Structures of Three Types of Local Quantum Channels Based on Quantum Correlations

Abstract

In a bipartite quantum system, quantum states are classified as classically correlated (CC) and quantum correlated (QC) states, the later are important resources of quantum information and computation protocols. Since correlations of quantum states may vary under a quantum channel, it is necessary to explore the influence of quantum channels on correlations of quantum states. In this paper, we discuss CC-preserving, QC-breaking and strongly CC-preserving local quantum channels of the form $\Phi _1\otimes \Phi _2$ and obtain the structures of these three types of local quantum channels. Moreover, we obtain a necessary and sufficient condition for a quantum state to be transformed into a CC state by a specific local channel $\Phi _1\otimes \Phi _2$ in terms of the structure of the input quantum state. Lastly, as applications of the obtained results, we present a classification of local quantum channels $\Phi _1\otimes \Phi _2$ and describe the quantum states which are transformed as CC ones by the corresponding local quantum channel.

Appendix: Proofs of Lemmas

Proof of Lemma 2.3 $(1)\Rightarrow (2):$ Suppose that $\Phi (\mathcal {M}_n)$ is commutative, then the operators in $\Phi (\mathcal {M}_n)$ are all normal and pairwise commute. Assume that $\Phi (\mathcal {M}_n)=\text{ span }\{A_1,A_2,\ldots ,A_m\},\ m=\dim (\Phi (\mathcal {M}_n)),$ where $[A_i,A_j]=0$ for $i\ne j$, then there exists an orthonormal basis $\{|k\rangle \}_{k=1}^n$ for $\mathbb {C}^n$ such that $A_j=\sum _{k=1}^n \lambda _{kj}|k\rangle \langle k|$ for all $j=1,2,\ldots ,m$. For any $A\in \mathcal {M}_n$, there exists a unique sequence $\{c_j(A)\}_{j=1}^m$ of complex numbers such that

$$\begin{aligned} \Phi (A)=\sum _{j=1}^mc_j(A)A_j=\sum _{k=1}^n\sum _{j=1}^m c_j(A)\lambda _{kj}|k\rangle \langle k|=\sum _{k=1}^nf_k(A)|k\rangle \langle k|, \end{aligned}$$

where $f_k(A)=\sum _{j=1}^mc_j(A)\lambda _{kj}$ for all $k$. Clearly, $f_k(A)=\langle k|\Phi (A)|k\rangle =\text{ tr }(\Phi ^\dag (|k\rangle \langle k|)A)$, where $\Phi ^\dag$ denotes the dual map of $\Phi$ with respect to the Hilbert-Schmidt inner product on $\mathcal {M}_n$. Thus,

$$\begin{aligned} \Phi (A)=\sum _{k=1}^n\text{ tr }(\Phi ^\dag (|k\rangle \langle k|)A)|k\rangle \langle k|,\ \ \forall A\in \mathcal {M}_n. \end{aligned}$$

It is easy to see that $\{\Phi ^\dag (|k\rangle \langle k|)\}$ is a POVM for ${\mathbb {C}}^n$. This shows that $\Phi$ is a measurement map.

$(2)\Rightarrow (1)$ and $(1)\Rightarrow (3)$: Evidently.

$(3)\Rightarrow (1):$ Let (3) hold. Then $[\Phi (\sigma _1), \Phi (\sigma _2)]=0$ for all positive operators $\sigma _1,\sigma _2$ on ${\mathbb {C}}^n$. By using the spectral theorem we see that $[\Phi (\sigma _1), \Phi (\sigma _2)]=0$ for all Hermitian operators $\sigma _1,\sigma _2$ on ${\mathbb {C}}^n$. So, $[\Phi (\sigma _1), \Phi (\sigma _2)]=0$ for all operators $\sigma _1,\sigma _2$ on ${\mathbb {C}}^n$. $\square$

Proof of Lemma 4.1

To show this, without loss of generality, we may assume that $\Phi (A) = tUAU^\dag + \frac{1-t}{n}\text{ tr }(A)I_n$ for all $A\in \mathcal {M}_n$ where $U$ is a unitary matrix, $-\frac{1}{n-1}\le t\le 1$ and $t\ne 0$. Clearly, for any $Y\in \mathcal {M}_n$, the matrix $X=\frac{1}{t}(U^\dag YU-\frac{1-t}{n}\text{ tr }(Y)I_n)$ satisfies $\Phi (X)=Y$. This implies that $\Phi$ is surjective. On the other hand, let $\Phi (A)=\Phi (B)$, that is, $tU(A-B)U^\dag +\frac{1-t}{n}\text{ tr }(A-B)I_n=0.$ Then we have $A-B=\frac{t-1}{tn}\text{ tr }(A-B)I_n$ since $t\ne 0$. Now we take trace operation and get that $(t-1)\text{ tr }(A-B)=t\cdot \text{ tr }(A-B)$. This shows that $\text{ tr }(A-B)=0$ and thus $\text{ tr }(A)=\text{ tr }(B)$. Since $\Phi (A)=\Phi (B)$, we get that $tUAU^\dag =tUBU^\dag$ and so $A=B$. Hence, $\Phi$ is injective. $\square$

Proof of Lemma 4.2

The “if part” is clear. We only need to prove the “only if part”. Let $\sum _{i=1}^k X_i\otimes Y_i=0$. Then for any pure states $|c\rangle , |d\rangle \in \mathbb {C}^m$ we have

$$\begin{aligned} tr _2\left( \left( \sum _{i=1}^k X_i\otimes Y_i\right) (I_n\otimes |d\rangle \langle c|)\right) =\sum _{i=1}^k \langle c|Y_i|d\rangle X_i=0. \end{aligned}$$

Since $\{X_i\}_{i=1}^{k}$ is a linearly independent family, $\langle c|Y_i|d\rangle =0$ for all $i$ and all $|c\rangle , |d\rangle \in \mathbb {C}^m$. That is, $Y_i=0$ for all $i\in \{1,2,\ldots , k\}.$ $\square$

Proof of Lemma 4.3

For any $\rho =\sum _iA_i\otimes B_i$, there exist $C_i, D_i$ such that $\Phi _1(C_i)=A_i$ and $\Phi _2(D_i)=B_i$ for all $i$. Put $\sigma =\sum _iC_i\otimes D_i$, then $(\Phi _1\otimes \Phi _2)(\sigma )=\rho$ and thus $\Phi _1\otimes \Phi _2$ is surjective. On the other hand, for any $\sigma _1,\sigma _2$ in $\mathcal {M}_n\otimes \mathcal {M}_m$, we write $\sigma _1=\sum _{i,j}E_{ij}\otimes A_{ij}, \sigma _2=\sum _{i,j}E_{ij}\otimes B_{ij},$ where $E_{ij}=|i\rangle \langle j|$ and $\{|i\rangle \}$ is an orthonormal basis for ${\mathbb {C}}^n$. Suppose that $(\Phi _1\otimes \Phi _2)(\sigma _1)=(\Phi _1\otimes \Phi _2)(\sigma _2)$, then

$$\begin{aligned} \sum _{i,j}\Phi _1(E_{ij})\otimes [\Phi _2(A_{ij})-\Phi _2(B_{ij})]=0. \end{aligned}$$

Since $\{E_{ij}\}$ is a Hamel basis for $\mathcal {M}_n$ and $\Phi _1$ is a linear isomorphism, $\{\Phi _1(E_{ij})\}$ is a linearly independent set. By Lemma 4.2, we have $\Phi _2(A_{ij})-\Phi _2(B_{ij})=0$ for all $i,j$, so $A_{ij}=B_{ij}$ for all $i,j$ since $\Phi _2$ is also a bijection. Therefore, $\sigma _1=\sigma _2$. Thus, $\Phi _1\otimes \Phi _2$ is a linear bijection. $\square$

Proof of Lemma 4.4

Necessity Suppose that $\Phi _1\otimes \Phi _2$ is strongly CC-preserving. Then $\Phi _1$ and $\Phi _2$ are commutativity preserving on states and then commutativity preserving. Assume that $\Phi _2$ is not commutativity preserving in both directions. Then there exist $\rho _2,\sigma _2\in \mathcal {D}({\mathbb {C}}^m)$ such that $[\rho _2,\sigma _2]\ne 0$ but $[\Phi _2(\rho _2),\Phi _2(\sigma _2)]=0$. Choose $\rho _1,\sigma _1\in \mathcal {D}({\mathbb {C}}^n)$ such that $\rho _1\ne \sigma _1$ and $[\rho _1,\sigma _1]=0$, and let $\rho =\frac{1}{2}\rho _1\otimes \rho _2+\frac{1}{2}\sigma _1\otimes \sigma _2.$ Then $\rho$ is not a CC state (Lemma 3.1) and

$$\begin{aligned} (\Phi _1\otimes \Phi _2)(\rho )=\frac{1}{2}\Phi _1(\rho _1)\otimes \Phi _2(\rho _2) +\frac{1}{2}\Phi _1(\sigma _1)\otimes \Phi _2(\sigma _2). \end{aligned}$$

Since $[\Phi _1(\rho _1),\Phi _1(\sigma _1)]=0$ and $[\Phi _2(\rho _2),\Phi _2(\sigma _2)]=0$, it follows from Lemma 3.1 that $(\Phi _1\otimes \Phi _2)(\rho )$ is a CC state, while $\rho$ is not a CC state, a contradiction. Thus, $\Phi _2$ is commutativity preserving in both directions. Similarly, one can show that $\Phi _1$ is commutativity preserving in both directions.

Sufficiency. Suppose that $\Phi _1$ and $\Phi _2$ are commutativity preserving in both directions. First, let us check that both $\Phi _1$ and $\Phi _2$ are bijective. To do this, we assume that $\Phi _1(X)=0$. Then $[\Phi _1(X),\Phi _1(A)]=0$ for all $A\in \mathcal {M}_n$. Since $\Phi _1$ is commutativity preserving in both directions, we conclude that $X=cI_n$. Because that $\Phi _1$ is trace-preserving, we see that $c=0$ and then $X=0$. This shows that $\Phi _1$ is injective and so bijective since $\dim {\mathcal {M}_n}=n^2<\infty$. Similarly, $\Phi _2$ is bijective. Let $\rho \in CC({\mathbb {C}}^n\otimes {\mathbb {C}}^m)$. Then by using Lemma 3.2, we can find two commuting families of normal operators $\{C_i\}$ and $\{D_i\}$ such that $\rho =\sum _iC_i\otimes D_i$. Thus, $(\Phi _1\otimes \Phi _2)(\rho )=\sum _i\Phi _1(C_i)\otimes \Phi _2(D_i)$. Since $\Phi _1$ and $\Phi _2$ are commutativity preserving and $\dag$-preserving, we see that $\{\Phi _1(C_i)\}$ and $\{\Phi _2(D_i)\}$ are commuting families of normal operators. By using Lemma 3.2 again, we conclude that $(\Phi _1\otimes \Phi _2)(\rho )$ is a CC state. Let $\rho \in \mathcal {D}({\mathbb {C}}^n\otimes {\mathbb {C}}^m)$ and $(\Phi _1\otimes \Phi _2)(\rho )$ be a CC state. It follows from Lemma 3.2 that $(\Phi _1\otimes \Phi _2)(\rho )=\sum _iA_i\otimes B_i$ for some commuting families $\{A_i\}$ and $\{B_i\}$ of normal operators. Since $\Phi _1$ and $\Phi _2$ are bijective (Lemma 4.1) and commutativity preserving in both directions, we can find two commuting families of normal operators $\{C_i\}$ and $\{D_i\}$ such that $\Phi _1(C_i)=A_i$ and $\Phi _2(D_i)=B_i$ for all $i$. Therefore, $(\Phi _1\otimes \Phi _2)(\sum _iC_i\otimes D_i)=\sum _iA_i\otimes B_i$ and thus $\rho =\sum _iC_i\otimes D_i$ since $\Phi _1\otimes \Phi _2$ are injective. Hence, $\rho \in CC({\mathbb {C}}^n\otimes {\mathbb {C}}^m)$ (Lemma 3.2). Thus, $\Phi _1\otimes \Phi _2$ is strongly CC-preserving. $\square$