Cloning by positive maps in von Neumann algebras

Abstract

We investigate cloning in the general operator algebra framework in arbitrary dimension assuming only positivity instead of strong positivity of the cloning operation, generalizing thus results obtained so far under that stronger assumption. The weaker positivity assumption turns out quite natural when considering cloning in the general $C^*$-algebra framework.

1 Introduction

Cloning and broadcasting of quantum states has recently become an important topic in Quantum Information Theory. Since its first appearance in [5, 14] in the form of a no-cloning theorem it has been investigated in various settings. The most interesting ones are the Hilbert space setup considered in [3, 8], and the setup of generic probabilistic models considered in [1, 2]. A common feature of these approaches consists in restricting attention to the finite-dimensional models; moreover, in the Hilbert space setup the map defining cloning or broadcasting is assumed to be completely positive. In [6] cloning and broadcasting are investigated in the general operator algebra framework, i.e. instead of the full algebra of all linear operators on a finite-dimensional Hilbert space an arbitrary von Neumann algebra on a Hilbert space of arbitrary dimension is considered; moreover, the cloning (broadcasting) operation is assumed to be a Schwarz (called also strongly positive) map instead of completely positive. The present paper can be viewed as a supplement to [6]; we follow the same approach weakening further the assumption on positivity of the cloning operation, namely, we assume only that it is positive. This weaker assumption is all we can hope for while considering the general problem of cloning in $C^*$-algebras (cf. [7]), thus the main theorem of the present paper (Theorem 8) is of importance for cloning in $C^*$-algebras. However, in our approach interesting results are obtained only for cloning, the problem of broadcasting in such a setup is still an open question.

It is probably worth mentioning that although cloning and broadcasting have their origins in quantum information theory, they are nevertheless purely mathematical objects concerning states on some $C^*$- or $W^*$-algebras, and thus their investigation is of independent mathematical interest. This is the approach taken in the present paper—we do not refer to any physical notions, however it is still possible (and hoped for) that the results obtained can find some physical applications.

The main results of this work are as follows. It is shown that all states cloneable by an operation are extreme points of the set of all states broadcastable by this operation, and a description of some algebra associated with the cloneable states is given. Moreover, for an arbitrary subset $\Gamma$ of the normal states of a von Neumann algebra it is proved that the states in $\Gamma$ are cloneable if and only if they have mutually orthogonal supports—the result obtained in [6] under the assumption that the cloning operation is a Schwarz map. Finally, the problem of uniqueness of the cloning operation is considered.

2 Preliminaries and notation

Let $\fancyscript{M}$ be an arbitrary von Neumann algebra with identity ${\small 1}\!\!1$ acting on a Hilbert space $\mathcal {H}$. The predual $\fancyscript{M}_*$ of $\fancyscript{M}$ is a Banach space of all normal, i.e. continuous in the $\sigma$-weak topology linear functionals on $\fancyscript{M}$.

A state on $\fancyscript{M}$ is a bounded positive linear functional $\rho :\fancyscript{M}\rightarrow \mathbb {C}$ of norm one. For a normal state $\rho$ its support, denoted by $\mathrm{s }(\rho )$, is defined as the smallest projection in $\fancyscript{M}$ such that $\rho (\mathrm{s }(\rho ))=\rho ({\small 1}\!\!1)$. In particular, we have

$$\begin{aligned} \rho (\mathrm{s }(\rho )x)=\rho (x\mathrm{s }(\rho ))=\rho (x),\qquad x\in \fancyscript{M}, \end{aligned}$$

and if $\rho (\mathrm{s }(\rho )x\mathrm{s }(\rho ))=0$ for $\mathrm{s }(\rho )x\mathrm{s }(\rho )\geqslant 0$ then $\mathrm{s }(\rho )x\mathrm{s }(\rho )=0$.

Let $\{\rho _{\theta }:\theta \in \Theta \}$ be a family of normal states on a von Neumann algebra $\fancyscript{M}$. Define the support of this family by

$$\begin{aligned} e=\bigvee _{\theta \in \Theta }\mathrm{s }(\rho _{\theta }). \end{aligned}$$

The family $\{\rho _{\theta }:\theta \in \Theta \}$ is said to be faithful if for each positive element $x\in \fancyscript{M}$ from the equality $\rho _{\theta }(x)=0$ for all $\theta \in \Theta$ it follows that $x=0$. It is seen that the faithfulness of this family is equivalent to the relation $e={\small 1}\!\!1$; moreover, if $\rho _{\theta }(exe)=0$ for all $\theta \in \Theta$ and $exe\geqslant 0$ then $exe=0$.

By a $W^*$-algebra of operators acting on a Hilbert space we shall mean a $C^*$-subalgebra of $\mathbb {B}(\mathcal {H})$ with identity, closed in the weak-operator topology. A typical example (and in fact the only one utilized in the paper) is the algebra $p\fancyscript{M}p$, where $p$ is a projection in $\fancyscript{M}$. For arbitrary $\mathcal {R}\subset \mathbb {B}(\mathcal {H})$ we denote by $W^*(\mathcal {R})$ the smallest $W^*$-algebra of operators on $\mathcal {H}$ containing $\mathcal {R}$.

A projection $p$ in a $W^*$-algebra $\fancyscript{M}$ is said to be minimal if it majorizes no other nonzero projection in $\fancyscript{M}$. A $W^*$-algebra $\fancyscript{M}$ is said to be atomic if the supremum of all minimal projections in $\fancyscript{M}$ equals the identity of $\fancyscript{M}$.

For $x,y\in \mathbb {B}(\mathcal {H})$ we define the Jordan product $x\circ y$ as follows

$$\begin{aligned} x\circ y=\frac{xy+yx}{2}. \end{aligned}$$

(The same symbol “$\circ$” will also be used for a linear functional $\varphi$ and a map $T$ on $\fancyscript{M}$, namely, $\varphi \circ T$ will stand for the functional defined as $(\varphi \circ T)(x)=\varphi (T(x)),\,x\in \fancyscript{M}$, however, it should not cause any confusion.) Let $\mathcal {A}\subset \mathbb {B}(\mathcal {H})$ be a linear space. $\mathcal {A}$ is said to be a $JW^*$-algebra if it is weak-operator closed, contains an identity $p$, i.e. $pa=ap=a$ for each $a\in \mathcal {A}$, and is closed with respect to the Jordan product, i.e. for any $a,b\in \mathcal {A}$ we have $a\circ b\in \mathcal {A}$.

Let $\fancyscript{M}$ and $\mathcal {N}$ be $W^*$-algebras. A linear map $T:\fancyscript{M}\rightarrow \mathcal {N}$ is said to be normal if it is continuous in the $\sigma$-weak topologies on $\fancyscript{M}$ and $\mathcal {N}$, respectively. It is called unital if it maps the identity of $\fancyscript{M}$ to the identity of $\mathcal {N}$.

Let $T$ be a positive map on $\fancyscript{M}$. [4, Proposition 3.2.4] yields that for any $x=x^*\in \fancyscript{M}$ we have

$$\begin{aligned} \Vert T\Vert T(x^2)\geqslant T(x)^2. \end{aligned}$$

For arbitrary $x\in \fancyscript{M}$ we obtain, applying the inequality above to the selfadjoint elements $x+x^*$ and $i(x-x^*)$, the following Schwarz inequality

$$\begin{aligned} \Vert T\Vert T(x^*\circ x)\geqslant T(x)^*\circ T(x), \end{aligned}$$

(cf. [11, Lemma 7.3]).

Let $T$ be a normal positive unital map on a $W^*$-algebra $\fancyscript{M}$. By analogy with the case where $T$ is completely positive, we define the multiplicative domain of $T$ as

$$\begin{aligned} \mathcal {A}=\{x\in \fancyscript{M}:T(x^*\circ x)=T(x)^*\circ T(x)\}. \end{aligned}$$

From [10, Theorem 1] it follows that $\mathcal {A}$ is a $JW^*$-subalgebra of $\fancyscript{M}$.

Let $\fancyscript{M}$ be a von Neumann algebra, and consider the tensor product $\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M}$. We have obvious counterparts $\Pi _{1,2}:(\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M})_*\rightarrow \fancyscript{M}_*$ of the partial trace on $(\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M})_*$ defined as

$$\begin{aligned} (\Pi _1\widetilde{\rho })(x)=\widetilde{\rho }(x\otimes {\small 1}\!\!1),\;(\Pi _2\widetilde{\rho })(x)=\widetilde{\rho }({\small 1}\!\!1\otimes x),\;\widetilde{\rho }\in (\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M})_*,\;x\in \fancyscript{M}. \end{aligned}$$

The main objects of our interest are the following two operations of broadcasting and cloning of states.

A linear map $K_*:\fancyscript{M}_*\rightarrow (\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M})_*$ sending states to states, and such that its dual $K:\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M}\rightarrow \fancyscript{M}$ is a unital positive map will be called a channel. (This terminology is almost standard, because by a “channel” is usually meant the map $K$ as above, however, with some additional assumption of complete- or at least two-positivity.) A state $\rho \in \fancyscript{M}_*$ is broadcast by channel $K_*$ if $(\Pi _i K_*)(\rho )=\rho ,\,i=1,2$; in other words, $\rho$ is broadcast by $K_*$ if for each $x\in \fancyscript{M}$

$$\begin{aligned} \rho (K(x\otimes {\small 1}\!\!1))=\rho (K({\small 1}\!\!1\otimes x))=\rho (x). \end{aligned}$$

A family of states is said to be broadcastable if there is a channel $K_*$ that broadcasts each member of this family.

A state $\rho \in \fancyscript{M}_*$ is cloned by channel $K_*$ if $K_*\rho =\rho \otimes \rho$. A family of states is said to be cloneable if there is a channel $K_*$ that clones each member of this family.

3 Broadcasting

The discussion in this section has an auxiliary character and is in main part a repetition for positive maps of the reasoning from [6] performed there for Schwarz maps. Its main purpose is to analyze some properties of broadcasting channels employed in Sect. 4.

Let $\fancyscript{M}$ be a von Neumann algebra, and let $\Gamma \subset \fancyscript{M}_*$ be a broadcastable family of states. Then there is a channel $K_*$ which broadcasts the states in $\Gamma$. Denote by $\mathcal {B}(K_*)$ the set of all normal states broadcastable by $K_*$. We have $\Gamma \subset \mathcal {B}(K_*)$, thus our main object of interest will be the set $\mathcal {B}(K_*)$. In the rest of this section we assume that we are given a fixed channel $K_*$. Define maps $L,R:\fancyscript{M}\rightarrow \fancyscript{M}$ as

$$\begin{aligned} L(x)=K(x\otimes {\small 1}\!\!1),\qquad R(x)=K({\small 1}\!\!1\otimes x),\qquad x\in \fancyscript{M}. \end{aligned}$$

Then $L$ and $R$ are unital normal positive maps on $\fancyscript{M}$. Observe that for a state $\rho$ broadcast by $K_*$ we have, for each $x\in \fancyscript{M}$,

$$\begin{aligned} (\rho \circ L)(x)=\rho (K(x\otimes {\small 1}\!\!1))=\rho (x),\qquad (\rho \circ R)(x)=\rho (K({\small 1}\!\!1\otimes x))=\rho (x), \end{aligned}$$

i.e. $\rho \circ L=\rho \circ R=\rho$. Consequently,

$$\begin{aligned} \mathcal {B}(K_*)=\{\rho \text { --- normal state}: \rho \circ L=\rho \circ R=\rho \}. \end{aligned}$$

Set

$$\begin{aligned} p=\bigvee _{\rho \in \mathcal {B}(K_*)}\mathrm{s }(\rho ). \end{aligned}$$

(1)

Then $\mathcal {B}(K_*)$ is a faithful family of states on the algebra $p\fancyscript{M}p$.

Define maps $K^{(p)}:\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M}\rightarrow p\fancyscript{M}p$ and $L^{(p)},R^{(p)}:\fancyscript{M}\rightarrow p\fancyscript{M}p$ by

$$\begin{aligned}&K^{(p)}(\widetilde{x})=pK(\widetilde{x})p,\qquad \widetilde{x}\in \fancyscript{M}\,\overline{\otimes }\,\fancyscript{M},\\&L^{(p)}(x)=pL(x)p,\qquad R^{(p)}(x)=pR(x)p,\qquad x\in \fancyscript{M}. \end{aligned}$$

Then $K^{(p)},\,L^{(p)},\,R^{(p)}$ are normal positive maps of norm one such that for each $x\in \fancyscript{M}$

$$\begin{aligned} K^{(p)}(x\otimes {\small 1}\!\!1)=L^{(p)}(x),\qquad K^{(p)}({\small 1}\!\!1\otimes x)=R^{(p)}(x), \end{aligned}$$

and

$$\begin{aligned} K^{(p)}({\small 1}\!\!1\otimes {\small 1}\!\!1)=L^{(p)}({\small 1}\!\!1)=R^{(p)}({\small 1}\!\!1)=p. \end{aligned}$$

Moreover, we have

$$\begin{aligned} p-pL(p)p=p({\small 1}\!\!1-L(p))p\geqslant 0, \end{aligned}$$

and for each $\rho \in \mathcal {B}(K_*)$ the $L$-invariance of $\rho$ yields

$$\begin{aligned} \rho (p-pL(p)p)=0, \end{aligned}$$

which means that

$$\begin{aligned} pL(p)p=p, \end{aligned}$$

(2)

since $\mathcal {B}(K_*)$ is faithful on $p\fancyscript{M}p$. The same relation holds also for $R$, thus

$$\begin{aligned} L^{(p)}(p)=R^{(p)}(p)=p. \end{aligned}$$

Consequently,

$$\begin{aligned} K^{(p)}({\small 1}\!\!1\otimes {\small 1}\!\!1)=L^{(p)}({\small 1}\!\!1)=L^{(p)}(p)=R^{(p)}({\small 1}\!\!1)=R^{(p)}(p)=p. \end{aligned}$$

Another description of $\mathcal {B}(K_*)$ is given by the following lemma.

Lemma 1

The following formula holds

$$\begin{aligned} \mathcal {B}(K_*)=\{\rho \text { --- normal state}: \rho \circ L^{(p)}=\rho \circ R^{(p)}=\rho \}. \end{aligned}$$

Proof

Assume that $\rho \in \mathcal {B}(K_*)$. Then, since $\mathrm{s }(\rho )\leqslant p$, we have for each $x\in \fancyscript{M}$

$$\begin{aligned} \rho (x)=\rho (L(x))=\rho (pL(x)p)=\rho (L^{(p)}(x)), \end{aligned}$$

and the same holds for $R^{(p)}$.

Conversely, if $\rho \circ L^{(p)}=\rho \circ R^{(p)}=\rho$, then

$$\begin{aligned} \rho ({\small 1}\!\!1)=\rho (L^{(p)}({\small 1}\!\!1))=\rho (p), \end{aligned}$$

showing that $\mathrm{s }(\rho )\leqslant p$, so for each $x\in \fancyscript{M}$

$$\begin{aligned} \rho (L(x))=\rho (pL(x)p)=\rho (L^{(p)}(x))=\rho (x), \end{aligned}$$

and by the same token $\rho \circ R=\rho$, which means that $\rho \in \mathcal {B}(K_*)$. $\square$

For a map $T$ on $\fancyscript{M}$ denote by $\mathcal {F}(T)$ its fixed-point space, i.e.

$$\begin{aligned} \mathcal {F}(T)=\{x\in \fancyscript{M}:T(x)=x\}. \end{aligned}$$

Let $\mathcal {A}$ be the multiplicative domain of $K^{(p)}$.

Lemma 2

The following relations hold

• (i) For each $x\in \mathcal {F}(L^{(p)})$ we have $x\otimes {\small 1}\!\!1\in \mathcal {A}$,

• (ii) For each $x\in \mathcal {F}(R^{(p)})$ we have ${\small 1}\!\!1\otimes x\in \mathcal {A}$.

Proof

It is enough to prove (i) since a proof of (ii) is analogous. Let $x\in \mathcal {F}(L^{(p)})$. Then $L^{(p)}(x)=x$ and $x=px=xp$. We have $\Vert L^{(p)}\Vert =1$, and the Schwarz inequality for the map $L^{(p)}$ yields

$$\begin{aligned} x^*\circ x=L^{(p)}(x)^*\circ L^{(p)}(x)\leqslant L^{(p)}(x^*\circ x), \end{aligned}$$

hence

$$\begin{aligned} p(x^*\circ x)p=x^*\circ x\leqslant L^{(p)}(x^*\circ x)=pL^{(p)} (x^*\circ x)p, \end{aligned}$$

or in other words

$$\begin{aligned} p(L^{(p)}(x^*\circ x)-x^*\circ x)p\geqslant 0. \end{aligned}$$

For an arbitrary $\rho \in \mathcal {B}(K_*)$ we have on account of the $L^{(p)}$-invariance of $\rho$

$$\begin{aligned} \rho (p(L^{(p)}(x^*\circ x)-x^*x)p)&=\rho (L^{(p)}(x^*\circ x)-x^*\circ x)\\&=\rho (L^{(p)}(x^*\circ x))-\rho (x^*\circ x)=0, \end{aligned}$$

and since the family $\mathcal {B}(K_*)$ is faithful on the algebra $p\fancyscript{M}p$ we obtain

$$\begin{aligned} p(L^{(p)}(x^*\circ x)-x^*\circ x)p= 0, \end{aligned}$$

which amounts to the equality

$$\begin{aligned} L^{(p)}(x^*\circ x)=x^*\circ x. \end{aligned}$$

Taking into account the definition of $K^{(p)}$ we get

$$\begin{aligned}&K^{(p)}(x^*\otimes {\small 1}\!\!1)\circ K^{(p)}(x\otimes {\small 1}\!\!1)=L^{(p)}(x^*)\circ L^{(p)}(x) =x^*\circ x=L^{(p)}(x^*\circ x)\\&\quad =pK(x^*\circ x\otimes {\small 1}\!\!1)p=K^{(p)}(x^*\circ x\otimes {\small 1}\!\!1)=K^{(p)}((x^*\otimes {\small 1}\!\!1)\circ (x\otimes {\small 1}\!\!1)), \end{aligned}$$

showing that $x\otimes {\small 1}\!\!1$ belongs to $\mathcal {A}$. $\square$

To simplify the notation let us agree on the following convention. For a positive map $T:\fancyscript{M}\rightarrow p\fancyscript{M}p$ such that

$$\begin{aligned} T({\small 1}\!\!1)=T(p)=p \end{aligned}$$

denote by $T_p$ the restriction $T|p\fancyscript{M}p$, so that $T_p:p\fancyscript{M}p\rightarrow p\fancyscript{M}p$. Now the positive unital maps from $p\fancyscript{M}p$ to $p\fancyscript{M}p$ will also be denoted with the use of index $p$, thus $T_p$ will stand for a positive map on the algebra $p\fancyscript{M}p$ such that $T_p(p)=p$. To justify this abuse of notation let us define for such a map $T_p$ the map $T$ as follows

$$\begin{aligned} T(x)=T_p(pxp),\qquad x\in \fancyscript{M}. \end{aligned}$$

(3)

It is clear that we have $T|p\fancyscript{M}p=T_p$, so for the consistency of our notation we only need the relation

$$\begin{aligned} T(x)=T(pxp),\qquad x\in \fancyscript{M}, \end{aligned}$$

which is a consequence of the following well-known fact whose proof can be found e.g. in [9, Lemma 2]).

Lemma 3

Let $T:\fancyscript{M}\rightarrow \fancyscript{M}$ be positive, and let $e$ be a projection in $\fancyscript{M}$ such that

$$\begin{aligned} T({\small 1}\!\!1)=T(e)=e. \end{aligned}$$

Then for each $x\in \fancyscript{M}$

$$\begin{aligned} T(x)=T(ex)=T(xe)=eT(x)=T(x)e. \end{aligned}$$

In the sequel while dealing with maps denoted by the same capital letter with or without index $p$ we shall always assume that they are connected by relation (3). If $T$, $T_p$, $V$ and $V_p$ are maps as above then it is easily seen that

$$\begin{aligned} (TV)_p=T_pV_p, \end{aligned}$$

in particular, for each positive integer $m$ we have

$$\begin{aligned} (T^m)_p=(T_p)^m. \end{aligned}$$

The same convention will be adopted also for states with supports contained in $p$, i.e. if $\varphi$ is a state on $\fancyscript{M}$ such that $\mathrm{s }(\varphi )\leqslant p$, then $\varphi _p$ will denote its restriction to $p\fancyscript{M}p$, and for an arbitrary state $\varphi _p$ on $p\fancyscript{M}p$ the state $\varphi$ will be defined as

$$\begin{aligned} \varphi (x)=\varphi _p(pxp),\qquad x\in \fancyscript{M}. \end{aligned}$$

Now we fix attention on the algebra $p\fancyscript{M}p$. In accordance with our convention, we define maps $L_p^{(p)},R_p^{(p)}:p\fancyscript{M}p\rightarrow p\fancyscript{M}p$ as

$$\begin{aligned} L_p^{(p)}=L^{(p)}|p\fancyscript{M}p,\qquad R_p^{(p)}=R^{(p)}|p\fancyscript{M}p. \end{aligned}$$

Clearly, $L_p^{(p)}$ and $R_p^{(p)}$ are normal positive unital maps such that for $\rho \in \mathcal {B}(K_*)$ the $\rho _p$ are $L_p^{(p)}$- and $R_p^{(p)}$-invariant. It is obvious that $\mathcal {F}(L_p^{(p)})=\mathcal {F}(L^{(p)})$ and $\mathcal {F}(R_p^{(p)})=\mathcal {F}(R^{(p)})$. Let $\mathfrak {S}_p$ be the semigroup of normal positive maps on $p\fancyscript{M}p$ generated by $L_p^{(p)}$ and $R_p^{(p)}$. Then $\mathcal {B}(K_*)_p$ defined as $\mathcal {B}(K_*)_p=\{\rho _p:\rho \in \mathcal {B}(K_*)\}$ is a faithful family of $\mathfrak {S}_p$-invariant normal states on $p\fancyscript{M}p$. Denote by $\mathcal {F}(\mathfrak {S}_p)$ the fixed-point space of $\mathfrak {S}_p$, i.e.

$$\begin{aligned} \mathcal {F}(\mathfrak {S}_p)=\{x\in p\fancyscript{M}p: S_p(x)=x\text { for each } S_p\in \mathfrak {S}_p\}. \end{aligned}$$

From the ergodic theorem for $W^*$-algebras (cf. [13]) we infer that $\mathcal {F}(\mathfrak {S}_p)$ is a $JW^*$-algebra, and there exists a normal faithful projection $\mathbb {E}_p$ from $p\fancyscript{M}p$ onto $\mathcal {F}(\mathfrak {S}_p)$ such that

$$\begin{aligned} \mathbb {E}_pS_p=S_p\mathbb {E}_p=\mathbb {E}_p,\qquad \text {for each}\quad S_p\in \mathfrak {S}_p, \end{aligned}$$

(4)

and

$$\begin{aligned} \rho _p\circ \mathbb {E}_p=\rho _p,\qquad \text {for each}\quad \rho \in \mathcal {B}(K_*). \end{aligned}$$

Furthermore, $\mathbb {E}_p$ is positive and has the following property reminiscent of an analogous property of conditional expectation: for any $x\in p\fancyscript{M}p$, $y\in \mathcal {F}(\mathfrak {S}_p)$ we have

$$\begin{aligned} \mathbb {E}_p(x\circ y)=(\mathbb {E}_px)\circ y. \end{aligned}$$

Moreover, if $\varphi _p$ is an arbitrary $\mathbb {E}_p$-invariant normal state on $p\fancyscript{M}p$ then from relation (4) we see that $\varphi _p$ is $\mathfrak {S}_p$-invariant. Conversely, if $\varphi _p$ is an arbitrary $\mathfrak {S}_p$-invariant normal state on $p\fancyscript{M}p$ then another consequence of the ergodic theorem is that $\varphi _p$ is also $\mathbb {E}_p$-invariant (this follows from the fact that for each $x\in p\fancyscript{M}p$, $\mathbb {E}_px$ lies in the $\sigma$-weak closure of the convex hull of $\{S_px:S_p\in \mathfrak {S}_p\}$). Consequently, we have the following equivalence for a normal state $\varphi _p$ on $p\fancyscript{M}p$: $\varphi _p$ is $\mathfrak {S}_p$-invariant if and only if it is $\mathbb {E}_p$-invariant.

Now we want to transfer these results from the algebra $p\fancyscript{M}p$ to the algebra $\fancyscript{M}$. Each element $S_p$ of $\mathfrak {S}_p$ has the form

$$\begin{aligned} S_p=(L_p^{(p)})^{r_1}(R_p^{(p)})^{r_2}\dots (L_p^{(p)})^{r_{m-1}}(R_p^{(p)})^{r_m}, \end{aligned}$$

where the integers $r_1,\dots ,r_m$ satisfy $r_1,r_m\geqslant 0$ and $r_2,\dots r_{m-1}>0$, $m=1,2,\dots$. Consequently,

$$\begin{aligned} S_p&=((L^{(p)})^{r_1})_p((R^{(p)})^{r_2})_p\dots ((L^{(p)})^{r_{m-1}})_p ((R^{(p)})^{r_m})_p\\&=((L^{(p)})^{r_1}(R^{(p)})^{r_2}\dots (L^{(p)})^{r_{m-1}} (R^{(p)})^{r_m})_p, \end{aligned}$$

showing that $S$ defined in accordance with our convention as

$$\begin{aligned} S(x)=S_p(pxp),\qquad x\in \fancyscript{M}, \end{aligned}$$

is an element of the semigroup $\mathfrak {S}$ generated by the maps $L^{(p)}$ and $R^{(p)}$. Thus we have

$$\begin{aligned} \mathfrak {S}_p=\{S_p:S\in \mathfrak {S}\}. \end{aligned}$$

It is easily seen that $\mathcal {F}(\mathfrak {S})=\mathcal {F}(\mathfrak {S}_p)$, where $\mathcal {F}(\mathfrak {S})$ denotes the fixed-points of $\mathfrak {S}$. Again in accordance with our convention, we define a map $\mathbb {E}:\fancyscript{M}\rightarrow \mathcal {F}(\mathfrak {S})$ by the formula

$$\begin{aligned} \mathbb {E}x=\mathbb {E}_p(pxp), \qquad x\in \fancyscript{M}. \end{aligned}$$

(5)

Then $\mathbb {E}$ is a normal positive projection onto the $JW^*$-algebra $\mathcal {F}(\mathfrak {S})$ such that $\mathbb {E}({\small 1}\!\!1)=p$. In the following proposition we obtain a characterization of $\mathcal {B}(K_*)$ in terms of the projection $\mathbb {E}$.

Proposition 4

Let $\varphi$ be a normal state on $\fancyscript{M}$. The following conditions are equivalent

• (i) $\varphi$ belongs to $\mathcal {B}(K_*)$,

• (ii) $\varphi$ is $\mathfrak {S}$-invariant,

• (iii) $\varphi =\varphi \circ \mathbb {E}$.

Proof

(i)$\Longrightarrow$(ii). It follows from Lemma 1.

(ii)$\Longrightarrow$(iii). We have

$$\begin{aligned} \varphi ({\small 1}\!\!1)=\varphi (L^{(p)}({\small 1}\!\!1))=\varphi (p), \end{aligned}$$

which means that $\mathrm{s }(\varphi )\leqslant p$. Consider the state $\varphi _p$. We have for each $x\in \fancyscript{M}$

$$\begin{aligned} \varphi _p(L_p^{(p)}(pxp))=\varphi (L^{(p)}(pxp))=\varphi (pxp)=\varphi _p(pxp), \end{aligned}$$

showing that $\varphi _p$ is $L_p^{(p)}$-invariant. In the same way it is shown that $\varphi _p$ is $R_p^{(p)}$-invariant, thus $\varphi _p$ is $\mathfrak {S}_p$-invariant. Since the $\mathfrak {S}_p$-invariance of $\varphi _p$ is equivalent to its $\mathbb {E}_p$-invariance, we have for each $x\in \fancyscript{M}$

$$\begin{aligned} \varphi (x)=\varphi (pxp)=\varphi _p(pxp)=\varphi _p(\mathbb {E}_p(pxp))=\varphi (\mathbb {E}_p(pxp))=\varphi (\mathbb {E}x). \end{aligned}$$

(iii)$\Longrightarrow$(i). Observe first that we have

$$\begin{aligned} \mathbb {E}L^{(p)}=\mathbb {E}=\mathbb {E}R^{(p)}. \end{aligned}$$

Indeed, taking into account (4), the fact that $L^{(p)}$ has its range contained in $p\fancyscript{M}p$, and Lemma 3 we obtain for each $x\in \fancyscript{M}$

$$\begin{aligned} \mathbb {E}(L^{(p)}(x))&=\mathbb {E}_p(p(L^{(p)}(x))p)=\mathbb {E}_p((L^{(p)}(x))\\&=\mathbb {E}_p(L_p^{(p)}(pxp))=\mathbb {E}_p(pxp)=\mathbb {E}x, \end{aligned}$$

and similarly for the second equality. Now we have

$$\begin{aligned} \varphi (L^{(p)}(x))=\varphi (\mathbb {E}(L^{(p)}(x)))=\varphi (\mathbb {E}x)=\varphi (x), \end{aligned}$$

and the same for $R^{(p)}$, which shows that $\varphi \in \mathcal {B}(K_*)$. $\square$

It turns out that the map $K^{(p)}$ has a special form on the tensor product algebra $\mathcal {F}(\mathfrak {S})\overline{\otimes }\mathcal {F}(\mathfrak {S})$ (this is the weak closure of the algebra of operators $\big \{\sum _{i=1}^mx_i\otimes y_i:x_i,y_i\in \mathcal {F}(\mathfrak {S})\big \}$ acting on $\mathcal {H}\otimes \mathcal {H}$).

Proposition 5

For each $x,y\in \mathcal {F}(\mathfrak {S})$ we have

$$\begin{aligned} K^{(p)}(x\otimes y)=x\circ y. \end{aligned}$$

(6)

Proof

Considering the semigroups $\big \{\big (R_p^{(p)}\big )^n:n=0,1,\dots \big \}$ and $\big \{\big (T_p^{(p)}\big )^n:n=0,1,\dots \big \}$ generated by $R_p^{(p)}$ and $T_p^{(p)}$ we immediately notice that the fixed-point spaces of these semigroups are equal to $\mathcal {F}(R_p^{(p)})$ and $\mathcal {F}(T_p^{(p)})$, respectively, and the above-mentioned ergodic theorem shows that $\mathcal {F}(R_p^{(p)})$ and $\mathcal {F}(T_p^{(p)})$ are $JW^*$-algebras. Moreover,

$$\begin{aligned} \mathcal {F}(\mathfrak {S})=\mathcal {F}(\mathfrak {S}_p) =\mathcal {F}(R_p^{(p)})\cap \mathcal {F}(T_p^{(p)}). \end{aligned}$$

Let $x,y\in \mathcal {F}(\mathfrak {S})$. Then by virtue of Lemma 2 we have $x\otimes {\small 1}\!\!1, {\small 1}\!\!1\otimes y\in \mathcal {A}$, and thus

$$\begin{aligned} K^{(p)}(x\otimes y)&=K^{(p)}((x\otimes {\small 1}\!\!1)\circ ({\small 1}\!\!1\otimes y))=K^{(p)}(x\otimes {\small 1}\!\!1)\circ K^{(p)}({\small 1}\!\!1\otimes y)\\&=R^{(p)}(x)\circ T^{(p)}(y)=x\circ y, \end{aligned}$$

showing the claim. $\square$

The next result is well-known in the case $p={\small 1}\!\!1$ (cf. [13, Lemma1]). Its proof for arbitrary $p$ is similar, so we omit it.

Lemma 6

For each $\rho \in \mathcal {B}(K_*)$ we have $s(\rho )\in \mathcal {F}(\mathfrak {S})$.

4 Cloning

Let $\fancyscript{M}$ be a von Neumann algebra, and let $K_*:\fancyscript{M}_*\rightarrow (\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M})_*$ be a channel. Denote by $\mathcal {C}(K_*)$ the set of all states cloneable by $K_*$, and put

$$\begin{aligned} e=\bigvee _{\rho \in \mathcal {C}(K_*)}\mathrm{s }(\rho ). \end{aligned}$$

The cloneable states and an associated algebra are described by

Theorem 7

The following conditions hold true:

• 1. The states in $\mathcal {C}(K_*)$ have mutually orthogonal supports, and are extreme points of $\mathcal {B}(K_*)$.

• 2. The algebra $e\mathcal {F}(\mathfrak {S})e$ is an atomic abelian $W^*$-subalgebra of $\mathcal {F}(\mathfrak {S})$, generated by $\{\mathrm{s }(\rho ):\rho \in \mathcal {C}(K_*)\}$, and such that $e\mathcal {F}(\mathfrak {S})e\subset \mathcal {F}(\mathfrak {S})'$ — the commutant of $\mathcal {F}(\mathfrak {S})$.

Proof

• 1. Since $\mathcal {C}(K_*)\subset \mathcal {B}(K_*)$ we may use the analysis of the preceding section. In particular, we adopt the setup and notation introduced there. For each $\rho \in \mathcal {C}(K_*)$ we have $K_*\rho =\rho \otimes \rho$, so taking into account Proposition 5 we obtain the equality

  $$\begin{aligned} \rho (x)\rho (y)&=\rho \otimes \rho (x\otimes y)=(K_*\rho )(x\otimes y)=\rho (K(x\otimes y))\nonumber \\&=\rho (pK(x\otimes y)p)=\rho (K^{(p)}(x\otimes y))=\rho (x\circ y) \end{aligned}$$

  (7)

  for all $x,y\in \mathcal {F}(\mathfrak {S})$. The equality above yields that for each $z\in \mathcal {F}(\mathfrak {S})$ and any $\rho \in \mathcal {C}(K_*)$ we have

  $$\begin{aligned} \rho (z^2)=\rho (z)^2. \end{aligned}$$

  (8)

  Let $x$ be an arbitrary selfadjoint element of $\mathcal {F}(\mathfrak {S})$. For each $\rho \in \mathcal {C}(K_*)$ we have by (8)

  $$\begin{aligned} \rho \big ((x-\rho (x){\small 1}\!\!1)^2\big )=\rho \big (x^2-2\rho (x)x-\rho (x)^2{\small 1}\!\!1\big ) =\rho (x^2)-\rho (x)^2=0, \end{aligned}$$

  which yields the equality

  $$\begin{aligned} \mathrm{s }(\rho )\big (x-\rho (x){\small 1}\!\!1\big )^2\mathrm{s }(\rho )=0, \end{aligned}$$

  i.e.

  $$\begin{aligned} \mathrm{s }(\rho )x=\rho (x)\mathrm{s }(\rho ). \end{aligned}$$

  (9)

  Since for an element $x$ of a $JW^*$-algebra $x+x^*$ and $x-x^*$ are also elements of this algebra the equality above holds for all $x\in \mathcal {F}(\mathfrak {S})$ as well. Let $\rho$ and $\varphi$ be two distinct states from $\mathcal {C}(K_*)$. Then by (9)

  $$\begin{aligned} \mathrm{s }(\rho )\mathrm{s }(\varphi )=\rho (\mathrm{s }(\varphi ))\mathrm{s }(\rho )\qquad \text {and}\qquad \mathrm{s }(\varphi )\mathrm{s }(\rho )=\varphi (\mathrm{s }(\rho ))\mathrm{s }(\varphi ), \end{aligned}$$

  which after taking adjoints yields the equality

  $$\begin{aligned} \mathrm{s }(\rho )\mathrm{s }(\varphi )=\mathrm{s }(\varphi )\mathrm{s }(\rho ). \end{aligned}$$

  Consequently,

  $$\begin{aligned} \rho (\mathrm{s }(\varphi ))\mathrm{s }(\rho )=\varphi (\mathrm{s }(\rho ))\mathrm{s }(\varphi )=\mathrm{s }(\varphi )\mathrm{s }(\rho ), \end{aligned}$$

  showing that either $\mathrm{s }(\rho )=\mathrm{s }(\varphi )$ or $\mathrm{s }(\rho )$ and $\mathrm{s }(\varphi )$ are orthogonal. If $\mathrm{s }(\rho )=\mathrm{s }(\varphi )$ then on account of (9) we would have

  $$\begin{aligned} \rho (x)\mathrm{s }(\rho )=\mathrm{s }(\rho )x=\mathrm{s }(\varphi )x=\varphi (x)\mathrm{s }(\varphi )=\varphi (x)\mathrm{s }(\rho ) \end{aligned}$$

  for each $x\in \mathcal {F}(\mathfrak {S})$, i.e.

  $$\begin{aligned} \rho |\mathcal {F}(\mathfrak {S})=\varphi |\mathcal {F}(\mathfrak {S}). \end{aligned}$$

  (10)

  Let $\mathbb {E}$ be the projection onto $\mathcal {F}(\mathfrak {S})$ defined by formula (5). We have $\rho =\rho \circ \mathbb {E}$ and $\varphi =\varphi \circ \mathbb {E}$, thus equality (10) yields

  $$\begin{aligned} \rho =\rho \circ \mathbb {E}=\varphi \circ \mathbb {E}=\varphi \end{aligned}$$

  contrary to the assumption that $\rho$ and $\varphi$ are distinct. Consequently, $\rho$ and $\varphi$ have orthogonal supports. Now take an arbitrary $\rho \in \mathcal {C}(K_*)$, and assume that

  $$\begin{aligned} \rho =\lambda \varphi _1+(1-\lambda )\varphi _2, \end{aligned}$$

  for some $0<\lambda <1$ and $\varphi _1,\varphi _2\in \mathcal {B}(K_*)$. Then

  $$\begin{aligned} 1=\rho (\mathrm{s }(\rho ))=\lambda \varphi _1(\mathrm{s }(\rho ))+(1-\lambda )\varphi _2(\mathrm{s }(\rho )), \end{aligned}$$

  showing that $\varphi _1(\mathrm{s }(\rho ))=\varphi _2(\mathrm{s }(\rho ))=1$, which means that $\mathrm{s }(\varphi _1),\mathrm{s }(\varphi _2)\leqslant \mathrm{s }(\rho )$. From equality (9) we obtain for $x\in \mathcal {F}(\mathfrak {S})$

  $$\begin{aligned} \varphi _{1,2}(x)=\varphi _{1,2}(\mathrm{s }(\rho )x)=\rho (x)\varphi _{1,2}(\mathrm{s }(\rho ))=\rho (x), \end{aligned}$$

  giving the relation

  $$\begin{aligned} \rho |\mathcal {F}(\mathfrak {S})=\varphi _{1,2}|\mathcal {F}(\mathfrak {S}). \end{aligned}$$

  We have $\rho =\rho \circ \mathbb {E}$ and $\varphi _{1,2}=\varphi _{1,2}\circ \mathbb {E}$, and thus

  $$\begin{aligned} \rho =\rho \circ \mathbb {E}=\varphi _{1,2}\circ \mathbb {E}=\varphi _{1,2}, \end{aligned}$$

  showing that $\rho$ is an extreme point of $\mathcal {B}(K_*)$.

• 2. From equality (9) we obtain that $\mathrm{s }(\rho )\in \mathcal {F}(\mathfrak {S})'$ for all $\rho \in \mathcal {C}(K_*)$, and that

  $$\begin{aligned} ex=\sum _{\rho \in \mathcal {C}(K_*)}\rho (x)\mathrm{s }(\rho ), \end{aligned}$$

  for all $x\in \mathcal {F}(\mathfrak {S})$, which means that $e\mathcal {F}(\mathfrak {S})=e\mathcal {F}(\mathfrak {S})e$ is a $W^*$-algebra generated by $\{\mathrm{s }(\rho ):\rho \in \mathcal {C}(K_*)\}$. By virtue of Lemma 6 we have $s(\rho )\in \mathcal {F}(\mathfrak {S})$ for each $\rho \in \mathcal {C}(K_*)$ thus $e\mathcal {F}(\mathfrak {S})e$ is a subalgebra of $\mathcal {F}(\mathfrak {S})$, and $e\mathcal {F}(\mathfrak {S})e\subset \mathcal {F}(\mathfrak {S})'$. Finally, $\mathrm{s }(\rho )$ is a minimal projection in $e\mathcal {F}(\mathfrak {S})e$. Indeed, for any projection $f\in \mathcal {F}(\mathfrak {S})$ equality (8) yields

  $$\begin{aligned} \rho (f)=0\text { or }1. \end{aligned}$$

  Now if $q$ is a projection in $e\mathcal {F}(\mathfrak {S})e$ such that $q\leqslant \mathrm{s }(\rho )$ and $q\ne \mathrm{s }(\rho )$ we cannot have $\rho (q)=1$, thus $\rho (q)=0$, and the faithfulness of $\rho$ on the algebra $\mathrm{s }(\rho )\fancyscript{M}\mathrm{s }(\rho )$ yields $q=0$. Consequently, algebra $e\mathcal {F}(\mathfrak {S})e$ being generated by minimal projections is atomic. $\square$

Theorem 8

Let $\Gamma$ be an arbitrary subset of normal states on a von Neumann algebra $\fancyscript{M}$. The following conditions are equivalent

• (i) $\Gamma$ is cloneable,

• (ii) The states in $\Gamma$ have mutually orthogonal supports.

Proof

The implication (i)$\Longrightarrow$(ii) follows from Theorem 7.

To prove (ii)$\Longrightarrow$(i) assume that the states from $\Gamma$ have orthogonal supports $\{e_i\}$, that is $\Gamma =\{\rho _i\}$ and $\mathrm{s }(\rho _i)=e_i$. Define a channel $K_*:\fancyscript{M}_*\rightarrow (\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M})_*$ as follows

$$\begin{aligned} K_*\varphi =\sum _i\varphi (e_i)\,\rho _i\otimes \rho _i+\varphi (e^{\bot })\,\widetilde{\omega },\qquad \varphi \in \fancyscript{M}_*, \end{aligned}$$

(11)

where

$$\begin{aligned} e=\sum _ie_i, \end{aligned}$$

and $\widetilde{\omega }$ is a fixed normal state on $\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M}$. Since $\rho _i(e^{\bot })=0$, we have

$$\begin{aligned} K_*\rho _i=\sum _j\rho _i(e_j)\,\rho _j\otimes \rho _j+ \rho _i(e^{\bot })\,\widetilde{\omega }=\rho _i\otimes \rho _i, \end{aligned}$$

showing that $K_*$ clones the $\rho _i$. $\square$

The above theorem yields an interesting corollary. Namely, a usual assumption about a channel is its complete (or at least two-) positivity, the assumption which we have tried to avoid in this work. It turns out that a stronger form of positivity gives the same cloneable states.

Corollary

Let $\Gamma$ be an arbitrary subset of normal states on a von Neumann algebra $\fancyscript{M}$. The following conditions are equivalent

• (i) $\Gamma$ is cloneable by a completely positive channel,

• (ii) $\Gamma$ is cloneable by a positive channel.

Proof

We need only to show the implication (ii)$\Longrightarrow$(i). From Theorem 8 (and with its notation) we know that the supports $e_i$ of states $\rho _i$ in $\Gamma$ are mutually orthogonal. Define a channel $K_*$ by formula (11). Then $K_*$ clones the $\rho _i$. Its dual map $K:\fancyscript{M}\,\overline{\otimes }\,\fancyscript{M}\,\rightarrow \,\fancyscript{M}$ has the form

$$\begin{aligned} K(\widetilde{x})=\sum _i\rho _i\otimes \rho _i(\widetilde{x})\,e_i +\widetilde{\omega } (\widetilde{x})\,e^{\bot },\qquad \widetilde{x}\in \fancyscript{M}\,\overline{\otimes }\,\fancyscript{M}, \end{aligned}$$

thus its range lies in the abelian von Neumann algebra generated by the projections $e_i$ and $e^{\bot }$. Consequently, on account of [12, Corollary IV.3.5] $K$ is completely positive. $\square$

Finally, let us say a few words about the uniqueness of the cloning operation.

Proposition 9

Let $\Gamma =\{\rho _i\}$ be an arbitrary subset of normal states on a von Neumann algebra $\fancyscript{M}$ such that the states in $\Gamma$ have mutually orthogonal supports. Put $e_i=\mathrm{s }(\rho _i)$, and assume that

$$\begin{aligned} \sum _ie_i={\small 1}\!\!1. \end{aligned}$$

Let the cloning channel $K_*'$ be defined as in Theorem 8, i.e.

$$\begin{aligned} K_*'\varphi =\sum _i\varphi (e_i)\,\rho _i\otimes \rho _i,\qquad \varphi \in \fancyscript{M}_*. \end{aligned}$$

Then for each channel $K_*$ that clones $\Gamma$ we have

$$\begin{aligned} K_*'=(\mathbb {E}\otimes \mathbb {E})_*K_*, \end{aligned}$$

where $\mathbb {E}$ is the projection from $\fancyscript{M}$ onto $\mathcal {F}(\mathfrak {S})$ defined by means of the dual $K$ of $K_*$ as in Sect. 3 (it turns out that in our case $\mathbb {E}$ is actually a conditional expectation).

Proof

Let $K_*$ be a channel cloning $\Gamma$. Since $\Gamma \subset \mathcal {C}(K_*)$ we get

$$\begin{aligned} e=\bigvee _{\omega \in \mathcal {C}(K_*)}\mathrm{s }(\omega )={\small 1}\!\!1. \end{aligned}$$

Theorem 7 asserts that the $\mathrm{s }(\omega )$ for $\omega \in \mathcal {C}(K_*)$ are minimal projections in $e\mathcal {F}(\mathfrak {S})e=\mathcal {F}(\mathfrak {S})$, thus the equality

$$\begin{aligned} \bigvee _ie_i={\small 1}\!\!1=\bigvee _{\omega \in \mathcal {C}(K_*)}\mathrm{s }(\omega ) \end{aligned}$$

yields $\Gamma =\mathcal {C}(K_*)$.

Denoting by $K'$ the dual of $K_*'$ we have for all $x,y\in \fancyscript{M}$

$$\begin{aligned} K'(x\otimes y)=\sum _i\rho _i(x)\rho _i(y)\,e_i. \end{aligned}$$

(12)

From Theorem 7 it follows that $\mathcal {F}(\mathfrak {S})$ is an abelian atomic $W^*$-algebra generated by $\{e_i\}$, i.e.

$$\begin{aligned} \mathcal {F}(\mathfrak {S})=\Big \{\sum _i\alpha _ie_i: \alpha _i\in \mathbb {C},\,\sup _i|\alpha _i|<\infty \Big \}. \end{aligned}$$

For the projection $\mathbb {E}$ defined as in Sect. 3, and arbitrary $x\in \fancyscript{M}$ we have

$$\begin{aligned} \mathbb {E}x=\sum _i\alpha _ie_i, \end{aligned}$$

for some coefficients $\alpha _i\in \mathbb {C}$ depending on $x$. Consequently,

$$\begin{aligned} \rho _j(x)=\rho _j(\mathbb {E}x)=\sum _i\alpha _i\rho _j(e_i)=\alpha _j, \end{aligned}$$

thus

$$\begin{aligned} \mathbb {E}x=\sum _i\rho _i(x)e_i. \end{aligned}$$

(From the formula above it immediately follows that $\mathbb {E}$ is a conditional expectation.) Consequently, we obtain

$$\begin{aligned} \mathbb {E}x\mathbb {E}y=\sum _i\rho _i(x)\rho _i(y)e_i, \end{aligned}$$

(13)

for all $x,y\in \fancyscript{M}$.

By (6) we get for any $x,y\in \fancyscript{M}$

$$\begin{aligned} K(\mathbb {E}\otimes \mathbb {E}(x\otimes y))=K(\mathbb {E}x\otimes \mathbb {E}y)=\mathbb {E}x\circ \mathbb {E}y=\mathbb {E}x\mathbb {E}y, \end{aligned}$$

(14)

and formulas (12), (13), and (14) yield

$$\begin{aligned} K'(x\otimes y)=K(\mathbb {E}\otimes \mathbb {E}(x\otimes y)), \end{aligned}$$

for any $x,y\in \fancyscript{M}$, which means that

$$\begin{aligned} K_*'=(\mathbb {E}\otimes \mathbb {E})_*K_*, \end{aligned}$$

finishing the proof. $\square$