},\ldots,c_{p})$ is called the $U$-CP nilpotent type of $\psi.$

##  5\. Quantum U-channels and quantum $U$-states

The $U$-states and the quantum $U$-channel, which are the $S$-space versions
of the states and quantum channel, respectively, are introduced in this
section. Together, we introduce $U$-separable and $U$-entangled states and
present the $U$-PPT criterion for $U$-separability of $U$-states.

######  Definition 5.1.

Let $\phi:M_{A}\to M_{B}$ be a linear map and $U_{A}$ and $U_{B}$ be the
fundamental unitaries in $M_{A}$ and $M_{B},$ respectively. Then

  1. (1)

$\phi$ is a quantum channel if it is CP and trace preserving, that is,
$\text{Tr}(\phi(V))=\text{Tr}(V)$ where $V\in M_{A}.$

  2. (2)

a linear map $\psi$ from $B(\mathcal{H}_{1})$ to $B(\mathcal{H}_{2})$ defined
by $\psi(V):=U_{2}\phi(U_{1}^{*}V)$ is a quantum $(U_{A},U_{B})$-channel if it
is $(U_{A},U_{B})$-CP and trace preserving.

######  Remark 5.2.

It is well known that $\phi$ is a quantum channel if and only if there exist
$m\times n$-matrices $R_{1},\ldots,R_{l}$ such that

$\displaystyle\phi(V)=\sum_{i=1}^{l}R_{i}^{*}VR_{i}\>\>\>\>\mbox{and}\>\>\>\>\sum_{i=1}^{l}R_{i}R_{i}^{*}=I$

where $V\in M_{A}.$ Indeed, if $\phi$ is a quantum channel, then it is a CP
map and trace preserving. Therefore by Kraus decomposition (3.1), there exist
$m\times n$-matrices $R_{1},\ldots,R_{l}$ such that
$\phi(V)=\sum_{i=1}^{l}R_{i}^{*}VR_{i},$ and if $\phi$ is a trace preserving
map, then $\phi^{*}(V)=\sum_{i=1}^{l}R_{i}VR_{i}^{*}$ is unital
($\text{Tr}(X)=\langle I_{X},X\rangle=\text{Tr}(\phi(X))=\langle
I_{X},\phi(X)\rangle=\langle\phi^{*}(I_{X}),X\rangle$) which implies
$\sum_{i=1}^{l}R_{i}R_{i}^{*}=I.$

Similarly, if $\psi$ is a quantum $(U_{A},U_{B})$-channel, then by Kraus
$U$-decomposition (3.2) we have
$\psi(V)=\sum_{i=1}^{l}R_{i}^{\\#_{A,B}}VR_{i},$ where
$R_{i}^{\\#_{A,B}}=U_{B}R^{*}_{i}U_{A}^{*}.$ Since $\psi$ is trace preserving,
it means $\psi^{*}$ is unital and we obtain
$I_{B}=\psi^{*}(I_{A})=\sum_{i=1}^{l}R_{i}R_{i}^{\\#_{A,B}}.$ Moreover,

$\displaystyle\sum_{i}R_{i}U_{B}^{*}R_{i}^{\\#_{A,B}}=R_{i}U_{B}^{*}U_{B}R^{*}_{i}U_{A}^{*}=U_{A}^{*}.$

A quantum state $\rho\in M_{n}(\mathbb{C})$ is a positive semi-definite matrix
with $\text{Tr}(\rho)=1.$

######  Definition 5.3.

Let $U$ be a fundamental unitary in $M_{n}(\mathbb{C}),$ then a matrix
$\rho\in M_{n}(\mathbb{C})$ is called a quantum $U$-state if the following
conditions hold:

  1. (1)

$\rho$ is $U$-positive, that is, $U^{*}\rho$ is positive and

  2. (2)

$\text{Tr}(U^{*}\rho)=1.$

######  Example 5.4.

Let $U$ be a fundamental unitary in $M_{l}(\mathbb{C}),$ where
$l\in\mathbb{N}.$ Define $\rho\in M_{l}(\mathbb{C})$ as
$\rho=|Ue\rangle\langle e|$ where $e\in\mathbb{C}^{l}$ with $\|e\|=1.$ Then

$\displaystyle U^{*}\rho=U^{*}|Ue\rangle\langle e|=|U^{*}Ue\rangle\langle
e|=|e\rangle\langle e|.$

It follows that $U^{*}\rho$ is positive and also note that
$\text{Tr}(U^{*}\rho)=\text{Tr}(|e\rangle\langle e|)=\langle e,e\rangle=1.$
Hence $\rho$ is a quantum $U$-state.

######  Proposition 5.5.

A quantum $(U_{A},U_{B})$-channel $\psi:M_{A}\to M_{B}$ maps quantum
$U_{A}$-states into quantum $U_{B}$-states.

###### Proof.

Let $V$ be a quantum $U_{A}$-state, that is, $V$ is $U_{A}$-positive and
$\text{Tr}(U_{A}^{*}V)=1.$ Since $\psi$ is a quantum $(U_{A},U_{B})$-channel,
we have

$\displaystyle\psi(V)=\sum_{i=1}^{l}R_{i}^{\\#_{A,B}}VR_{i}=\sum_{i=1}^{l}U_{B}R^{*}_{i}U_{A}^{*}VR_{i},$

for some $m\times n$-matrices $R_{1},\ldots,R_{l}.$ Since $V$ is
$U_{A}$-positive, we have $U_{A}^{*}V\geq 0.$ Therefore
$U_{B}^{*}\psi(V)=\sum_{i=1}^{l}R^{*}_{i}U_{A}^{*}VR_{i}\geq 0,$ that is,
$\psi(V)$ is $U_{B}$-positive. Furthermore, we obtain

$\displaystyle\text{Tr}(U_{B}^{*}\psi(V))$
$\displaystyle=\text{Tr}(\sum_{i=1}^{l}R^{*}_{i}U_{A}^{*}VR_{i})=\text{Tr}(\sum_{i=1}^{l}U_{A}^{*}VR_{i}R^{*}_{i})=\text{Tr}(U_{A}^{*}V\sum_{i=1}^{l}R_{i}R_{i}^{*})$
$\displaystyle=\text{Tr}(U_{A}^{*}V)=1,$

which proves that $\psi(V)$ is a quantum $U_{B}$-state. ∎

A bipartite quantum state $\rho\in M_{A}\otimes M_{B}$ is a product state if
$\rho=\rho_{A}\otimes\rho_{B}$ with $\rho_{A}\in M_{A}^{+}$ and $\rho_{B}\in
M_{B}^{+}$ and is separable if it is a convex combination of product states.
Moreover, it is entangled if it is not separable. We define
$\tau:=t\otimes\text{id}:M_{A}\otimes M_{B}\to M_{A}\otimes M_{B}$ where $t$
is the transpose on $M_{A}.$ We call the $\tau$ map the partial transpose or
the blockwise transpose and a bipartite quantum state $\rho$ is positive
partial transpose (PPT) if $\rho^{\tau}:=t\otimes\text{id}(\rho)$ is positive.
The positive partial transpose criterion says that if $\rho$ is separable,
then $\rho$ is positive partial transpose.

######  Definition 5.6.

Let $U_{A}$ and $U_{B}$ be the fundamental unitaries in $M_{A}$ and $M_{B},$
respectively. Let $U_{A}\otimes U_{B}$ be the fundamental unitary in
$M_{A}\otimes M_{B}$ and $\rho\in M_{A}\otimes M_{B}$ be a bipartite quantum
$U_{A}\otimes U_{B}$-state. Then

  1. (1)

$\rho$ is a product $U_{A}\otimes U_{B}$-state if
$\rho=\rho_{A}\otimes\rho_{B}$ where $\rho_{A}\in M_{A}^{U+}$ and $\rho_{B}\in
M_{B}^{U+}.$

  2. (2)

$\rho$ is $U_{A}\otimes U_{B}$-separable if it is a convex combination of
product $U_{A}\otimes U_{B}$-states.

  3. (3)

$\rho$ is $U_{A}\otimes U_{B}$-entangled if it is not $U_{A}\otimes
U_{B}$-separable.

  4. (4)

$\rho$ is $U_{A}\otimes U_{B}$-positive partial transpose if the partial
transpose $\rho^{\tau}$ is $U^{t}_{A}\otimes U_{B}$-positive, that is,
$(\overline{U}_{A}\otimes U_{B}^{*})(\rho^{\tau})$ is positive.

######  Proposition 5.7.

If a bipartite quantum $U_{A}\otimes U_{B}$-state $\rho\in M_{A}\otimes M_{B}$
is $U_{A}\otimes U_{B}$-separable, then $\rho$ is $U_{A}\otimes
U_{B}$-positive partial transpose.

###### Proof.

Consider that $\rho$ is $U_{A}\otimes U_{B}$-separable, it means we can write
it as a convex combination of product $U_{A}\otimes U_{B}$-states, that is,

$\displaystyle\rho$ $\displaystyle=\sum_{i=1}^{l}p_{i}(U_{A}\otimes
U_{B})(|z_{i}\rangle\langle z_{i}|)=\sum_{i=1}^{l}p_{i}(U_{A}\otimes
U_{B})(|x_{i}\rangle\otimes|y_{i}\rangle)(\langle x_{i}|\otimes\langle
y_{i}|)$ $\displaystyle=\sum_{i=1}^{l}p_{i}(U_{A}\otimes
U_{B})(|x_{i}\rangle\langle x_{i}|\otimes|y_{i}\rangle\langle
y_{i}|)=\sum_{i=1}^{l}p_{i}U_{A}(|x_{i}\rangle\langle x_{i}|)\otimes
U_{B}(|y_{i}\rangle\langle y_{i}|),$

with $\sum_{i=1}^{l}p_{i}=1,$ and
$|z_{i}\rangle=|x_{i}\rangle\otimes|y_{i}\rangle\in M_{A}\otimes M_{B}.$ Since
$(U_{A}(|x_{i}\rangle\langle
x_{i}|))^{t}=|\overline{x_{i}}\rangle\langle\overline{x_{i}}|\overline{U_{A}^{*}},$
we obtain

$\displaystyle\rho^{\tau}=t\otimes\text{id}(\rho)=\sum_{i=1}^{l}p_{i}|\overline{x_{i}}\rangle\langle\overline{x_{i}}|\overline{U_{A}^{*}}\otimes
U_{B}(|y_{i}\rangle\langle y_{i}|).$

Since
$\overline{U_{A}}|\overline{x_{i}}\rangle\langle\overline{x_{i}}|\overline{U_{A}^{*}}$
is a positive matrix in $M_{A},$ $(\overline{U}_{A}\otimes
U_{B}^{*})(\rho^{\tau})$ is positive. ∎

##  6\. U-entanglement breaking maps

In this section, we consider the special class of quantum channels which can
be simulated by a classical channel in the following sense: The sender makes a
measurement on the input state $\rho$, and send the outcome $k$ via a
classical channel to the receiver who then prepares an agreed upon state
$R_{k}.$ Such channels can be written in the form

$\displaystyle\phi(\rho)=\sum_{k}R_{k}\text{Tr}(E_{k}\rho),$

where each $R_{k}$ is a density matrix (density matrices, also called density
operators, which conceptually take the role of the state vectors, that is,
$R_{k}$ is a positive semi-definite matrix with $\text{Tr}(R_{k})=1$) and the
${E_{k}}$ form a positive operator valued measure ($\\{E_{k}\\}_{k}$ form a
positive operator valued measure means for each $k,$ $E_{k}$ is positive semi-
definite and $\sum_{k}E_{k}=id_{A}$). We call this the “Holevo form” because
it was introduced by Holevo in [13]. In this context, it is natural to
consider the class of channels which break entanglement.

######  Definition 6.1.

Let $\phi:M_{A}\to M_{B}$ be a quantum channel. If $(id_{n}\otimes\phi)(S)$ is
always separable for all bipartite quantum states $S\in
M_{n}(\mathbb{C})\otimes M_{A},$ then we call it an entanglement breaking map.

Let $U_{A}$ and $U_{B}$ be the fundamental unitaries in $M_{A}$ and $M_{B},$
respectively. The family $\\{F_{k}\\}_{k}$ is a $U_{A}$-positive operator
valued measure if each $F_{k}U_{A}$ is positive semi-definite and
$\sum_{k}F_{k}U_{A}=id_{A}$ (or $\sum_{k}F_{k}=U_{A}^{*}$) and $D$ is called
$U_{A}$-density matrix if $D$ is a $U_{A}$-positive semi-definite matrix, that
is, $U_{A}^{*}D$ is positive semi-definite matrix with
$\text{Tr}(U_{A}^{*}D)=1.$

######  Definition 6.2.

Let $\psi:M_{A}\to M_{B}$ be a $(U_{A},U_{B})$-quantum channel.

  1. (1)

$\psi$ is said to be $(U_{A},U_{B})$-entanglement breaking if
$(id_{n}\otimes\psi)(S)$ is $I_{n}\otimes U_{B}$-separable for any
$I_{n}\otimes U_{A}$-density matrix $S\in M_{n}(\mathbb{C})\otimes M_{A}.$

  2. (2)

$\psi$ is in $(U_{A},U_{B})$-Holevo form if it can be expressed as

$\displaystyle\psi(\rho)=\sum_{k}D_{k}\text{Tr}(F_{k}\rho),$

where $D_{k}$ is a $U_{B}$-density matrix, that is, $U_{B}^{*}D_{k}$ is
positive semi-definite matrix and $\text{Tr}(U_{B}^{*}D_{k})=1$ and $F_{k}$ is
a $U_{A}$-positive operator valued measure in $M_{A},$ that is $F_{k}U_{A}$ is
positive semi-definite and $\sum_{k}F_{k}U_{A}=id_{A}.$

######  Theorem 6.3.

Let $\psi:M_{A}\to M_{B}$ be a $(U_{A},U_{B})$-quantum channel. Then the
following statements are equivalent:

  1. (1)

$\psi$ is $(U_{A},U_{B})$-entanglement breaking;

  2. (2)

$\psi$ is in $(U_{A},U_{B})$-Holevo form .

###### Proof.

$(1)\implies(2):$ Suppose $\psi$ is $(U_{A},U_{B})$-entanglement breaking. The
map $\phi$ given by $\phi(V)=U^{*}_{B}\psi(U_{A}V)$ is a quantum channel and
we have for each $n\in\mathbb{N},$

$\displaystyle
id_{n}\otimes\phi=id_{n}\otimes(U^{*}_{B}\psi(U_{A}))=(I_{n}\otimes
U^{*}_{B})(id_{n}\otimes\psi)(I_{n}\otimes U_{A}).$ (6.1)

Let $S\in M_{n}(\mathbb{C})\otimes M_{A}$ be a density matrix. One can easily
verify that $(I_{n}\otimes U_{A})S$ is a $(I_{n}\otimes U_{A})$-density
matrix, that is, $(I_{n}\otimes U_{A}^{*})(I_{n}\otimes U_{A})S$ is positive
and $\text{Tr}((I_{n}\otimes U_{A}^{*})(I_{n}\otimes U_{A})S)=1$ which
trivially hold as $(I_{n}\otimes U_{A}^{*})(I_{n}\otimes U_{A})S=S.$ Since
$(id_{n}\otimes\psi)(I_{n}\otimes U_{A})S$ is $(I_{n}\otimes
U_{B})$-separable, $(id_{n}\otimes\phi)(S)$ is separable. This implies that
$\phi$ is an entanglement breaking map. Now using [12, Theorem 4], we can
write $\phi$ in the Holevo form, that is,

$\displaystyle\phi(\rho)=\sum_{k}R_{k}\text{Tr}(E_{k}\rho),$

where each $R_{k}$ is a density matrix and $\\{E_{k}\\}_{k}$ is a positive
operator valued measure with $\sum_{k}E_{k}=id_{A}.$ Observe that

$\displaystyle\psi(\rho)=U_{B}\phi(U_{A}^{*}\rho)=\sum_{k}U_{B}R_{k}\text{Tr}(E_{k}U_{A}^{*}\rho)=\sum_{k}D_{k}\text{Tr}(F_{k}\rho),$

where $D_{k}:=U_{B}R_{k}$ and $F_{k}:=E_{k}U_{A}^{*}.$ Note that $D_{k}$ is a
$U_{B}$-density matrix since $U_{B}^{*}D_{k}=U_{B}^{*}U_{B}R_{k}=R_{k}$ and
$R_{k}$ is already a density matrix in $M_{B}$ and also $\\{F_{k}\\}_{k}$ is a
$U_{A}$-positive operator valued measure in $M_{A}$ as
$E_{k}U_{A}^{*}U_{A}=E_{k}$ is positive semi-definite and
$\sum_{k}E_{k}U_{A}^{*}U_{A}=id_{A}.$

$(2)\implies(1):$ Assume that $\psi$ has the $(U_{A},U_{B})$-Holevo form, it
means $\psi(\rho)=\sum_{k}D_{k}\text{Tr}(F_{k}\rho),$ where $D_{k}$ is a
$U_{B}$-density matrix and $\\{F_{k}\\}_{k}$ is a $U_{A}$-positive operator
valued measure in $M_{A}.$ Define $\phi$ by
$\phi(\rho)=U^{*}_{B}\psi(U_{A}\rho),$ where $\rho\in M_{A}.$ We obtain

$\displaystyle\phi(\rho)=U^{*}_{B}\psi(U_{A}\rho)=U^{*}_{B}\psi(U_{A}\rho)$
$\displaystyle=U^{*}_{B}\sum_{k}D_{k}\text{Tr}(F_{k}U_{A}\rho)$
$\displaystyle=\sum_{k}U^{*}_{B}D_{k}\text{Tr}(F_{k}U_{A}\rho).$

Since $D_{k}$ is a $U_{B}$-density matrix and $\\{F_{k}\\}_{k}$ is a
$U_{A}$-positive operator valued measure in $M_{A},$ $\phi$ has a Holevo form
and by [12, Theorem 4] $\phi$ is an entanglement breaking map and hence
Equation (6.1) implies that $\psi$ is a $(U_{A},U_{B})$-entanglement breaking
map. ∎

######  Remark 6.4.

Let $\phi,\psi:M_{A}\to M_{B}$ be linear maps such that
$\psi(\rho)=U_{B}\phi(U_{A}^{*}\rho),$ where $\rho\in M_{A}.$ As we know
$\phi$ is positive if and only if $\psi$ is a $(U_{A},U_{B})$-positive map.
Suppose $\phi$ is a quantum channel, that is, $\psi$ is a
$(U_{A},U_{B})$-quantum channel. Note that $\theta\circ\phi$ is a CP map for
any CP map $\theta:M_{B}\to M_{C}$ if and only if $\omega\circ\psi$ is
$(U_{A},U_{C})$-CP for any $(U_{B},U_{C})$-CP $\omega:M_{B}\to M_{C}.$
Therefore, it follows from Theorem 6.3 that $\phi$ is an entanglement breaking
map if and only if $\psi$ is a $(U_{A},U_{B})$-entanglement breaking map.

##  7\. Examples of fundamental unitary and $U$-CP maps

In this section, we provide concrete examples of completely $U$-positive maps
and examples of $3\otimes 3$ quantum $U$-states which are $U$-entangled and
$U$-separable. It is easy to observe that the $2\times 2$ identity matrix $I$
and the Pauli matrices

$\sigma_{x}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\quad\sigma_{y}=\begin{pmatrix}0&-\iota\\\
\iota&0\end{pmatrix},\quad\sigma_{z}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}$

form a basis for $M_{2}(\mathbb{C})$. That is, for any $A\in
M_{2}(\mathbb{C}),$ we have $A=aI+b\sigma_{x}+c\sigma_{y}+d\sigma_{z}$ where
$a,b,c,d\in\mathbb{C}$. Any fundamental unitary on the $2$-dimensional complex
$S$-space has the form

$U=\begin{pmatrix}a&b\\\
-e^{\iota\phi}\overline{b}&e^{\iota\phi}\overline{a}\end{pmatrix}$ (7.1)

where $\phi\in\mathbb{R}$ and $a,b\in\mathbb{C}$ such that
$|a|^{2}+|b|^{2}=1.$ For example, if we choose $a=1$ and $b=0$, then we have
the unitary

$\begin{pmatrix}1&0\\\ 0&e^{\iota\phi}\end{pmatrix}$

which is called a Phase Gate (see [17]) that represents a rotation about the
$z$-axis by an angle $\phi$ on the Bloch sphere.

If we define an $S$-space with respect to the fundamental unitary $U$ as in
(7.1), then $U^{*}A=aU^{*}+b\sigma_{x}^{U}+c\sigma_{y}^{U}+d\sigma_{z}^{U},$
where $\sigma_{x}^{U}=U^{*}\sigma_{x},$ $\sigma_{y}^{U}=U^{*}\sigma_{y},$ and
$\sigma_{z}^{U}=U^{*}\sigma_{z},$ and we call these matrices $U$-Pauli
matrices.

Let $U_{1}=\begin{pmatrix}1&0\\\ 0&\iota\end{pmatrix}$ and
$U_{2}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\\ 1&1\end{pmatrix}$ be two
unitaries which are not symmetries, where $U_{1}$ is the Phase gate for
$\phi=\frac{\pi}{2}.$

  1. (1)

Consider the $S$-space $(\mathbb{C}^{2},U_{1}).$ For any $A\in
M_{2}(\mathbb{C}),$ we have

$U^{*}_{1}A=\left[a\begin{pmatrix}1&0\\\ 0&-\iota\end{pmatrix}-\iota
b\begin{pmatrix}0&\iota\\\ 1&0\end{pmatrix}-\iota c\begin{pmatrix}0&1\\\
\iota&0\end{pmatrix}+d\begin{pmatrix}1&0\\\ 0&\iota\end{pmatrix}\right]$

and

$(U^{*}_{1}A)^{*}=\left[\overline{a}\begin{pmatrix}1&0\\\
0&\iota\end{pmatrix}+\overline{b}\begin{pmatrix}0&\iota\\\
1&0\end{pmatrix}+\overline{c}\begin{pmatrix}0&1\\\
\iota&0\end{pmatrix}+\overline{d}\begin{pmatrix}1&0\\\
0&-\iota\end{pmatrix}\right].$

Comparing $U^{*}_{1}A$ and $(U^{*}_{1}A)^{*}$, one may easily find out that
$A$ is $U_{1}$-self adjoint if and only if $a=\overline{d},$ $-\iota
c=\overline{c}$ and $-\iota b=\overline{b}$ , that is, $A$ has the form

$\displaystyle A=\begin{pmatrix}a+d&b-\iota c\\\ b+\iota
c&a-d\end{pmatrix}=\begin{pmatrix}a+\overline{a}&b+\overline{c}\\\
b-\overline{c}&a-\overline{a}\end{pmatrix}=\begin{pmatrix}2\Re(a)&b+\overline{c}\\\
b-\overline{c}&2\iota\Im(a)\end{pmatrix}$

and $U_{1}^{*}A$ has the form

$\displaystyle U_{1}^{*}A=\begin{pmatrix}a+d&b-\iota c\\\ c-\iota
b&-\iota(a-d)\end{pmatrix}=\begin{pmatrix}a+\overline{a}&b+\overline{c}\\\
c+\overline{b}&\iota(a-\overline{a})\end{pmatrix}=\begin{pmatrix}2\Re(a)&b+\overline{c}\\\
c+\overline{b}&2\Im(a)\end{pmatrix}$

where $a,b,c\in\mathbb{C}$. Further, $U^{*}_{1}A$ is positive, that is, $A$ is
$U_{1}$-positive if and only if

$0\leq\Re(a)\quad\mbox{and}\quad
4\Re(a)\Im(a)\geq(b+\overline{c})(\overline{b}+{c})$

Also, $U^{*}_{1}A$ is a quantum state, that is, $A$ is a quantum $U_{1}$-state
if and only if

$\Re(a)+\Im(a)=\frac{1}{2}.$

In particular, if $a=\frac{1}{2}\in\mathbb{R},$ $b=t$ and $c=-t$ for all
$t\geq 0$, then all the above relations are trivially satisfied. In other
words, for $t\geq 0$,

$A=\rho_{t}=\begin{pmatrix}1&0\\\ 2t&0\end{pmatrix}$

provides a one parameter family of quantum $U_{1}$-states in
$M_{2}(\mathbb{C})$. Similarly, the following provides a one parameter family
of quantum $U_{1}\otimes U_{1}$-states

$\frac{1}{16}\begin{pmatrix}1&0&0&0\\\ 2t&0&0&0\\\ 2t&0&0&0\\\
4t^{2}&0&0&0\end{pmatrix},$

where $t\geq 0$.

Since $M_{2}(\mathbb{C})$ is a unital $*$-algebra, any $*$-homomorphism $\pi$
from $M_{2}(\mathbb{C})$ into $M_{2}(\mathbb{C})$ has the form
$\pi(A)=W^{*}AW$ for some unitary matrix $W\in M_{2}(\mathbb{C}).$ If $\phi$
is a $U_{1}$-CP map defined on $M_{2}(\mathbb{C}),$ then by Theorem 2.2 there
exist a $*$-homomorphism $\pi$ on $M_{2}(\mathbb{C})$ and a matrix $V\in
M_{2}(\mathbb{C})$ such that

$\phi(A)=V^{\\#}\pi(A)V,$

where $V^{\\#}=U_{1}V^{*}U_{1}^{*}.$ For example, if we consider
$V=\begin{pmatrix}\alpha&0\\\ 0&\beta\end{pmatrix}$ and a unitary
$W=\begin{pmatrix}\gamma&0\\\ 0&\delta\end{pmatrix},$ then we get $U_{1}$-CP
$\phi$ in the following form:

$\displaystyle\phi(A)$
$\displaystyle=V^{\\#}\pi(A)V=(U_{1}V^{*}U_{1}^{*})(W^{*}AW)V=\begin{pmatrix}\overline{\alpha}&0\\\
0&\overline{\beta}\end{pmatrix}\begin{pmatrix}a_{11}&\overline{\gamma}a_{12}\delta\\\
\overline{\delta}a_{21}\gamma&a_{22}\end{pmatrix}\begin{pmatrix}\alpha&0\\\
0&\beta\end{pmatrix}$ $\displaystyle=\begin{pmatrix}\overline{\alpha}\alpha
a_{11}&\overline{\alpha}\overline{\gamma}\delta\beta a_{12}\\\
\overline{\beta}\overline{\delta}\gamma\alpha a_{21}&\overline{\beta}\beta
a_{22}\end{pmatrix},$

where $A=\begin{pmatrix}a_{11}&a_{12}\\\ a_{21}&a_{22}\end{pmatrix}\in
M_{2}(\mathbb{C})$. Furthermore, if $|\alpha|=|\beta|=1,$ then $\phi(A)$ is of
the form

$\phi(A)=\begin{pmatrix}a_{11}&\overline{\alpha}\overline{\gamma}\delta\beta
a_{12}\\\ \overline{\beta}\overline{\delta}\gamma\alpha
a_{21}&a_{22}\end{pmatrix}.$

  2. (2)

Consider the $S$-space $(\mathbb{C}^{2},U_{2}).$ For any $A\in
M_{2}(\mathbb{C}),$ we obtain

$U^{*}_{2}A=\frac{1}{\sqrt{2}}\left[a\begin{pmatrix}1&1\\\
-1&1\end{pmatrix}+b\begin{pmatrix}1&1\\\ 1&-1\end{pmatrix}-\iota
c\begin{pmatrix}-1&1\\\ -1&-1\end{pmatrix}+d\begin{pmatrix}1&-1\\\
-1&-1\end{pmatrix}\right].$

Comparing $U^{*}_{2}A$ and $(U^{*}_{2}A)^{*}$, one may easily find out that
$A$ is $U_{2}$-self adjoint if and only if $b$ and $d$ are reals and
$c=-\iota\overline{a}$, that is, $A$ has the form

$\begin{pmatrix}a+d&-\overline{a}+b\\\ \overline{a}+b&a-d\end{pmatrix}$

where $a\in\mathbb{C}$ and $b,d\in\mathbb{R}$. Further, $U^{*}_{2}A$ is
positive, that is, $A$ is $U_{2}$-positive if and only if

$-(b+d)\leq 2\Re(a)\quad\mbox{and}\quad b^{2}+d^{2}\leq
2((\Re(a))^{2}-(\Im(a))^{2}).$

Also, $U^{*}_{2}A$ is a quantum state, that is, $A$ is a quantum $U_{2}$-state
if and only if

$\Re(a)=\frac{\sqrt{2}}{4},\quad-(b+d)\leq\frac{\sqrt{2}}{2}\quad\mbox{and}\quad
b^{2}+d^{2}\leq\frac{1}{4}-2(\Im(a))^{2}.$

In particular, if $a=\sqrt{2}/4\in\mathbb{R}$ and $b=d=t/4$, with
$-\sqrt{2}\leq t\leq\sqrt{2}$, then all the above relations are trivially
satisfied. In other words, for $-\sqrt{2}\leq t\leq\sqrt{2}$,

$\rho_{t}=\frac{1}{4}\begin{pmatrix}t+\sqrt{2}&t-\sqrt{2}\\\
t+\sqrt{2}&-t+\sqrt{2}\end{pmatrix}$

provides a one parameter family of quantum $U_{2}$-states in
$M_{2}(\mathbb{C})$. Similarly, the following provides a one parameter family
of quantum $U_{2}\otimes U_{2}$-states

$\frac{1}{16}\begin{pmatrix}t^{2}+2\sqrt{2}t+2&t^{2}-2&t^{2}-2&t^{2}-2\sqrt{2}t+2\\\
t^{2}+2\sqrt{2}t+2&-t^{2}+2&t^{2}-2&-t^{2}+2\sqrt{2}t-2\\\
t^{2}+2\sqrt{2}t+2&t^{2}-2&-t^{2}+2&-t^{2}+2\sqrt{2}t-2\\\
t^{2}+2\sqrt{2}t+2&-t^{2}+2&-t^{2}+2&t^{2}-2\sqrt{2}t+2\end{pmatrix},$

where $-\sqrt{2}\leq t\leq\sqrt{2}$.

Also, similar to the earlier example, we get any $U_{2}$-CP map $\phi$ in the
following form:

$\displaystyle\phi(A)$
$\displaystyle=V^{\\#}\pi(A)V=(U_{2}V^{*}U_{2}^{*})(W^{*}AW)V$
$\displaystyle=\frac{1}{{2}}\begin{pmatrix}\overline{\alpha}+\overline{\beta}&\overline{\alpha}-\overline{\beta}\\\
\overline{\alpha}-\overline{\beta}&\overline{\alpha}+\overline{\beta}\end{pmatrix}\begin{pmatrix}a_{11}&\overline{\gamma}a_{12}\delta\\\
\overline{\delta}a_{21}\gamma&a_{22}\end{pmatrix}\begin{pmatrix}\alpha&0\\\
0&\beta\end{pmatrix}$
$\displaystyle=\frac{1}{{2}}\begin{pmatrix}(\overline{\alpha}+\overline{\beta})\alpha
a_{11}+(\overline{\alpha}-\overline{\beta})\overline{\delta}\gamma\alpha
a_{21}&(\overline{\alpha}+\overline{\beta})\overline{\gamma}\delta\beta
a_{12}+(\overline{\alpha}-\overline{\beta})\beta a_{22}\\\
(\overline{\alpha}-\overline{\beta})\alpha
a_{11}+(\overline{\alpha}+\overline{\beta})\overline{\delta}\gamma\alpha
a_{21}&(\overline{\alpha}-\overline{\beta})\overline{\gamma}\delta\beta
a_{12}+(\overline{\alpha}+\overline{\beta})\beta a_{22}\end{pmatrix},$

where $A=\begin{pmatrix}a_{11}&a_{12}\\\ a_{21}&a_{22}\end{pmatrix}\in
M_{2}(\mathbb{C})$. Also if $|\alpha|=|\beta|=1,$ then $\phi(A)$ is of the
form

$\phi(A)=\frac{1}{{2}}\begin{pmatrix}(1+\overline{\beta}\alpha)a_{11}+(1-\overline{\beta}\alpha)\overline{\delta}\gamma
a_{21}&(\overline{\alpha}\beta+1)\overline{\gamma}\delta
a_{12}+(\overline{\alpha}\beta-1)a_{22}\\\
(1-\overline{\beta}\alpha)a_{11}+(1+\overline{\beta}\alpha)\overline{\delta}\gamma
a_{21}&(\overline{\alpha}\beta-1)\overline{\gamma}\delta
a_{12}+(\overline{\alpha}\beta+1)a_{22}\end{pmatrix}.$

  3. (3)

Let $\mathbb{C}^{3}$ be a $3$-dimensional $S$-space with an indefinite metric
induced by $U_{3}$, where $U_{3}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1&0\\\
1&1&0\\\ 0&0&\sqrt{2}\end{pmatrix}.$ It is easy to observe that the matrices

$\displaystyle\mu_{1}=\begin{pmatrix}1&0&0\\\ 1&0&0\\\
0&0&0\end{pmatrix},\quad\mu_{2}=\begin{pmatrix}0&1&0\\\ 0&1&0\\\
0&0&0\end{pmatrix},\quad\mu_{3}=\begin{pmatrix}0&0&1\\\ 0&0&1\\\
0&0&0\end{pmatrix},$ $\displaystyle\mu_{4}=\begin{pmatrix}-1&0&0\\\ 1&0&0\\\
0&0&0\end{pmatrix},\quad\mu_{5}=\begin{pmatrix}0&-1&0\\\ 0&1&0\\\
0&0&0\end{pmatrix},\quad\mu_{6}=\begin{pmatrix}0&0&-1\\\ 0&0&1\\\
0&0&0\end{pmatrix},$ $\displaystyle\mu_{7}=\begin{pmatrix}0&0&0\\\ 0&0&0\\\
\sqrt{2}&0&0\end{pmatrix},\quad\mu_{8}=\begin{pmatrix}0&0&0\\\ 0&0&0\\\
0&\sqrt{2}&0\end{pmatrix},\quad\mu_{9}=\begin{pmatrix}0&0&0\\\ 0&0&0\\\
0&0&\sqrt{2}\end{pmatrix}$

form a basis for $M_{3}(\mathbb{C})$. Thus, for any $A\in M_{3}(\mathbb{C})$,
we have $A=\sum\limits_{i=1}^{9}a_{i}\mu_{i},$ where $a_{i}\in\mathbb{C}$.
Then, we get

$\displaystyle A=\begin{pmatrix}a_{1}-a_{4}&a_{2}-a_{5}&a_{3}-a_{6}\\\
a_{1}+a_{4}&a_{2}+a_{5}&a_{3}+a_{6}\\\
a_{7}\sqrt{2}&a_{8}\sqrt{2}&a_{9}\sqrt{2}\end{pmatrix}.$ (7.2)

Since

$\displaystyle U_{3}^{*}A=\sqrt{2}\begin{pmatrix}a_{1}&a_{2}&a_{3}\\\
a_{4}&a_{5}&a_{6}\\\ a_{7}&a_{8}&a_{9}\end{pmatrix},$

after comparing $U_{3}^{*}A$ and $(U_{3}^{*}A)^{*}$, one may easily find out
that $A$ is $U_{3}$-self adjoint if and only if $a_{1},$ $a_{5}$ and $a_{9}$
are reals and $a_{2}=\overline{a_{4}},$ $a_{3}=\overline{a_{7}}$ and
$a_{6}=\overline{a_{8}}$, that is, $U_{3}^{*}A$ has the form

$U_{3}^{*}A=\sqrt{2}\begin{pmatrix}a_{1}&a_{2}&a_{3}\\\
\overline{a_{2}}&a_{5}&a_{6}\\\
\overline{a_{3}}&\overline{a_{6}}&a_{9}\end{pmatrix}.$

Further, $U_{3}^{*}A$ is positive, that is, $A$ is $U_{3}$-positive if and
only if the following conditions hold:

$\displaystyle a_{1}\geq 0,$ (7.3) $\displaystyle a_{1}a_{5}-|a_{2}|^{2}\geq
0$ (7.4) and $\displaystyle
a_{1}a_{5}a_{9}-a_{1}|a_{6}|^{2}-|a_{2}|^{2}a_{9}-|a_{3}|^{2}a_{5}+2\Re(a_{2}\overline{a_{3}}a_{6})\geq
0.$ (7.5)

Also, $U_{3}^{*}A$ is a quantum state, that is, $A$ is a quantum $U_{3}$-state
if and only if

$a_{1}+a_{5}+a_{9}=\frac{1}{\sqrt{2}}.$

In particular, if we choose $a_{i}=\dfrac{1}{3\sqrt{2}}$ in (7.2), then the
matrix $A=\dfrac{1}{3}\begin{pmatrix}0&0&0\\\ \sqrt{2}&\sqrt{2}&\sqrt{2}\\\
1&1&1\end{pmatrix}$ is a $U_{3}$-state, where

$\displaystyle U_{3}^{*}A=\frac{1}{3}\begin{pmatrix}1&1&1\\\ 1&1&1\\\
1&1&1\end{pmatrix}.$

Using this example we give the following quantum separable $U_{3}\otimes
U_{3}$-state:

$\displaystyle\frac{1}{9}\begin{pmatrix}0&0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0&0\\\
2&2&2&2&2&2&2&2&2\\\
\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}\\\
0&0&0&0&0&0&0&0&0\\\
\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}&\sqrt{2}\\\
1&1&1&1&1&1&1&1&1\end{pmatrix}.$

In [7], Choi gave the following entangled state which has positive partial
transpose:

$\displaystyle\begin{pmatrix}1&0&0&0&1&0&0&0&1\\\ 0&2&0&1&0&0&0&0&0\\\
0&0&\frac{1}{2}&0&0&0&1&0&0\\\ 0&1&0&\frac{1}{2}&0&0&0&0&0\\\
1&0&0&0&1&0&0&0&1\\\ 0&0&0&0&0&2&0&1&0\\\ 0&0&1&0&0&0&2&0&0\\\
0&0&0&0&0&1&0&\frac{1}{2}&0\\\ 1&0&0&0&1&0&0&0&1\end{pmatrix}.$

Consider

$\displaystyle C:=\frac{2}{21}\begin{pmatrix}1&0&0&0&1&0&0&0&1\\\
0&2&0&1&0&0&0&0&0\\\ 0&0&\frac{1}{2}&0&0&0&1&0&0\\\
0&1&0&\frac{1}{2}&0&0&0&0&0\\\ 1&0&0&0&1&0&0&0&1\\\ 0&0&0&0&0&2&0&1&0\\\
0&0&1&0&0&0&2&0&0\\\ 0&0&0&0&0&1&0&\frac{1}{2}&0\\\
1&0&0&0&1&0&0&0&1\end{pmatrix}.$

Note that

$\displaystyle U_{3}\otimes
U_{3}=\frac{1}{2}\begin{pmatrix}1&-1&0&-1&1&0&0&0&0\\\ 1&1&0&-1&-1&0&0&0&0\\\
0&0&\sqrt{2}&0&0&-\sqrt{2}&0&0&0\\\ 1&-1&0&1&-1&0&0&0&0\\\
1&1&0&1&1&0&0&0&0\\\ 0&0&\sqrt{2}&0&0&\sqrt{2}&0&0&0\\\
0&0&0&0&0&0&\sqrt{2}&-\sqrt{2}&0\\\ 0&0&0&0&0&0&\sqrt{2}&\sqrt{2}&0\\\
0&0&0&0&0&0&0&0&2\end{pmatrix}.$

One may easily check that

$\displaystyle A=(U_{3}\otimes
U_{3})C=\frac{1}{21}\begin{pmatrix}2&-3&0&-\frac{3}{2}&2&0&0&0&2\\\
0&1&0&\frac{1}{2}&0&0&0&0&0\\\
0&0&\frac{1}{\sqrt{2}}&0&0&-2\sqrt{2}&\sqrt{2}&-\sqrt{2}&0\\\
0&-1&0&-\frac{1}{2}&0&0&0&0&0\\\ 2&3&0&\frac{3}{2}&2&0&0&0&2\\\
0&0&\frac{1}{\sqrt{2}}&0&0&2\sqrt{2}&\sqrt{2}&\sqrt{2}&0\\\
0&0&\sqrt{2}&0&0&-\sqrt{2}&2\sqrt{2}&-\frac{1}{\sqrt{2}}&0\\\
0&0&\sqrt{2}&0&0&\sqrt{2}&2\sqrt{2}&\frac{1}{\sqrt{2}}&0\\\
2&0&0&0&2&0&0&0&2\end{pmatrix}$

is a $U_{3}\otimes U_{3}$-entangled state.

### Acknowledgement

Azad Rohilla thanks Harsh Trivedi for a research visit during February-March
2024 at The LNM Institute of Information Technology. Harsh Trivedi is
supported by MATRICS-SERB Research Grant, File No: MTR/2021/000286, by SERB,
Department of Science & Technology (DST), Government of India. Harsh Trivedi
acknowledges the DST-FIST program (Govt. of India) FIST - No.
SR/FST/MS-I/2018/24.

## References

  * [1] J.-P. Antoine and S. Ōta, _Unbounded GNS representations of a ∗-algebra in a Kreĭn space_, Lett. Math. Phys. 18 (1989), no. 4, 267–274.
  * [2] B. V. R. Bhat and N. Mallick, _Nilpotent completely positive maps_ , Positivity, 2014, 18, pp.567-577.
  * [3] J. Bognár, _Indefinite inner product spaces_ , Springer-Verlag, New York-Heidelberg, 1974, Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 78.
  * [4] P. J. M. Bongaarts, _Maxwell’s equations in axiomatic quantum field theory. I. Field tensor and potentials_ , J. Mathematical Phys. 18 (1977), no. 7, 1510–1516.
  * [5] H.-J. Borchers, _On the structure of the algebra of field operators. II_ , Comm. Math. Phys. 1 (1965), 49–56.
  * [6] M. D. Choi, _Completely positive linear maps on complex matrices_ , Linear algebra and its applications. 1975 Jun 1; 10(3):285-90.
  * [7] M. D. Choi and R. V. Kadison, _Positive linear maps in Operator Algebras and Applications_ Kingston, 1980, Proc. Sympos. Pure Math, 1980.
  * [8] S. Dey and H. Trivedi, _KSGNS construction for $\tau$-maps on S-modules and $\mathfrak{K}$-families_, Oper. Matrices 11 (2017), no. 3, 679–696.
  * [9] S. Dey and H. Trivedi, _Bures Distance and Transition Probability for $\alpha$-CPD-Kernels_, Complex Anal. Oper. Theory 13, 2171–2190 (2019).
  * [10] P. A. M. Dirac, _The physical interpretation of quantum mechanics_ , Proc. Roy. Soc. London. Ser. A. 180 (1942), 1–40.
  * [11] J. Heo, _Quantum J-channels on Krein spaces_. (English summary) Quantum Inf. Process. 22 (2023), no. 1, Paper No. 16, 18 pp.
  * [12] M. Horodecki, P. W. Shor, and M. B. Ruskai, _Entanglement breaking channels_ Reviews in Mathematical Physics 15, no. 06 (2003): 629-641.
  * [13] A. S. Holevo, _Coding Theorems for Quantum Channels_ , Russian Math. Surveys 53, 1295–1331 (1999).
  * [14] W. Pauli, _On Dirac’s new method of field quantization_ , Rev. Modern Phys. 15 (1943), 175–207.
  * [15] F. Philipp, F. H. Szafraniec, and C. Trunk, _Selfadjoint operators in S-spaces_ , J. Funct. Anal. 260 (2011), no. 4, 1045–1059.
  * [16] F.-S. Raúl; F. Raúl, _J-states and quantum channels between indefinite metric spaces._ (English summary) Quantum Inf. Process., 21 (2022), no. 4, Paper No. 139, 32 pp.
  * [17] E. G. Rieffel and P. H. Wolfgang, _Quantum computing: A gentle introduction_ , MIT press, 2011.
  * [18] W. F. Stinespring, _Positive functions on $C^{*}$-algebras_, Proc. Amer. Math. Soc. 6 (1955), 211–216.
  * [19] F. H. Szafraniec, _Two-sided weighted shifts are ‘almost Krein’ normal_ , Spectral theory in inner product spaces and applications, Oper. Theory Adv. Appl., vol. 188, Birkhäuser Verlag, Basel, 2009, pp. 245–250.