Channel polarization of two-dimensional-input quantum symmetric channels

Being attracted by the property of classical polar code, researchers are trying to find its analogue in quantum fields, which is called quantum polar code. The first step and the key to design quantum polar code is to find out for the quantity which can measure the quality of quantum channels, whether there is a polarization phenomenon which is similar to classical channel polarization. Coherent information is believed to be the quantum analogue of classical mutual information and the quantity to measure the capacity of quantum channel. In this paper, we define a class of quantum channels called quantum symmetric channels and prove that for quantum symmetric channels, under the similar channel combining and splitting process as in the classical channel polarization, the maximum single-letter coherent information of the coordinate channels will polarize. That is to say, there is a channel polarization phenomenon in quantum symmetric channels.

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Authors and Affiliations

1. Harbin Institute of Technology, Shenzhen, Shenzhen, 518055, China

   Zhengzhong Yi, Zhipeng Liang & Xuan Wang

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All authors conceived the work, analyzed the results, and wrote the manuscript.

Corresponding author

Correspondence to Xuan Wang.

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The authors have no competing interests to declare that are relevant to the content of this article.

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Appendix A: Proof of Proposition 5

Proof

Proving \(\mathcal {E}^{\otimes N}\) is a QSC is to prove each row of its BTPM is a permutation of the first row and each column of its BTPM is a permutation of the first column. Here, we will use the same method, which is used to prove Theorem 8 in Sect. 3.1, to prove it.

Suppose the input and output of \(\mathcal {E}^{\otimes N}\) are \(|Q_{1}^{N}\rangle \) and \(|V_{1}^{N}\rangle \), respectively, where \(Q_{1}^{N}=(Q_{1},\dots ,Q_{N})\in \{0,1\}^{N}\) and \(V_{1}^{N}=(V_{1},\dots ,V_{N})\in \{0,1\}^{N}\). Then, the basis transition probability is

Since \(\mathcal {E}\) is a QSC, we have

Hence, we have

which means each row of the BTPM of \(\mathcal {E}^{\otimes N}\) is a permutation of the first row and each column of its BTPM is a permutation of the first column. Thus, \(\mathcal {E}^{\otimes N}\) is a QSC. \(\square \)

Appendix B: A particular rule

Before proving Theorem 9, we make a particular rule which will be used in the second step of the proof.

This rule is used to label the operator elements of a channel through a one-to-one relationship between operator elements and output states. First, we fixed the input state \(|Q_{1}^{N}\rangle \) of the quantum combined channel \(\mathcal {E}_{N}\) to \(|0_{1}^{N}\rangle \), and then, arbitrary operator element \(F_{k} \in \left\{ F_{k}\right\} _{k=0, \ldots , 2^{N}-1}\) of the N-copy channel \(\mathcal {E}^{\otimes N}\) uniquely corresponds to a output state \(|V_{1}^{N}\rangle \), \(V_{1}^{N}\in \mathcal {Y}^{N}\), namely

By Definition 3, we have

where the subscript k of \(F_{k}\) is the decimal number of the binary sequence \(b_{1} b_{2} \ldots b_{N}\).

To further understanding this rule, we take 2-copy channel \(\mathcal {E}^{\otimes 2}\), for example, and primal channel \(\mathcal {E}\) is bit flip channel whose operator elements are \(\{ E_{0}=\sqrt{p}X,E_{1}=\sqrt{1-p}I\}\). It is easy to obtain that four operator elements of \(\mathcal {E}^{\otimes 2}\) are \(F_{0}=pX^{1}\otimes X^{2}\), \(F_{1}=\sqrt{p(1-p)}X^{1}\otimes I^{2}\), \(F_{2}=\sqrt{p(1-p)}I^{1}\otimes X^{2}\) and \(F_{3}=(1-p)I^{1}\otimes I^{2}\), respectively. Assume that the input state of primal channel \(\mathcal {E}\) will only take value from \(|0\rangle =\left( \begin{array}{l}1 \\ 0\end{array}\right) \) or \(|1\rangle =\left( \begin{array}{l}0 \\ 1\end{array}\right) \). Then, the input space \(\{|Q_{1}^{2}\rangle \}\) of the quantum combined channel \(\mathcal {E}_{2}\) must be \(\{|Q_{1}^{2}\rangle \}=\{|00\rangle ,|01\rangle ,|10\rangle ,|11\rangle \}\), and the output space \(\{|V_{1}^{2}\rangle \}\) of the quantum combined channel \(\mathcal {E}_{2}\) must be \(\{|V_{1}^{2}\rangle \}=\{|00\rangle ,|01\rangle ,|10\rangle ,|11\rangle \}\), which means different operator elements \(F_{k},\ 0\le k \le 3\), will map the input space \(\{|Q_{1}^{2}\rangle \}\) to the same output space \(\{|V_{1}^{2}\rangle \}\). Thus, we fixed the input state to \(|00\rangle \), and a one-to-one relationship between operator element \(F_{k}\) \((0\le k\le 3)\) and output state \(|V_{1}^{2}\rangle \) of the channel \(\mathcal {E}^{\otimes 2}\) is established; namely, \(F_{0}\) corresponds to \(|11\rangle \), \(F_{1}\) corresponds to \(|10\rangle \), \(F_{2}\) corresponds to \(|01\rangle \), and \(F_{3}\) corresponds to \(|00\rangle \).

By using Theorem 8 and Eq. (B4), we have

for all \(Q_{1}^{N}\in \mathcal {X}^{N}\) and \(V_{1}^{N}\in \mathcal {Y}^{N}\).

Appendix C: Proof of Theorem 9

In this section, we prove Theorem 9 that the quantum coordinate channels \(\{\mathcal {E}_{N}^{(i)}\}\) are QQSCs. At the second step of the proof, we use the particular rule that we make in “Appendix B.”

Proof

In Sect. 2.5, we define quantum coordinate channel \(\mathcal {E}_{N}^{(i)}\), \(1\le i \le N\), whose input is \(\rho ^{Q_{i}}\) and output is \(\rho ^{V_{1}^{N},R_{1}^{i-1}}\).

1. The first step of the proof: obtain the general form of density operator  \(\rho ^{V_{1}^{N},R_{1}^{i-1}}\) of quantum joint system \(V_{1}^{N},R_{1}^{i-1}\).

Assume that each input state \(\rho ^{Q_{i}}\) of the quantum combined channel \(\mathcal {E}_{N}\) is \(\rho ^{Q_{i}}=q|0\rangle \langle 0|+(1-q)|1\rangle \langle 1|\). Then, we have

where \(Pr\left( |Q_{1}^{N}\rangle \langle Q_{1}^{N}|\right) =\prod _{i=1}^{N}Pr(|Q_{i}\rangle \langle Q_{i}|)\), alphabet \(\mathcal {X}=\{0,1\}\) and \(\mathcal {X}^{N}\) is the N-power extension alphabet of \(\mathcal {X}\). Introduce reference system \(\rho ^{R_{1}^{N}}=\rho ^{R_{1}} \otimes \cdots \otimes \rho ^{R_{N}}\) to purify \(\rho ^{Q_{i}^{N}}\), where \(\rho ^{R_{1}} = \cdots = \rho ^{R_{N}}=\rho ^{Q_{1}} = \cdots = \rho ^{Q_{N}}\). We have

Then, the density operator \(\rho ^{Q_{1}^{N},R_{1}^{N}}\) of the joint system \(Q_{1}^{N},R_{1}^{N}\) is

We use a unitary operator \(U_N\) which only acts on system \(Q_{1}^{N}\) to represent the encoding process \(|Q_{1}^{N}\rangle \rightarrow |C_{1}^{N}\rangle \), and we have

where \(C_{1}^{N}=Q_{1}^{N} G_{N}\), \(\tilde{C}_{1}^{N}=\tilde{Q}_{1}^{N} G_{N}\), and \(G_{N}\) is generator matrix.

The channel \(\mathcal {E}^{\otimes N}\), whose operator elements are \(\{ F_{k}\}_{k=0, \ldots , 2^{N}-1}\), follows the encoding process \(|Q_{1}^{N}\rangle \rightarrow |C_{1}^{N}\rangle \). Then, the density operator \(\rho ^{V_{1}^{N},R_{1}^{N}}\) of the output of the channel \(\mathcal {E}^{\otimes N}\) is

Notice that the channel \(\mathcal {E}^{\otimes N}\) is the last layer of the channel \(\mathcal {E}_{N}\), so the density operator \(\rho ^{V_{1}^{N},R_{1}^{N}}\) is also the output of the channel \(\mathcal {E}_{N}\). Then, we perform partial trace over the system \(R_{i}^{N}\) and obtain

Equation (C12) guarantees that

must be a unit vector, since it is easy to verify \(\sum _{Q_{1}^{i-1}\in \mathcal {X}^{i-1}}Pr\left( |Q_{1}^{i-1}\rangle \langle Q_{1}^{i-1}|\right) =1\). Divide Eq. (C12) into two parts: \(Q_{i}=0\) and \(Q_{i}=1\), we have

where

and

For \(\rho _{V_{1}^{N},R_{1}^{i-1}}^{(0)}\) and \(\rho _{V_{1}^{N},R_{1}^{i-1}}^{(1)}\), we exchange summation order and obtain

and

2. The second step of the proof: prove that the density operator \(\rho ^{V_{1}^{N},R_{1}^{i-1}}\) can be diagonalized with respect to a set of basis \(\{ |m^{\prime }\rangle \}_{m=0,\ldots ,2^{N}-1}\).

We will prove that density operators \(\rho _{V_{1}^{N},R_{1}^{i-1}}^{(0)}\) and \(\rho _{V_{1}^{N},R_{1}^{i-1}}^{(1)}\) can be diagonalized with respect to a same set of basis \(\{ |m^{\prime }\rangle \}_{m=0,\ldots ,2^{N}-1}\), namely

We consider \(\rho _{V_{1}^{N},R_{1}^{i-1}}^{(0)}\) only, since the proof method of \(\rho _{V_{1}^{N},R_{1}^{i-1}}^{(1)}\) is the same as that of \(\rho _{V_{1}^{N},R_{1}^{i-1}}^{(0)}\). We first prove that the vector \(|m^{\prime }\rangle \) can be written as

Since for all \(Q_{i+1}^{N}\in \mathcal {X}^{N-i}\), operation elements \(\{F_{k}\}_{k=0,\ldots ,2^{N}-1}\) will map vector

to a same set of orthogonal basis \(\{ |m^{\prime }\rangle \}_{m=0,\ldots ,2^{N}-1}\), which contains \(2^{N}\) basis vectors. Thus, without losing generality, we can let \(Q_{i+1}^{N}=0_{i+1}^{N}\). Using Eq. (B4), Eq. (B6) and Theorem 8, we have

which proves Eq. (C21).

Observe Eq. (C23), there is a one-to-one relationship between \(F_{k}\) and \(V_{1}^{N}\); thus, sum over all \(F_{k}\) is sum over all \(V_{1}^{N}\) and Eq. (C17) can be rewritten as

Here, we use the fact that \(Pr_{N}\left( |V_{1}^{N}\rangle ||Q_{1}^{N}\rangle \right) =Pr_{N}\left( |a_{1}^{N} G_{N} \cdot V_{1}^{N}\rangle ||Q_{1}^{N} \oplus a_{1}^{N}\rangle \right) \) which is according to Theorem 8, so let \(a_{1}^{N}=Q_{1}^{i-1},0,0_{i+1}^{N}\); we have

For Eq. (C24), we exchange summation order and obtain

where

and

\(Pr_{N}^{(i)}\left( |m^{\prime }\rangle ||0\rangle \right) \) is the transition probability which means the probability of the input state \(|0\rangle \langle 0|\) changing into \(|m^{\prime }\rangle \langle m^{\prime }|\). Using Eq. (C25), then Eq. (C28) can be rewritten as

Using the same method, Eq. (C20) can be easily proved, and we have

Thus, the basis transition probabilities can be uniformly expressed as

3. The third step of the proof: use Arikan’s method to prove the basis transition probability matrix is symmetric.

Next, we will prove that the basis transition probability matrix is symmetric. We will refer to the proof method which Arikan used to prove that classical coordinate channels \(\{W_{N}^{(i)}\}\) are symmetric if the primal binary-input discrete memoryless channel W is symmetric.

By Theorem 8, we have

for arbitrary \((a_{1}^{i-1},1,a_{i+1}^{N})\in \mathcal {X}^{N}\), and thus, Eq. (C31) can be rewritten as

Substitute Eq. (C21) into \(Pr_{N}^{(i)}\left( |m^{\prime }\rangle ||Q_{i}\rangle \right) \), and connect with Eq. (C33), we have

Here, we take \(a_{1}^{N}= (a_{1}^{i-1},1,a_{i+1}^{N})\), and the proof is completed. Equation (C34) means arbitrary row of the BTPM of the quantum coordinate channel \(\mathcal {E}_{N}^{(i)}\) is a permutation of the other row. \(\square \)

Appendix D: Proof of Proposition 11

In this section, we prove Proposition 11 that we can derive \(Pr_{N}^{(i)}\left( |m\rangle ||Q_{i}\rangle \right) \) from the TPM of classical coordinate channels \(W_{N}^{(i)}\).

Proof

According to Arikan’s theorem [6], the transition probabilities of classical coordinate channels \(\{W_{N}^{(i)}\}\) are

and Arikan has proved that classical combined channel \(W_{N}\) and classical coordinate channels \(\{W_{N}^{(i)}\}\) are all symmetric, which satisfies

and

for all \(u_{1}^{i-1}\in \mathcal {X}^{i-1}\).

Using Theorem 8, Proposition 10 and Eq. (D36), we have

for all \(V_{1}^{N}=y_{1}^{N} \in \mathcal {Y}^{N}\) and \(Q_{1}^{N}=u_{1}^{N} \in \mathcal {X}^{N}\).

Substitute Eqs. (D38) and (D37) into Eq. (51), we have

Equation (D39) means we can derive \(Pr_{N}^{(i)}\left( |m\rangle ||Q_{i}\rangle \right) \) from the TPM of classical coordinate channels \(W_{N}^{(i)}\), which completes the proof. \(\square \)

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