#  A geometric characterization of
invertible quantum measurement maps

Kan He  Faculty of Mathematics, Institute of Mathematics, Taiyuan University
of Technology, Taiyuan, 030024, P. R. China<EMAIL_ADDRESS>,  Jin-
Chuan Hou  Faculty of Mathematics, Institute of Mathematics, Taiyuan
University of Technology, Taiyuan, 030024, P. R. China
<EMAIL_ADDRESS>and  Chi-Kwong Li  Department of Mathematics, College
of William Mary, Williamsburg, VA 23187-8795, USA; Faculty of Mathematics,
Institute of Mathematics, Taiyuan University of Technology, Taiyuan, 030024,
P. R. China<EMAIL_ADDRESS>
###### Abstract.

A geometric characterization is given for invertible quantum measurement maps.
Denote by ${\mathcal{S}}(H)$ the convex set of all states (i.e., trace-1
positive operators) on Hilbert space $H$ with dim$H\leq\infty$, and
$[\rho_{1},\rho_{2}]$ the line segment joining two elements
$\rho_{1},\rho_{2}$ in ${\mathcal{S}}(H)$. It is shown that a bijective map
$\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ satisfies
$\phi([\rho_{1},\rho_{2}])\subseteq[\phi(\rho_{1}),\phi(\rho_{2})]$ for any
$\rho_{1},\rho_{2}\in{\mathcal{S}}$ if and only if $\phi$ has one of the
following forms

$\rho\mapsto\frac{M\rho M^{*}}{{\rm tr}(M\rho M^{*})}\quad\hbox{ or
}\quad\rho\mapsto\frac{M\rho^{T}M^{*}}{{\rm tr}(M\rho^{T}M^{*})},$

where $M$ is an invertible bounded linear operator and $\rho^{T}$ is the
transpose of $\rho$ with respect to an arbitrarily fixed orthonormal basis.

2002 Mathematical Subject Classification. 47B49, 47L07, 47N50

Key words and phrases. Quantum states, Quantum measurement, Segment preserving
maps

This work is partially supported by National Natural Science Foundation of
China (11171249, 11271217, 11201329), a grant to International Cooperating
Research from Shanxi (2011081039). Li was also supported by a USA NSF grant
and a HK RCG grant.

##  1\. Introduction and the main result

In the mathematical framework of the theory of quantum information, a state is
a positive operator of trace 1 acting on a complex Hilbert space $H$. Denote
by ${\mathcal{S}}(H)$ the set of all states on $H$, that is, of all positive
operators with trace 1. It is clear that ${\mathcal{S}}(H)$ is a closed convex
subset of ${\mathcal{T}}(H)$, the Banach space of all trace-class operators on
$H$ endowed with the trace-norm $\|\cdot\|_{\rm Tr}$. In quantum information
science and quantum computing, 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 entanglement
witnesses, which are special types of positive maps; see [11]. In this
connection, it is helpful to know the characterizations of maps leaving
invariant some important subsets or quantum properties. Such questions have
attracted the attention of many researchers; for example, see [1, 2, 4, 6, 8,
9, 10].

In this paper, we characterize invertible maps
$\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ that satisfies

$\phi([\rho_{1},\rho_{2}])\subseteq[\phi(\rho_{1}),\phi(\rho_{2})]\quad\hbox{
for any }\rho_{1},\rho_{2}\in{\mathcal{S}}(H),$

where $[\rho_{1},\rho_{2}]=\\{t\rho_{1}+(1-t)\rho_{2}:t\in[0,1]\\}$ denotes
the closed line segment joining two states $\rho_{1},\rho_{2}$. In other
words, we characterize maps on states such that for any
$\rho_{1},\rho_{2}\in{\mathcal{S}}(H)$ and $0\leq t\leq 1$, there is some $s$
with $0\leq s\leq 1$ such that

$\phi(t\rho_{1}+(1-t)\rho_{2})=s\phi(\rho_{1})+(1-s)\phi(\rho_{2}).$

This question is motivated by the study of affine isomorphisms on
${\mathcal{S}}(H)$; see [2]. Recall that an affine isomorphism on
${\mathcal{S}}(H)$ is a bijective map
$\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ satisfying

$\phi(t\rho_{1}+(1-t)\rho_{2})=t\phi(\rho_{1})+(1-t)\phi(\rho_{2})\quad\hbox{
for all }t\in[0,1]\hbox{ and }\rho_{1},\rho_{2}\in{\mathcal{S}}(H).$

Evidently, we have the implications (c) $\Rightarrow$ (b) $\Rightarrow$ (a)
for a bijective map $\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ for the
following conditions.

  * (a)

$\phi([\rho_{1},\rho_{2}])\subseteq[\phi(\rho_{1}),\phi(\rho_{2})]$ for any
$\rho_{1},\rho_{2}\in{\mathcal{S}}(H)$.

  * (b)

$\phi([\rho_{1},\rho_{2}])=[\phi(\rho_{1}),\phi(\rho_{2})]$ for any
$\rho_{1},\rho_{2}\in{\mathcal{S}}(H)$.

  * (c)

$\phi$ is an affine isomorphism.

It was shown in [2] that an affine isomorphism
$\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ has the form

$None$ $\rho\mapsto U\rho U^{*}\quad\hbox{ or }\quad\rho\mapsto
U\rho^{T}U^{*},$

where $U$ is a unitary operator and $\rho^{T}$ is the transpose of $\rho$ with
respect to a certain orthonormal basis for $H$. Note that unitary similarity
transforms correspond to evolutions of quantum systems, and many maps that
leave invariant subsets or quantum properties of the states have the form
described in (1.1). One may be tempted to conjecture that maps on states
satisfying (a) or (b) above also have the forms described in (1.1). However,
this is not true as shown by our results. It turns out that the maps
satisfying condition (a) and (b) are closely related to quantum measurements.

Recall that in quantum mechanics a fine-grained quantum measurement is
described by a collection $\\{M_{m}\\}$ of measurement operators acting on the
state space $H$ satisfying $\sum_{m}M_{m}^{*}M_{m}=I$. 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 $M_{j}$ is fixed, we get a
measurement map $\phi_{j}$ defined by $\phi_{j}(\rho)=\frac{M_{j}\rho
M_{j}^{*}}{{\rm tr}(M_{j}\rho M_{j}^{*})}$ from the convex subset
${\mathcal{S}}_{M}(H)=\\{\rho:M_{j}\rho M_{j}^{*}\not=0\\}$ of the (convex)
set ${\mathcal{S}}(H)$ of states into ${\mathcal{S}}(H)$. If $M_{j}$ is
invertible, then $\phi_{j}:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ is
bijective and will be called an invertible measurement map. Observe that a
measurement map $\phi_{j}$ satisfies (a), (b), and is not of the standard form
(1.1) in general.

In this paper, we show that, up to the transpose, bijective maps on states
satisfying (a) or (b) are precisely invertible measurement maps. The following
is our main result.

Theorem 1. Let ${\mathcal{S}}(H)$ be the convex set of all states on Hilbert
space $H$ with $2\leq\dim H\leq\infty$. The following statements are
equivalent for a bijective map
$\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$.

  * (a)

$\phi([\rho_{1},\rho_{2}])\subseteq[\phi(\rho_{1}),\phi(\rho_{2})]$ for any
$\rho_{1},\rho_{2}\in{\mathcal{S}}(H)$.

  * (b)

$\phi([\rho_{1},\rho_{2}])=[\phi(\rho_{1}),\phi(\rho_{2})]$ for any
$\rho_{1},\rho_{2}\in{\mathcal{S}}(H)$.

  * (c)

There is an invertible bounded linear operator $M\in{\mathcal{B}}(H)$ such
that $\phi$ has the form

$\rho\mapsto\frac{M\rho M^{*}}{{\rm tr}(M\rho M)}\quad{\rm
or}\quad\rho\mapsto\frac{M\rho^{T}M^{*}}{{\rm tr}(M\rho^{T}M^{*})},$

where $\rho^{T}$ is the transpose of $\rho$ with respect to an orthonormal
basis.

It is interesting to note that condition (a) is much weaker than condition
(b). For example, condition (a) does not even ensure that
$\phi([\rho_{1},\rho_{2}])$ is a convex (connected) subset of
$[\phi(\rho_{1}),\phi(\rho_{2})]$. It turns out that the two conditions (a)
and (b) are equivalent for a bijective map, and the map must be a measurement
map or the composition of the transpose map with a measurement map.

The proof of Theorem 1 is done in the next few sections. In Section 2, we will
establish the equivalence of (a) and (b) using a result of P$\breve{a}$les
[12]. Then we verify the equivalence of (b) and (c). We treat the finite
dimensional case in Section 3. Using the result in Section 3, we complete the
proof for the infinite dimensional case in Section 4.

##  2\. The equivalence of the first two conditions

The implication of (b) $\Rightarrow$ (a) is clear. We consider the implication
(a)$\Rightarrow$(b).

Assume (a) holds. We will prove that
$\phi([\rho,\sigma])=[\phi(\rho),\phi(\sigma)]$ for any quantum states
$\rho,\sigma$. If $\rho=\sigma$, it is trivial. Suppose $\rho\neq\sigma$.

Note that $\rho,\sigma\in{\mathcal{S}}(H)$ are linearly dependent if and only
if $\rho=\sigma$. So, if $\rho,\sigma$ are linearly independent, then
$\phi(\rho),\phi(\sigma)$ are linearly independent as
$\phi(\rho)\neq\phi(\sigma)$ by the injectivity of $\phi$. Let
${\mathcal{HT}}(H)$ be the real linear space of all self-adjoint trace-class
operators on $H$. As $\phi$ is injective, we must have
$\phi(]\rho,\sigma[)\subset]\phi(\rho),\psi(\sigma)[$ for any
$\rho,\sigma\in{\mathcal{S}}(H)$, where
$]\rho,\sigma_{2}[=[\rho,\sigma]\setminus\\{\rho,\sigma\\}$ is the open line
segment joining $\rho,\sigma$. So by P$\breve{a}$les’ result [12, Theorem 2],
there exists a real linear map
$\psi:{\mathcal{HT}}(H)\rightarrow{\mathcal{HT}}(H)$, a real linear functional
$f:{\mathcal{HT}}(H)\rightarrow{\mathbb{R}}$, an operator
$B\in{\mathcal{H}T}(H)$ and a real number $c$ such that

$None$ $\phi(\rho)=\frac{\psi(\rho)+B}{f(\rho)+c}\quad\mbox{\rm and}\quad
f(\rho)+c>0$

hold for all $\rho\in{\mathcal{S}}(H)$. Thus, for any
$\rho,\sigma\in{\mathcal{S}}(H)$ with $\rho\not=\sigma$ and any $t\in[0,1]$,
there exists $s\in[0,1]$ such that

$\phi(t\rho+(1-t)\sigma)=s\phi(\rho)+(1-s)\phi(\sigma)=s\frac{\psi(\rho)+B}{f(\rho)+c}+(1-s)\frac{\psi(\sigma)+B}{f(\sigma)+c}.$

On the other hand, by the linearity of $\psi$ and $f$, we have

$\displaystyle\phi(t\rho+(1-t)\sigma)$ $\displaystyle=$
$\displaystyle\frac{\psi(t\rho+(1-t)\sigma)+B}{f(t\rho+(1-t)\sigma)+c}$
$\displaystyle=$ $\displaystyle
t\frac{\psi(\rho)+B}{f(t\rho+(1-t)\sigma)+c}+(1-t)\frac{\psi(\sigma)+B}{f(t\rho+(1-t)\sigma)+c}.$

Write $\lambda_{t,\rho,\sigma}=f(t\rho+(1-t)\sigma)+c$, we get

$(\frac{s}{f(\rho)+c}-\frac{t}{\lambda_{t,\rho,\sigma}})(\psi(\rho)+B)+(\frac{1-s}{f(\sigma)+c}-\frac{1-t}{\lambda_{t,\rho,\sigma}})(\psi(\sigma)+B)=0.$

As $\rho\neq\sigma$, $\phi(\rho)$ and $\phi(\sigma)$ are linearly independent.
This implies that $\psi(\rho)+B$ and $\psi(\sigma)+B$ are linearly
independent, too. It follows that

$\frac{t}{f(t\rho+(1-t)\sigma)+c}=\frac{s}{f(\rho)+c}\ {\rm and}\
\frac{1-t}{f(t\rho+(1-t)\sigma)}=\frac{1-s}{f(\sigma)+c}.$

Clearly, $s$ is continuously dependent of $t$ such that $\lim_{t\rightarrow
0}s=0$ and $\lim_{t\rightarrow 1}s=1$. Hence we must have
$\phi([\rho,\sigma])=[\phi(\rho),\phi(\sigma)]$. Thus, condition (b) holds. ∎

Denote by ${\mathcal{P}ur}(H)=\\{x\otimes x:x\in H,\|x\|=1\\}$ the set of pure
states in ${\mathcal{S}}(H)$. The following lemma is useful for our future
discussion.

Lemma 2.1. If condition (b) of Theorem 1 holds, then $\phi$ preserves pure
states in both directions, that is,
$\phi({\mathcal{P}ur}(H))={\mathcal{P}ur}(H)$ .

Proof It is clear that ${\mathcal{S}}(H)$ is a convex set and its extreme
point set is the set ${\mathcal{P}ur}(H)$ of all pure states (rank-1
projections). For any $P\in{\mathcal{P}ur}(H)$, if
$\phi^{-1}(P)\not\in{\mathcal{P}ur}(H)$, then there are two states
$Q,R\in{\mathcal{S}}(H)$ such that $Q\not=R$ and $\phi^{-1}(P)=tQ+(1-t)R$. As
$\phi([\rho,\sigma])\subseteq[\phi(\rho),\phi(\sigma)]$ for any $\rho,\sigma$,
there is some $s\in[0,1]$ such that
$P=\phi(\phi^{-1}(P))=\Phi(tQ+(1-t)R)=s\phi(Q)+(1-s)\phi(R)$. Since
$\phi(Q)\neq\phi(R)$, this contradicts the fact that $P$ is extreme point. So
$\phi^{-1}$ sends pure states to pure states. Similarly, since
$\phi([\rho,\sigma])\supseteq[\phi(\rho),\phi(\sigma)]$ for any states
$\rho,\sigma$, one can show that $\phi$ maps pure states into pure states. ∎

##  3\. Proof of Theorem 1: finite dimensional case

In this section we assume that $\dim H=n<\infty$. In such a case, we may
regard ${\mathcal{HT}}(H)$ the same as ${\bf H}_{n}$, the real linear space of
$n\times n$ Hermitian matrices. Since the implication (c) $\Rightarrow$ (b),
we needs only prove the implication (b) $\Rightarrow$ (c). We divide the proof
of this implication into several assertions. Assume (b) holds.

Assertion 3.1. $\phi(\frac{I}{n})$ is invertible.

Let $\phi(\frac{I}{n})=T$. In order to prove $T$ is invertible, we show that
$\phi$ maps invertible states to invertible states. Note that $\phi$ has the
form of Eq.(2.1), that is, for any $\rho\in{\mathcal{S}}(H)$,
$\phi(\rho)=\frac{\psi(\rho)+B}{f(\rho)+c}$. Since ${\bf H}_{n}$ is finite
dimensional, the linear map $\psi$ and the linear functional $f$ are bounded.
So $\phi$ is continuous. $\phi^{-1}$ is also continuous as $\phi$ preserves
line segment and hence has the form of Eq.(2.1). Thus $\phi$ maps open sets to
open sets. Denote by $G({\mathcal{S}}(H))$ the subset of all invertible
states. $G({\mathcal{S}}(H))$ is an open subset of ${\mathcal{S}}(H)$. In
fact, $G({\mathcal{S}}(H))$ is the maximal open set of all interior points of
${\mathcal{S}}(H)$. To see this, assume that a state $\rho$ is not invertible;
then there are mutually orthogonal rank-one projections $P_{i}$ ($i=1,2,\ldots
n$), an integer $1\leq k<n$ and scalars $t_{i}>0$ with $\sum_{i=1}^{k}t_{i}=1$
such that $\rho=\Sigma^{k}_{i=1}t_{i}P_{i}$. For any $\varepsilon>0$ small
enough so that $\frac{\varepsilon}{2k}<\min\\{t_{1},t_{2}\ldots,t_{k}\\}$, let

$\rho_{\varepsilon}=\Sigma^{k}_{i=1}(t_{i}-\frac{\varepsilon}{2k})P_{i}+\Sigma^{n}_{j=k+1}(\frac{\varepsilon}{2(n-k)})P_{j}.$

Then $\rho_{\varepsilon}$ is an invertible state and

$\|\rho-\rho_{\varepsilon}\|_{{\rm
tr}}\leq\Sigma^{k}_{i=1}\frac{\varepsilon}{2k}+\Sigma^{n}_{j=k+1}\frac{\varepsilon}{2(n-k)}=\varepsilon.$

It follows that for any state $\rho$ and any $\varepsilon>0$, there is an
invertible state $\sigma$ such that
$\rho\in\\{\tau\in{\mathcal{S}}(H):\|\tau-\sigma\|_{\rm Tr}<\varepsilon\\}$.
So the trace norm closure of $G({\mathcal{S}}(H))$ equals ${\mathcal{S}}(H)$.
Thus $G({\mathcal{S}}(H))$ is the set of all interior points of
${\mathcal{S}}(H)$. Since $\phi$ preserves the open sets, we have
$\phi(G({\mathcal{S}}(H)))\subseteq G({\mathcal{S}}(H))$. So $\phi$ preserves
the invertible states. In particular, $\phi(\frac{I}{n})$ is invertible. ∎

By Assertion 1, there is an invertible operator $R\in{\mathcal{B}}(H)$ such
that $\phi(\frac{I}{n})=RR^{*}$. Let $S=R^{-1}$; then the map
$\tilde{\phi}:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ defined by

$\rho\mapsto\frac{S\phi(\rho)S^{*}}{{\rm tr}(S\phi(\rho)S^{*})}$

is bijective, sends line segments to line segments in both directions, i.e.,
$\tilde{\phi}([\rho,\sigma])=[\tilde{\phi}(\rho),\tilde{\phi}(\sigma)]$, and
satisfies $\tilde{\phi}(\frac{I}{n})=\frac{I}{n}$.

Assertion 3.2. $\tilde{\phi}$ maps orthogonal rank one projections to
orthogonal rank one projections.

If $\\{P_{1},\dots,P_{n}\\}$ is an orthogonal set of rank one projections
satisfying $P_{1}+\cdots+P_{n}=I$, then there are $t_{i}\in[0,1]$
$(i=1,...,n)$ with $\Sigma_{i=1}^{n}t_{i}=1$ such that

$\frac{I}{n}=\tilde{\phi}(\frac{I}{n})=\tilde{\phi}(\frac{(P_{1}+\cdots+P_{n})}{n})=t_{1}\tilde{\phi}(P_{1})+\cdots+t_{n}\tilde{\phi}(P_{n})\geq
t_{i}\tilde{\phi}(P_{i})$

for each $i=1,\dots,n$. Because $\tilde{\phi}(P_{i})$ is a rank one orthogonal
projection and $I/n-t_{i}\tilde{\phi}(P_{i})$ is positive semidefinite, we see
that $1/n\geq t_{i}$ for $i=1,\dots,n$. Taking trace, we have

$1={\rm tr}(I/n)=\sum_{i=1}^{n}t_{i}.$

Thus, $t_{1}=\cdots=t_{n}=1/n$. So, $I=\sum_{i=1}^{n}\tilde{\phi}(P_{i})$.
This implies that $\\{\tilde{\phi}(P_{1}),\ldots,\tilde{\phi}(P_{n})\\}$ is an
orthogonal set of rank one projections. Hence, $\tilde{\phi}$ sends orthogonal
rank one projections to orthogonal rank one projections. ∎

By [12, Theorem 2] again, $\tilde{\phi}$ has the form of Eq.(2.1), that is,

$None$ $\tilde{\phi}(\rho)=\frac{\psi(\rho)+B}{f(\rho)+c}$

holds for any $\rho\in{\mathcal{S}}(H)$, where
$\psi:{\mathbf{H}}_{n}({\mathbb{C}})\rightarrow{\mathbf{H}}_{n}({\mathbb{C}})$
is a real linear map, ${\mathbf{H}}_{n}({\mathbb{C}})$ is the real linear
space of all $n\times n$ hermitian matrices,
$B\in{\mathbf{H}}_{n}({\mathbb{C}})$,
$f:{\mathbf{H}}_{n}({\mathbb{C}})\rightarrow{\mathbb{R}}$ is a real linear
functional and $c$ is a real constant with $f(\rho)+c>0$ for all
$\rho\in{\mathcal{S}}(H)$.

Next we consider the two cases of dim$H>2$ and dim$H=2$ respectively.

Assertion 3.3. Assume dim$H>2$. The functional $f$ in Eq.(3.1) 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 $\\{e_{i}\\}^{n}_{i=1}$, 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 $\tilde{\phi}$
preserves the rank one projections in both directions, there is a rank one
projection $Q_{i}=x_{i}\otimes x_{i}$ such that

$x_{i}\otimes x_{i}=Q_{i}=\tilde{\phi}(P_{i})=\frac{\psi(e_{i}\otimes
e_{i})+B}{f(e_{i}\otimes e_{i})+c}.$

So

$\psi(e_{i}\otimes e_{i})+B=(f(e_{i}\otimes e_{i})+c)(x_{i}\otimes x_{i}).$

As $\tilde{\phi}(\frac{I}{n})=\frac{I}{n}$ and
$\frac{I}{n}=\frac{1}{n}\sum^{n}_{i=1}e_{i}\otimes e_{i}$, we have

$\frac{I}{n}=\tilde{\phi}(\frac{1}{n}\sum^{n}_{i=1}e_{i}\otimes
e_{i})=\frac{\psi(\sum^{n}_{i=1}\frac{1}{n}e_{i}\otimes
e_{i})+B}{f(\sum^{n}_{i=1}\frac{1}{n}e_{i}\otimes
e_{i})+c}=\frac{\sum^{n}_{i=1}\frac{1}{n}\psi(e_{i}\otimes
e_{i})+n\frac{1}{n}B}{\sum^{n}_{i=1}\frac{1}{n}f(e_{i}\otimes
e_{i})+n\frac{1}{n}c}.$

Then

$None$ $\frac{I}{n}=\frac{\frac{1}{n}(\sum^{n}_{i=1}\psi(e_{i}\otimes
e_{i})+B)}{\frac{1}{n}(\sum^{n}_{i=1}f(e_{i}\otimes
e_{i})+c)}=\frac{\sum^{n}_{i=1}(\psi(e_{i}\otimes
e_{i})+B)}{\sum^{n}_{i=1}(f(e_{i}\otimes e_{i})+c)}.$

On the other hand, by Assertion 3.2, we have

$\frac{I}{n}=\tilde{\phi}(\frac{I}{n})=\frac{1}{n}\sum^{n}_{i=1}\tilde{\phi}(e_{i}\otimes
e_{i})=\frac{1}{n}\sum^{n}_{i=1}\frac{\psi(e_{i}\otimes
e_{i})+B}{f(e_{i}\otimes e_{i})+c}.$

Thus we get

$None$ $I=\sum^{n}_{i=1}\frac{\psi(e_{i}\otimes e_{i})+B}{f(e_{i}\otimes
e_{i})+c}.$

Let $A_{i}=\psi(e_{i}\otimes e_{i})+B$ and $a_{i}=f(e_{i}\otimes e_{i})+c$.
Then Eq.(3.2) and Eq.(3.3) imply that

$I=n(\frac{A_{1}+A_{2}+\ldots+A_{n}}{a_{1}+a_{2}+\ldots+a_{n}})=\frac{A_{1}}{a_{1}}+\frac{A_{2}}{a_{2}}+\ldots+\frac{A_{n}}{a_{n}}.$

Note that $A_{i}=a_{i}Q_{i}$, where $Q_{i}=\tilde{\phi}(e_{i}\otimes
e_{i})=x_{i}\otimes x_{i}$. Therefore, we get that

$I=n(\frac{a_{1}Q_{1}+a_{2}Q_{2}+\ldots+a_{n}Q_{n}}{a_{1}+a_{2}+\ldots+a_{n}})=\frac{a_{1}Q_{1}}{a_{1}}+\frac{a_{2}Q_{2}}{a_{2}}+\ldots+\frac{a_{n}Q_{n}}{a_{n}}.$

It follows that

$n(\frac{a_{1}Q_{1}+a_{2}Q_{2}+\ldots+a_{n}Q_{n}}{a_{1}+a_{2}+\ldots+a_{n}})=Q_{1}+Q_{2}+\ldots+Q_{n}.$

Since $\\{Q_{i}\\}_{i=1}^{n}$ is an orthogonal set of rank one projections, we
see that

$\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}=a_{1}=a_{2}=\ldots=a_{n}.$

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 $z\in[x,y]^{\perp}$. It follows from
the above argument 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 $\rho$. ∎

Assertion 3.4. Assume dim$H>2$. $\phi$ has the form stated in Theorem 1 (c).

Every state is a convex combination of some pure states, i.e. convex
combination of some rank one projections. Therefore, by Assertion 3.3, we have

$\tilde{\phi}(\rho)=\frac{\psi(\rho)+B}{\alpha+c}$

holds for all $\rho$. Then by the linearity of $\psi$, it is clear that
$\tilde{\phi}$ is an affine isomorphism, i.e., for any states $\rho,\sigma$
and scalar $\lambda$ with $0\leq\lambda\leq 1$,
$\tilde{\phi}(\lambda\rho+(1-\lambda)\sigma)=\lambda\tilde{\phi}(\rho)+(1-\lambda)\tilde{\phi}(\sigma).$
By a result due to Kadison (Ref. [2, Theorem 8.1]), $\tilde{\phi}$ has the
standard form, that is, there exists a unitary operator $U\in{\mathcal{B}}(H)$
such that $\tilde{\phi}$ has the form

$\tilde{\phi}(\rho)=U\rho U^{*}\ \mbox{\rm for all }\rho\quad\hbox{ or
}\quad\rho\mapsto U\rho^{T}U^{*}\ \mbox{for all }\ \rho.$

Now recalled that $\tilde{\phi}$ is defined by
$\tilde{\phi}(\rho)=S\phi(\rho)S^{*}/{\rm tr}(S\phi(\rho)S^{*})$. If
$\tilde{\phi}$ takes the first form, then we have

$\phi(\rho)={\rm
tr}(S\phi(\rho)S^{*})S^{-1}\tilde{\phi}(\rho)(S^{*})^{-1}={\rm
tr}(S\phi(\rho)S^{*})S^{-1}U\rho U^{*}(S^{*})^{-1}.$

As $1={\rm tr}(\phi(\rho))={\rm tr}(S\phi(\rho)S^{*}){\rm tr}(S^{-1}U\rho
U^{*}(S^{*})^{-1})$, so

${\rm tr}(S\phi(\rho)S^{*})=\frac{1}{{\rm tr}(S^{-1}U\rho
U^{*}(S^{*})^{-1})}.$

Letting $M=S^{-1}U$, we get $\phi(\rho)=\frac{M\rho M^{*}}{{\rm tr}(M\rho
M^{*})}$ for all $\rho$, that is, $\phi$ has the first form stated in (c) of
Theorem 1.

Similarly, if $\tilde{\phi}$ takes the second form, then $\phi$ takes the
second form stated in (c) of Theorem 1. ∎

Assertion 3.5. Condition (c) of Theorem 1 holds for the case of dim$H=2$.

Assume that dim$H=2$. Denote by ${\mathcal{S}}_{2}=\mathcal{S}(H)$ the convex
set of $2\times 2$ positive matrices with the trace 1. Then the map
$\tilde{\phi}:{\mathcal{S}}_{2}\rightarrow{\mathcal{S}}_{2}$ is a bijective
map preserving segment in both directions satisfying
$\tilde{\phi}(\frac{1}{2}I_{2})=\frac{1}{2}I_{2}$. Let us identify
${\mathcal{S}}_{2}$ with the unit ball
$({\mathbb{R}}^{3})_{1}=\\{(x,y,z)^{T}\in{\mathbb{R}}^{3}:x^{2}+y^{2}+z^{2}\leq
1\\}$ of ${\mathbb{R}}^{3}$ by the following way. Let
$\pi:({\mathbb{R}}^{3})_{1}\rightarrow{\mathcal{S}}_{2}$ be the map defined by

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

$\pi$ 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
rank one projection, and $0=(0,0,0)^{T}$ if and only if the corresponding
matrix is $\pi(0)=\frac{1}{2}I$. The map
$\tilde{\phi}:{\mathcal{S}}_{2}\rightarrow{\mathcal{S}}_{2}$ induces a map
$\hat{\phi}:({\mathbb{R}}^{3})_{1}\rightarrow({\mathbb{R}}^{3})_{1}$ by the
following equation

$\tilde{\phi}(\rho)=\frac{1}{2}I+\pi(\hat{\phi}(\pi^{-1}(\rho))).$

Since $\tilde{\phi}$ is a segment preserving bijective map and $\pi$ is an
affine isomorphism, the map $\hat{\phi}$ is a bijective map preserving segment
in both directions, that is, $\hat{\phi}([u,v])=[\hat{\phi}(u),\hat{\phi}(v)]$
for $u,v\in({\mathbb{R}}^{3})_{1}$. So $\hat{\phi}$ maps the surface of
$({\mathbb{R}}^{3})_{1}$ onto the surface of $({\mathbb{R}}^{3})_{1}$. Since
$\tilde{\phi}(\frac{1}{2}I)=\frac{1}{2}I$, we have that
$\hat{\phi}((0,0,0)^{T})=(0,0,0)^{T}$.

Applying the P$\breve{a}$les’ result [12, Theorem 2] 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 $r\in{\mathbb{R}}$ such that
$f((x,y,z)^{T})+r>0$ and

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

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 $r>0$.
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 $x^{2}_{0}+y^{2}_{0}+z^{2}_{0}=1$ 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

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

and thus

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

Similarly

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

By the linearity of $L$ we have
$r_{1}x_{0}+r_{2}y_{0}+r_{3}z_{0}+r=-r_{1}x_{0}-r_{2}y_{0}-r_{3}z_{0}+r$.
Hence $r_{1}x_{0}+r_{2}y_{0}+r_{3}z_{0}=0$, a contradiction. So, we have
$f=0$, and thus $\hat{\phi}=\frac{L}{r}$ is linear. Now it is clear that
$\tilde{\phi}$ is an affine isomorphism as $\pi$ is an affine isomorphism.
Applying a similar argument to the proof of Assertion 3.4 and the Kadison’s
result, one sees that $\tilde{\phi}$ has the standard form. Thus, Theorem 1
(c) holds. ∎

By Assertions 3.4 and 3.5, we get the proof of Theorem 1 for finite-
dimensional case.

##  4\. Proof: infinite dimensional case

In this section we give a proof of our main result for infinite dimensional
case. Similar to the previous section, we need only establish the implication
(b) $\Rightarrow$ (c). We begin with two lemmas.

Let $V_{1},V_{2}$ be linear spaces on a field ${\mathbb{F}}$,
$\upsilon:{\mathbb{F}}\rightarrow{\mathbb{F}}$ a nonzero ring automorphism. A
map $A:V_{1}\rightarrow V_{2}$ is called a $\upsilon$-linear operator if
$A(\lambda x)=\upsilon(\lambda)Ax$ for all $x\in V_{1}$. The following lemma
is similar to [7, Lemma 2.3.1].

Lemma 4.1 Let $V_{1},V_{2}$ be linear spaces on a field ${\mathbb{F}}$,
$\tau,\upsilon:{\mathbb{F}}\rightarrow{\mathbb{F}}$ nonzero ring auto-
isomorphisms. Suppose $A:V_{1}\rightarrow V_{2}$ is a $\tau$-linear
transformation, $B:V_{1}\rightarrow V_{2}$ is a $\upsilon$-linear
transformation, and $\dim{\rm span}({\rm ran}(B))\geq 2$. If $\ker
B\subseteq\ker A$ and $Ax$ and $Bx$ are linearly dependent for all $x\in V$,
then $\tau=\upsilon$ and $A=\lambda B$ for some scalar $\lambda$.

Proof As $\ker B\subseteq\ker A$, for every $x\in V_{1}$, there is some scalar
$\lambda_{x}$ such that $Ax=\lambda_{x}Bx$. If $Bx\not=0$, then there exists
$y\in V_{1}$ such that $Bx,By$ are linearly independent. Then
$\lambda_{x+y}(Bx+By)=A(x+y)=\lambda_{x}Bx+\lambda_{y}By$. This implies that
$\lambda_{x}=\lambda_{x+y}=\lambda_{y}$. Moreover, for any
$\alpha\in{\mathbb{F}}$, we have $\lambda_{\alpha x}=\lambda_{x}$. If $Bx=0$,
then $Ax=0$. Thus it follows that there exists a scalar $\lambda$ such that
$Ax=\lambda Bx$ holds for all $x\in V_{1}$. So, $A=\lambda B$ and
$\tau=\upsilon$. $\Box$

Lemma 4.2 Let ${\mathcal{S}}(H)$ be the set of all states on Hilbert space $H$
with $\dim H=\infty$, and $\phi:{\mathcal{S}}(H)\rightarrow{\mathcal{S}}(H)$ a
bijective map. If $\phi$ satisfies that, for any $t\in[0,1]$ and
$\rho,\sigma\in{\mathcal{S}}(H)$, there is $s\in[0,1]$ such that

$\phi(t\rho+(1-t)\sigma)=s\phi(\rho)+(1-s)\phi(\sigma),$

then, $\phi$ is continuous and there is an invertible bounded linear or
conjugate linear operator $T$ such that

$\phi(x\otimes x)=\frac{Tx\otimes Tx}{\|Tx\|^{2}}{\ for\ all\ unit\ vectors\
}x\in H.$

Proof We complete the proof by checking several assertions. First we restate
Lemma 2.1 as:

Assertion 4.1. $\phi$ preserves pure states (rank one projections) in both
directions.

Assertion 4.2. For any $x_{i}\otimes x_{i}\in{\mathcal{P}ur}(H)$ with
$\\{x_{1},x_{2}\ldots,x_{n}\\}$ linearly independent, let

$F(x_{1},\ldots,x_{n})=C(x_{1},\ldots,x_{n})\cup F_{0}(x_{1},\ldots,x_{n}),$

where $C(x_{1},\ldots,x_{n})={\rm cov}\\{x_{i}\otimes x_{i}:i=1,2\ldots,n\\}$
is the convex hull of $\\{x_{i}\otimes x_{i}\\}_{i=1}^{n}$,

$\begin{array}[]{rl}F_{0}(x_{1},\ldots,x_{n})=&\\{Z\in{\mathcal{S}}(H)\setminus
C(x_{1},\ldots,x_{n}):\mbox{\rm there exists some }\\\
&W\in{\mathcal{S}}(H)\setminus C(x_{1},\ldots,x_{n})\ \mbox{\rm such that
}[Z,W]\cap C(x_{1},\ldots,x_{n})\not=\emptyset\\}.\end{array}$

Let $H_{0}={\rm span}\\{x_{1},\ldots,x_{n}\\}$. Then we have

$None$ $F(x_{1},\ldots,x_{n})={\mathcal{S}}(H_{0})\oplus\\{0\\}.$

Obviously, $C(x_{1},\ldots,x_{n})\subset{\mathcal{S}}(H_{0})\oplus\\{0\\}$. If
$Z\in F_{0}(x_{1},\ldots,x_{n})$, then there exists some
$W\in{\mathcal{S}}(H)\setminus C(x_{1},\ldots,x_{n})$, $t_{i}>0$ with
$\sum_{i=1}^{n}t_{i}=1$ and $t\in(0,1)$ such that

$\sum_{i=1}^{n}t_{i}x_{i}\otimes x_{i}=tZ+(1-t)W.$

Let $P_{0}\in{\mathcal{B}}(H)$ be the projection from $H$ onto $H_{0}$. As
$\sum_{i=1}^{n}t_{i}x_{i}\otimes x_{i}-tZ=(1-t)W\geq 0$ and
$(I-P_{0})\sum_{i=1}^{n}t_{i}x_{i}\otimes
x_{i}=\sum_{i=1}^{n}t_{i}x_{i}\otimes x_{i}(I-P_{0})=0$, we see that
$(I-P_{0})Z=Z(I-P_{0})=0$, which implies that $P_{0}ZP_{0}=Z$ and hence
$Z\in{\mathcal{S}}(H_{0})\oplus\\{0\\}$.

Conversely, assume that $Z\in{\mathcal{S}}(H_{0})\oplus\\{0\\}$. Since
$C(x_{1},\ldots,x_{n})\subset{\mathcal{S}}(H_{0})\oplus\\{0\\}$, we may assume
that $Z$ is not a convex combination of $\\{x_{i}\otimes x_{i}\\}_{i=1}^{n}$.
Because $\\{x_{i}\\}_{i=1}^{n}$ is a linearly independent set, there exists an
operator $S\in{\mathcal{B}}(H_{0})$ such that $\\{e_{i}=Sx_{i}\\}_{i=1}^{n}$
is an orthonormal basis of $H_{0}$. Then, consider

$S(\sum_{i=1}^{n}a_{i}x_{i}\otimes
x_{i}-Z)S^{*}=\sum_{i=1}^{n}a_{i}Sx_{i}\otimes
Sx_{i}-SZS^{*}=\sum_{i=1}^{n}a_{i}e_{i}\otimes e_{i}-SZS^{*}.$

It is clear that for sufficient large $a_{i}>0$,
$\sum_{i=1}^{n}a_{i}e_{i}\otimes e_{i}-SZS^{*}\geq 0$, and hence,
$W=\sum_{i=1}^{n}a_{i}x_{i}\otimes x_{i}-Z\geq 0$. This entails that

$\frac{\sum_{i=1}^{n}a_{i}x_{i}\otimes
x_{i}}{\sum_{i=1}^{n}a_{i}}=\frac{1}{\sum_{i=1}^{n}a_{i}}Z+\frac{{\rm
tr}(W)}{\sum_{i=1}^{n}a_{i}}(\frac{W}{{\rm tr}(W)}),$

that is, $Z\in F_{0}(x_{1},\ldots,x_{n})\subset F(x_{1},\ldots,x_{n})$. This
finishes the proof of Eq.(4.1). ∎

Assertion 4.3. For any finite-dimensional subspace $H_{0}\subset H$, there
exists a subspace $H_{1}$ with $\dim H_{1}=\dim H_{0}$ such that

$\phi({\mathcal{S}}(H_{0})\oplus\\{0\\})={\mathcal{S}}(H_{1})\oplus\\{0\\}.$

Assume that $\dim H_{0}=n$. Choose an orthonormal basis
$\\{x_{i}\\}_{i=1}^{n}$ of $H_{0}$. Then by Assertion 4.1, there are unit
vectors $u_{i}\in H$ such that $\phi(x_{i}\otimes x_{i})=u_{i}\otimes u_{i}$.
It is clear that $\\{u_{i}\\}_{i=1}^{n}$ is a linearly independent set. Let
$H_{1}={\rm span}\\{u_{i}\\}_{i=1}^{n}$. Then $\dim H_{1}=n$, and by Eq.(4.1)
in Assertion 4.2, we have
$F(x_{1},\ldots,x_{n})={\mathcal{S}}(H_{0})\oplus\\{0\\}$,
$F(u_{1},\ldots,u_{n})={\mathcal{S}}(H_{1})\oplus\\{0\\}$. Since the bijection
$\phi$ preserves segments and pure states in both directions, it is easily
checked that $\phi(F(x_{1},\ldots,x_{n}))=F(u_{1},\ldots,u_{n})$, and the
conclusion of Assertion 4.3 follows. ∎

Assertion 4.4. For any finite dimensional subspace $\Lambda\subset H$, there
exists a subspace $H_{\Lambda}\subset H$ with $\dim H_{\Lambda}=\dim\Lambda$
and an invertible linear or conjugate linear operator
$M_{\Lambda}:\Lambda\rightarrow H_{\Lambda}$ such that

$\phi(P_{\Lambda}\rho P_{\Lambda})=\frac{Q_{\Lambda}M_{\Lambda}\rho
M_{\Lambda}^{*}Q_{\Lambda}}{{\rm tr}(M_{\Lambda}\rho M_{\Lambda}^{*})}$

for all $\rho\in{\mathcal{S}}(\Lambda)$, where $P_{\Lambda}$ and $Q_{\Lambda}$
are respectively the projections onto $\Lambda$ and $H_{\Lambda}$. Moreover,
the $M_{\Lambda}$ can be chosen so that
$M_{\Lambda_{1}}=M_{\Lambda_{2}}|_{\Lambda_{1}}$ whenever
$\Lambda_{1}\subseteq\Lambda_{2}$.

Let $H_{0}$ be a finite dimensional subspace of $H$ and let
$\\{e_{1},e_{2},\ldots,e_{n}\\}$ be an orthonormal basis of $H_{0}$. By
Assertion 4.1 there exist unit vectors $\\{u_{1},u_{2},\ldots,u_{n}\\}$ such
that $\phi(e_{i}\otimes e_{i})=u_{i}\otimes u_{i}$. Let $H_{1}={\rm
span}\\{u_{1},u_{2},\ldots,u_{n}\\}$. By Assertion 4.1 again, $\dim
H_{1}=n=\dim H_{0}$. It follows from Assertion 4.3 that, for any
$\rho\in{\mathcal{S}}(H)$, $P_{0}\rho P_{0}=\rho$ implies that
$P_{1}\phi(\rho)P_{1}=\phi(\rho)$. Thus $\phi$ induces a bijective map
$\phi_{0}:{\mathcal{S}}(H_{0})\rightarrow{\mathcal{S}}(H_{1})$ by
$\phi_{0}(\rho)=\phi(P_{0}\rho P_{0})|_{H_{1}}$. Applying Theorem 1 for finite
dimensional case just proved in Section 2, we obtain that there is an
invertible bounded linear operator $M:H_{0}\rightarrow H_{1}$ such that
$\phi_{0}$ has the form

$\rho\mapsto\frac{M\rho M^{*}}{{\rm tr}(M^{*}M\rho)}\quad{\rm
or}\quad\rho\mapsto\frac{M\rho^{T}M^{*}}{{\rm tr}(M^{*}M\rho^{T})},$

where $\rho^{T}$ is the transpose of $\rho$ with respect to the orthonormal
basis $\\{e_{1},e_{2},\ldots,e_{n}\\}$. In the last case, we let
$J:H_{0}\rightarrow H_{0}$ be the conjugate linear operator defined by
$J(\sum_{i=1}^{n}\xi_{i}e_{i})=\sum_{i=1}^{n}\bar{\xi_{i}}e_{i}$, and let
$M^{\prime}=MJ$. Then, $M^{\prime}:H_{0}\rightarrow H_{1}$ is invertible
conjugate linear and
$\phi_{0}(\rho)=\frac{M^{\prime}\rho{M^{\prime}}^{*}}{{\rm
tr}({M^{\prime}}^{*}M^{\prime}\rho)}$ for all $\rho\in{\mathcal{S}}(H_{0})$.
Therefore, the first part of the Assertion 4.4 is true.

Let $\Lambda_{i}$, $i=1,2$, are finite dimensional subspaces of $H$ and
$M_{i}$s are associated operators as that obtained above way. If
$\Lambda_{1}\subseteq\Lambda_{2}$, then, for any unit vector
$x\in\Lambda_{1}$, we have $\frac{M_{1}x\otimes
M_{1}x}{\|M_{1}x\|^{2}}=\phi(x\otimes x)=\frac{M_{2}x\otimes
M_{2}x}{\|M_{2}x\|^{2}}$. It follows that $M_{1}x$ and $M_{2}x$ are linearly
dependent. By Lemma 4.2 we see that $M_{2}|_{\Lambda_{1}}=\lambda M_{1}$ for
some scalar $\lambda$. As $\frac{(\lambda M)\rho(\lambda M)^{*}}{{\rm
tr}((\lambda M)^{*}(\lambda M)\rho)}=\frac{M\rho M^{*}}{{\rm
tr}(M^{*}M\rho)}$, we may choose $M_{2}$ so that $M_{2}|_{\Lambda_{1}}=M_{1}$.
∎

Assertion 4.5. There exists a linear or conjugate linear bijective
transformation $T:H\rightarrow H$ such that

$\phi(x\otimes x)=\frac{Tx\otimes Tx}{\|Tx\|^{2}}$

for every unit vector $x\in H$ and $T|_{\Lambda}=M_{\Lambda}$ for every finite
dimensional subspace $\Lambda$ of $H$.

For any $x\in H$, there is finite dimensional subspace $\Lambda$ such that
$x\in\Lambda$. Let $Tx=M_{\Lambda}x$. Then, by Assertion 4.4, $T:H\rightarrow
H$ is well defined, linear or conjugate linear. And by Assertion 4.1, $T$ is
bijective.

Note that P$\breve{a}$les’ result (Theorem 2 in [12]) holds true for the
infinite dimensional case. Since $\phi$ preserves segment, by [12, Theorems
1-2], there exists a linear operator
$\Gamma:{\mathcal{HT}}(H)\rightarrow{\mathcal{HT}}(H)$, a linear functional
$g:{\mathcal{HT}}(H)\rightarrow{\mathbb{R}}$, a scalar $b\in{\mathbb{R}}$ and
some operator $B\in\mathcal{HT}(B)$ such that

$None$ $\phi(\rho)=\frac{\Gamma\rho+B}{g(\rho)+b}$

for all $\rho\in{\mathcal{S}}(H)$, where ${\mathcal{HT}}(H)$ denotes the set
of all self-adjoint Trace-class operators in ${\mathcal{B}}(H)$ and
$g(\rho)+b>0$ for all $\rho\in{\mathcal{S}}(H)$. ∎

Assertion 4.6. The functions $g$, $\Gamma$ in Eq.(4.2) are bounded and hence
$\phi$ is continuous.

Note that, for any $\rho_{1},\rho_{2}\in{\mathcal{S}}(H)$ and any $t\in(0,1)$,
there exists some $s(t)\in(0,1)$ such that

$\phi(t\rho_{1}+(1-t)\rho_{2})=s(t)\phi(\rho_{1})+(1-s(t))\phi(\rho_{2}).$

Combining this with Eq.(4.2), one gets

$None$
$\frac{t\Gamma\rho_{1}+(1-t)\Gamma\rho_{2}+B}{tg(\rho_{1})+(1-t)g(\rho_{2})+b}=s(t)\frac{\Gamma\rho_{1}+B}{g(\rho_{1})+b}+(1-s(t))\frac{\Gamma\rho_{2}+B}{g(\rho_{2})+b}.$

Note that different states are linearly independent. Comparing the
coefficients of $\Gamma\rho_{1}$ in Eq.(4.3), one sees that

$None$ $s(t)=\frac{t(g(\rho_{1})+b)}{tg(\rho_{1})+(1-t)g(\rho_{2})+b}.$

It follows that $s(t)\rightarrow 1$ when $t\rightarrow 1$. If
$\rho=\sum_{i=1}^{n}t_{i}\rho_{i}\in{\mathcal{S}}(H)$ with
$\rho_{i}\in{\mathcal{S}}(H)$, one can get some $p_{i}$ so that
$\phi(\rho)=\phi(\sum_{i=1}^{n}t_{i}\rho_{i})=\sum_{i=1}^{n}p_{i}\phi(\rho_{i})$,
where $\sum_{i=1}^{n}t_{i}=\sum_{i=1}^{n}p_{i}=1$. Similarly we can check that

$None$ $p_{i}=\frac{t_{i}(g(\rho_{i})+b)}{\sum_{i=1}^{n}t_{i}g(\rho_{i})+b}.$

Suppose that $\rho,\rho_{i}\in{\mathcal{S}}(H)$ with
$\rho=\sum_{i=1}^{\infty}t_{i}\rho_{i}$, where $t_{i}>0$ and
$\sum_{i=1}^{\infty}t_{i}=1$. Then

$None$
$\begin{array}[]{rl}\phi(\rho)=&\phi(\sum_{i=1}^{\infty}t_{i}\rho_{i})\\\
=&\phi((\sum_{j=1}^{k}t_{j})\sum_{i=1}^{k}(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})\rho_{i}+(1-\sum_{j=1}^{k}t_{j})\sum_{i=k+1}^{\infty}(\frac{t_{i}}{1-\sum_{j=1}^{k}t_{j}})\rho_{i})\\\
=&s_{k}\phi(\sum_{i=1}^{k}(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})\rho_{i})+(1-s_{k})\phi(\sum_{i=k+1}^{\infty}(\frac{t_{i}}{1-\sum_{j=1}^{k}t_{j}})\rho_{i}).\end{array}$

Thus there exist scalars $q_{i}^{(k)}>0$ with $\sum_{i=1}^{k}q_{i}^{(k)}=1$
such that

$s_{k}\phi(\sum_{i=1}^{k}(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})\rho_{i})=\sum_{i=1}^{k}s_{k}q_{i}^{(k)}\phi(\rho_{i}).$

According to Eq.(4.4), Eq.(4.5), and keeping in mind that $g$ is a linear
functional, a simple calculation reveals that

$None$
$\begin{array}[]{rl}s_{k}=&\frac{(\sum_{j=1}^{k}t_{j})(g(\sum_{i=1}^{k}(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})\rho_{i})+b)}{(\sum_{j=1}^{k}t_{j})(g(\sum_{i=1}^{k}(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})\rho_{i}))+(1-\sum_{j=1}^{k}t_{j})g(\sum_{i=k+1}^{\infty}(\frac{t_{i}}{1-\sum_{j=1}^{k}t_{j}})\rho_{i}))+b}\\\
=&\frac{(\sum_{j=1}^{k}t_{j})(g(\sum_{i=1}^{k}(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})\rho_{i})+b)}{g(\rho)+b},\end{array}$
$None$
$q_{i}^{(k)}=\frac{(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})(g(\rho_{i})+b)}{\sum_{i=1}^{k}(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})g(\rho_{i})+b}=\frac{(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})(g(\rho_{i})+b)}{g(\sum_{i=1}^{k}(\frac{t_{i}}{\sum_{j=1}^{k}t_{j}})\rho_{i})+b},$

and

$None$ $s_{k}q_{i}^{(k)}=\frac{t_{i}(g(\rho_{i})+b)}{g(\rho)+b}.$

Observe that $s_{k}q_{i}^{(k)}$ is independent to $k$. Since
$\sum_{i=1}^{k}t_{i}\rightarrow 1$ as $k\rightarrow\infty$, we must have
$s_{k}\rightarrow 1$ as $k\rightarrow\infty$. Eqs.(4.6)-(4.9) imply that

$\sum_{i=1}^{\infty}\frac{t_{i}(g(\rho_{i})+b)}{g(\rho)+b}=1$

and

$None$
$\phi(\sum_{i=1}^{\infty}t_{i}\rho_{i})=\sum_{i=1}^{\infty}(\frac{t_{i}(g(\rho_{i})+b)}{g(\rho)+b})\phi(\rho_{i}).$

In particular, we have

$None$
$g(\sum_{i=1}^{\infty}t_{i}\rho_{i})=\sum_{i=1}^{\infty}t_{i}g(\rho_{i}).$

We assert that $\sup\\{g(\rho):\rho\in{\mathcal{S}}(H)\\}<\infty$. Assume that
$\sup\\{g(\rho):\rho\in{\mathcal{S}}(H)\\}=\infty$. Then, for any positive
integer $i$, there exists $\rho_{i}\in{\mathcal{S}}(H)$ satisfying that
$g(\rho_{i})>2^{i}$. Let
$\rho_{0}=\sum_{i=1}^{\infty}\frac{1}{2^{i}}\rho_{i}$,
$\sigma_{k}=\sum_{i=1}^{k}\frac{1}{2^{i}}\rho_{i}$, then
$\sigma_{k}\rightarrow\rho_{0}$, and

$g(\sigma_{k})=\sum_{i=1}^{k}\frac{1}{2^{i}}g(\rho_{i})\geq\sum_{i=1}^{k}1=k.$

Since $g(\rho_{i})\geq 0$, by Eq.(4.11), we have $g(\rho_{0})\geq
g(\sigma_{k})\geq k$ for every $k$, contradicting to the fact that
$g(\rho_{0})<\infty$. Now the fact $g(\rho)+b>0$ for all $\rho$ entails that
there exists a positive number $c$ such that
$\sup\\{|g(\rho)|:\rho\in{\mathcal{S}}(H)\\}=c$. Thus $g$ is continuous on
${\mathcal{HT}}(H)$ and

$None$ $\|g\|=c<\infty.$

Since

$\|\Gamma\rho\|\leq\|\Gamma\rho+B\|+\|B\|\leq\|\Gamma\rho+B\|_{\rm
Tr}+\|B\|=g(\rho)+b+\|B\|\leq c+|b|+\|B\|$

holds for all $\rho\in{\mathcal{S}}(H)$, it follows that $\Gamma$ is
$\|\cdot\|_{\rm tr}$-$\|\cdot\|$ continuous from ${\mathcal{HT}}(H)$ into
itself. Hence, if $\rho_{n},\rho\in{\mathcal{S}}(H)$ and $\|\cdot\|_{\rm
tr}$-$\lim_{n\rightarrow\infty}\rho_{n}=\rho$, then
$\|\cdot\|$-$\lim_{n\rightarrow\infty}\phi(\rho_{n})=\phi(\rho)$. However,
convergence under trace-norm topology and convergence under uniform-norm
topology are the same for states [15]. Hence we have $\|\cdot\|_{\rm
tr}$-$\lim_{n\rightarrow\infty}\phi(\rho_{n})=\phi(\rho)$, i.e., $\phi$ is
continuous under the trace-norm topology. ∎

Assertion 4.7. The operator $T$ in Assertion 4.5 is bounded.

For any finite dimensional subspace $\Lambda\subset H$, let $M_{\Lambda}$ be
the invertible linear or conjugate linear operator stated in Assertion 4.4.
Then for any $\rho\in{\mathcal{S}}(H)$ with range in $\Lambda$, we have
$\frac{\Gamma\rho+B}{g(\rho)+b}=\frac{Q_{\Lambda}M_{\Lambda}\rho
M_{\Lambda}^{*}Q_{\Lambda}}{{\rm tr}(M_{\Lambda}\rho M_{\Lambda}^{*})}$. Thus

$\Gamma\rho+B=\lambda_{\rho}Q_{\Lambda}M_{\Lambda}\rho
M_{\Lambda}^{*}Q_{\Lambda},$

where

$\lambda_{\rho}=\frac{g(\rho)+b}{{\rm tr}(M_{\Lambda}\rho M_{\Lambda}^{*})}.$

For any $\sigma\in{\mathcal{S}}(H)$ with range in $\Lambda$ and
$\sigma\not=\rho$, and for any $0<t<1$, by considering $t\rho+(1-t)\sigma$ one
gets

$\lambda_{\rho}=\lambda_{t\rho+(1-t)\sigma}=\lambda_{\sigma}.$

This implies that there exists a scalar $d>0$ such that $\lambda_{\rho}=d$ for
all $\rho$ with range in $\Lambda$. Use Assertion 4.4 again, it is clear that
$d$ is not dependent to $\Lambda$. Thus, the equation

${\rm tr}(M_{\Lambda}\rho M_{\Lambda}^{*})=d^{-1}(g(\rho)+b)$

holds for all finite rank $\rho\in{\mathcal{S}}(H)$. In particular, for any
unit vector $x\in\Lambda$, by Assertion 4.6, $\|g\|<\infty$ and we have

$\|M_{\Lambda}x\|^{2}=d^{-1}(g(x\otimes x)+b)\leq d^{-1}(\|g\|+|b|)<\infty,$

which implies that $\|M_{\Lambda}\|\leq\sqrt{d^{-1}(\|g\|+|b|)}.$ It follows
that, for any unit vector $x\in H$, we have
$\|Tx\|\leq\sqrt{d^{-1}(\|g\|+|b|)}$ and hence
$\|T\|\leq\sqrt{d^{-1}(\|g\|+|b|)}$.

The proof is finished.∎

Now we are in a position to give a proof of the main theorem for infinite
dimensional case.

Proof of Theorem 1: infinite dimensional case. Similar to the finite
dimensional case, we need only to show (b) $\Rightarrow$ (c).

Assume (b). By Lemma 4.2, there is a bounded invertible linear or conjugate
linear operator $T$ such that $\phi(x\otimes x)=\frac{Tx\otimes
Tx}{\|Tx\|^{2}}=\frac{Tx\otimes xT^{*}}{\|Tx\|^{2}}$ for all unit vectors
$x\in H$. Let $\rho$ be any finite rank state. Then there exists a finite
dimensional subspace $\Lambda$ of $H$ such that the range of $\rho$ is
contained in $\Lambda$. By Assertion 4.4 in the proof of Lemma 4.3, we have
$\phi(\rho)=\frac{(Q_{\Lambda}M_{\Lambda})\rho(Q_{\Lambda}M_{\Lambda})^{*}}{{\rm
tr}((Q_{\Lambda}M_{\Lambda})\rho(Q_{\Lambda}M_{\Lambda})^{*})}=\frac{T\rho
T^{*}}{{\rm tr}(T\rho T^{*})}$. Since the set of finite-rank states is dense
in ${\mathcal{S}}(H)$ and, by Lemma 4.3, $\phi$ is continuous, we get that
$\phi(\rho)=\frac{T\rho T^{*}}{{\rm tr}(T^{*}T\rho)}$ for all states $\rho$ as
desired, completing the proof.∎

Acknowldegement

We thank Wen-ling Huang for helpful discussions on the segment preserving
bijections $\hat{\phi}$ in Assertion 3.5.

## References

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