Contractive freedoms of ensembles and quantum channels for infinite-dimensional systems

Abstract

In this paper, we reveal that the unitary freedom in ensemble of pure states with a given state and the unitary freedom in operator-sum representation for a given channel are no longer valid for infinite-dimensional systems. The replacements of them are, respectively, the contractive freedom in ensemble and the contractive freedom in operator-sum representation. Precisely, we show that, (1) two ensembles $\{|\phi _i\rangle , p_i\}$ and $\{|\psi _j\rangle , q_j\}$ determine the same state if and only if there exists some contractive matrix $V=(v_{ij})$ such that $\sqrt{p_i}|\phi _i\rangle =\sum _j v_{ij}\sqrt{q_j}|\psi _j\rangle$ for each i; (2) two sequences $\{A_i\}$ and $\{B_j\}$ of Kraus operators in operator-sum representations determine the same quantum channel if and only if there exists some contractive matrix $V=(v_{ij})$ such that $A_i=\sum _jv_{ij}B_j$ for each i. All possible quantum ensembles of pure states with any given state and all possible operator-sum representations of any given channel are described.

Appendices

Appendix A: Proof of the contractive freedom in the ensemble

The purpose of “Appendix A” is to prove Theorem 1, the main result in Sect. 2. Before doing this, we need several lemmas.

Lemma A.1

Let H and K be two separable complex Hilbert spaces. Suppose $\{|g_{i}\rangle \}$ is an orthonormal set of K, $\{|x_{i}\rangle \}$ is a sequence of vectors of H. If $\sum _{i}|x_{i}\rangle \otimes |g_{i}\rangle =0$, then $|x_{i}\rangle =0$ for all i.

Proof

For any j, because $\sum _{i}|x_{i}\rangle \otimes |g_{i}\rangle =0,$ we have

$$\begin{aligned} 0=\langle x_j|\otimes \langle g_j|\sum _{i}|x_i\rangle \otimes |g_i\rangle =\sum _{i}\langle x_j|x_{i}\rangle \langle g_{j}|g_{i}\rangle =\langle x_j|x_{j}\rangle . \end{aligned}$$

So $|x_{j}\rangle =0$, as required. $\square$

Recall that $A\in {\mathcal {B}}(H)$ is positive, denoted by $A\ge 0$, if $\langle \psi |A|\psi \rangle \ge 0$ for all vectors $|\psi \rangle \in H$.

Lemma A.2

Let $\{|{\psi _{j}}\rangle \}_{j=1}^{M}$ with $M\le \infty$ be a sequences of vectors in a complex Hilbert space H. Then, for any positive semi-definite $M\times M$ matrix $A=(a_{ij})$, we have

$$\begin{aligned} \sum _{i=1}^{M}\sum _{j=1}^M a_{ij}|{\psi _{j}}\rangle \langle {\psi _{i}}|\ge 0. \end{aligned}$$

Proof

Since $A=(a_{ij})\ge 0$, there exists $M\times M$ matrix $B=(b_{ij})$ such that $A=B^\dag B$. Let $|{\phi _{i}}\rangle =\sum _{j=1}^{M} b_{ij}|{\psi _{j}}\rangle$ for each $i=1,2,\ldots$. Then,

$$\begin{aligned} \begin{aligned} \sum _{i=1}^{M}\sum _{j=1}^M a_{ij}|{\psi _{j}}\rangle \langle {\psi _{i}}|&=\sum _{i=1}^{M}\sum _{j=1}^M\sum _{l=1}^M {\bar{b}}_{li}b_{lj}|{\psi _{j}}\rangle \langle {\psi _{i}}| \\&= \sum _{l=1}^{M} \left( \sum _{j=1}^Mb_{lj}| {\psi _{j}}\rangle \right) \left( \sum _{i=1}^M{\bar{b}}_{li}\langle {\psi _{i}}|\right) \\&=\sum _{l=1}^M |\phi _l\rangle \langle \phi _l| \ge 0. \end{aligned} \end{aligned}$$

$\square$

Let H be the state space with $\dim H\le \infty$. For a state $\rho$, let $\rho =\sum _{i=1}^{N}\lambda _{i}|e_{i}\rangle \langle e_{i}|$ be the spectral decomposition of $\rho$, where $\{|e_{i}\rangle \}_{i=1}^N$ with $N\le \infty$ is an orthonormal set consisting of all eigenvectors belonging to nonzero eigenvalues $\{\lambda _{i}\}_{i=1}^N$ of the state $\rho$ with $\lambda _i\ge \lambda _{i+1}>0$. Then, ${\mathcal {E}}=\{|e_{i}\rangle , \lambda _{i} \}_{i=1}^N=\{|{\widetilde{e}}_{i}\rangle \}_{i=1}^N$ is a $\rho$-ensemble which we call an eigen-ensemble of $\rho$. Clearly, rank$(\rho )=N$, the dimension of the range of $\rho$ as an operator.

Theorem A.3

Let H be a separable complex Hilbert space.

• (a) Assume that $\{|\widetilde{\phi _{i}}\rangle \}_{i=1}^{M_1}$ and $\{|\widetilde{\psi _{j}}\rangle \}_{j=1}^{M_2}$ are two ensembles of pure states in system H, where $1\le M_1,M_2\le \infty$. If there exists an $M_1\times M_2$ contractive matrix $V=(v_{ij})$, so that $|\widetilde{\phi _{i}}\rangle =\sum _{j=1}^{M_2} v_{ij}|\widetilde{\psi _{j}}\rangle$ for each $i=1,2,\ldots M_1$, then $\sum _{i=1}^{M_1}| \widetilde{\phi _{i}}\rangle \langle \widetilde{\phi _{i}}|=\sum _{j=1}^{M_2}|\widetilde{\psi _{j}} \rangle \langle \widetilde{\psi _{j}}|$; that is, $\{|\widetilde{\phi _{i}}\rangle \}_{i=1}^{M_1}$ and $\{|\widetilde{\psi _{j}}\rangle \}_{j=1}^{M_2}$ are ensembles of pure states with the same quantum state.

• (b) Let $\{|\widetilde{\phi _{i}}\rangle \}_{i=1}^M$ be any ensemble of pure states with the quantum state $\rho \in {{\mathcal {S}}}(H)$, where $1\le M \le +\infty$. Then, there exist an $M\times M$ isometric matrix $V=(v_{ij})$ and an eigen-ensemble $\{|\widetilde{e_{j}}\rangle \}_{j=1}^N$ with quantum state $\rho$ such that $|{\widetilde{\phi }}_{i}\rangle =\sum _{j=1}^N v_{ij}|{\widetilde{e}}_{j}\rangle$ for each $i=1,2,\ldots$. Particularly, if rank$(\rho )<\infty$, V can be chosen as a unitary matrix.

• (c) Let $\{|\widetilde{\psi _{j}}\rangle \}_{j=1}^{M}$ be a sequence of vectors and $W=(w_{ij})$ be an $N\times M$ contractive matrix. Let $|\widetilde{\phi _{i}}\rangle =\sum _{j=1}^{M} w_{ij}|\widetilde{\psi _{j}}\rangle$ for each $i=1,2,\ldots N$. Then, $\mathrm{Tr}(\sum _{i=1}^N |\widetilde{\phi _{i}}\rangle \langle \widetilde{\phi _{i}}|)\le \mathrm{Tr}(\sum _{j=1}^M |\widetilde{\psi _{j}}\rangle \langle \widetilde{\psi _{j}}|)$.

Proof

(a) As $V=(v_{ij})$ is contractive, that is, $\Vert V\Vert \le 1$, we must have $V^\dag V\le I$, the $M_2\times M_2$ identity matrix. Let $B=(b_{ij})=V^\dag V$; then $b_{ij}=\sum _{l=1}^{M_1}{\bar{v}}_{ li}v_{lj}$. It follows that

$$\begin{aligned} \begin{aligned} \rho _\phi&=\sum _{l=1}^{M_1}|\widetilde{\phi _{l}}\rangle \langle \widetilde{\phi _{l}}|=\sum _{l=1}^{M_1}\sum _{j=1}^{M_2}v_{lj}| \widetilde{\psi _{j}}\rangle \sum _{i=1}^{M_2} {\overline{v}}_{li}\langle \widetilde{\psi _{i}}| \\&=\sum _{i,j=1}^{M_2}\sum _{l=1}^{M_1}v_{lj}{\overline{v}}_{li}| \widetilde{\psi _{j}}\rangle \langle \widetilde{\psi _{i}}|\\&= \sum _{i,j=1}^{M_2}b_{ij}|\widetilde{\psi _{j}}\rangle \langle \widetilde{\psi _{i}}|. \end{aligned} \end{aligned}$$

Let $A=I-B$; then $A=(a_{ij})\ge 0$ and $a_{ij}=\delta _{ij}-b_{ij}$, where $\delta _{ii}=1$ and $\delta _{ij}=0$ for $i\not =j$. By Lemma A.2, we see that

$$\begin{aligned} \begin{aligned} \rho _\psi -\rho _\phi&=\sum _{j=1}^{M_2}|\widetilde{\psi _{j}}\rangle \langle \widetilde{\psi _{j}}|-\sum _{l=1}^{M_1}|\widetilde{\phi _{l}} \rangle \langle \widetilde{\phi _{l}}|\\&= \sum _{i,j=1}^{M_2} (\delta _{ij}-b_{ij})|\widetilde{\psi _{j}} \rangle \langle \widetilde{\psi _{i}}|=\sum _{i,j=1}^{M_2} a_{ij}|\widetilde{\psi _{j}}\rangle \langle \widetilde{\psi _{i}}|\ge 0. \end{aligned} \end{aligned}$$

This forces $\rho _\phi =\rho _\psi$ because it is easily checked that, for any two states $\rho _1$ and $\rho _2$, $\rho _1\le \rho _2$ if and only if $\rho _1=\rho _2$. In fact, there exists an orthonormal basis $\{|e_j\rangle \}_{j}$ and a sequence $\{\lambda _j\}$ of nonnegative numbers such that $\rho _2=\sum _j\lambda _j|e_j\rangle \langle e_j|$. Write $\rho _1=(c_{ij})$ associated with this basis. Then, $\rho _1\le \rho _2$ implies that $c_{ii}=\langle e_i|\rho _1|e_i\rangle \le \langle e_i|\rho _2|e_i\rangle =\lambda _i$. As $\sum _i c_{ii}=\mathrm{Tr} (\rho _1)=\mathrm{Tr} (\rho _2)=\sum _i\lambda _i=1$, we must have $c_{ii}=\lambda _i$. Thus, $\rho _2-\rho _1$ is a positive semi-definite matrix with zero diagonal entries, and hence, $\rho _2-\rho _1=0$.

(b) Let $\{|\widetilde{\phi _{i}}\rangle \}_{i=1}^M$ be any ensemble of pure states with the quantum state $\rho$, where $1\le M \le +\infty$. Consider an auxiliary system with the associated Hilbert space K of dimension M and fix an orthonormal basis $\{|g_{j}\rangle \}_{j=1}^M$ of K. Then, $|\Psi \rangle = \sum _{i=1}^M|\widetilde{\phi _{i}}\rangle \otimes |g_{i}\rangle$ is a purification of $\rho$, that is, $\text {Tr}_K|\Psi \rangle \langle \Psi |=\rho$. By the Schmidt decomposition for pure states in bipartite systems, there exists a sequence $\{\xi _j\}_{j=1}^{N}$ with $\xi _j\ge \xi _{j+1}>0$ for each j, an orthonormal sequence $\{|e_j\rangle \}_{j=1}^{N}$ in H and an orthonormal sequence $\{|f_j\rangle \}_{j=1}^{N}$ in K with $1\le N\le \infty$ such that $|\Psi \rangle =\sum _{j=1}^{N}\xi _j|e_{j}\rangle \otimes |f_{j}\rangle$. Since $\rho =\mathrm{Tr}_K|\Psi \rangle \langle \Psi |=\sum _{j=1}^{N}\xi _j^2|e_j\rangle \langle e_j|$, we see that the rank of $\rho$ is N and $\{|e_j\rangle ,\xi ^2_j\}_{j=1}^N$ is an eigen-ensemble with $\rho$. Let $|{\widetilde{e}}_{j}\rangle =\xi _j|e_j\rangle$; then

$$\begin{aligned} |\Psi \rangle =\sum _{j=1}^N|{\widetilde{e}}_{j}\rangle \otimes |f_{j}\rangle . \end{aligned}$$

If $N<+\infty$ (particularly, if $M<\infty$), we can extend $\{|f_{j}\rangle \}_{j=1}^N$ to an orthonormal basis $\{|f_{j}\rangle \}_{j=1}^M$ of K, and let the correspondent $|{\widetilde{e}}_{j}\rangle =|0\rangle$ for $N<j\le M$. Since any two orthonormal bases are unitarily related, we see that there exists an $M\times M$ unitary matrix $U=(u_{ij})$ such that $|f_{l}\rangle =\sum _{i=1}^M u_{il}|g_{i}\rangle$. It follows that

$$\begin{aligned} \begin{aligned} \sum _{i=1}^M|\widetilde{\phi _{i}}\rangle \otimes |g_{i}\rangle&= |\Psi \rangle =\sum _{j=1}^M|{\widetilde{e}}_{j}\rangle \otimes |f_{j}\rangle \\&=\sum _{j=1}^M|{\widetilde{e}}_{j}\rangle \otimes \sum _{i=1}^M u_{ij}|g_{i}\rangle \\&=\sum _{i=1}^M \left( \sum _{j=1}^M u_{ij}|{\widetilde{e}}_{j} \rangle \right) \otimes |g_{i}\rangle . \end{aligned} \end{aligned}$$

So we have $\sum _{i=1}^M(\sum _{j=1}^M u_{ij}|{\widetilde{e}}_{j} \rangle -|\widetilde{\phi _{i}}\rangle ) \otimes |g_{i}\rangle =0$. By Lemma A.1, we get

$$\begin{aligned} |\widetilde{\phi _{i}}\rangle =\sum _{j=1}^M u_{ij}|{\widetilde{e}}_{j}\rangle =\sum _{j=1}^N u_{ij}|{\widetilde{e}}_{j}\rangle , \end{aligned}$$

$i=1,2,\ldots ,M$.

If $N=+\infty$, we have $M=+\infty$ as $M\ge N$. So there exists an isometric operator $V=(v_{ij})$ on K such that $|f_{j}\rangle =\sum _{i=1}^\infty v_{ij}|g_{i}\rangle$, $j=1,2,\ldots$. It follows that

$$\begin{aligned} \begin{aligned} \sum _{i=1}^\infty |\widetilde{\phi _{i}}\rangle \otimes |g_{i}\rangle = |\Psi \rangle&=\sum _{j=1}^\infty |{\widetilde{e}}_{j}\rangle \otimes |f_{j}\rangle \\&=\sum _{j=1}^\infty |{\widetilde{e}}_{j}\rangle \otimes \sum _{i=1}^\infty v_{ij}|g_{i}\rangle \\&=\sum _{i=1}^\infty \left( \sum _{j=1}^\infty v_{ij}|{\widetilde{e}}_{j}\rangle \right) \otimes |g_{i}\rangle . \end{aligned} \end{aligned}$$

Hence, by Lemma A.1, we get $|{\widetilde{\phi }}_{i} \rangle =\sum _{j=1}^\infty v_{ij}|{\widetilde{e}}_{j}\rangle$ for each $i=1,2,\dots$. Equivalently,

$$\begin{aligned} \begin{pmatrix} |{\widetilde{\phi }}_{1}\rangle \\ |{\widetilde{\phi }}_{2}\rangle \\ \vdots \\ |{\widetilde{\phi }}_{i}\rangle \\ \vdots \end{pmatrix} =V \begin{pmatrix} |{\widetilde{e}}_{1}\rangle \\ |{\widetilde{e}}_{2}\rangle \\ \vdots \\ |{\widetilde{e}}_{j}\rangle \\ \vdots \end{pmatrix} \end{aligned}$$

with V an isometric matrix.

(c) Clearly, we may assume that $\{|\widetilde{\psi _{j}} \rangle \}_{j=1}^{M}$ is a quantum ensemble of pure states. Let $\rho =\sum _{i=1}^M |\widetilde{\psi _{i}}\rangle \langle \widetilde{\psi _{i}}|$ and $\{|\widetilde{e_{j}}\rangle \}_{j=1}^P$ be an eigen-ensemble with the state $\rho$. Then, by (b) and its proof, there exists a $P\times M$ isometric matrix $V=(v_{ij})$ such that $|{\widetilde{\psi }}_{j}\rangle =\sum _{k=1}^P v_{jk}|{\widetilde{e}}_{k}\rangle$ for each $j=1,2,\ldots , M$. Let ${{\mathcal {E}}}= \{|\widetilde{\psi _{j}}\rangle \}_{j=1}^M$, ${{\mathcal {E}}}_0= \{|\widetilde{e_{k}}\rangle \}_{k=1}^P$ and ${{\mathcal {E}}}_1=\{|\widetilde{\phi _{i}}\rangle \}_{i=1}^N$. Then, we have ${{\mathcal {E}}}_1= W{{\mathcal {E}}}= WV{{\mathcal {E}}}_0=R{{\mathcal {E}}}_0$, where $R=WV=(r_{ik})$ is an $N\times P$ contractive matrix. It follows that, for any k, one has $\sum _{i=1}^N |r_{ik}|^2\le 1$. Then,

$$\begin{aligned} \begin{aligned} \mathrm{Tr}\left( \sum _{i=1}^N |\widetilde{\phi _{i}}\rangle \langle \widetilde{\phi _{i}}|\right)&= \sum _{i=1}^N \mathrm{Tr} \left( \sum _{k,h=1}^P r_{ik}{\bar{r}}_{ih}|\widetilde{e_{k}}\rangle \langle \widetilde{e_{h}}|\right) \\&=\sum _{i=1}^N \sum _{k=1}^P |r_{ik}|^2\Vert |\widetilde{e_{k}}\rangle \Vert ^2 \\&= \sum _{k=1}^P\left( \sum _{i=1}^N |r_{ik}|^2\right) \Vert | \widetilde{e_{k}}\rangle \Vert ^2 \\&\le \sum _{k=1}^P \Vert |\widetilde{e_{k}}\rangle \Vert ^2 =1=\mathrm{Tr}\left( \sum _{j=1}^M |\widetilde{\psi _{j}}\rangle \langle \widetilde{\psi _{j}}|\right) , \end{aligned} \end{aligned}$$

as desired. $\square$

Now we are at a position to give a proof of the Contractive freedom in ensembles.

Proof of Theorem 1

The implications “(2)$\Rightarrow$(3)$\Rightarrow$(4)” are obvious. “(4)$\Rightarrow$(1)” is an immediate consequence of Theorem A.3 (a). The remain is to check the implication “(1)$\Rightarrow$(2).”

Assume that $\{|\widetilde{\phi _{i}}\rangle \}_{i=1}^{M_1}$ and $\{|\widetilde{\psi _{i}}\rangle \}_{i=1}^{M_2}$ are two ensembles of pure states with a quantum state $\rho$, that is, $\rho =\sum _{i=1}^{M_1}|\widetilde{\phi _{i}}\rangle \langle \widetilde{\phi _{i}}|=\sum _{i=1}^{M_2}|\widetilde{\psi _{i}} \rangle \langle \widetilde{\psi _{i}}|$. With no loss of the generality, assume that $1\le M_2\le M_1\le +\infty$. Making use of (b) in Theorem A.3, there exists an $M_1\times M_1$ isometric matrix $V_1$ and an $M_2\times M_2$ isometric matrix $V_2$ such that

$$\begin{aligned} \begin{pmatrix} |{\widetilde{\phi }}_{1}\rangle \\ |{\widetilde{\phi }}_{2}\rangle \\ \vdots \\ |{\widetilde{\phi }}_{i}\rangle \\ \vdots \end{pmatrix}=V_1 \begin{pmatrix} |{\widetilde{e}}_{1}\rangle \\ |{\widetilde{e}}_{2}\rangle \\ \vdots \\ |{\widetilde{e}}_{j}\rangle \\ \vdots \end{pmatrix} \quad \mathrm{and} \quad \begin{pmatrix} |{\widetilde{\psi }}_{1}\rangle \\ |{\widetilde{\psi }}_{2}\rangle \\ \vdots \\ |{\widetilde{\psi }}_{i}\rangle \\ \vdots \end{pmatrix}=V_2 \begin{pmatrix} |{\widetilde{e}}_{1}^\prime \rangle \\ |{\widetilde{e}}_{2}^\prime \rangle \\ \vdots \\ |{\widetilde{e}}_{j}^\prime \rangle \\ \vdots \end{pmatrix} \end{aligned}$$

(A1)

for some eigen-ensembles $\{|{\widetilde{e}}_{i}\rangle \}_{i=1}^{M_1}$ and $\{|{\widetilde{e}}_{i}^\prime \rangle \}_{j=1}^{M_2}$ of $\rho$ with $|{\widetilde{e}}_{l}\rangle =0$ and $|{\widetilde{e}}_{h}^\prime \rangle =0$ whenever $l,h> N$, where N is the rank of $\rho$.

Case 1 $N<\infty$.

By Theorem A.3 (b) both $V_1$ and $V_2$ can be chosen as unitary matrices. As $M_2\le M_1\le \infty$, if $M_2\not =M_1$, then $M_2<\infty$. Thus, we can add, respectively $\{ |{\widetilde{e}}_{M_2+1}^\prime \rangle , |{\widetilde{e}}_{M_2+2}^\prime \rangle ,\ldots \}$ and $\{|{\widetilde{\psi }}_{M_2+1}\rangle , |{\widetilde{\psi }}_{M_2+2}\rangle ,\ldots \}$ into $\{|{\widetilde{e}}_{i}^\prime \rangle \}_{i=1}^{M_2}$ and $\{|{\widetilde{\psi }}_{i}\rangle \}_{i=1}^{M_2}$ with $|{\widetilde{e}}_{l}^\prime \rangle =0$ and $|{\widetilde{\psi }}_{l}\rangle =0$ if $l>M_2$. Clearly, $V_2$ can be extended to an $M_1\times M_1$ unitary matrix so that Eq. (A1) still holds with $V_1, V_2$ unitary and $M_2=M_1$.

Also, it is easily checked that there exists an $M_1\times M_1$ unitary matrix V such that

$$\begin{aligned} \begin{pmatrix} |{\widetilde{e}}_{1}\rangle \\ |{\widetilde{e}}_{2}\rangle \\ \vdots \\ |{\widetilde{e}}_{i}\rangle \\ \vdots \end{pmatrix}=V \begin{pmatrix} |{\widetilde{e}}_{1}^\prime \rangle \\ |{\widetilde{e}}_{2}^\prime \rangle \\ \vdots \\ |{\widetilde{e}}_{j}^\prime \rangle \\ \vdots \end{pmatrix} \end{aligned}$$

(A2)

as both $\{|{\widetilde{e}}_1\rangle , |{\widetilde{e}}_2\rangle , \ldots , |{\widetilde{e}}_N\rangle \}$ and $\{|{\widetilde{e}}_1^\prime \rangle , |{\widetilde{e}}_2^\prime \rangle , \ldots , |{\widetilde{e}}_N^\prime \rangle \}$ are eigen-ensembles of $\rho$. It follows from Eq. (A1) that

$$\begin{aligned} \begin{aligned} \begin{pmatrix} |{\widetilde{\phi }}_{1}\rangle \\ |{\widetilde{\phi }}_{2}\rangle \\ \vdots \\ |{\widetilde{\phi }}_{i}\rangle \\ \vdots \end{pmatrix}&= V_1 \begin{pmatrix} |{\widetilde{e}}_{1}\rangle \\ |{\widetilde{e}}_{2}\rangle \\ \vdots \\ |{\widetilde{e}}_{i}\rangle \\ \vdots \end{pmatrix}=V_1V\begin{pmatrix} |{\widetilde{e}}_{1}^\prime \rangle \\ |{\widetilde{e}}_{2}^\prime \rangle \\ \vdots \\ |{\widetilde{e}}_{j}^\prime \rangle \\ \vdots \end{pmatrix} \\&= V_1VV_2^\dag \begin{pmatrix} |{\widetilde{\psi }}_{1}\rangle \\ |{\widetilde{\psi }}_{2}\rangle \\ \vdots \\ |{\widetilde{\psi }}_{j}\rangle \\ \vdots \end{pmatrix}= W \begin{pmatrix} |{\widetilde{\psi }}_{1}\rangle \\ |{\widetilde{\psi }}_{2}\rangle \\ \vdots \\ |{\widetilde{\psi }}_{j}\rangle \\ \vdots \end{pmatrix}. \end{aligned} \end{aligned}$$

(A3)

with $W=V_1VV_2^\dag$. Note that, W is unitary in this case.

Particularly, we have shown that the last assertion in Theorem 1 concerning $\min \{M_1,M_2\}<\infty$ is true.

Case 2 $N=\infty$.

In this case, $N=M_1=M_2=\infty$. By Theorem A.3 (b), Eq. (A1) is true with $V_1$ and $V_2$ isometric matrices. Since Eq. (A2) is also valid for the case $N=\infty$ and Eq. (A3) is true with $V_1$ and $V_2$ isometric matrices, we see that $|{\widetilde{\phi }}_{i}\rangle =\sum _j w_{ij}|{\widetilde{\psi }}_{j}\rangle$ for each i with $W=V_1VV_2^\dag$. Clearly, W is contractive and

$$\begin{aligned} \begin{aligned} \begin{pmatrix} |{\widetilde{\psi }}_{1}\rangle \\ |{\widetilde{\psi }}_{2}\rangle \\ \vdots \\ |{\widetilde{\psi }}_{j}\rangle \\ \vdots \end{pmatrix}&= V_2 \begin{pmatrix} |{\widetilde{e}}_{1}^\prime \rangle \\ |{\widetilde{e}}_{2}^\prime \rangle \\ \vdots \\ |{\widetilde{e}}_{j}^\prime \rangle \\ \vdots \end{pmatrix}=V_2V^\dag \begin{pmatrix} |{\widetilde{e}}_{1}\rangle \\ |{\widetilde{e}}_{2}\rangle \\ \vdots \\ |{\widetilde{e}}_{i}\rangle \\ \vdots \end{pmatrix} \\&= V_2V^\dag V_1^\dag \begin{pmatrix} |{\widetilde{\phi }}_{1}\rangle \\ |{\widetilde{\phi }}_{2}\rangle \\ \vdots \\ |{\widetilde{\phi }}_{i}\rangle \\ \vdots \end{pmatrix}=W^\dag \begin{pmatrix} |{\widetilde{\phi }}_{1}\rangle \\ |{\widetilde{\phi }}_{2}\rangle \\ \vdots \\ |{\widetilde{\phi }}_{i}\rangle \\ \vdots \end{pmatrix} \end{aligned} \end{aligned}$$

(A4)

gives that $|{\widetilde{\psi }}_{j}\rangle =\sum _i {\overline{w}}_{ij}|{\widetilde{\phi }}_{i}\rangle$ for each j, completing the proof. $\square$

Appendix B: Proof of the contractive freedom in the operator-sum representation

The purpose of “Appendix B” is to give a proof of the main result Theorem 6 in Sect. 3.

Proof

The implications “(2)$\Rightarrow$(3)$\Rightarrow$(4)” are obvious.

“(4)$\Rightarrow$(1)”: Assume that there exists a $\max \{M,N\}\times \max \{M,N\}$ contractive matrix $V=(v_{ij})$ such that $A_i=\sum _j v_{ij} B_j$ for each i. For any pure state $|\psi \rangle \in H$, let $\rho _i=\Phi _i(|\psi \rangle \langle \psi |)$, $i=1,2$. Then, $\rho _1=\sum _i A_i|\psi \rangle \langle \psi |A_i^\dag$ and $\rho _2=\sum _j B_j|\psi \rangle \langle \psi |B_j^\dag$ are two states. Since $A_i|\psi \rangle =\sum _j v_{ij} B_j|\psi \rangle$ and $V=(v_{ij})$ is contractive, by Theorem 1, we have $\rho _1=\rho _2$. So, $\Phi _1(|\psi \rangle \langle \psi |)=\Phi _2(|\psi \rangle \langle \psi |)$ holds for all pure states $|\psi \rangle$, which entails $\Phi _1=\Phi _2$.

“(1)$\Rightarrow$(2)”: Assume that $\Phi _1=\Phi _2$. Note that, the last assertion of Theorem 6 is just [21, Theorem 2.4] and thus is true. So we may assume that $M=N=\infty$.

For any positive integer n and any pure state $|\psi \rangle \in H^{(n)}=H\oplus H\oplus \cdots \oplus H$, the direct sum of n copies of H, $\Phi _1=\Phi _2$ implies that

$$\begin{aligned} \sum _{i=1}^\infty A_i^{(n)}|\psi \rangle \langle \psi |A_i^{(n)^\dag }= \sum _{j=1}^\infty B_j^{(n)}|\psi \rangle \langle \psi |B_j^{(n)^\dag }\in {{\mathcal {S}}}(K^{(n)}), \end{aligned}$$

where $A^{(n)}$ stands for the direct sum of n copies of A. So, $\{A_i^{(n)}|\psi \rangle \}_{i=1}^M$ and $\{B_j^{(n)}|\psi \rangle \}_{j=1}^N$ are two ensembles of pure states with the same state. By the dual bi-contractive freedom in ensembles (Theorem 1), there exists an $M\times M$ contractive matrix $W_\psi =(w_{ij}(\psi ))$, such that

$$\begin{aligned} A_i^{(n)}|\psi \rangle =\sum _{j=1}^\infty w_{ij}(\psi )B_j^{(n)}|\psi \rangle , \; i=1, 2,\dots , \end{aligned}$$

and

$$\begin{aligned} B_j^{(n)}|\psi \rangle =\sum _{j=1}^\infty {\overline{w}}_{ij}(\psi )A_i^{(n)}|\psi \rangle , \; i=1, 2,\dots . \end{aligned}$$

Equivalently, we have shown that, for any $|\psi \rangle \in H^{(n)}$, there is a contractive operator $W_\psi \in {\mathcal {B}}(l^2)$ such that

$$\begin{aligned} \mathbf{A }^{(n)}|\psi \rangle =(W_\psi \otimes I) \mathbf{B }^{(n)}|\psi \rangle , \end{aligned}$$

(B1)

and

$$\begin{aligned} \mathbf{B }^{(n)}|\psi \rangle =(W_\psi ^\dag \otimes I) \mathbf{A }^{(n)}|\psi \rangle , \end{aligned}$$

(B2)

where

$$\begin{aligned} \mathbf{A }^{(n)}= \begin{pmatrix} A_1^{(n)}\\ A_2^{(n)}\\ \vdots \\ A_i^{(n)}\\ \vdots \end{pmatrix} \qquad \mathrm{and} \qquad \mathbf{B }^{(n)}= \begin{pmatrix} B_1^{(n)}\\ B_2^{(n)}\\ \vdots \\ B_j^{(n)}\\ \vdots \end{pmatrix}. \end{aligned}$$

It follows that, for any vectors $|\psi _1\rangle , \dots , |\psi _n\rangle \in H$, by letting $|\psi \rangle =|\psi _1\rangle \oplus |\psi _2\rangle \oplus \cdots \oplus |\psi _n\rangle \in H^{(n)}$ and applying Eqs. (B1) and (B2),

$$\begin{aligned} \mathbf{A }|\psi _k\rangle =(W_\psi \otimes I) \mathbf{B }|\psi _k\rangle \end{aligned}$$

(B3)

and

$$\begin{aligned} \mathbf{B }|\psi _k\rangle =(W_\psi ^\dag \otimes I) \mathbf{A }|\psi _k\rangle \end{aligned}$$

(B4)

hold for each $k=1,2,\dots ,n.$ In other words, Eqs. (B3) and (B4) ensure that, for any finite set $F\subset H$, there exists a contractive operator $W_F\in {{\mathcal {B}}}(l^2)$ such that

$$\begin{aligned} \mathbf{A }|\psi \rangle =(W_F\otimes I) \mathbf{B }|\psi \rangle \quad \text{ for } \text{ every } \ |\psi \rangle \in F \end{aligned}$$

(B5)

and

$$\begin{aligned} \mathbf{B }|\psi \rangle =(W_F^\dag \otimes I) \mathbf{A }|\psi \rangle \quad \text{ for } \text{ every } \ |\psi \rangle \in F. \end{aligned}$$

(B6)

Let ${{\mathcal {F}}}=\{F: F\subset H \, \hbox {is finite}\}$. For $F_1,F_2\in {{\mathcal {F}}}$, we say $F_1\preceq F_2$ if $F_1\subseteq F_2$. Then, $({{\mathcal {F}}}, \preceq )$ is a directed set and consequently, $\{W_F\}_{F\in {\mathcal {F}}}$ is a net in ${{\mathcal {B}}}_1(l^2)$, the unit ball of ${{\mathcal {B}}}(l^2)$. It follows from Eqs. (B5) and (B6) that

$$\begin{aligned} \mathrm{WOT-}\lim _F(W_F\otimes I) \mathbf{B }=\mathbf{A } \end{aligned}$$

and

$$\begin{aligned} \mathrm{WOT-}\lim _F (W_F^\dag \otimes I) \mathbf{A }=\mathbf{B } \end{aligned}$$

as operators in ${{\mathcal {B}}}(H, K^{(\infty )})$, where WOT-$\lim$ stands for taking limits under the weak operator topology (briefly, WOT). On the other hand, since the closed unit ball ${{\mathcal {B}}}_1(l^2)$ of ${{\mathcal {B}}}(l^2)$ is WOT compact [22, Theorem 5.1.3.], there exists a subnet $\{W_{F_\alpha }\}$ of $\{W_F\}$ so that WOT-$\lim _{F_\alpha } W_{F_\alpha }=V$ for some $V\in {{\mathcal {B}}}_1(l^2)$. Therefore, we have

$$\begin{aligned} \begin{aligned} (V\otimes I)\mathbf{B }&= \mathrm{WOT-}\lim _{F_\alpha } (W_{F_\alpha }\otimes I)\mathbf{B } \\&=\mathrm{WOT-}\lim _F (W_{F }\otimes I)\mathbf{B }=\mathbf{A }, \end{aligned} \end{aligned}$$

that is,

$$\begin{aligned} \mathbf{A }=(V\otimes I)\mathbf{B }. \end{aligned}$$

(B7)

Since the conjugation is continuous under the weak operator topology, we see that

$$\begin{aligned} \begin{aligned} (V^\dag \otimes I)\mathbf{A }&= \mathrm{WOT-} \lim _{F_\alpha } (W_{F_\alpha }^\dag \otimes I)\mathbf{A } \\&=\mathrm{WOT-} \lim _F (W_{F }^\dag \otimes I)\mathbf{A }=\mathbf{B }, \end{aligned} \end{aligned}$$

that is,

$$\begin{aligned} \mathbf{B }=(V^\dag \otimes I)\mathbf{A }. \end{aligned}$$

(B8)

Obviously, Eqs. (B7) and (B8) together are equivalent to the statement (2). So (2) is true and the proof completes. $\square$