# Quantum measurements and maps preserving strict convex combinations and pure
states

Lihua Yang, Jinchuan Hou  Department of Mathematics, Taiyuan University of
Technology , Taiyuan 030024, P. R. China<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, a characterization of maps between quantum states that preserve
pure states and strict convex combinations is obtained. Based on this
characterization, a structural theorem for maps between multipartite quantum
states that preserve separable pure states and strict convex combinations is
established. Then these results are applied to characterize injective (local)
quantum measurements and answer some conjectures proposed in
[J.Phys.A:Math.Theor. 45 (2012) 205305].

PACS. 03.65.Aa, 03.65.Ud, 03.65.Db, 03.67.-a

Key words and phrases. quantum measurement, pure states, separable pure
states, strict convex combination preserving maps

This work is partially supported by Natural Science Foundation of China
(11171249, 11271217), and a grant from International Cooperation Program in
Sciences and Technology of Shanxi (2011081039).

##  1\. Introduction

In the theory of quantum mechanics, a state is a positive operator of trace 1
acting on a complex Hilbert space $H$. Denote by ${\mathcal{S}}(H)$ and
$\mathcal{P}ur(H)$ respectively the set of all states and the set of all pure
states ($i.e.$ rank-1 projections) on $H$. In quantum information theory we
deal, in general, with multipartite systems. The underlying space $H$ of a
multipartite composite quantum system is a tensor product of underlying spaces
$H_{i}$ of its subsystems, that is $H=H_{1}\otimes H_{2}\otimes\cdots\otimes
H_{n}$. If $n=2$, the system is called a bipartite system. The definition of
multipartite separability was introduced in [13] as a natural extension of the
notion of separability in bipartite case [14]. Let us denote the set of all
states in an $n$-partite system by ${\mathcal{S}}(H_{1}\otimes\cdots\otimes
H_{n})$. In the case $\dim H<\infty$, a state
$\rho\in{\mathcal{S}}(H_{1}\otimes\cdots\otimes H_{n})$ is said to be (fully)
separable if it admits a representation of the form

$\rho=\sum_{i}p_{i}\rho^{(1)}_{i}\otimes\cdots\otimes\rho^{(n)}_{i},$

where $p_{i}>0$ with $\sum_{i}p_{i}=1$ and
$\rho_{i}^{(k)}\in{\mathcal{S}}(H_{k})$. Otherwise, $\rho$ is said to be
entangled. Denote by respectively ${\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2}\otimes\cdots\otimes H_{n})$ and ${\mathcal{P}ur}_{\rm sep}(H_{1}\otimes
H_{2}\otimes\cdots\otimes H_{n})$ the set of all separable states and the set
of all separable pure states on $H_{1}\otimes H_{2}\otimes\cdots\otimes
H_{n}$. It is obvious that

$\begin{array}[]{rl}&{\mathcal{P}ur}_{\rm sep}(H_{1}\otimes
H_{2}\otimes\cdots\otimes
H_{n})={\mathcal{P}ur}(H_{1})\otimes{\mathcal{P}ur}(H_{2})\otimes\cdots\otimes{\mathcal{P}ur}(H_{n})\\\
=&\\{P_{1}\otimes P_{2}\otimes\cdots\otimes
P_{n}:P_{i}\in{\mathcal{P}ur}(H_{i}),i=1,2,\ldots,n\\}.\end{array}$

The theory of maps on the set of states plays an important role in quantum
computation and quantum information science. It is important to understand,
characterize, and construct different classes of maps on states. For instance,
all quantum channels and quantum operations are completely positive linear
maps; in quantum error correction, one has to construct the recovery map for a
given channel; to study the entanglement of states, one constructs NCP (non
completely positive) positive maps and entanglement witnesses. Many
researchers pay their attention to the problem of characterizing the maps on
the states; ref. [1, 2, 3, 6].

The present paper is motivated by some conjectures proposed in [8] where the
bijective maps $\Phi:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})\rightarrow{\mathcal{S}}_{\rm sep}(H_{1}\otimes H_{2})$ that preserve
strict convex combinations were studied.

Recall that a map $\phi$ between convex sets is said to be (strict) convex
combination preserving if, for any $\rho,\sigma\in{\mathcal{S}}(H)$ and
$t\in[0,1]$ ($t\in(0,1)$), there is some $s$ with $0\leq s\leq 1$ ($0<s<1$)
such that $\phi(t\rho+(1-t)\sigma)=s\phi(\rho)+(1-s)\phi(\sigma)$. It is
obvious that $\phi$ preserves (strict) convex combination if and only if
$\phi([\rho,\sigma])\subseteq[\phi(\rho),\phi(\sigma)]$
($\phi((\rho,\sigma))\subseteq(\phi(\rho),\phi(\sigma))$), where $[A,B]$
stands for the closed (open) line segment joint $A$ and $B$, that is,
$[A,B]=\\{tA+(1-t)B:0\leq t\leq 1\\}$ ($(A,B)=[A,B]\setminus\\{A,B\\}$, here
we define $(\rho,\rho)=\\{\rho\\}$). The maps preserve strict convex
combinations are closely related to quantum measurements. In quantum mechanics
a fine-grained quantum measurement is described by a collection $\\{M_{m}\\}$
of measurement operators acting on the Hilbert space $H$ corresponding to the
system satisfying $\sum_{m}M_{m}^{*}M_{m}=I$; ref. for example, [4]. Let
$M_{j}$ be a measurement operator. If the state of the quantum system is
$\rho\in{\mathcal{S}}(H)$ before the measurement, then the state after the
measurement is $\frac{M_{j}\rho M_{j}^{*}}{{\rm Tr}(M_{j}\rho M_{j}^{*})}$
whenever $M_{j}\rho M_{j}^{*}\not=0$. If we fix an $M_{j}=M$ is fixed, we get
a measurement map $\phi$ defined by $\phi(\rho)=\frac{M\rho M^{*}}{{\rm
Tr}(M\rho M^{*})}$ from the convex subset ${\mathcal{S}}_{M}(H)=\\{\rho:M\rho
M^{*}\not=0\\}$ of the (convex) set ${\mathcal{S}}(H)$ into
${\mathcal{S}}(H)$, which is a map that preserves the strict convex
combinations and sends pure states to pure states. If $M$ is invertible
(injective), then $\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ is
bijective (injective) and will be called an invertible (injective) measurement
map.

The problem of characterizing the strict convex combination maps between
quantum states was firstly attacked by [5]. It is shown in [5] that a
bijective map $\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$, $\dim H\geq
2$, is (strict) convex combination preserving if and only if $\phi$ is an
invertible quantum measurement map or the composition of transpose and an
invertible quantum measurement map. Note that
$\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ is bijective and (strict)
convex combination preserving imply that $\phi$ preserves pure states in both
directions, that is, $\phi({\mathcal{P}ur}(H))={\mathcal{P}ur}(H)$. Let
$\psi:{\mathcal{S}}_{\rm sep}(H_{1}\otimes H_{2})\rightarrow{\mathcal{S}}_{\rm
sep}(H_{1}\otimes H_{2})$ be a bijective map. In [8], based on the work of
[5], it is shown that, if $\Phi$ is (strict) convex combination preserving and
if

$None$ $\begin{array}[]{rl}&\psi(P_{1}\otimes P_{2})\ \mbox{ is a product
state for any}\ P_{i}\in{\mathcal{S}}(H_{i})\ \mbox{\rm with}\\\ &{\rm
rank}P_{i}=1\ \mbox{ and }{\rm rank}P_{j}=2,\ \ (1\leq i\not=j\leq
2),\end{array}$

then $\psi$ is a composition of an invertible local quantum measurement (i.e.,
the map of the form $\rho\mapsto\frac{(S\otimes T)\rho(S\otimes T)^{*}}{{\rm
Tr}((S\otimes T)\rho(S\otimes T)^{*})}$ with $S,T$ invertible) and some of the
following maps: the transpose, the partial transpose and the swap. In the
sequel we say a map is essentially a (local) quantum measurement if it is a
(local) quantum measurement or a composition of a (local) quantum measurement
with any one of the following maps: the transpose, the partial transpose, and
the swap.

It is conjectured in [8] that if a bijective map $\psi:{\mathcal{S}}_{\rm
sep}(H_{1}\otimes H_{2})\rightarrow{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})$ is (strict) convex combination preserving, then it sends in fact
product states to product states, and thus the additional assumption Eq.(1.1)
in the result just mentioned above is superfluous.

The purpose of the present paper is to answer these conjectures affirmatively
for finite dimensional systems. To do this, we need first to characterize
general maps between states that preserve the strict convex combinations and
send pure states to pure states. Let $H$ and $K$ be complex Hilbert spaces of
dimension $\geq 2$ with $\dim H<\infty$. We show in Section 2 that a map
$\psi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(K)$ preserves strict convex
combinations and pure states if and only if it has one of the following three
forms: (1) $\psi$ is contractive to a pure state, i.e, there exists a pure
state $Q\in{\mathcal{P}ur}(K)$ such that $\psi(\rho)=Q$ for all
$\rho\in{\mathcal{S}}(H)$; (2) there exist distinct pure states
$Q_{i}\in{\mathcal{P}ur}(K)$, $i=1,2$ such that
$\psi({\mathcal{P}ur}(H))=\\{Q_{1},Q_{2}\\}$, and
$\psi({\mathcal{S}}(H))\subseteq[Q_{1},Q_{2}]$; (3) $\dim H\leq\dim K$ and
there exists an injective operator $M\in\mathcal{B}(H,K)$ such that
$\psi(\rho)=\frac{M\rho M^{*}}{{\rm Tr}(M\rho M^{*})}$ for all
$\rho\in{\mathcal{S}}(H)$, or $\psi(\rho)=\frac{M\rho^{t}M^{*}}{{\rm
Tr}(M\rho^{t}M^{*})}$ for all $\rho\in{\mathcal{S}}(H)$, where $A^{t}$ is the
transpose of $A$ with respect to an arbitrarily fixed orthonormal basis of
$H$, that is, $\psi$ is essentially an injective quantum measurement (see
Theorem 2.5). Note that, by our result, if $\psi$ is strict convex combination
preserving and pure state preserving and if $\psi$ is not continuous, then
$\psi$ must have the form (2), and an example of such map is given (Remark
2.7). Based on the results obtained in Section 2, we are able to give a
structure theorem of maps $\psi:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})\rightarrow{\mathcal{S}}_{\rm sep}(K_{1}\otimes K_{2})$ that preserve
strict convex combinations and separable pure states in Section 3, where
$2\leq\dim H_{i}<\infty$ and $\dim K_{i}\geq 2$, $i=1,2$. We show that such
maps can have ten possible forms (Theorem 3.2). Consequently, if the range of
$\psi$ is non-collinear or a singleton, then $\psi$ sends product states to
product states (Corollary 3.3); and moreover, if the range of $\psi$ also
contains a state $\sigma$ so that both of its reductions ${\rm
Tr}_{1}(\sigma)$ and ${\rm Tr}_{2}(\sigma)$ have rank $\geq 2$, then $\psi$ is
essentially an injective local quantum measurement (Corollary 3.4). These
results particularly answer the conjectures in [8] mentioned above. Section 4
is a brief discussion of the same topic for multipartite systems. The similar
structure theorem is valid for maps $\psi:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2}\otimes\cdots\otimes H_{n})\rightarrow{\mathcal{S}}_{\rm
sep}(K_{1}\otimes K_{2}\otimes\cdots\otimes K_{n})$ that preserve strict
convex combinations and separable pure states, but with more complicated
expressions. Particularly, if the range of $\psi$ is non-collinear or a
singleton, then $\psi$ sends product states to product states; and moreover,
if the range of $\psi$ contains a state $\sigma$ so that each reduction ${\rm
Tr}^{i}(\sigma)$ has rank $\geq 2$, then $\psi$ is essentially a local
injective quantum measurement (Theorem 4.1, Corollary 4.2). Section 5 is a
short conclusion.

##  2\. Maps preserving pure states and strict convex combinations

In this section we characterize the maps between the convex sets of quantum
states that send pure states to pure states and preserve the strict convex
combinations.

We start by giving a simple lemma which is easily checked.

Lemma 2.1. Let $\\{Q_{1},Q_{2},\cdots,Q_{r}\\}$ be a linearly independent set
of rank one projections acting on a Hilbert space $H$. If
$\sum_{i=1}^{r}t_{i}Q_{i}=P_{0}$ is a projection for some
$t_{i}>0$,$i=1,2,\ldots,r$, then $\\{Q_{i}\\}_{i=1}^{r}$ are orthogonal and
$t_{i}=1$ for all $i$.

Let ${\bf H}_{m}$ be the real linear space of all $m\times m$ Hermitian
matrices and let ${\mathcal{P}}_{m}$ be the set of all rank-1 $m\times m$
projection matrices. The next lemma comes from [3] which can be viewed as a
characterization of linear preservers of pure states. Also, ref. [9] for
infinite dimensional case.

Lemma 2.2. Suppose $\phi:{\bf H}_{m}\rightarrow{\bf H}_{n}$ is a linear map
satisfying $\phi({\mathcal{P}}_{m})\subseteq{\mathcal{P}}_{n}$. Then one of
the following holds:

(i) There is $Q\in{\mathcal{P}}_{n}$ such that $\phi(A)={\rm Tr}(A)Q$ for all
$A\in{\bf H}_{m}$.

(ii) $m\leq n$ and there is a $U\in M_{n\times m}$ with $U^{*}U=I_{m}$ such
that $\phi(A)=UAU^{*}$ for all $A\in{\bf H}_{m}$, or $\phi(A)=UA^{t}U^{*}$ for
all $A\in{\bf H}_{m}$.

The following lemma is the main result in [12], which gives a characterization
of strict convex combination preserving maps in terms of linear ones.

Lemma 2.3. Let $X$ and $Y$ be real linear spaces and $D\subseteq X$ a nonempty
convex subset. Assume that $\phi:D\rightarrow Y$ is a strict convex
combination preserving map such that $\phi(D)$ is non-collinear (i.e.,
$\phi(D)$ contains a nondegenerate triangle). Then, there exist a linear
transformation $A:X\rightarrow Y$, a linear functional
$f:X\rightarrow{\mathbb{R}}$, a vector $y_{0}\in Y$, and a scalar
$b\in{\mathbb{R}}$ such that

$f(x)+b>0\quad\mbox{for all }\quad x\in D$

and

$\phi(x)=\frac{Ax+y_{0}}{f(x)+b}\quad\mbox{for all }\quad x\in D.$

By using of Lemma 2.1 and Lemma 2.3 we can prove the following lemma, which is
also crucial for proving our main result.

Lemma 2.4. Let $H$ be a complex Hilbert space with $2\leq\dim H=r<\infty$ and
$\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ be a map preserving pure
state and strict convex combinations. If $\phi(\frac{1}{r}I)=\frac{1}{r}I$ and
${\rm ran}\phi$ is non-collinear, then $\phi$ is affine.

Proof. As $\phi$ preserves pure states and strict convex combinations, by
Lemma 2.1, $\phi(\frac{1}{r}I)=\frac{1}{r}I$ implies that $\phi$ maps
orthogonal pure states to orthogonal pure states. Also, by Lemma 2.3, $\phi$
is strict convex combination preserving and ${\rm ran}\phi$ is non-collinear
together imply that $\phi$ has the form
$\phi(\rho)=\frac{\Gamma(\rho)+D}{f(\rho)+d}$ for any
$\rho\in{\mathcal{S}}(H)$, where
$\Gamma:\mathcal{B}_{sa}(H)\rightarrow\mathcal{B}_{sa}(H)$ is a linear
transformation, $f:\mathcal{B}_{\rm sa}(H)\rightarrow\mathbb{R}$ is a linear
functional, $D\in\mathcal{B}_{\rm sa}(H)$ and $d\in\mathbb{R}$ with
$f(\rho)+d>0$ for all $\rho\in{\mathcal{S}}(H)$, ${\mathcal{B}}_{\rm sa}(H)$
is the real linear space of all self-adjoint operators in ${\mathcal{B}}(H)$.
Since $\dim H<\infty$, $\Gamma$ and $f$ are continuous. It follows that $\phi$
is continuous. To prove the lemma, we consider two cases of $\dim H>2$ and
$\dim H=2$ respectively.

Case 1. $\dim H>2$.

We will show that $f$ is a constant on ${\mathcal{S}}(H)$, that is, there is a
real number $a$ such that $f(\rho)=a$ for all $\rho\in{\mathcal{S}}(H)$.

For any normalized orthogonal basis $\\{e_{i}\\}_{i=1}^{r}$ of $H$, let
$P_{i}=e_{i}\otimes e_{i}$. We first claim that $f(e_{i}\otimes
e_{i})=f(e_{j}\otimes e_{j})$ for any $i$ and $j$. Since $\phi$ preserves pure
states, there is a pure state $Q_{i}=x_{i}\otimes x_{i}$ such that

$x_{i}\otimes x_{i}=Q_{i}=\phi(P_{i})=\frac{\Gamma(e_{i}\otimes
e_{i})+D}{f(e_{i}\otimes e_{i})+d}.$

So

$\Gamma(e_{i}\otimes e_{i})+D=(f(e_{i}\otimes e_{i})+d)(x_{i}\otimes x_{i}).$

As $\phi(\frac{I}{r})=\frac{I}{r}$ and
$\frac{I}{r}=\frac{1}{r}\sum_{i=1}^{r}e_{i}\otimes e_{i}$, we have

$\frac{I}{r}=\phi(\frac{1}{r}\sum_{i=1}^{r}e_{i}\otimes
e_{i})=\frac{\Gamma(\sum_{i=1}^{r}\frac{1}{r}e_{i}\otimes
e_{i})+D}{f(\sum_{i=1}^{r}\frac{1}{r}e_{i}\otimes
e_{i})+d}=\frac{\sum_{i=1}^{r}\frac{1}{r}\Gamma(e_{i}\otimes
e_{i})+r\frac{1}{r}D}{\sum_{i=1}^{r}\frac{1}{r}f(e_{i}\otimes
e_{i})+r\frac{1}{r}d},$

that is,

$\frac{I}{r}=\frac{\frac{1}{r}(\sum_{i=1}^{r}(\Gamma(e_{i}\otimes
e_{i})+D))}{\frac{1}{r}(\sum_{i=1}^{r}(f(e_{i}\otimes
e_{i})+d))}=\frac{\sum_{i=1}^{r}(\Gamma(e_{i}\otimes
e_{i})+D)}{\sum_{i=1}^{r}(f(e_{i}\otimes e_{i})+d)}.$

Let $A_{i}=\Gamma(e_{i}\otimes e_{i})+D$ and $a_{i}=f(e_{i}\otimes e_{i})+d$.
Then $A_{i}=a_{i}Q_{i}$ and the above equation becomes to

$I=r(\frac{A_{1}+A_{2}+\cdots+A_{r}}{a_{1}+a_{2}+\cdots+a_{r}})=r(\frac{a_{1}Q_{1}+a_{2}Q_{2}+\cdots+a_{r}Q_{r}}{a_{1}+a_{2}+\cdots+a_{r}}).$

Applying Lemma 2.1, we see that $\frac{ra_{i}}{a_{1}+a_{2}+\cdots+a_{r}}=1$
for each $i=1,2,\ldots,r$ and hence

$a_{1}=a_{2}=\cdots=a_{r}=\frac{a_{1}+a_{2}+\cdots+a_{r}}{r}.$

This implies that there is some scalar $a$ such that $f(e_{i}\otimes e_{i})=a$
holds for all $i$. Now for arbitrary unit vectors $x$, $y\in H$, as $\dim
H>2$, there is a unit vector $z\in H$ such that $z\in[x,y]^{\bot}$. It follows
from what proved above that $f(x\otimes x)=f(z\otimes z)=f(y\otimes y)$. So
$f(x\otimes x)=a$ for all unit vectors $x\in H$. Since each state is a convex
combination of pure states, by the linearity of $f$, we get that $f(\rho)=a$
holds for every state $\rho$. Therefore, we have

$\phi(\rho)=\frac{\Gamma(\rho)+D}{a+d}$

holds for all $\rho$. Then by the linearity of $\Gamma$, it is clear that
$\phi$ is affine, i.e., for any states $\rho$, $\sigma$ and scalar $\lambda$
with $0\leq\lambda\leq 1$,
$\phi(\lambda\rho+(1-\lambda)\sigma)=\lambda\phi(\rho)+(1-\lambda)\phi(\sigma)$.

Case 2. $\dim H=2$.

By fixing an orthonormal basis of $H$ we may identify $\mathcal{S}(H)$ with
$\mathcal{S}_{2}$, the convex set of $2\times 2$ positive matrices with the
trace 1. Then $\phi:\mathcal{S}_{2}\rightarrow\mathcal{S}_{2}$ is a map
preserving pure states and strict convex combinations satisfying
$\phi(\frac{1}{2}I_{2})=\frac{1}{2}I_{2}$. Let us identify $\mathcal{S}_{2}$
with the Bloch ball representation
$(\mathbb{R}^{3})_{1}=\\{(x,y,z)^{t}\in\mathbb{R}^{3}:x^{2}+y^{2}+z^{2}\leq
1\\}$ by the following way. Let
$\pi:(\mathbb{R}^{3})_{1}\rightarrow\mathcal{S}_{2}$ be the map defined by

$(x,y,z)^{t}\longmapsto\frac{1}{2}I_{2}+\frac{1}{2}\left(\begin{array}[]{cc}z&x-iy\\\
x+iy&-z\end{array}\right).$

$\pi$ is a bijective affine isomorphism. Note that $v=(x,y,z)^{t}$ satisfies
$x^{2}+y^{2}+z^{2}=1$ if and only if the corresponding matrix $\pi(v)$ is a
pure state, and $0=(0,0,0)^{t}$ if and only if the corresponding matrix is
$\pi(0)=\frac{1}{2}I$. The map
$\phi:\mathcal{S}_{2}\rightarrow\mathcal{S}_{2}$ induces a map
$\hat{\phi}:(\mathbb{R}^{3})_{1}\rightarrow(\mathbb{R}^{3})_{1}$ by the
following equation

$\phi(\rho)=\pi(\hat{\phi}(\pi^{-1}(\rho))).$

Since $\phi$ is pure state and strict convex combination preserving and
continuous, and $\pi$ is an affine isomorphism, it is easily checked that the
map $\hat{\phi}$ is strict convex combination preserving and maps the surface
of $(\mathbb{R}^{3})_{1}$ into the surface of $(\mathbb{R}^{3})_{1}$. Since
$\phi(\frac{1}{2}I)=\frac{1}{2}I$, we have that
$\hat{\phi}((0,0,0)^{t})=(0,0,0)^{t}$. It is also clear that the range of
$\hat{\phi}$ is non-collinear.

Now applying Lemma 2.3 to $\hat{\phi}$, there exists a linear transformation
$L:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$, a linear functional
$f:\mathbb{R}^{3}\rightarrow\mathbb{R}$, a vector $u_{0}\in\mathbb{R}^{3}$ and
a scalar $s\in\mathbb{R}$ such that $f((x,y,z)^{t})+s>0$ and

$\hat{\phi}((x,y,z)^{t})=\frac{L((x,y,z)^{t})+u_{0}}{f((x,y,z)^{T})+s}$

for each $(x,y,z)^{T}\in(\mathbb{R}^{3})_{1}$. Since
$\hat{\phi}((0,0,0)^{T})=(0,0,0)^{t}$, we have $u_{0}=0$ and $s>0$.
Furthermore, the linearity of $f$ implies that there are real scalars $r_{1}$,
$r_{2}$, $r_{3}$ such that $f((x,y,z)^{t})=r_{1}x+r_{2}y+r_{3}z$. We claim
that $r_{1}=r_{2}=r_{3}=0$ and hence $f=0$. If not, then there is a vector
$(x_{0},y_{0},z_{0})^{T}$ satisfying $x_{0}^{2}+y_{0}^{2}+z_{0}^{2}=1$ such
that $f((x_{0},y_{0},z_{0})^{T})=r_{1}x_{0}+r_{2}y_{0}+r_{3}z_{0}\neq 0$. It
follows that

$1=\|\hat{\phi}((x_{0},y_{0},z_{0})^{t})\|=\|\frac{L((x_{0},y_{0},z_{0})^{t})}{r_{1}x_{0}+r_{2}y_{0}+r_{3}z_{0}+s}\|,$

and thus

$\|L((x_{0},y_{0},z_{0})^{t})\|=r_{1}x_{0}+r_{2}y_{0}+r_{3}z_{0}+s.$

Similarly

$\|L((-x_{0},-y_{0},-z_{0})^{t})\|=-r_{1}x_{0}-r_{2}y_{0}-r_{3}z_{0}+s.$

By the linearity of $L$ we have
$r_{1}x_{0}+r_{2}y_{0}+r_{3}z_{0}+s=-r_{1}x_{0}-r_{2}y_{0}-r_{3}z_{0}+s$.
Hence $r_{1}x_{0}+r_{2}y_{0}+r_{3}z_{0}=0$, a contradiction. So, we have
$f=0$, and thus $\hat{\phi}=\frac{L}{s}$ is affine. Now it is clear that
$\phi$ is affine as $\pi$ is an affine isomorphism. $\Box$

The following is the main result of this section.

Theorem 2.5. Let $H,K$ be complex Hilbert spaces with $2\leq\dim H<\infty$ and
${\mathcal{S}}(H),{\mathcal{S}}(K)$ the convex sets of all states on $H,K$,
respectively. Let $\psi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(K)$ be a map.
Then $\psi$ preserves pure states and strict convex combinations (that is,
$\psi({\mathcal{P}ur}(H))\subseteq{\mathcal{P}ur}(K)$ and
$\psi((\rho,\sigma))\subseteq(\psi(\rho),\psi(\sigma))$ for any
$\rho,\sigma\in{\mathcal{S}}(H)$) if and only if one of the following holds:

(1) There exists $\sigma_{0}\in{\mathcal{P}ur}(K)$ such that
$\psi(\rho)=\sigma_{0}$ for all $\rho\in{\mathcal{S}}(H)$.

(2) There exist distinct pure states $Q_{i}\in{\mathcal{P}ur}(K)$, $i=1,2$
such that $\psi({\mathcal{P}ur}(H))=\\{Q_{1},Q_{2}\\}$, and a map
$h:{\mathcal{S}}(H)\rightarrow[0,1]$ such that, for any
$\rho_{1},\rho_{2}\in{\mathcal{S}}(H)$ and any $t\in(0,1)$,
$h(t\rho_{1}+(1-t)\rho_{2})=sh(\rho_{1})+(1-s)h(\rho_{2})$ for some
$s\in(0,1)$, and $\psi(\rho)=h(\rho)Q_{1}+(1-h(\rho))Q_{2}$ for all
$\rho\in{\mathcal{S}}(H)$.

(3) $\dim H\leq\dim K$ and there exists an injective operator
$M\in\mathcal{B}(H,K)$ such that $\psi(\rho)=\frac{M\rho M^{*}}{{\rm Tr}(M\rho
M^{*})}$ for all $\rho\in{\mathcal{S}}(H)$, or
$\psi(\rho)=\frac{M\rho^{t}M^{*}}{{\rm Tr}(M\rho^{t}M^{*})}$ for all
$\rho\in{\mathcal{S}}(H)$, where $A^{t}$ is the transpose of $A$ with respect
to an arbitrarily fixed orthonormal basis of $H$.

We remark here that the form (3) can be restated as:

(3′) $\dim H\leq\dim K$ and there exists an injective linear or conjugate
linear operator $M:H\rightarrow K$ such that $\psi(\rho)=\frac{M\rho
M^{*}}{{\rm Tr}(M\rho M^{*})}$ for all $\rho\in{\mathcal{S}}(H)$.
This statement is more convenient some times.

The following corollary is immediate, which essentially gives a
characterization of injective quantum measurement maps or the transpose of an
injective quantum measurement. We say that a map $\psi$ is open line segment
preserving if $\psi((\rho,\sigma))=(\psi(\rho),\psi(\sigma))$ for any
$\rho,\sigma$.

Corollary 2.6. Let $H,K$ be complex Hilbert spaces with $2\leq\dim H<\infty$
and $\psi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(K)$ be a map. Then the
following statements are equivalent.

(1) $\psi$ is strict convex combination preserving with
$\psi({\mathcal{P}ur}(H))\subseteq{\mathcal{P}ur}(K)$ and non-collinear range.

(2) $\psi$ is open line segment preserving with
$\psi({\mathcal{P}ur}(H))\subseteq{\mathcal{P}ur}(K)$ and non-collinear range.

(3) $\dim H\leq\dim K$ and there exists an injective operator
$M\in\mathcal{B}(H,K)$ such that $\psi(\rho)=\frac{M\rho M^{*}}{{\rm Tr}(M\rho
M^{*})}$ for all $\rho\in{\mathcal{S}}(H)$, or
$\psi(\rho)=\frac{M\rho^{t}M^{*}}{{\rm Tr}(M\rho^{t}M^{*})}$ for all
$\rho\in{\mathcal{S}}(H)$, where $A^{t}$ is the transpose of $A$ with respect
to an arbitrarily fixed orthonormal basis of $H$.

Particularly, if $\psi$ is bijective, then, by [5, Lemma 2.1], we have
$\psi(\mathcal{P}ur(H))={\mathcal{P}ur}(K)$. Also the surjectivity of $\psi$
implies the surjectivity of $M$. Thus the above corollary is a generalization
of the main result in [5] for finite dimensional case.

Remark 2.7. If the map $\psi$ has the form (1) or (3) of Theorem 2.5, then
$\psi$ is continuous. However, if $\psi$ has the form (2), $\psi$ is not
continuous and may have erratic behavior. For example, Assume $H$ is of
dimension 2. Let $Q_{1},Q_{2}$ be two distinct pure states on $K$. Divide
${\mathcal{P}ur}(H)$ into two disjoint parts
${\mathcal{P}ur}(H)={\mathcal{P}}_{1}\cup{\mathcal{P}}_{2}$ with the property
$P\in{\mathcal{P}}_{1}\Leftrightarrow P^{\bot}\in{\mathcal{P}}_{2}$ and define
$\psi(P)=Q_{1}$ if $P\in{\mathcal{P}}_{1}$; $\psi(P)=Q_{2}$ if
$P\in{\mathcal{P}}_{2}$; $\psi(tP_{1}+(1-t)P_{2})=\frac{1}{2}(Q_{1}+Q_{2})$ if
$P_{i}\in{\mathcal{P}}_{i}$, $i=1,2$, and $t\in(0,1)$, where $Q_{1},Q_{2}$ are
any distinct pure states on $K$. Then,
$\psi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(K)$ is strict convex
combination preserving and
$\psi({\mathcal{P}ur}(H))\subset{\mathcal{P}ur}(K)$. $\psi$ has the form (2)
in Theorem 2.5.

Now let us start to prove the main result of this section.

Proof of Theorem 2.5.

If $\psi$ has the form (1) or (2) or (3), it is clear that $\psi$ is pure
states and strict convex combination preserving. Conversely, assume that
$\psi$ is pure states and strict convex combination preserving. We will show
that $\psi$ has one of the forms stated in (1), (2) and (3).

Assume $\dim H=m<\infty$.

As $\frac{1}{m}I\in{\mathcal{S}}(H)$, where $I$ is the identity on $H$,
$\psi(\frac{1}{m}I)$ is positive with trace 1. So
$\psi(\frac{1}{m}I)=\frac{RR^{*}}{{\rm Tr}(RR^{*})}$ for some bounded linear
operator $R$ from $H$ into $K$.

Claim 1. $\dim{\rm ran}R\leq m$ and ${\rm ran}\psi(\rho)\subseteq{\rm ran}R$
holds for all $\rho\in\mathcal{S}(H)$.

Firstly we will show that ${\rm ran}\psi(P)\subseteq{\rm ran}R$ holds for any
pure state $P\in\mathcal{S}(H)$. Let $P=P_{1}$. There exist pure states
$\\{P_{2},\cdots,P_{m}\\}$ such that $\\{P_{1},P_{2},\cdots,P_{m}\\}$ is an
orthogonal set satisfying $P_{1}+P_{2}+\cdots+P_{m}=I$. Since $\psi$ is strict
convex combination preserving, there are $p_{i}\in(0,1)$ $(i=1,2,\cdots,m)$
with $\sum_{i=1}^{m}p_{i}=1$ such that

$\frac{RR^{*}}{{\rm
Tr}(RR^{*})}=\psi(\frac{1}{m}I)=\psi(\frac{1}{m}\sum_{i=1}^{m}P_{i})=\sum_{i=1}^{m}p_{i}\psi(P_{i}).$

Note that $\psi(P_{i})$s are rank-1 projections by the assumption. It follows
that $\dim{\rm ran}R\leq\sum_{i=1}^{m}\dim{\rm ran}\psi(P_{i})\leq m$ and, for
any $i$, we have $0\leq\psi(P_{i})\leq p_{i}^{-1}\frac{RR^{*}}{{\rm
Tr}(RR^{*})}$. Hence ${\rm ran}\psi(P_{i})\subseteq{\rm ran}R$ for all $i$.
Particularly, ${\rm ran}\psi(P)={\rm ran}\psi(P_{1})\subseteq{\rm ran}R$.

For any $\rho\in{\mathcal{S}}(H)$, let $\rho=\sum_{i=1}^{m}t_{i}P_{i}$ be its
spectral resolution. As $t_{i}\geq 0$, $\sum_{i}t_{i}=1$ and $\psi$ is strict
convex combination preserving, there are $s_{i}\in[0,1]$ with
$\sum_{i}s_{i}=1$ and $s_{i}=0$ if $t_{i}=0$ such that
$\psi(\rho)=\sum_{i=1}^{m}s_{i}\psi(P_{i})$. Now it is clear that ${\rm
ran}\psi(\rho)\subseteq{\rm ran}R$ because ${\rm ran}\psi(P_{i})\subseteq{\rm
ran}R$ for all $i=1,2,\ldots,m$.

As a result, if $R$ is of rank-1, then it is clear that $\psi$ has the form
(1) in Theorem 2.5.

So, in the sequel, we assume that rank$(R)=r\geq 2$. Thus we can define a map
$\phi:\mathcal{S}(H)\rightarrow\mathcal{S}(H)$ by

$None$ $\phi(\rho)=\frac{R^{[-1]}\psi(\rho)R^{[-1]*}}{{\rm
Tr}(R^{[-1]}\psi(\rho)R^{[-1]*})},$

where $R^{[-1]}$ is the Moore-Penrose generalized inverse of $R$. It is clear
that $\phi(\mathcal{P}ur(H))\subseteq\mathcal{P}ur(H)$ and $\phi$ preserves
strict convex combination. Write $\phi(\frac{1}{m}I_{m})=\frac{Q_{0}}{{\rm
Tr}(Q_{0})}$, where $Q_{0}=R^{[-1]}R$. This implies that $Q_{0}$ is a
projection with ${\rm rank}Q_{0}=r\leq m$. It follows that, there exists an
orthonormal set $\\{e_{1},\ldots,e_{r}\\}\subset H$ such that
$\sum_{i=1}^{r}\phi(P_{i})=Q_{0}$, where $P_{i}=e_{i}\otimes e_{i}$. Let
$H_{1}={\rm span}\\{e_{1},\ldots,e_{r}\\}$ and $K_{1}=Q_{0}(H)$. Then $\dim
H_{1}=\dim K_{1}=r$ and
$\widetilde{\phi}=\phi|_{{\mathcal{S}}(H_{1})}:{\mathcal{S}}(H_{1})\rightarrow{\mathcal{S}}(K_{1})$
is a strict convex combination preserver sending pure states to pure states.
As
$\widetilde{{\phi}}(\frac{1}{r}I_{H_{1}})=\sum_{i=1}^{r}q_{i}\phi(P_{i})=SS^{*}$
is an invertible state on $K_{1}$, it induces a strict convex combination
preserver
$\widehat{{\phi}}:{\mathcal{S}}(H_{1})\rightarrow{\mathcal{S}}(H_{1})$
satisfying
$\widehat{\phi}({\mathcal{P}ur}(H_{1}))\subseteq{\mathcal{P}ur}(H_{1})$ and
$\widehat{\phi}(\frac{1}{r}I_{H_{1}})=\frac{1}{r}I_{H_{1}}$, where
$\widehat{\phi}$ is defined by

$None$
$\widehat{\phi}(\rho)=\frac{S^{-1}\widetilde{\phi}(\rho){S^{-1}}^{*}}{{\rm
Tr}(S^{-1}\widetilde{\phi}(\rho){S^{-1}}^{*})}.$

It follows that $\widehat{\phi}$ maps orthogonal pure states to orthogonal
ones.

Claim 2. If $r\geq 3$, then there is a unitary operator $U:H_{1}\rightarrow
H_{1}$ such that either $\widehat{\phi}(\rho)=U\rho U^{*}$ for every
$\rho\in{\mathcal{S}}(H_{1})$ or $\widehat{\phi}(\rho)=U\rho^{t}U^{*}$ for
every $\rho\in{\mathcal{S}}(H_{1})$.

If $r\geq 3$, then $\widehat{\phi}({\mathcal{S}}(H_{1}))$ contains at least 3
rank one projections which are orthogonal to each other and hence non-
collinear. So, by Lemma 2.4, $\widehat{\phi}$ is affine, that is
$\widehat{\phi}(t\rho+(1-t)\sigma)=t\widehat{\phi}(\rho)+(1-t)\widehat{\phi}(\sigma)$
holds for any $\rho,\sigma\in{\mathcal{S}}(H_{1})$ and $t\in[0,1]$. Thus
$\widehat{\phi}$ is injective and can be extended to an injective linear map
from ${\mathcal{B}}_{\rm sa}\cong{\bf H}_{r}$ into ${\mathcal{B}}_{\rm
sa}\cong{\bf H}_{r}$.

Now by Lemma 2.2, there is a unitary operator $U:H_{1}\rightarrow H_{1}$ such
that either $\widehat{\phi}(\rho)=U\rho U^{*}$ for every
$\rho\in{\mathcal{S}}(H_{1})$ or $\widehat{\phi}(\rho)=U\rho^{t}U^{*}$ for
every $\rho\in{\mathcal{S}}(H_{1})$.

If $P$ and $Q$ are projections and $PQ=0$, we say that $P$ and $Q$ are
orthogonal, denoted by $P\bot Q$. $P^{\bot}$ stands for $I-P$.

Claim 3. If $r=2$, then either

(i) there exist
${\mathcal{P}}_{1},{\mathcal{P}}_{2}\subseteq{\mathcal{P}ur}(H_{1})$
satisfying ${\mathcal{P}}_{1}\cup{\mathcal{P}}_{2}={\mathcal{P}ur}(H_{1})$ and
$P,I_{2}-P$ can not be in the same ${\mathcal{P}}_{i}$; and there exist
$Q_{1},Q_{2}\in{\mathcal{P}ur}(H_{1})$ with $Q_{1}\bot Q_{2}$ such that
$\widehat{\phi}({\mathcal{P}}_{i})=\\{Q_{i}\\}$, $i=1,2$, and
$\widehat{\phi}({\mathcal{S}}(H_{1})\subseteq[Q_{1},Q_{2}]$.

(ii) there exists a unitary operator $U:H_{1}\rightarrow H_{1}$ such that
either $\widehat{\phi}(\rho)=U\rho U^{*}$ for every
$\rho\in{\mathcal{S}}(H_{1})$ or $\widehat{\phi}(\rho)=U\rho^{t}U^{*}$ for
every $\rho\in{\mathcal{S}}(H_{1})$.

If there exist $Q_{1},Q_{2},Q_{3}\in\widehat{\phi}({\mathcal{S}}(H_{1}))$ such
that they are non-collinear, then Lemma 2.4 is applicable. It follows that
$\widehat{\phi}$ is affine and can be extended to a linear or conjugate linear
map, still denoted by $\widehat{\phi}$ from ${\bf H}_{2}$ into ${\bf H}_{2}$,
which is unital and rank-1 projection preserving. Therefore, use Lemma 2.2 we
see that (ii) is true.

Assume that $\widehat{\phi}({\mathcal{S}}(H_{1})$ is collinear; then there
exist $P_{1},P_{2}\in{\mathcal{S}}(H_{1})$ such that
$\widehat{\phi}(P_{i})=Q_{i}$, $Q_{1}+Q_{2}=I_{2}$ and
$\widehat{\phi}({\mathcal{S}}(H_{1})\subseteq[Q_{1},Q_{2}]$. It entails that
$\widehat{\phi}({\mathcal{P}ur}(H_{1}))=\\{Q_{1},Q_{2}\\}$. Let
${\mathcal{P}}_{i}=\widehat{\phi}^{-1}(\\{Q_{i}\\})$, $i=1,2$. Then,
${\mathcal{P}}_{1}\cap{\mathcal{P}}_{2}=\emptyset$ and
${\mathcal{P}}_{1}\cup{\mathcal{P}}_{2}={\mathcal{P}ur}(H_{1})$. If
$P\in{\mathcal{P}}_{1}$, then

$\frac{1}{2}\widehat{\phi}(P)+\frac{1}{2}\widehat{\phi}(P^{\bot})=\widehat{\phi}(\frac{1}{2}(P+P^{\bot}))=\frac{1}{2}I_{2}.$

So $\widehat{\phi}(P^{\bot})=Q_{2}$, and $P^{\bot}\in{\mathcal{P}}_{2}$. Now,
it is clear that (i) holds.

Claim 4. If $r=m\geq 3$, then $\psi(\rho)=\frac{M\rho M^{*}}{{\rm Tr}(M\rho
M^{*})}$ for all $\rho\in{\mathcal{S}}(H)$; or
$\psi(\rho)=\frac{M\rho^{t}M^{*}}{{\rm Tr}(M\rho^{t}M^{*})}$ for all
$\rho\in{\mathcal{S}}(H)$, where $M:H\rightarrow K$ is an injective linear
operator. So, $\psi$ has the form (3) of Theorem 2.5.

In fact, in this case we have $\widehat{\phi}=\widetilde{\phi}=\phi$. By Claim
2, there is a unitary $U:H\rightarrow H$ such that $\phi(\rho)=U\rho U^{*}$
for all $\rho$ or $\phi(\rho)=U\rho^{t}U^{*}$ for all $\rho$. Let $M=RU$. Then
$M:H\rightarrow K$ is injective and the claim holds.

By use of Claim 4, the following claim is obvious.

Claim 5. If $r=m=2$, then either $\psi$ has the form (3) or has the form (2)
of Theorem 2.5.

Claim 6. If $2=r<m$, then $\psi$ has the form (2).

As $r=2$, $\widehat{\phi}$ has two possible forms (i) and (ii) stated in Claim
3.

If $\widehat{\phi}$ has the form (i), then there exist distinct pure states
$Q_{1},Q_{2}$ on $K$ such that $\psi({\mathcal{P}ur}(H))=\\{Q_{1},Q_{2}\\}$.
It is clear that, in this case, we have $\psi(\rho)\in[Q_{1},Q_{2}]$ for every
$\rho\in{\mathcal{S}}(H)$, that is, $\psi$ is of the form (2) stated in
Theorem 2.5.

We assert that the case (ii) does not occur. If $\widehat{\phi}$ takes form
(ii), then there exists an orthogonal set of pure states
$\\{P_{1}=e_{1}\otimes e_{1},\ldots,P_{m}=e_{m}\otimes e_{m}\\}$ such that
ran$(\psi)={\mathcal{S}}(K_{1})$, where $K_{1}={\rm span}\\{u_{1},u_{2}\\}$
with $u_{i}\otimes u_{i}=\psi(e_{i}\otimes e_{i})$, $i=1,2$. Thus $\psi$ is
continuous when restricted on ${\mathcal{S}}(H_{1})$ with $H_{1}={\rm
span}\\{e_{1},e_{2}\\}$. Note that $\psi(P_{3})=\psi(P_{1})$ or $\psi(P_{2})$,
say $\psi(P_{3})=\psi(P_{1})=Q_{1}$. Let $P(\alpha,\beta)=(\alpha e_{1}+\beta
e_{2})\otimes(\alpha e_{1}+\beta e_{2})\in{\mathcal{P}ur}(H_{1})$, where
$\alpha,\beta\in{\mathbb{C}}$ satisfying $|\alpha|^{2}+|\beta|^{2}=1$. Then,
$P(\alpha,\beta)$ is continuous and hence $\psi(P(\alpha,\beta))$ is
continuous in $\alpha,\beta$. Since
$\\{P(\alpha,\beta),P(\bar{\beta},-\bar{\alpha}),P_{3},\ldots,P_{m}\\}$ is
still a complete orthogonal set of rank-1 projections, one of them must be
$\psi(P_{3})=Q_{1}$ and another be $Q_{2}$. It follows that the range of
$\psi(P(\alpha,\beta))$ can take at most two distinct value and thus must be a
constant function. However, $\psi(P(1,0))=\psi(P_{1})=Q_{1}$ and
$\psi(P(0,1))=\psi(P_{2})=Q_{2}$, a contradiction. So, this case can not
occur, finishing the proof of Claim 6.

Claim 7. The case of $3\leq r<m$ can not occur.

On the contrary, suppose $m>r\geq 3$. Then by Lemma 2.3, $\phi$ is continuous.
Choose any orthogonal set of rank one projections $\\{P_{i}=x_{i}\otimes
x_{i}\\}_{i=1}^{m}$ satisfying $\sum_{i=1}^{m}P_{i}=I$. Then there exists
$p_{i}>0$ with $\sum_{i=1}^{m}p_{i}=1$ such that
$\phi(\frac{1}{m}I)=\phi(\frac{1}{m}\sum_{i=1}^{m}P_{i})=\sum_{i=1}^{m}p_{i}\phi(P_{i})=\frac{1}{r}Q_{0}$.
It follows that $m$ rank-1 projections $\phi(P_{1}),\cdots,\phi(P_{m})$ are
linearly dependent. Without loss of generality, assume
$\\{\phi(P_{1}),\cdots,\phi(P_{r})\\}$ is linearly independent. Then for any
$j>r$, there exists $i_{j}\leq r$ such that $\phi(P_{j})=\phi(P_{i_{j}})$. So
there exist $q_{i}>0$, $(1\leq i\leq r)$, such that
$\phi(\frac{1}{m}I)=\sum_{i=1}^{r}q_{i}\phi(P_{i})$. By Lemma 2.1, we obtain
that $q_{i}=\frac{1}{r}$ and $\\{\phi(P_{i})\\}_{i=1}^{r}$ is an orthogonal
set of rank-1 projections. Consequently we obtain that, for any two orthogonal
rank-1 projections $Q_{1}=y_{1}\otimes y_{1}$, $Q_{2}=y_{2}\otimes y_{2}$ on
$H$, either $\phi(Q_{1})=\phi(Q_{2})$ or $\phi(Q_{1})\bot\phi(Q_{2})$. Let
$Q^{\prime}_{1}=(\alpha y_{1}+\beta y_{2})\otimes(\alpha y_{1}+\beta y_{2})$,
$Q^{\prime}_{2}=(\bar{\beta}y_{1}-\bar{\alpha}y_{2})\otimes(\bar{\beta}y_{1}-\bar{\alpha}y_{2})$,
where $|\alpha|^{2}+|\beta|^{2}=1$. Clearly $Q^{\prime}_{1}\bot
Q^{\prime}_{2}$. We assert that:

$None$
$\phi(Q_{1})=\phi(Q_{2})\Rightarrow\phi(Q^{\prime}_{1})=\phi(Q^{\prime}_{2})\quad\mbox{\rm
and}\quad\phi(Q_{1})\bot\phi(Q_{2})\Rightarrow\phi(Q^{\prime}_{1})\bot\phi(Q^{\prime}_{2}).$

To see this, let $f(\alpha,\beta)=\phi(Q^{\prime}_{1})$,
$g(\alpha,\beta)=\phi(Q^{\prime}_{2})$ and let
$h(\alpha,\beta)=\|f(\alpha,\beta)-g(\alpha,\beta)\|$. As $\phi$ is
continuous, we see that $f,g$ and $h$ are continuous in $\alpha,\beta$. Also
note that $h(\alpha,\beta)\in\\{0,1\\}$ for any $(\alpha,\beta)$. Hence, if
$\phi(Q_{1})=\phi(Q_{2})$, then $h(1,0)=0$, which forces
$h(\alpha,\beta)\equiv 0$ and consequently, $f(\alpha,\beta)=g(\alpha,\beta)$
for all $\alpha,\beta$; if $\phi(Q_{1})\bot\phi(Q_{2})$, then
$h(\alpha,\beta)\equiv h(0,1)=1$, which implies that $f(\alpha,\beta)\bot
g(\alpha,\beta)$ for all $\alpha,\beta$. So the assertion (2.4) is true.

Now for the chosen orthogonal set $\\{P_{i}\\}_{i=1}^{m}$, as $3\leq r<m$, by
what proved above, we can rearrange the order of $\\{\phi(P_{i})\\}_{i=1}^{m}$
so that $\\{\phi(P_{1}),\ldots,\phi(P_{r})\\}$ is an orthogonal set. Then
$\phi(P_{1})+\cdots+\phi(P_{r})=Q_{0}$, $\phi(P_{r+j})$ equals to
$\phi(P_{i})$ for some $i$ with $1\leq i\leq r$.

Assume that there exist two distinguished projections in
$\\{\phi(P_{r+1}),\ldots,\phi(P_{m})\\}$, say
$\phi(P_{r+1})\not=\phi(P_{r+2})$. Let $P^{\prime}_{r+1}=(\alpha x_{r+1}+\beta
x_{r+2})\otimes(\alpha x_{r+1}+\beta x_{r+2})$ and
$P^{\prime}_{r+2}=(\bar{\beta}x_{r+1}-\bar{\alpha}x_{r+2})\otimes(\bar{\beta}x_{r+1}-\bar{\alpha}x_{r+2})$
with $|\alpha|^{2}+|\beta|^{2}=1$. By Eq.(2.4), $f(\alpha,\beta)\bot
g(\alpha,\beta)$, where $f(\alpha,\beta)=\phi(P^{\prime}_{r+1})$ and
$g(\alpha,\beta)=\phi(P^{\prime}_{r+2})$. Since
$\\{P_{1},\ldots,P_{r},P^{\prime}_{r+1},P^{\prime}_{r+2},\ldots,P_{m}\\}$ is
still orthogonal, we see that
$f(\alpha,\beta)\in\\{\phi(P_{1}),\ldots,\phi(P_{r})\\}$. The continuity of
$f$ then implies that $f(\alpha,\beta)\equiv\phi(P_{i_{0}})$ for some $1\leq
i_{0}\leq r$. Similarly, $g(\alpha,\beta)\equiv\phi(P_{i_{1}})$ for some
$1\leq i_{1}\leq r$. Since $f$ and $g$ has the same range, we must have
$i_{0}=i_{1}$, but this contradicts to $f(1,0)\bot g(1,0)$.

Therefore, we may assume that $\phi(P_{r+1})=\cdots=\phi(P_{m})=\phi(P_{1})$.
Now Let $P^{\prime}_{1}=(\alpha x_{1}+\beta x_{2})\otimes(\alpha x_{1}+\beta
x_{2})$ and
$P^{\prime}_{2}=(\bar{\beta}x_{1}-\bar{\alpha}x_{2})\otimes(\bar{\beta}x_{1}-\bar{\alpha}x_{2})$
with $|\alpha|^{2}+|\beta|^{2}=1$. Denote
$f(\alpha,\beta)=\phi(P^{\prime}_{1})$ and
$g(\alpha,\beta)=\phi(P^{\prime}_{2})$. Then by Eq.(2.4) again we have
$f(\alpha,\beta)\bot g(\alpha,\beta)$. Note that
$f(1,0)=\phi(P_{1})=\phi(P_{r+1})\in\\{f(\alpha,\beta),g(\alpha,\beta)\\}$ for
any $(\alpha,\beta)$. This entails
$\\{f(\alpha,\beta),g(\alpha,\beta)\\}=\\{\phi(P_{1}),\phi(P_{2})\\}$ for any
$(\alpha,\beta)$. So it follows from the continuity of $f$ and $g$ that
$f(\alpha,\beta)\equiv\phi(P_{1})$ and $g(\alpha,\beta)\equiv\phi(P_{2})$,
contradicting to the fact that $f$ and $g$ has the same range. So the claim is
true.

Combining Claims 1-7, we see that $\psi$ preserves pure state and strict
convex combinations will imply that $\psi$ takes one of the form (1), (2) and
(3), completing the proof of Theorem 2.5. $\Box$

##  3\. Maps preserving separable pure states and strict convex combinations:
bipartite systems

This section is devoted to giving a structure theorem of maps preserve
separable pure states and strict convex combinations for bipartite systems.
Using this structure theorem we are able to answer the conjectures for finite
dimensional case mentioned in the introduction section that were proposed in
[8] .

The following simple lemma may be found in [10].

Lemma 3.1 Let $H$ be a complex Hilbert space of any dimension and
$T\in{\mathcal{B}}(H)$ a finite rank operator. Then $\frac{1}{{\rm
rank}T}\|T\|_{\rm Tr}^{2}\leq\|T\|_{2}^{2}\leq\|T\|_{\rm Tr}^{2}$, where
$\|T\|_{\rm Tr}$ and $\|T\|_{2}$ are respectively the trace-norm and the
Hilbert-Schmidt norm of $T$.

Recall that ${\mathcal{S}}_{\rm sep}(H_{1}\otimes H_{2})$ and
$\mathcal{P}ur(H_{1})\otimes\mathcal{P}ur(H_{2})$ stand respectively for the
convex set of all separable states and the set of all separable pure states in
bipartite system $H_{1}\otimes H_{2}$. Now let us present the main result of
this section.

Theorem 3.2 Let $H_{1},H_{2},K_{1},K_{2}$ be complex Hilbert spaces with
$2\leq\dim H_{i}<\infty$, $i=1,2$. Let $\psi:{\mathcal{S}}_{\rm
sep}(H_{1}\otimes H_{2})\rightarrow{\mathcal{S}}_{\rm sep}(K_{1}\otimes
K_{2})$ be a map. If $\psi$ preserves separable pure states and strict convex
combinations, then one of the following statements holds.

(1) There exists $R_{1}\otimes
R_{2}\in\mathcal{P}ur(K_{1})\otimes\mathcal{P}ur(K_{2})$ such that

$\psi(A\otimes B)=R_{1}\otimes R_{2}$

for all $A\in{\mathcal{S}}(H_{1})$ and $B\in{\mathcal{S}}(H_{2})$.

(2) $\dim H_{1}\leq\dim K_{1}$, there exist $R_{2}\in\mathcal{P}ur(K_{2})$ and
injective $M_{1}\in\mathcal{B}(H_{1},K_{1})$ such that

$\psi(A\otimes B)=\frac{M_{1}AM_{1}^{*}}{{\rm Tr}(M_{1}AM_{1}^{*})}\otimes
R_{2}\ \mbox{\rm for all }A\in{\mathcal{S}}(H_{1})\ \mbox{\rm and}\
B\in{\mathcal{S}}(H_{2})$

or

$\psi(A\otimes B)=\frac{M_{1}A^{t}M_{1}^{*}}{{\rm
Tr}(M_{1}A^{t}M_{1}^{*})}\otimes R_{2}\ \mbox{\rm for all
}A\in{\mathcal{S}}(H_{1})\ \mbox{\rm and}\ B\in{\mathcal{S}}(H_{2}).$

(3) $\dim H_{2}\leq\dim K_{2}$, there exist $R_{1}\in\mathcal{P}ur(K_{1})$ and
injective $M_{2}\in\mathcal{B}(H_{2},K_{2})$ such that

$\psi(A\otimes B)=R_{1}\otimes\frac{M_{2}BM_{2}^{*}}{{\rm
Tr}(M_{2}BM_{2}^{*})}\ \mbox{\rm for all }A\in{\mathcal{S}}(H_{1})\ \mbox{\rm
and}\ B\in{\mathcal{S}}(H_{2})$

or

$\psi(A\otimes B)=R_{1}\otimes\frac{M_{2}B^{t}M_{2}^{*}}{{\rm
Tr}(M_{2}B^{t}M_{2}^{*})}\ \mbox{\rm for all }A\in{\mathcal{S}}(H_{1})\
\mbox{\rm and}\ B\in{\mathcal{S}}(H_{2}).$

(4) $\dim H_{2}\leq\dim K_{1}$, there exist $R_{2}\in\mathcal{P}ur(K_{2})$ and
injective $M_{1}\in\mathcal{B}(H_{2},K_{1})$ such that

$\psi(A\otimes B)=\frac{M_{1}BM_{1}^{*}}{{\rm Tr}(M_{1}BM_{1}^{*})}\otimes
R_{2}\ \mbox{\rm for all }A\in{\mathcal{S}}(H_{1})\ \mbox{\rm and}\
B\in{\mathcal{S}}(H_{2})$

or

$\psi(A\otimes B)=\frac{M_{1}B^{t}M_{1}^{*}}{{\rm
Tr}(M_{1}B^{t}M_{1}^{*})}\otimes R_{2}\ \mbox{\rm for all
}A\in{\mathcal{S}}(H_{1})\ \mbox{\rm and}\ B\in{\mathcal{S}}(H_{2}).$

(5) $\dim H_{1}\leq\dim K_{2}$, there exist $R_{1}\in\mathcal{P}ur(K_{1})$ and
injective $M_{2}\in\mathcal{B}(H_{1},K_{2})$ such that

$\psi(A\otimes B)=R_{1}\otimes\frac{M_{2}AM_{2}^{*}}{{\rm
Tr}(M_{2}AM_{2}^{*})}\ \mbox{\rm for all }A\in{\mathcal{S}}(H_{1})\ \mbox{\rm
and}\ B\in{\mathcal{S}}(H_{2})$

or

$\psi(A\otimes B)=R_{1}\otimes\frac{M_{2}A^{t}M_{2}^{*}}{{\rm
Tr}(M_{2}A^{t}M_{2}^{*})}\ \mbox{\rm for all }A\in{\mathcal{S}}(H_{1})\
\mbox{\rm and}\ B\in{\mathcal{S}}(H_{2}).$

(6) $\dim H_{i}\leq\dim K_{i}$, $i=1,2$, there exist injective
$M_{1}\in\mathcal{B}(H_{1},K_{1})$ and $M_{2}\in\mathcal{B}(H_{2},K_{2})$ such
that

$\psi(A\otimes B)=\frac{M_{1}\Psi_{1}(A)M_{1}^{*}}{{\rm
Tr}(M_{1}\Psi_{1}(A)M_{1}^{*})}\otimes\frac{M_{2}\Psi_{2}(B)M_{2}^{*}}{{\rm
Tr}(M_{2}\Psi_{2}(B)M_{2}^{*})}$

for all $A\in{\mathcal{S}}(H_{1})\ \mbox{\rm and}\ B\in{\mathcal{S}}(H_{2})$,
where $\Psi_{i}:{\mathcal{B}}(H_{i})\rightarrow{\mathcal{B}}(H_{i})$, $i=1,2$,
is the identity, or the transpose.

(7) $\dim H_{1}\leq\dim K_{2}$ and $\dim H_{2}\leq\dim K_{1}$, there exist
injective $M_{1}\in\mathcal{B}(H_{2},K_{1})$ and
$M_{2}\in\mathcal{B}(H_{1},K_{2})$ such that

$\psi(A\otimes B)=\frac{M_{1}\Psi_{2}(B)M_{1}^{*}}{{\rm
Tr}(M_{1}\Psi_{2}(B)M_{1}^{*})}\otimes\frac{M_{2}\Psi_{1}(A)M_{2}^{*}}{{\rm
Tr}(M_{2}\Psi_{1}(A)M_{2}^{*})}$

for all $A\in{\mathcal{S}}(H_{1})\ \mbox{\rm and}\ B\in{\mathcal{S}}(H_{2})$,
where $\Psi_{i}:{\mathcal{B}}(H_{i})\rightarrow{\mathcal{B}}(H_{i})$, $i=1,2$,
is the identity, or the transpose.

(8) $\min\\{\dim H_{1},\dim H_{2}\\}\leq\dim K_{1}$, there exist
$R_{2}\in{\mathcal{P}ur}(K_{2})$ and a strict convex combination preserving
map $\varphi_{1}:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})\rightarrow{\mathcal{S}}(K_{1})$ such that

$\psi(A\otimes B)=\varphi_{1}(A\otimes B)\otimes R_{2}$

for all $A\in\mathcal{S}(H_{1})$ and $B\in\mathcal{S}(H_{2})$. Moreover,
$\varphi_{1}$ satisfies that for each $P\otimes
Q\in\mathcal{P}ur(H_{1})\otimes\mathcal{P}ur(H_{2})$, $\varphi_{1}(P\otimes
Q)=\frac{M_{P}QM_{P}^{*}}{{\rm
Tr}(M_{P}QM_{P}^{*})}=\frac{N_{Q}PN_{Q}^{*}}{{\rm Tr}(N_{Q}PN_{Q}^{*})}$ for
some injective linear or conjugate linear (may not synchronously) operators
$M_{P}:H_{2}\rightarrow K_{1}$ and $N_{Q}:H_{1}\rightarrow K_{1}$.

(9) $\min\\{\dim H_{1},\dim H_{2}\\}\leq\dim K_{2}$, there exist
$R_{1}\in\mathcal{P}ur(K_{1})$ and a strict convex combination preserving map
$\varphi_{2}:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})\rightarrow{\mathcal{S}}(K_{2})$ such that

$\psi(A\otimes B)=R_{1}\otimes\varphi_{2}(A\otimes B)$

for all $A\in\mathcal{S}(H_{1})$ and $B\in\mathcal{S}(H_{2})$. Moreover,
$\varphi_{2}$ satisfies that, for each $P\otimes
Q\in\mathcal{P}ur(H_{1})\otimes\mathcal{P}ur(H_{2})$, $\varphi_{2}(P\otimes
Q)=\frac{M_{P}QM_{P}^{*}}{{\rm
Tr}(M_{P}QM_{P}^{*})}=\frac{N_{Q}PN_{Q}^{*}}{{\rm Tr}(N_{Q}PN_{Q}^{*})}$, for
some injective linear or conjugate linear (may not synchronously) operators
$M_{P}:H_{2}\rightarrow K_{2}$ and $N_{Q}:H_{1}\rightarrow K_{2}$.

(10) There exist $P_{i}^{\prime}\in{\mathcal{P}ur}(K_{1})$ and
$Q_{i}^{\prime}\in{\mathcal{P}ur}(K_{2})$, $i=1,2$ such that
$\psi({\mathcal{P}ur}(H_{1})\otimes{\mathcal{P}ur}(H_{2}))=\\{P_{1}^{\prime}\otimes
Q_{1}^{\prime},P_{2}^{\prime}\otimes Q_{2}^{\prime}\\}$, and
ran$(\psi)\subseteq[P_{1}^{\prime}\otimes Q_{1}^{\prime},P_{2}^{\prime}\otimes
Q_{2}^{\prime}]$.
Where the transpose is taken with respect to an arbitrarily fixed orthonormal
basis.

Proof. Suppose $\dim H_{1}=m$, $\dim H_{2}=n$. If the range of $\psi$ is
collinear, then it is clear that either (1) holds or (10) holds. So, in the
sequel, we assume that the range of $\psi$ is non-collinear. Denote by
$\mathcal{B}_{\rm sa}(H)$ the real linear space of all self-adjoint operators
on the Hilbert space $H$. Consider the partial traces ${\rm
Tr_{1}}:\mathcal{B}_{\rm sa}(K_{1}\otimes K_{2})\rightarrow\mathcal{B}_{\rm
sa}(K_{2})$ and ${\rm Tr_{2}}:\mathcal{B}_{\rm sa}(K_{1}\otimes
K_{2})\rightarrow\mathcal{B}_{\rm sa}(K_{1})$ on $\mathcal{B}_{\rm
sa}(K_{1}\otimes K_{2})\equiv\mathcal{B}_{\rm
sa}(K_{1})\otimes\mathcal{B}_{\rm sa}(K_{2})$ defined by ${\rm
Tr_{1}}(A\otimes B)=({\rm Tr}A)B$ and ${\rm Tr_{2}}(A\otimes B)=({\rm Tr}B)A$.
Clearly, ${\rm Tr_{1}}$ and ${\rm Tr_{2}}$ are linear maps. Define two maps
$\phi_{1}:(\mathcal{S}(H_{1}),\mathcal{S}(H_{2}))\rightarrow\mathcal{S}(K_{1})$
and
$\phi_{2}:(\mathcal{S}(H_{1}),\mathcal{S}(H_{2}))\rightarrow\mathcal{S}(K_{2})$
by

$\phi_{1}(A,B)={\rm Tr_{2}}(\psi(A\otimes B))\ {\rm and}\ \phi_{2}(A,B)={\rm
Tr_{1}}(\psi(A\otimes B)).$

Notice that

$\psi(P\otimes Q)=\phi_{1}(P,Q)\otimes\phi_{2}(P,Q)\ {\rm for\ all}\
P\in\mathcal{P}ur(H_{1})\ {\rm and}\ Q\in\mathcal{P}ur(H_{2}).$

Fix a $Q\in\mathcal{P}ur(H_{2})$; then the maps
$\phi_{1}(\cdot,Q):\mathcal{S}(H_{1})\rightarrow\mathcal{S}(K_{1})$ and
$\phi_{2}(\cdot,Q):\mathcal{S}(H_{1})\rightarrow\mathcal{S}(K_{2})$ are both
strict convex combination preserving and
$\phi_{1}(\mathcal{P}ur(H_{1}),Q)\subseteq\mathcal{P}ur(K_{1})$ while
$\phi_{2}(\mathcal{P}ur(H_{1}),Q)\subseteq\mathcal{P}ur(K_{2})$. Therefore,
applying Theorem 2.5 to $\phi_{1}(\cdot,Q)$ and $\phi_{2}(\cdot,Q)$,
respectively, we get that, for $i=1,2$, either

(i) there exists pure state $R_{iQ}\in{\mathcal{P}ur}(K_{i})$ such that
$\phi_{i}(A,Q)=R_{iQ}$ for all $A\in\mathcal{S}(H_{1})$;
or

(ii) there are pure states
$R_{1Q}^{(i)},R_{2Q}^{(i)}\in{\mathcal{P}ur}(K_{i})$ and a strict convex
combination preserving map $h_{iQ}:{\mathcal{S}}(H_{1})\rightarrow[0,1]$ such
that $\phi_{i}({\mathcal{P}ur}(H_{1}),Q)=\\{R_{1Q}^{(i)},R_{2Q}^{(i)}\\}$ and
$\phi_{i}(A,Q)=h_{iQ}(A)R_{1Q}^{(i)}+(1-h_{iQ}(A))R_{2Q}^{(i)}$ for all
$A\in\mathcal{S}(H_{1})$;
or

(iii) there exists an injective linear or conjugate linear operator
$M_{iQ}:H_{1}\rightarrow K_{i}$ such that
$\phi_{i}(A,Q)=\frac{M_{iQ}AM_{iQ}^{*}}{{\rm Tr}(M_{iQ}AM_{iQ}^{*})}$ for all
$A\in\mathcal{S}(H_{1})$.

As $\psi$ is strict convex combination preserving and ${\rm ran}\psi$ is non-
collinear, Lemma 2.3 is applicable. Thus we have
$\psi(\rho)=\frac{\Gamma(\rho)+D}{f(\rho)+d}$, where $\Gamma:\mathcal{B}_{\rm
sa}(H_{1}\otimes H_{2})\rightarrow\mathcal{B}_{\rm sa}(K_{1}\otimes K_{2})$ is
a linear map, $D\in\mathcal{B}_{\rm sa}(K_{1}\otimes K_{2})$ is an operator,
$f:\mathcal{B}_{\rm sa}(H_{1}\otimes H_{2})\rightarrow\mathbb{R}$ is a linear
functional and $d\in\mathbb{R}$ with $f(\rho)+d>0$ for all
$\rho\in\mathcal{S}_{\rm sep}(H_{1}\otimes H_{2})$. As both $H_{1}$ and
$H_{2}$ are finite dimensional, $\Gamma$, $f$ are continuous. It follows that
$\psi$ is continuous. So both $\phi_{1}$ and $\phi_{2}$ are continuous on
$\mathcal{S}(H_{1})$. These facts will be used frequently.

Let us first consider the map $\phi_{1}$.

Claim 1. Either $\phi_{1}(\cdot,Q)$ has the form (i) for all
$Q\in\mathcal{P}ur(H_{2})$ or $\phi_{1}(\cdot,Q)$ has the form (iii) for all
$Q\in\mathcal{P}ur(H_{2})$.

As mentioned above, for any fixed pure state $Q\in{\mathcal{P}ur}(H_{2})$,
$\phi_{1}(\cdot,Q)$ takes one of the forms (i)-(iii). As $\phi_{1}$ is
continuous, by Remark 2.7 we see that $\phi_{1}(\cdot,Q)$ can not have the
form (ii) for any $Q\in\mathcal{P}ur(H_{2})$. Thus, for any $Q$,
$\phi_{1}(\cdot,Q)$ takes the form (i) or the form (iii).

Furthermore, we will show that either $\phi_{1}(\cdot,Q)$ has the form (i) for
all $Q\in\mathcal{P}ur(H_{2})$ or $\phi_{1}(\cdot,Q)$ has the form (iii) for
all $Q\in\mathcal{P}ur(H_{2})$.

To do this, for any $A\in{\mathcal{S}}(H_{1})$ so that ${\rm rank}(A)\geq 2$,
define $F_{A}:\mathcal{P}ur(H_{2})\rightarrow\mathbb{R}$ by
$F_{A}(Q)=\|\phi_{1}(A,Q)\|_{2}$, where $\|\cdot\|_{2}$ is the Hilbert-Schmidt
norm. Notice that when $\phi_{1}(\cdot,Q)$ takes the form (i), then
$F_{A}(Q)=\|\phi_{1}(A,Q)\|_{2}=1$; when $\phi_{1}(\cdot,Q)$ takes the form
(iii), then $\phi_{1}(A,Q)=\frac{M_{1Q}AM_{1Q}^{*}}{{\rm
Tr}(M_{1Q}AM_{1Q}^{*})}$ and
$F_{A}(Q)=\|\phi_{1}(A,Q)\|_{2}=\frac{\|M_{1Q}AM_{1Q}^{*}\|_{2}}{\|M_{1Q}A^{\frac{1}{2}}\|_{2}^{2}}<1$
as ${\rm rank}A\geq 2$ and $M_{1Q}$ is injective.

If there exist two distinct $Q_{1},Q_{2}\in\mathcal{P}ur(H_{2})$, such that
$\phi_{1}(\cdot,Q_{1})$ has the form (i) while $\phi_{1}(\cdot,Q_{2})$ has the
form (iii), that is, $\phi_{1}(\cdot,Q_{1})=R_{1Q_{1}}$,
$\phi_{1}(\cdot,Q_{2})=\frac{M_{1Q_{2}}(\cdot)M_{1Q_{2}}^{*}}{{\rm
Tr}(M_{1Q_{2}}(\cdot)M_{1Q_{2}}^{*})}$. Let $Q_{1}=x\otimes x$ and
$Q_{2}=y\otimes y$ with unit vectors $x,y\in H_{2}\cong\mathbb{C}^{n}$. Note
that $x$ and $y$ are linearly independent. For any $t\in[0,1]$, define

$Q(t)=\frac{1}{\|x+t(y-x)\|^{2}}(x+t(y-x))\otimes(x+t(y-x))\in\mathcal{P}ur(H_{2}).$

Then, $Q(t)$ is continuous in $t$, $Q(0)=Q_{1}$, $Q(1)=Q_{2}$, and
$\phi_{1}(\cdot,Q(t))$ has the form (i) or (iii) for any $t$. Fix an
$A\in{\mathcal{S}}(H_{1})$ with rank$A\geq 2$ and let
$t_{0}=\max\\{t\in[0,1]:F_{A}(Q(t))=1\\}$. Then $F_{A}(Q(t_{0}))=1$ and so
$\phi_{1}(\cdot,Q(t_{0}))$ has the form (i). Thus
$\|\phi_{1}(A,Q(t_{0}))\|_{2}=1$ for any $A\in{\mathcal{S}}(H_{1})$. For any
$1\geq t>t_{0}$, $\phi_{1}(\cdot,Q(t))$ has the form (iii). Thus there exist
$\\{t_{n}\\}$, $t_{n}>t_{0}$, such that $\phi_{1}(\cdot,Q(t_{n}))$ has the
form (iii) and $t_{n}\rightarrow t_{0}$. Then by Lemma 3.1, for any given
sufficient small $\varepsilon>0$, there exist
$\\{A_{t_{n}}\\}\subseteq{\mathcal{S}}(H_{1})$ with ${\rm rank}A_{t_{n}}=2$,
such that
$\frac{1}{2}\leq\|\phi_{1}(A_{t_{n}},Q(t_{n}))\|_{2}^{2}\leq\frac{1}{2}+\varepsilon<1$.
The reason of the existence of $\\{A_{t_{n}}\\}$ for each $n$ is that, we can
find rank-2 operator $B_{t_{n}}\in$ran$\phi_{1}(\cdot,Q_{t_{n}})$ such that
$\frac{1}{2}\leq\|B_{t_{n}}\|_{2}^{2}\leq\frac{1}{2}+\varepsilon<1$, thus
there exists $A_{t_{n}}\in{\mathcal{S}}(H_{1})$ such that
$B_{t_{n}}=\frac{M_{t_{n}}A_{t_{n}}M_{t_{n}}^{*}}{{\rm
Tr}(M_{t_{n}}A_{t_{n}}M_{t_{n}}^{*})}$. As $M_{t_{n}}$ is injective, we see
that rank$A_{t_{n}}=2$. Now, since ${\mathcal{S}}(H_{1})$ is a compact set,
$\\{A_{t_{n}}\\}\subseteq{\mathcal{S}}(H_{1})$ has a convergent subsequence
$\\{A_{t_{n_{i}}}\\}\subseteq\\{A_{t_{n}}\\}$, say $A_{t_{n_{i}}}\rightarrow
A_{0}$ as $t_{n_{i}}\rightarrow t_{0}$. Then the continuity of $\psi$ entails
that
$\|\phi_{1}(A_{t_{n_{i}}},Q(t_{n_{i}}))\|_{2}^{2}\rightarrow\|\phi_{1}(A_{0},Q(t_{0}))\|_{2}^{2}$.
But this is a contradiction because $\|\phi_{1}(A_{0},Q(t_{0}))\|_{2}^{2}=1$
and
$\|\phi_{1}(A_{t_{n_{i}}},Q(t_{n_{i}}))\|_{2}^{2}\leq\frac{1}{2}+\varepsilon<1$.
Thus either $\phi_{1}(\cdot,Q)$ has the form (i) for all
$Q\in\mathcal{P}ur(H_{2})$ or $\phi_{1}(\cdot,Q)$ has the form (iii) for all
$Q\in\mathcal{P}ur(H_{2})$.

Similarly, we have

Claim 1′. Either $\phi_{2}(\cdot,Q)$ has the form (i) for all
$Q\in\mathcal{P}ur(H_{2})$ or $\phi_{2}(\cdot,Q)$ has the form (iii) for all
$Q\in\mathcal{P}ur(H_{2})$.

Claim 2. One of the following holds:

(a) For all $Q\in\mathcal{P}ur(H_{2})$, both $\phi_{1}(\cdot,Q)$ and
$\phi_{2}(\cdot,Q)$ have the form (i).

(b) For all $Q\in\mathcal{P}ur(H_{2})$, $\phi_{1}(\cdot,Q)$ has the form (i)
and $\phi_{2}(\cdot,Q)$ has the form (iii).

(c) For all $Q\in\mathcal{P}ur(H_{2})$, $\phi_{1}(\cdot,Q)$ has the form (iii)
and $\phi_{2}(\cdot,Q)$ has the form (i).

We need only to check that, for all $Q\in\mathcal{P}ur(H_{2})$,
$\phi_{1}(\cdot,Q)$ and $\phi_{2}(\cdot,Q)$ can not have the form (iii)
simultaneously. Suppose, on the contrary, there exists
$Q_{0}\in\mathcal{P}ur(H_{2})$ such that both $\phi_{1}(\cdot,Q_{0})$ and
$\phi_{2}(\cdot,Q_{0})$ are of the form (iii). Then there exist injective
linear or conjugate linear operators $M_{1Q_{0}}:H_{1}\rightarrow K_{1}$ and
$M_{2Q_{0}}:H_{1}\rightarrow K_{2}$ such that
$\phi_{i}(A,Q_{0})=\frac{M_{iQ_{0}}AM_{iQ_{0}}^{*}}{{\rm
Tr}(M_{iQ_{0}}AM_{iQ_{0}}^{*})}$ for all $A\in{\mathcal{S}}(H_{1})$, $i=1,2$.
Thus, we must have $\dim H_{1}\leq\min\\{\dim K_{1},\dim K_{2}\\}$ and

$\begin{array}[]{rl}\psi(P\otimes
Q_{0})=&\phi_{1}(P,Q_{0})\otimes\phi_{2}(P,Q_{0})=(\frac{M_{1Q_{0}}PM_{1Q_{0}}^{*}}{{\rm
Tr}(M_{1Q_{0}}PM_{1Q_{0}}^{*})})\otimes(\frac{M_{2Q_{0}}PM_{2Q_{0}}^{*}}{{\rm
Tr}(M_{2Q_{0}}PM_{2Q_{0}}^{*})})\\\ =&\frac{(M_{1Q_{0}}\otimes
M_{2Q_{0}})(P\otimes P)(M_{1Q_{0}}\otimes M_{2Q_{0}})^{*}}{{\rm
Tr}((M_{1Q_{0}}\otimes M_{2Q_{0}})(P\otimes P)(M_{1Q_{0}}\otimes
M_{2Q_{0}})^{*})}\end{array}$

for all $P\in\mathcal{P}ur(H_{1})$. Particularly, take $P_{1}=e_{1}\otimes
e_{1}$, $P_{2}=e_{2}\otimes e_{2}$, $P_{3}=\frac{1}{2}(e_{1}\otimes
e_{1}+e_{1}\otimes e_{2}+e_{2}\otimes e_{1}+e_{2}\otimes e_{2})$ and
$P_{4}=\frac{1}{2}(e_{1}\otimes e_{1}-e_{1}\otimes e_{2}-e_{2}\otimes
e_{1}+e_{2}\otimes e_{2})$, where $e_{1},e_{2}\in H_{1}$ are unit vectors with
$e_{1}\perp e_{2}$. Then $P_{1}+P_{2}=P_{3}+P_{4}$ and so $P_{1}\otimes
Q_{0}+P_{2}\otimes Q_{0}=P_{3}\otimes Q_{0}+P_{4}\otimes Q_{0}$. As $\psi$ is
strict convex combination preserving, there exist $s_{1}\in(0,1)$ and
$s_{2}\in(0,1)$, such that

$\begin{array}[]{rl}&\psi(\frac{1}{2}P_{1}\otimes
Q_{0}+\frac{1}{2}P_{2}\otimes Q_{0})=s_{1}\psi(P_{1}\otimes
Q_{0})+(1-s_{1})\psi(P_{2}\otimes Q_{0})\\\ =&s_{1}\frac{(M_{1Q_{0}}\otimes
M_{2Q_{0}})(P_{1}\otimes P_{1})(M_{1Q_{0}}\otimes M_{2Q_{0}})^{*}}{{\rm
Tr}((M_{1Q_{0}}\otimes M_{2Q_{0}})(P_{1}\otimes P_{1})(M_{1Q_{0}}\otimes
M_{2Q_{0}})^{*})}+(1-s_{1})\frac{(M_{1Q_{0}}\otimes M_{2Q_{0}})(P_{2}\otimes
P_{2})(M_{1Q_{0}}\otimes M_{2Q_{0}})^{*}}{{\rm Tr}((M_{1Q_{0}}\otimes
M_{2Q_{0}})(P_{2}\otimes P_{2})(M_{1Q_{0}}\otimes
M_{2Q_{0}})^{*})}\end{array}$

and

$\begin{array}[]{rl}&\psi(\frac{1}{2}P_{3}\otimes
Q_{0}+\frac{1}{2}P_{4}\otimes Q_{0})=s_{2}\psi(P_{3}\otimes
Q_{0})+(1-s_{2})\psi(P_{4}\otimes Q_{0})\\\ =&s_{2}\frac{(M_{1Q_{0}}\otimes
M_{2Q_{0}})(P_{3}\otimes P_{3})(M_{1Q_{0}}\otimes M_{2Q_{0}})^{*}}{{\rm
Tr}((M_{1Q_{0}}\otimes M_{2Q_{0}})(P_{3}\otimes P_{3})(M_{1Q_{0}}\otimes
M_{2Q_{0}})^{*})}+(1-s_{2})\frac{(M_{1Q_{0}}\otimes M_{2Q_{0}})(P_{4}\otimes
P_{4})(M_{1Q_{0}}\otimes M_{2Q_{0}})^{*}}{{\rm Tr}((M_{1Q_{0}}\otimes
M_{2Q_{0}})(P_{4}\otimes P_{4})(M_{1Q_{0}}\otimes
M_{2Q_{0}})^{*})}.\end{array}$

It follows that there exist $a_{i}\neq 0$, $i=1,2,3,4$, such that

$a_{1}P_{1}\otimes P_{1}+a_{2}P_{2}\otimes P_{2}=a_{3}P_{3}\otimes
P_{3}+a_{4}P_{4}\otimes P_{4}.$

As $P_{4}=P_{1}+P_{2}-P_{3}$, then $0=a_{1}P_{1}\otimes
P_{1}+a_{2}P_{2}\otimes P_{2}-a_{3}P_{3}\otimes P_{3}-a_{4}P_{4}\otimes
P_{4}=P_{1}\otimes a_{1}P_{1}+P_{2}\otimes a_{2}P_{2}-P_{3}\otimes
a_{3}P_{3}-P_{1}\otimes a_{4}P_{4}-P_{2}\otimes a_{4}P_{4}+P_{3}\otimes
a_{4}P_{4}=P_{1}\otimes(a_{1}P_{1}-a_{4}P_{4})+P_{2}\otimes(a_{2}P_{2}-a_{4}P_{4})-P_{3}\otimes(a_{3}P_{3}-a_{4}P_{4})$.
As $P_{1}$, $P_{2}$ and $P_{3}$ are linearly independent, by [7], we have
$a_{1}P_{1}-a_{4}P_{4}=0$, $a_{2}P_{2}-a_{4}P_{4}=0$ and
$a_{3}P_{3}-a_{4}P_{4}=0$, which is a contradiction. So the claim is true.

Similarly, one can check that

Claim 3. One of the following holds:

(a′) For all $P\in\mathcal{P}ur(H_{1})$, both $\phi_{1}(P,\cdot)$ and
$\phi_{2}(P,\cdot)$ have the form (i).

(b′) For all $P\in\mathcal{P}ur(H_{1})$, $\phi_{1}(P,\cdot)$ has the form (i)
and $\phi_{2}(P,\cdot)$ has the form (iii).

(c′) For all $P\in\mathcal{P}ur(H_{1})$, $\phi_{1}(P,\cdot)$ has the form
(iii) and $\phi_{2}(P,\cdot)$ has the form (i).

Claim 4. (a) and (a′) can not hold simultaneously.

In fact, if (a) and (a′) hold, that is, for all $Q\in\mathcal{P}ur(H_{2})$, we
have $\phi_{i}(A,Q)=R_{iQ}$, and, for all $P\in\mathcal{P}ur(H_{1})$, we have
$\phi_{i}(P,B)=R_{iP}$. Fix $P_{0}\in\mathcal{P}ur(H_{1})$ and
$Q_{0}\in\mathcal{P}ur(H_{2})$. Then we get

$\phi_{i}(P,Q)=\phi_{i}(P,Q_{0})=\phi_{i}(P_{0},Q_{0})=R_{i}.$

Therefore, $\psi(P\otimes Q)=\phi_{1}(P,Q)\otimes\phi_{2}(P,Q)=R_{1}\otimes
R_{2}$ for all $P\otimes Q\in\mathcal{P}ur(H_{1})\otimes\mathcal{P}ur(H_{2})$.
We know that for any $A\otimes
B\in{\mathcal{S}}(H_{1})\otimes{\mathcal{S}}(H_{2})$, there exist
$\\{A_{i}\\}_{i=1}^{m}\subseteq\mathcal{P}ur(H_{1})$,
$\\{B_{j}\\}_{j=1}^{n}\subseteq\mathcal{P}ur(H_{2})$,
$\\{a_{i}\\}_{i=1}^{m}\subseteq[0,1]$, $\\{b_{j}\\}_{j=1}^{n}\subseteq[0,1]$
satisfying $\sum_{i=1}^{m}a_{i}=1$, $\sum_{j=1}^{n}b_{j}=1$, such that
$A=\sum_{i=1}^{m}a_{i}A_{i}$, $B=\sum_{j=1}^{n}b_{j}B_{j}$. Then, there exist
$c_{ij}\geq 0$ with $\sum_{i,j}c_{ij}=1$ such that $\psi(A\otimes
B)=\sum_{i,j}c_{ij}\psi(A_{i}\otimes B_{j})=R_{1}\otimes R_{2}$, this
contradicts to the assumption that the range of $\psi$ is non-collinear. So
the Claim 4 is true.

Claim 5. If (a) and (b′) hold, then $\psi$ has the form (3), that is, there
exist $R_{1}\in\mathcal{P}ur(K_{1})$ and injective linear or conjugate linear
operator $M_{2}:H_{2}\rightarrow K_{2}$ such that

$\psi(A\otimes B)=R_{1}\otimes\frac{M_{2}BM_{2}^{*}}{{\rm
Tr}(M_{2}BM_{2}^{*})}$

for all $A\in{\mathcal{S}}(H_{1})$ and $B\in{\mathcal{S}}(H_{2})$. It is clear
that $\dim H_{2}\leq\dim K_{2}$.

In this case, for any $P\otimes
Q\in\mathcal{P}ur(H_{1})\otimes\mathcal{P}ur(H_{2})$ we have

$\psi(P\otimes Q)=\phi_{1}(P,Q)\otimes\phi_{2}(P,Q)=R_{1Q}\otimes
R_{2Q}=R_{1P}\otimes\frac{M_{2P}QM_{2P}^{*}}{{\rm Tr}(M_{2P}QM_{2P}^{*})},$

which implies that $R_{1Q}=R_{1P}$ is independent of $P,Q$, and
$\frac{M_{2P}QM_{2P}^{*}}{{\rm Tr}(M_{2P}QM_{2P}^{*})}=R_{2Q}$. Thus for any
fixed $Q\in\mathcal{P}ur(H_{2})$, $\frac{M_{2P_{1}}QM_{2P_{1}}^{*}}{{\rm
Tr}(M_{2P_{1}}QM_{2P_{1}}^{*})}=\frac{M_{2P_{2}}QM_{2P_{2}}^{*}}{{\rm
Tr}(M_{2P_{2}}QM_{2P_{2}}^{*})}=R_{2Q}$ holds for any distinct
$P_{1}\in\mathcal{P}ur(H_{1})$ and $P_{2}\in\mathcal{P}ur(H_{1})$. Thus for
all $Q\in\mathcal{P}ur(H_{2})$, we have $\frac{M_{2P_{1}}QM_{2P_{1}}^{*}}{{\rm
Tr}(M_{2P_{1}}QM_{2P_{1}}^{*})}=\frac{M_{2P_{2}}QM_{2P_{2}}^{*}}{{\rm
Tr}(M_{2P_{2}}QM_{2P_{2}}^{*})}$, that is, $\frac{M_{2P}QM_{2P}^{*}}{{\rm
Tr}(M_{2P}QM_{2P}^{*})}$ is independent of $P$. So there exist
$R_{1}\in\mathcal{P}ur(K_{1})$ and injective linear or conjugate linear
operator $M_{2}:H_{2}\rightarrow K_{2}$ such that $\psi(P\otimes
Q)=R_{1}\otimes\frac{M_{2}QM_{2}^{*}}{{\rm Tr}(M_{2}QM_{2}^{*})}$ for all
separable pure states $P\otimes Q$.

Now, for any $B\in{\mathcal{S}}(H_{2})$ and $P\in{\mathcal{P}ur}(H_{1})$,
writing $B=\sum_{j=1}^{n}b_{j}B_{j}$ as in the proof of Claim 4, we have
$\psi(P\otimes
B)=\psi(P\otimes\sum_{j=1}^{n}b_{j}B_{j})=\sum_{j=1}^{n}b^{\prime}_{j}\psi(P\otimes
B_{j})=\sum_{j=1}^{n}b^{\prime}_{j}(R_{1}\otimes\frac{M_{2}B_{j}M_{2}^{*}}{{\rm
Tr}(M_{2}B_{j}M_{2}^{*})})=R_{1}\otimes\sum_{j=1}^{n}b^{\prime}_{j}\frac{M_{2}B_{j}M_{2}^{*}}{{\rm
Tr}(M_{2}B_{j}M_{2}^{*})}$ is a product state. But we already know that
$\phi_{2}(P,\cdot)=\frac{M_{2}(\cdot)M_{2}^{*}}{{\rm
Tr}(M_{2}(\cdot)M_{2}^{*})}$, so ${\rm Tr}_{1}(\psi(P\otimes
B))=\phi_{2}(P,B)=\frac{M_{2}BM_{2}^{*}}{{\rm Tr}(M_{2}BM_{2}^{*})}$. Thus
$\psi(P\otimes B)=R_{1}\otimes\frac{M_{2}BM_{2}^{*}}{{\rm
Tr}(M_{2}BM_{2}^{*})}$. Then, for any $A\in{\mathcal{S}}(H_{1})$, writing
$A=\sum_{i=1}^{m}a_{i}A_{i}$ as in the proof of Claim 4, we obtain
$\psi(A\otimes B)=\psi(\sum_{i=1}^{m}a_{i}A_{i}\otimes
B)=\sum_{i=1}^{m}a^{\prime}_{i}\psi(A_{i}\otimes
B)=\sum_{i=1}^{m}a^{\prime}_{i}(R_{1}\otimes\frac{M_{2}BM_{2}^{*}}{{\rm
Tr}(M_{2}BM_{2}^{*})})=R_{1}\otimes(\sum_{i=1}^{m}a^{\prime}_{i}\frac{M_{2}BM_{2}^{*}}{{\rm
Tr}(M_{2}BM_{2}^{*})}))$, which is a product states. Therefore, we must have
$\psi(A\otimes B)=R_{1}\otimes\frac{M_{2}BM_{2}^{*}}{{\rm
Tr}(M_{2}BM_{2}^{*})}$, where $M_{2}$ is linear or conjugate linear. In the
case that $M_{2}$ is a conjugate linear operator, it is well known that, there
exists a linear operator $N_{2}$ such that $\frac{M_{2}BM_{2}^{*}}{{\rm
Tr}(M_{2}BM_{2}^{*})}=\frac{N_{2}B^{t}N_{2}^{*}}{{\rm
Tr}(N_{2}B^{t}N_{2}^{*})}$ for all $B$, where the transpose is taken with
respect to an arbitrarily fixed orthonormal basis of $H_{2}$. So the claim is
true.

Similarly, one can show the following Claims 6-8.

Claim 6. If (a) and (c′) hold, then (4) holds.

Claim 7. If (b) and (a′) hold, then $\psi$ has the form (5).

Claim 8. If (c) and (a′) hold, then $\psi$ takes the form (2).

Claim 9. If (b) and (c′) hold, then $\psi$ has the form (7).

Suppose that (b) and (c′) hold; then
$\phi_{1}(P,Q)=\phi_{1}(P_{0},Q)=\frac{M_{1P_{0}}QM_{1P_{0}}^{*}}{{\rm
Tr}(M_{1P_{0}}QM_{1P_{0}}^{*})}$ and
$\phi_{2}(P,Q)=\phi_{2}(P,Q_{0})=\frac{M_{2Q_{0}}PM_{2Q_{0}}^{*}}{{\rm
Tr}(M_{2Q_{0}}PM_{2Q_{0}}^{*})}$. Thus, we obtain

$\psi(P\otimes
Q)=\phi_{1}(P,Q)\otimes\phi_{2}(P,Q)=\frac{M_{1P_{0}}QM_{1P_{0}}^{*}}{{\rm
Tr}(M_{1P_{0}}QM_{1P_{0}}^{*})}\otimes\frac{M_{2Q_{0}}PM_{2Q_{0}}^{*}}{{\rm
Tr}(M_{2Q_{0}}PM_{2Q_{0}}^{*})}$

for all $P\in\mathcal{P}ur(H_{1})$ and $Q\in\mathcal{P}ur(H_{2})$, where
$M_{1P_{0}}:H_{2}\rightarrow K_{1}$ and $M_{2Q_{0}}:H_{1}\rightarrow K_{2}$
are injective linear or conjugate linear operators. It follows that $\dim
H_{1}\leq\dim K_{2}$ and $\dim H_{2}\leq\dim K_{1}$. Let $M_{1}=M_{1P_{0}}$
and $M_{2}=M_{2Q_{0}}$. Then $\psi(P\otimes Q)=\frac{M_{1}QM_{1}^{*}}{{\rm
Tr}(M_{1}QM_{1}^{*})}\otimes\frac{M_{2}PM_{2}^{*}}{{\rm Tr}(M_{2}PM_{2}^{*})}$
for all separable pure states $P\otimes Q$. For any $A\otimes
B\in{\mathcal{S}}(H_{1}\otimes H_{2})$, write $A=\sum_{i=1}^{m}a_{i}A_{i}$ and
$B=\sum_{j=1}^{n}b_{j}B_{j}$ by the spectral theorem, where $a_{i},b_{j}\geq
0$ with $\sum_{i=1}^{m}a_{i}=1,\sum_{j=1}^{n}b_{j}=1$ and $A_{i}$s, $B_{j}$s
pure states. Thus, for any $Q\in{\mathcal{P}ur}(H_{2})$, $\psi(A\otimes
Q)=\psi(\sum_{i=1}^{m}a_{i}A_{i}\otimes
Q)=\sum_{i=1}^{m}a^{\prime}_{i}\psi(A_{i}\otimes
Q)=\sum_{i=1}^{m}a^{\prime}_{i}(\frac{M_{1}QM_{1}^{*}}{{\rm
Tr}(M_{1}QM_{1}^{*})}\otimes\frac{M_{2}A_{i}M_{2}^{*}}{{\rm
Tr}(M_{2}A_{i}M_{2}^{*})})=\frac{M_{1}QM_{1}^{*}}{{\rm
Tr}(M_{1}QM_{1}^{*})}\otimes(\sum_{i=1}^{m}a^{\prime}_{i}\frac{M_{2}A_{i}M_{2}^{*}}{{\rm
Tr}(M_{2}A_{i}M_{2}^{*})})$ is a product state. As
$\phi_{2}(\cdot,Q)=\frac{M_{2}(\cdot)M_{2}^{*}}{{\rm
Tr}(M_{2}(\cdot)M_{2}^{*})}$, we obtain that ${\rm Tr}_{1}(\psi(A\otimes
Q))=\phi_{2}(A,Q)=\frac{M_{2}AM_{2}^{*}}{{\rm Tr}(M_{2}AM_{2}^{*})}$. So we
must have $\psi(A\otimes Q)=\frac{M_{1}QM_{1}^{*}}{{\rm
Tr}(M_{1}QM_{1}^{*})}\otimes\frac{M_{2}AM_{2}^{*}}{{\rm
Tr}(M_{2}AM_{2}^{*})}$. It follows that $\psi(A\otimes
B)=\psi(A\otimes\sum_{j=1}^{n}b_{j}B_{j})=\sum_{j=1}^{n}b^{\prime}_{j}\psi(A\otimes
B_{j})=\sum_{j=1}^{n}b^{\prime}_{j}(\frac{M_{1}B_{j}M_{1}^{*}}{{\rm
Tr}(M_{1}B_{j}M_{1}^{*})}\otimes\frac{M_{2}AM_{2}^{*}}{{\rm
Tr}(M_{2}AM_{2}^{*})})=(\sum_{j=1}^{n}b^{\prime}_{j}\frac{M_{1}B_{j}M_{1}^{*}}{{\rm
Tr}(M_{1}B_{j}M_{1}^{*})})\otimes\frac{M_{2}AM_{2}^{*}}{{\rm
Tr}(M_{2}AM_{2}^{*})}$. On the other hand, we also have $\psi(P\otimes
B)=\psi(P\otimes\sum_{j=1}^{n}b_{j}B_{j})=\sum_{j=1}^{n}b^{\prime}_{j}\psi(P\otimes
B_{j})=\sum_{j=1}^{n}b^{\prime}_{j}(\frac{M_{1}B_{j}M_{1}^{*}}{{\rm
Tr}(M_{1}B_{j}M_{1}^{*})}\otimes\frac{M_{2}PM_{2}^{*}}{{\rm
Tr}(M_{2}PM_{2}^{*})})=(\sum_{j=1}^{n}b^{\prime}_{j}\frac{M_{1}B_{j}M_{1}^{*}}{{\rm
Tr}(M_{1}B_{j}M_{1}^{*})})\otimes\frac{M_{2}PM_{2}^{*}}{{\rm
Tr}(M_{2}PM_{2}^{*})}$. As
$\phi_{1}(P,\cdot)=\frac{M_{1}(\cdot)M_{1}^{*}}{{\rm
Tr}(M_{1}(\cdot)M_{1}^{*})}$, then ${\rm Tr}_{2}(\psi(P\otimes
B))=\phi_{1}(P,B)=\frac{M_{1}BM_{1}^{*}}{{\rm Tr}(M_{1}BM_{1}^{*})}$. Thus we
get $\psi(P\otimes B)=\frac{M_{1}BM_{1}^{*}}{{\rm
Tr}(M_{1}BM_{1}^{*})}\otimes\frac{M_{2}PM_{2}^{*}}{{\rm Tr}(M_{2}PM_{2}^{*})}$
and then $\psi(A\otimes B)=\psi(\sum_{i=1}^{m}a_{i}A_{i}\otimes
B)=\sum_{i=1}^{m}a_{i}\psi(A_{i}\otimes
B)=\sum_{i=1}^{m}a^{\prime}_{i}(\frac{M_{1}BM_{1}^{*}}{{\rm
Tr}(M_{1}BM_{1}^{*})}\otimes\frac{M_{2}A_{i}M_{2}^{*}}{{\rm
Tr}(M_{2}A_{i}M_{2}^{*})})=\frac{M_{1}BM_{1}^{*}}{{\rm
Tr}(M_{1}BM_{1}^{*})}\otimes(\sum_{i=1}^{m}a^{\prime}_{i}\frac{M_{2}A_{i}M_{2}^{*}}{{\rm
Tr}(M_{2}A_{i}M_{2}^{*})})$. Now it is clear that $\psi(A\otimes
B)=\frac{M_{1}BM_{1}^{*}}{{\rm
Tr}(M_{1}BM_{1}^{*})}\otimes\frac{M_{2}AM_{2}^{*}}{{\rm Tr}(M_{2}AM_{2}^{*})}$
for any $A\in{\mathcal{S}}(H_{1})$ and $B\in{\mathcal{S}}(H_{2})$. Hence the
claim is true.

Similarly, we have

Claim 10. If (c) and (b′) hold, then $\psi$ has the form (6).

Claim 11. If (b) and (b′) hold, then $\psi$ has the form (9).

Assume (b) and (b′) hold synchronously. Then for any $P\otimes
Q\in\mathcal{P}ur(H_{1})\otimes\mathcal{P}ur(H_{2})$ we have $\psi(P\otimes
Q)=\phi_{1}(P,Q)\otimes\phi_{2}(P,Q)=R_{1Q}\otimes\frac{M_{2Q}PM_{2Q}^{*}}{{\rm
Tr}(M_{2Q}PM_{2Q}^{*})}=R_{1P}\otimes\frac{M_{2P}QM_{2P}^{*}}{{\rm
Tr}(M_{2P}QM_{2P}^{*})}$. It follows that there exist
$R_{1}\in\mathcal{P}ur(K_{1})$ such that $R_{1Q}=R_{1P}=R_{1}$ and
$\frac{M_{2Q}PM_{2Q}^{*}}{{\rm
Tr}(M_{2Q}PM_{2Q}^{*})}=\frac{M_{2P}QM_{2P}^{*}}{{\rm Tr}(M_{2P}QM_{2P}^{*})}$
for all $P$, $Q$. Thus there exists a strict convex combination preserving map
$\varphi_{2}:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})\rightarrow{\mathcal{S}}(K_{2})$ such that, for each $P\otimes
Q\in{\mathcal{P}ur}(H_{1})\otimes{\mathcal{P}ur}(H_{2})$,
$\varphi_{2}(P\otimes Q)=\frac{M_{P}QM_{P}^{*}}{{\rm
Tr}(M_{P}QM_{P}^{*})}=\frac{N_{Q}PN_{Q}^{*}}{{\rm Tr}(N_{Q}PN_{Q}^{*})}$ for
some injective, may not synchronously, linear or conjugate linear operators
$M_{P}:H_{2}\rightarrow K_{2}$, $N_{Q}:H_{1}\rightarrow K_{2}$, and

$\psi(\rho)=R_{1}\otimes\varphi_{2}(\rho)$

for all $\rho\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes H_{2})$. In this case
$\max\\{\dim H_{1},\dim H_{2}\\}\leq\dim K_{2}$. So $\psi$ has the form (9)
and Claim 11 is true.

Similarly,

Claim 12. If (c) and (c′) hold, then $\psi$ takes the form (8).

Combining the claims 4-12, we complete the proof of Theorem 3.2. $\Box$

By Theorem 3.2, the following corollary is immediate, which gives an
affirmative answer to a conjecture in [8] without the injectivity assumption.

Corollary 3.3 Let $\psi:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})\rightarrow{\mathcal{S}}_{\rm sep}(K_{1}\otimes K_{2})$ be a map with
$2\leq\dim H_{i}<\infty$, $i=1,2$, and ${\rm ran}\psi$ non-collinear or a
singleton (i.e., contains only one element). If $\psi$ preserves separable
pure states and strict convex combinations, then it sends product states to
product states.

Now we give a characterization of injective local quantum measurements, which
reveals that, in almost all situations the maps preserving separable pure
states and strict convex combinations are essentially the injective local
quantum measurements.

Corollary 3.4 Let $H_{1},H_{2},K_{1},K_{2}$ be Hilbert spaces with $2\leq\dim
H_{i}<\infty$, $i=1,2$ and let $\psi:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})\rightarrow{\mathcal{S}}_{\rm sep}(K_{1}\otimes K_{2})$ be a map. Then
the following statements are equivalent.

(1) $\psi$ is strict convex combination preserving,
$\psi({\mathcal{P}ur}(H_{1})\otimes{\mathcal{P}ur}(H_{2}))\subseteq{\mathcal{P}ur}(K_{1})\otimes{\mathcal{P}ur}(K_{2})$
and the range of $\psi$ is non-collinear containing a state $\sigma$ so that
both reductions ${\rm Tr}_{1}(\sigma)$ and ${\rm Tr}_{2}(\sigma)$ have rank
$\geq 2$.

(2) $\psi$ is open line segment preserving,
$\psi({\mathcal{P}ur}(H_{1})\otimes{\mathcal{P}ur}(H_{2}))\subseteq{\mathcal{P}ur}(K_{1})\otimes{\mathcal{P}ur}(K_{2})$
and the range of $\psi$ is non-collinear containing a state $\sigma$ so that
both reductions ${\rm Tr}_{1}(\sigma)$ and ${\rm Tr}_{2}(\sigma)$ have rank
$\geq 2$.

(3) Either

(1∘) there exist injective operators $M_{1}\in\mathcal{B}(H_{1},K_{1})$ and
$M_{2}\in\mathcal{B}(H_{2},K_{2})$ such that

$\psi(\rho)=\frac{(M_{1}\otimes M_{2})\Phi(\rho)(M_{1}\otimes M_{2})^{*}}{{\rm
Tr}((M_{1}\otimes M_{2})\Phi(\rho)(M_{1}\otimes M_{2})^{*})}$

for all $\rho\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes H_{2})$;
or

(2∘) there exist injective operators $M_{1}\in\mathcal{B}(H_{2},K_{1})$ and
$M_{2}\in\mathcal{B}(H_{1},K_{2})$ such that

$\psi(\rho)=\frac{(M_{1}\otimes M_{2})\Phi(\Theta(\rho))(M_{1}\otimes
M_{2})^{*}}{{\rm Tr}((M_{1}\otimes M_{2})\Phi(\Theta(\rho))(M_{1}\otimes
M_{2})^{*})}$

for all $\rho\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes H_{2})$.
Here $\Phi$ is the identity, or the transpose, or the partial transpose of the
first system or the partial transpose of the second system with respect to an
arbitrarily fixed product basis, $\Theta$ is the swap.

Proof. (1)$\Leftrightarrow$(2)$\Leftarrow$(3) is obvious, we need to check the
(1)$\Rightarrow$(3).

Assume (1). It is clear that $\psi$ has one of the forms Theorem 3.2.(1)-(10)
as $\psi$ satisfies all the conditions of Theorem 3.2. Furthermore, the
assumption that there exists $\sigma$ in ${\rm ran}\psi$ so that rank${\rm
Tr}_{i}(\sigma)\geq 2$, $i=1,2$ forces that $\psi$ can only take the form (6)
or (7), that is, $\psi(A\otimes B)=\frac{M_{1}\Psi_{1}(A)M_{1}^{*}}{{\rm
Tr}(M_{1}\Psi_{1}(A)M_{1}^{*})}\otimes\frac{M_{2}\Psi_{2}(B)M_{2}^{*}}{{\rm
Tr}(M_{2}\Psi_{2}(B)M_{2}^{*})}$ for all $A\otimes B\in{\mathcal{S}}_{\rm
sep}(H_{1}\otimes H_{2})$, or $\psi(A\otimes
B)=\frac{M_{1}\Psi_{2}(B)M_{1}^{*}}{{\rm
Tr}(M_{1}\Psi_{2}(B)M_{1}^{*})}\otimes\frac{M_{2}\Psi_{1}(A)M_{2}^{*}}{{\rm
Tr}(M_{2}\Psi_{1}(A)M_{2}^{*})}$ for all $A\otimes B\in{\mathcal{S}}_{\rm
sep}(H_{1}\otimes H_{2})$, where $\Psi_{i}$ is the identity or the transpose
with respect to an arbitrarily fixed orthonormal basis, $i=1,2$. Thus either

(i) $\psi(A\otimes B)=\frac{(M_{1}\otimes M_{2})\Phi(A\otimes B)(M_{1}\otimes
M_{2})^{*}}{{\rm Tr}(M_{1}\otimes M_{2})\Phi(A\otimes B)(M_{1}\otimes
M_{2})^{*})}$ for all $A\otimes B\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})$; or

(ii) $\psi(A\otimes B)=\frac{(M_{1}\otimes M_{2})\Phi(\Theta(A\otimes
B))(M_{1}\otimes M_{2})^{*}}{{\rm Tr}(M_{1}\otimes M_{2})\Phi(\Theta(A\otimes
B))(M_{1}\otimes M_{2})^{*})}$ for all $A\otimes B\in{\mathcal{S}}_{\rm
sep}(H_{1}\otimes H_{2})$,
where $\Phi$ is the identity, or the transpose, or the partial transpose of
the first system or the partial transpose of the second system with respect to
an arbitrarily fixed product basis, $\Theta$ is the swap.

Let

$\Delta(\rho)=\frac{(M_{1}^{[-1]}\otimes
M_{2}^{[-1]})\psi(\Phi(\rho))(M_{1}^{[-1]}\otimes M_{2}^{[-1]})^{*}}{{\rm
Tr}((M_{1}^{[-1]}\otimes M_{2}^{[-1]})\psi(\Phi(\rho))(M_{1}^{[-1]}\otimes
M_{2}^{[-1]})^{*})}$

if $\psi$ has form (i), and let

$\Delta(\rho)=\frac{\Theta[(M_{1}^{[-1]}\otimes
M_{2}^{[-1]})\psi(\Phi(\rho))(M_{1}^{[-1]}\otimes M_{2}^{[-1]})^{*}]}{{\rm
Tr}((M_{1}^{[-1]}\otimes M_{2}^{[-1]})\psi(\Phi(\rho))(M_{1}^{[-1]}\otimes
M_{2}^{[-1]})^{*})}$

if $\psi$ takes the form (ii). Then $\Delta:{\mathcal{S}}_{\rm
sep}(H_{1}\otimes H_{2})\rightarrow{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})$ is a bijective map preserving separable pure states and strict convex
combinations. Furthermore $\Delta(A\otimes B)=A\otimes B$ for all $A\otimes
B\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes H_{2})$. Then by [8, Lemma 2.7], we
get $\Delta(\rho)=\rho$ for any $\rho\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})$. Now it is clear that $\psi$ has the form (1∘) or (2∘), finishing the
proof. $\Box$

The following result is a generalization of the main result in [8] by omitting
the additional assumption in Eq.(1.1) for finite dimensional case, and thus,
answer affirmatively a conjecture proposed in [8] , as mentioned in the
introduction section.

Theorem 3.5 Let $H_{i},K_{i},$ be complex Hilbert spaces with $2\leq\dim
H_{i}<\infty$, $i=1,2$, and $\psi:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2})\rightarrow{\mathcal{S}}_{\rm sep}(K_{1}\otimes K_{2})$ an bijective
map. Then $\Phi$ is convex combination preserving if and only if either

(1) there exist invertible operators $S\in\mathcal{B}(H_{1},K_{1})$ and
$T\in\mathcal{B}(H_{2},K_{2})$ such that

$\psi(\rho)=\frac{(S\otimes T)\Psi(\rho)(S\otimes T)^{*}}{{\rm Tr}((S\otimes
T)\Psi(\rho)(S\otimes T)^{*})}$

for all $\rho\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes H_{2})$;
or

(2) there exist invertible operators $S\in\mathcal{B}(H_{2},K_{1})$ and
$T\in\mathcal{B}(H_{1},K_{2})$ such that

$\psi(\rho)=\frac{(S\otimes T)\Psi(\Theta(\rho))(S\otimes T)^{*}}{{\rm
Tr}((S\otimes T)\Psi(\Theta(\rho))(S\otimes T)^{*})}$

for all $\rho\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes H_{2})$.
Here $\Psi$ is the identity, or the transpose, or the partial transpose of the
first system or the partial transpose of the second system with respect to an
arbitrarily fixed product basis, $\Theta$ is the swap.

Proof. We need only check the “only if” part. By the assumption, $\psi$ is
bijective and strict convex combination preserving from ${\mathcal{S}}_{\rm
sep}(H_{1}\otimes H_{2})$ onto ${\mathcal{S}}_{\rm sep}(K_{1}\otimes K_{2})$.
Particularly, the range of $\psi$ is non-collinear. By [8], we have
$\psi({\mathcal{P}ur}_{\rm sep}(H_{1}\otimes H_{2}))={\mathcal{P}ur}_{\rm
sep}(K_{1}\otimes K_{2})$, that is, $\psi$ preserves separable pure states in
both directions.

Then by Corollary 3.4, $\psi$ has either the form (1) or the form (2) with
$T,S$ injective. Since $\psi({\mathcal{P}ur}_{\rm sep}(H_{1}\otimes
H_{2}))={\mathcal{P}ur}_{\rm sep}(K_{1}\otimes K_{2})$, it is clear that both
$T$ and $S$ are invertible. Hence the theorem is true. $\Box$

##  4\.  Maps preserving separable pure states and strict convex combinations:
Multipartite systems

The results similar to that in Section 3 for bipartite case are valid for
multipartite cases, of course, with more complicated expressions. The proofs
are also similar. In this section we only list some of them, which have
relatively simple expressions and may have more applications. The meanings of
the notations used here are also similar to that in Section 3.

Suppose $\dim H_{i}=n_{i}$. For $1\leq r_{1}<\cdots<r_{p}\leq n$, define the
partial trace which is a linear map ${\rm
Tr}^{r_{1},\cdots,r_{p}}:{\mathcal{B}}_{\rm
sa}(\otimes_{i=1}^{n}H_{i})\rightarrow{\mathcal{B}}_{\rm
sa}(\otimes_{j=1}^{p}H_{r_{j}})$ as follows:

$\otimes_{i=1}^{n}A_{i}\longmapsto(\prod_{i\neq{r_{1},\cdots,r_{p}}}{\rm
Tr}A_{i})\otimes_{j=1}^{p}A_{r_{j}}.$

In particular, the linear map ${\rm Tr}^{r}:{\mathcal{B}}_{\rm
sa}(\otimes_{i=1}^{n}H_{i})\rightarrow{\mathcal{B}}_{\rm sa}(H_{r})$ is given
by ${\rm Tr}^{r}(\otimes_{i=1}^{n}A_{i})=(\prod_{i\neq r}{\rm
Tr}(A_{i}))A_{r}$. We call ${\rm Tr}^{r}(\rho)$ the reduction state of
$\rho\in{\mathcal{S}}(H_{1}\otimes H_{2}\otimes\cdots\otimes H_{n})$ in the
subsystem ${\mathcal{S}}(H_{r})$. A multipartite state
$\rho\in{\mathcal{S}}(\otimes_{i=1}^{n}H_{i})$ is called a product state if
$\rho=\otimes_{i=1}^{n}\rho_{i}$ for some $\rho_{i}\in{\mathcal{S}}(H_{i})$.

The following result corresponds to Corollary 3.3 and Corollary 3.4.

Theorem 4.1 Let $\psi:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2}\otimes\cdots\otimes H_{n})\rightarrow{\mathcal{S}}_{\rm
sep}(K_{1}\otimes K_{2}\otimes\cdots\otimes K_{n})$ be a strict convex
combination preserving map, with $2\leq\dim H_{i}<\infty$, $i=1,2,\cdots,n$,
and
$\psi(\mathcal{P}ur(H_{1})\otimes\mathcal{P}ur(H_{2})\otimes\cdots\otimes\mathcal{P}ur(H_{n}))\subseteq\mathcal{P}ur(K_{1})\otimes\mathcal{P}ur(K_{2})\otimes\cdots\otimes\mathcal{P}ur(K_{n})$.

(1) If the range of $\psi$ is non-collinear or a singleton, then $\psi$ maps
product states to product states.

(2) If the range of $\psi$ is non-collinear and contains a state $\sigma$ so
that its reduction state ${\rm Tr}^{i}(\sigma)$ has rank $\geq 2$ for each
$i=1,2,\ldots,n$, then there exist a permutation
$\pi:(1,\cdots,n)\mapsto(\pi(1),\cdots,\pi(n))$ of $(1,\cdots,n)$ and
injective linear or conjugate linear (may not simultaneously) operators
$M_{j}:H_{\pi(j)}\rightarrow K_{j}$, $j=1,\cdots,n$, such that

$\psi(\rho)=\frac{(M_{1}\otimes\cdots\otimes
M_{n})\Theta_{\pi}(\rho)(M_{1}^{*}\otimes\cdots\otimes M_{n}^{*})}{{\rm
Tr}((M_{1}\otimes\cdots\otimes
M_{n})\Theta_{\pi}(\rho)(M_{1}^{*}\otimes\cdots\otimes M_{n}^{*}))}$

for all $\rho\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes\cdots\otimes H_{n})$.
Here $\Theta_{\pi}:{\mathcal{B}}_{\rm sa}(H_{1}\otimes
H_{2}\otimes\cdots\otimes H_{n})\rightarrow{\mathcal{B}}_{\rm
sa}(H_{\pi(1)}\otimes H_{\pi(2)}\otimes\cdots\otimes H_{\pi(n)})$ is a linear
map determined by $\Theta_{\pi}(A_{1}\otimes A_{2}\otimes\cdots\otimes
A_{n})=A_{\pi(1)}\otimes A_{\pi(2)}\otimes\cdots\otimes A_{\pi(n)}$. It is
clear that $\dim H_{\pi(j)}\leq\dim K_{j}$.

The following result is corresponding to Theorem 3.5.

Corollary 4.2 Let $\psi:{\mathcal{S}}_{\rm sep}(H_{1}\otimes
H_{2}\otimes\cdots\otimes H_{n})\rightarrow{\mathcal{S}}_{\rm
sep}(K_{1}\otimes K_{2}\otimes\cdots\otimes K_{n})$ be a bijective map with
$2\leq\dim H_{i}<\infty$, $i=1,2,\cdots,n$. Then $\psi$ is strict convex
combination preserving if and only if there exist a permutation $\pi$ of
$(1,2,\cdots,n)$ and invertible linear or conjugate linear (may not
simultaneously) operators $M_{j}:H_{\pi(j)}\rightarrow K_{j}$, $j=1,\cdots,n$,
such that

$None$ $\psi(\rho)=\frac{(M_{1}\otimes\cdots\otimes
M_{n})\Theta_{\pi}(\rho)(M_{1}^{*}\otimes\cdots\otimes M_{n}^{*})}{{\rm
Tr}((M_{1}\otimes\cdots\otimes
M_{n})\Theta_{\pi}(\rho)(M_{1}^{*}\otimes\cdots\otimes M_{n}^{*}))}$

holds for all $\rho\in{\mathcal{S}}_{\rm sep}(H_{1}\otimes\cdots\otimes
H_{n})$. Here $\Theta_{\pi}:{\mathcal{B}}_{\rm sa}(H_{1}\otimes
H_{2}\otimes\cdots\otimes H_{n})\rightarrow{\mathcal{B}}_{\rm
sa}(H_{\pi(1)}\otimes H_{\pi(2)}\otimes\cdots\otimes H_{\pi(n)})$ is the
linear map determined by $\Theta_{\pi}(A_{1}\otimes A_{2}\otimes\cdots\otimes
A_{n})=A_{\pi(1)}\otimes A_{\pi(2)}\otimes\cdots\otimes A_{\pi(n)}$. It is
clear that $\dim H_{\pi(j)}=\dim K_{j}$.

##  5\. Conclusion

The quantum measurement map preserve strict convex combinations and sends pure
states to pure states. Similarly, each local quantum measurement map preserves
separable pure states and strict convex combinations. These facts make it
interesting to study the problem of characterizing the maps between states in
(multipartite) quantum systems that are (separable) pure state preserving and
strict convex combination preserving. These problems are basic and interesting
both in quantum information science and mathematics science, and their
solutions will present a geometric characterization of (local) quantum
measurements and help us to understand better the quantum measurement.

In the present paper, we give a characterization of the maps
$\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(K)$ with $2\leq\dim H<\infty$
that preserve pure states and strict convex combinations, and give a structure
theorem of the maps $\psi:{\mathcal{S}}(H_{1}\otimes H_{2}\otimes\cdots\otimes
H_{n})\rightarrow{\mathcal{S}}(K_{1}\otimes K_{2}\otimes\cdots\otimes K_{n})$
with $2\leq\dim H_{i}<\infty$ that preserve separable pure states and strict
convex combinations. From these results we get a characterization of injective
(local) quantum measurements. In almost all situations, for example, in the
case that the range of $\psi$ is non-collinear or a singleton, $\psi$ sends
product states to product states. In particular, if the range of $\psi$ is
non-collinear and contains an element with each reduction having rank $\geq
2$, then $\phi$ is essentially an injective local quantum measurement. Thus we
answer affirmatively two conjectures proposed in [8].

Finally we remark that, the main results obtained in [5, 8] hold for both
finite dimensional systems and infinite dimensional systems. However, in the
present paper we only deal with finite dimensional systems. We conjecture that
the results in this paper are also valid for infinite dimensional case but new
tools are needed to prove this.

Acknowledgement. The authors wish to give their thanks to the referees for
helpful comments and suggestions to improve the expression of this paper.

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