The Encoding and Decoding Complexities of Entanglement-Assisted Quantum Stabilizer Codes

TL;DR

This paper analyzes the encoding and decoding complexities of entanglement-assisted quantum stabilizer codes, providing practical implementation bounds and discussing the computational hardness of decoding in this context.

Contribution

It introduces a practical encoding circuit complexity for entanglement-assisted quantum stabilizer codes and extends decoding hardness results to these codes.

Findings

Encoding complexity for $[[n,k;c]]$ codes is $O(n(n-k+c)/ ext{log} n$.

Special case $c=0$ reduces to known $O(n^2/ ext{log} n)$ complexity.

Shared entanglement increases encoding complexity but offers benefits.

Abstract

Quantum error-correcting codes are used to protect quantum information from decoherence. A raw state is mapped, by an encoding circuit, to a codeword so that the most likely quantum errors from a noisy quantum channel can be removed after a decoding process. A good encoding circuit should have some desired features, such as low depth, few gates, and so on. In this paper, we show how to practically implement an encoding circuit of gate complexity $O(n(n-k+c)/\log n)$ for an $[[n,k;c]]$ quantum stabilizer code with the help of $c$ pairs of maximally-entangled states. For the special case of an $[[n,k]]$ stabilizer code with $c=0$, the encoding complexity is $O(n(n-k)/\log n)$, which is previously known to be $O(n^2/\log n)$. For $c>0,$ this suggests that the benefits from shared entanglement come at an additional cost of encoding complexity. Finally we discuss decoding of entanglement-assisted quantum stabilizer codes and extend previously known computational hardness results on decoding quantum stabilizer codes.

Equations

$$|0{⟩}^{\otimes n-k}|x⟩=X_{n-k+1}^{x_1}\cdots X_n^{x_k}|0{⟩}^{\otimes n}$$

$$\phi (g)\triangleq (u_1\cdots u_n|v_1\cdots v_n)$$

$${\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left[\kern 0.0pt\kern-2.5pt\kern-5.0pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\scriptstyle\mbox{~{}}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{k}\\$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I_{s\times s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O\\$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I_{c\times c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I_{c\times c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O\\$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I_{c\times c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I_{c\times c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt\crcr}}}}\right]$}}.}$$

$${H^{\prime}_{\text{raw}}=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left[\kern 0.0pt\kern-2.5pt\kern-5.0pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\scriptstyle\mbox{~{}}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{k}\\s\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I_{s\times s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O\\c\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I_{c\times c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O\\c\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I_{c\times c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt\crcr}}}}\right]$}},}$$

$$[{\bar{Z}}_i,{\bar{Z}}_j]$$

$$[{\bar{X}}_i,{\bar{X}}_j]$$

$$[{\bar{Z}}_i,{\bar{X}}_j]$$

$$\{{\bar{Z}}_i,{\bar{X}}_i\}$$

$$H_{(s+2c)\times 2n}^{\prime }=$$

$$\displaystyle~{}~{}\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left[\kern 0.0pt\kern-2.5pt\kern-5.55557pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\scriptstyle\mbox{~{}}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c+k}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c+k}\\s\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle A$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle B$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle C\\c\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{1}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{5}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{2}\\c\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{3}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{6}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{4}$\hfil\kern 5.0pt\crcr}}}}\right]$}}=\left[\begin{smallmatrix}\varphi({\bar{Z}}_{1})\\[-6.0pt] \vdots\\ \varphi({\bar{Z}}_{s})\\ \varphi({\bar{X}}_{s+1})\\[-6.0pt] \vdots\\ \varphi({\bar{X}}_{s+c})\\ \varphi({\bar{Z}}_{s+1})\\[-6.0pt] \vdots\\ \varphi({\bar{Z}}_{s+c})\\ \end{smallmatrix}\right].$$

$$B+C{A}^T+B^T+A{C}^T=O,$$

$$M_5=M_1{C}^T+M_2{A}^T,$$

$$M_6=M_3{C}^T+M_4{A}^T,$$

$$M_1{M}_2^T+M_2{M}_1^T=O,$$

$$M_3{M}_4^T+M_4{M}_3^T=O,$$

$$M_1{M}_4^T+M_2{M}_3^T=I.$$

$$H_{(s+2c)\times 2(n+c)}=$$

$$\displaystyle\quad\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left[\kern 0.0pt\kern-2.5pt\kern-5.55557pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\scriptstyle\mbox{~{}}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c+k}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{s}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c+k}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{c}\\s\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle A$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle B$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle C$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O\\c\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{1}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{5}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{2}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O\\c\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{3}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{6}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle M_{4}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I$\hfil\kern 5.0pt\crcr}}}}\right]$}},$$

$$\displaystyle\begin{array}[]{c}\text{1st block}\cr\vdots\cr\frac{n-k}{m}\text{th block}\cr\end{array}\ \left[\begin{matrix}I_{m\times m}&&&A_{1}\cr&\ddots&&\vdots\cr&&I_{m\times m}&~{}A_{\frac{n-k}{m}}\cr\end{matrix}\right],$$

$$I_{m\times m}$$

$$\to \left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]=D_{m\times m}.$$

$$[I{A}_i]=$$

$$\to$$

$${B_{0}=\left[\begin{matrix}O_{m\times m}&*&\cdots&*\cr*&O_{m\times m}&*\cdots&*\cr\vdots&*\atop{\vdots}&\ddots&\vdots\atop*\cr*&*&\cdots*&O_{m\times m}\cr\end{matrix}\right]\begin{array}[]{c}\text{1st block}\cr\cdot\cr\cdot\cr\cdot\cr\frac{n-k}{m}\text{th block}.\cr\end{array}}$$

$$\underset{\text{phase gates}}{\underbrace{O(n-k)}}+\underset{\text{CNOT gates}}{\underbrace{O\left(2^m\cdot \frac{n-k}{m}\right)}}+\underset{\text{CZ gates}}{\underbrace{O\left((m^2+n-k)\cdot \frac{n-k}{m}\right)}}$$

$$=O\left((m+2^m+m^2+n-k)\frac{n-k}{m}\right)=O\left((2^m+n-k)\frac{n-k}{m}\right).$$

$$\displaystyle\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left[\kern 0.0pt\kern-2.5pt\kern-5.0pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\scriptstyle\mbox{~{}}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{n-k}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{n-k}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\overbrace{}^{k}\\n-k\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle A$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle B$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle C\\~{}~{}~{}k\big{\{}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle I$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle\vrule$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle C^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle O$\hfil\kern 5.0pt\crcr}}}}\right]$}}\\[5pt]$$

$$2\cdot O\left((2^m+k)\cdot \frac{n-k}{m}\right)+O\left((2^m+n-k)\cdot \frac{n-k}{m}\right)+O(n-k)$$

$$=O\left((2^m+n+m)\cdot \frac{n-k}{m}\right)=O\left((2^m+n)\cdot \frac{n-k}{m}\right).$$

$$\displaystyle\left[\begin{array}[]{cc|cc}I&A&B&C\\ O&M_{1}&M_{1}C^{T}+M_{2}A^{T}&M_{2}\\ O&M_{3}&M_{3}C^{T}+M_{4}A^{T}&M_{4}\end{array}\right]$$

$$\to$$

$$=$$

$$M_1=[M_{11}{M}_{12}]$$

$$\{\begin{array}{l}{M}_{11}{M}_{21}^T+M_{12}{M}_{22}^T+M_{21}{M}_{11}^T+M_{22}{M}_{12}^T=O\\ M_{31}{M}_{41}^T+M_{32}{M}_{42}^T+M_{41}{M}_{31}^T+M_{42}{M}_{32}^T=O\\ M_{11}{M}_{41}^T+M_{12}{M}_{42}^T+M_{21}{M}_{31}^T+M_{22}{M}_{32}^T=I;\end{array}$$

$$\to$$

$$M_{12}$$

$$K_2=M_{21}{M}_{11}^T+M_{22}{M}_{12}^T$$

$$K_3=M_{32}+M_{31}{M}_{11}^{-1}{M}_{12}$$

$$K_4=M_{41}{M}_{11}^T+M_{42}{M}_{12}^T$$

$$\to$$

Full text

A good encoding circuit should have some desired features, such as low depth, few gates, and so on. In this paper, we show how to practically implement an encoding circuit of gate complexity $O(n(n-k+c)/\log n)$ for an $[[n,k;c]]$ quantum stabilizer code with the help of $c$ pairs of maximally-entangled states. For the special case of an $[[n,k]]$ stabilizer code with $c=0$, the encoding complexity is $O(n(n-k)/\log n)$, which is previously known to be $O(n^{2}/\log n)$. For $c>0,$ this suggests that the benefits from shared entanglement come at an additional cost of encoding complexity.

Finally we discuss decoding of entanglement-assisted quantum stabilizer codes and extend previously known computational hardness results on decoding quantum stabilizer codes.

I Introduction

Quantum computers are powerful. However they are hard to build because quantum states are vulnerable and physical gates are imperfect. How to handle quantum noises has been an important problem in quantum information processing. A possible method is to use quantum error-correcting codes [1, 2, 3], in which quantum information is encoded in a codespace so that the most likely errors can be treated.

The class of quantum stabilizer codes (QSCs) have similar features to classical linear codes and are convenient for practical implementations [4]. An $[[n,k]]$ QSC is a $k$-qubit subspace of the $n$-qubit state space. The mapping from the raw $k$-qubit space to the encoded space can be implemented by an encoding circuit consisting of elementary gates. Clearly a low-complexity encoding circuit is desired, since quantum coherence decays with time and quantum gates cannot be implemented perfectly. In addition, the syndrome measurement circuit for error correction is also closely related to the encoding circuit [5]. As a consequence, a low complexity encoding circuit can be potentially used in designing fault-tolerant procedures to achieve higher error threshold.

In this paper we study the gate complexity of an encoding circuit. To encode an $[[n,k]]$ QSC, Cleve and Gottesman showed that $O(n(n-k))$ gates are required [6], which is proportional to the dimension of the underlying check matrix. Aaronson and Gottesman argued that $O(n^{2}/\log n)$ gates are sufficient [7] by using the CNOT circuit decomposition algorithm in [8]. Herein we show that this complexity can be further reduced to $O(n(n-k)/\log n)$ by studying the structure of the check matrix and applying a variant of the decomposition algorithm in [8]. More generally, we show that the encoding complexity for an $[[n,k;c]]$ entanglement-assisted quantum stabilizer code (EAQSC) [9] is $O(n(n-k+c)/\log n)$. It is known that entanglement may increase the code rate [9], provide a larger minimum distance [10], or improve the decoding performance [11]. It is clear now that these benefits come at an additional cost of encoding compleity.

On the other hand, decoding a linear code is shown to be NP-complete by Berlekamp, McEliece, and van Tilborg [12]. In the quantum case, several decoding problems are also shown to be NP-hard [13, 14, 15]. Especially in the presence of degeneracy, the optimal decoding method finds the most probable equivalent coset of errors corresponding to an error syndrome and it is shown to be #P-complete by Iyer and Poulin [16]. In this paper we will discuss the decoding procedure of a general EAQSC and show that the corresponding decoding problem is also #P-complete.

This paper is organized as follows. In Section II we give the notation and the basics of quantum codes. The encoding complexity of $O(n(n-k+c)/\log n))$ is shown in Section III. In Section IV, we discuss the hardness of decoding an EAQSC. Then we conclude.

II The EAQSC Scheme

We first review basics of stabilizer codes. Let ${\cal G}_{n}$ be the $n$-fold Pauli group $\{i^{e}\bigotimes_{j=1}^{n}M_{j}:M_{j}\in\{I,X,Y,Z\},e\in\{0,1,2,3\}\}$. Let ${\cal S}\subset{\cal G}_{n}$ be an Abelian stabilizer group, generated by $n-k$ independent stabilizer generators ${\bar{Z}}_{1},\dots,{\bar{Z}}_{n-k}$, such that $-I\notin{\cal S}$. Then an $[[n,k]]$ stabilizer code defined by ${\cal S}$ is ${\cal C}({\cal S})\triangleq\{|v\rangle\in{\mathbb{C}}^{2^{n}}:~{}{g|v\rangle=|v\rangle}~{}~{}\forall~{}g\in{\cal S}\}$.

Let $Z_{i}\triangleq I^{\otimes(i-1)}\otimes Z\otimes I^{\otimes(n-i)}$ and similarly for $X_{i}$. For ${x}=(x_{1}\cdots x_{k})\in\{0,1\}^{n}$, consider the following initial state

[TABLE]

with $k$-qubit logical state $|x\rangle$ and $n-k$ ancillas in $|0\rangle$ before encoding. This state has stabilizers $Z_{1},\dots,Z_{n-k}$. A unitary encoding circuit $U$ for ${\cal C}({\cal S})$ maps $|x\rangle$ to $|\bar{x}\rangle\triangleq U|0\rangle^{\otimes n-k}|x\rangle$ with stabilizers ${\bar{Z}}_{i}=UZ_{i}U^{\dagger},~{}i=1,\dots,n-k$. The set of the $2^{k}$ vectors $\{|\bar{x}\rangle:x\in\{0,1\}^{n}\}$ forms a basis of ${\cal C}({\cal S})$. Also $|\bar{x}\rangle$ can be written as ${\bar{X}}_{n-k+1}^{x_{1}}\cdots{\bar{X}}_{n}^{x_{k}}|\bar{0}\rangle,$ where ${\bar{X}}_{j}=UX_{j}U^{\dagger},~{}j=n-k+1,\dots,n,$ are called seed generators and $|\bar{0}\rangle=U|0\rangle^{\otimes n}$ is a coded zero, which can be generated by $|\bar{0}\rangle=\frac{1}{\sqrt{2^{n-k}}}\prod_{i=1}^{n-k}(I+{\bar{Z}}_{i})|0\rangle^{\otimes n}=\frac{1}{\sqrt{2^{n-k}}}\sum_{g\in{\cal S}}g|0\rangle^{\otimes n}$ such that $g|\bar{0}\rangle=|\bar{0}\rangle$ for all $g\in{\cal S}$.

Define a homomorphism $\varphi:{\cal G}_{n}\to\{0,1\}^{2n}$ by

[TABLE]

for $g=\sigma_{1}\otimes\cdots\otimes\sigma_{n}\in{\cal G}_{n}$ such that $\sigma_{j}\mapsto(u_{j}|v_{j})$ by $I\mapsto(0|0),X\mapsto(1|0),Y\mapsto(1|1),Z\mapsto(0|1)$. An $(n-k)\times 2n$ binary check matrix $H=[I~{}A|B~{}C]$ is the matrix with rows $\varphi({\bar{Z}}_{i}),~{}i=1,\dots,n-k$. For example, the initial state has a check matrix $[O~{}O|I~{}O]$, where $O$ is the all-zero matrix of appropriate dimensions. Since ${\cal S}$ is an Abelian subgroup, $H$ has to satisfy the commutation condition $H\Lambda H^{T}=O$, where $\Lambda=\left[{O_{n\times n}\atop I_{n\times n}}\vrule~{}{I_{n\times n}\atop O_{n\times n}}\right].$ The encoding circuit can be implemented by the reverse of a series of elementary gates that transform $H$ to $[O~{}O|I~{}O]$ [6, 5, 7].

EAQSCs are a coding scheme that the sender (Alice) and the receiver (Bob) share some maximally-entangled Einstein-Podolsky-Rosen (EPR) pairs $|\Phi^{+}\rangle_{AB}$ [9, 17]. An initial basis state of the overall system is $|0\rangle^{\otimes s}\otimes|\Phi^{+}\rangle^{\otimes c}_{AB}\otimes|x_{1}\cdots x_{k}\rangle$, where Bob holds $c$ qubits (a half the $c$ EPR pairs) and Alice holds the remaining $n=s+c+k$ qubits prior to communication. After encoding, Alice sends her $n$ qubits to Bob through a noisy quantum channel and Bob’s $c$ qubits are assumed to be error-free. Every initial basis state is stabilized by a set of operators in ${\cal G}_{n+c}$ with corresponding check matrix (before encoding)

$H_{\text{raw}}=$

[TABLE]

Similarly, we can consider a unitary encoding circuit $U$ that maps $H_{\text{raw}}$ to some $H$ with $H\Lambda H^{T}=O$. Upon reception, Bob does a decoding on the total of $(n+c)$ qubits according to the check matrix $H$.

We can consider a simplified check matrix of $H_{\text{raw}}$ without the columns corresponding to Bob’s qubits:

[TABLE]

which is corresponding to simplified stabilizers $\{Z_{1},\dots,Z_{s},$ $X_{s+1},\dots,X_{s+c},$ $Z_{s+1},\dots,Z_{s+c}\}$ in ${\cal G}_{n}$. After encoding by $U$, we have simplified stabilizer generators ${\bar{Z}}_{i}=UZ_{i}U^{\dagger}$ and ${\bar{X}}_{i}=UX_{i}U^{\dagger}$ satisfying the following commutation relations

[TABLE]

Thus we have a non-Abelian subgroup ${\cal S}^{\prime}=\langle{\bar{Z}}_{1},\dots,{\bar{Z}}_{s},$ ${\bar{X}}_{s+1},\dots,{\bar{X}}_{s+c},$ ${\bar{Z}}_{s+1},\dots,{\bar{Z}}_{s+c}\rangle$ with corresponding simplified check matrix

[TABLE]

By (2) we have

[TABLE]

Without loss of generality, we assume $M_{1}=[M_{11}~{}M_{12}]$ with non-singular $M_{11}$. Now a standard form of the check matrix (with additional columns corresponding Bob’s qubits) is

[TABLE]

which satisfies $H\Lambda H^{T}=O$.

III Encoding complexity

We consider the decomposition of a stabilizer circuit [7] by Clifford gates {CNOT, H, P} and (possibly) some swap operations. (See, for example, [18] for the encoding cirucit of an EAQSC.) For convenience, we use an additional controlled-$Z$ gate (CZ), which can be decomposed as a CNOT and two H gates, in the circuit decomposition. Consider a check matrix of the form $H=[H_{X}|H_{Z}]$. We have the following gate operation rules:

A CNOT gate from qubit $i$ to qubit $j$ adds column $i$ to column $j$ in $H_{X}$ and adds column $j$ to column $i$ in $H_{Z}$. 2. 2.

A CZ gate from qubit $i$ to qubit $j$ adds column $i$ in $H_{X}$ to column $j$ in $H_{Z}$ and adds column $j$ in $H_{X}$ to column $i$ in $H_{Z}$. 3. 3.

A Hadamard gate on qubit $i$ swaps column $i$ in $H_{X}$ with column $i$ in $H_{Z}$. 4. 4.

A phase gate on qubit $i$ adds column $i$ in $H_{X}$ to column $i$ in $H_{Z}$.

Patel, Markov, and Hayes proposed an efficient reduction algorithm to decompose a CNOT circuit [8]. Motivated by their method, we propose the following algorithm for the reduction of a matrix of the form $[I\ A|B^{\prime}\ O]$ for our purpose.

Lemma 1**.**

For an $(n-k)\times 2n$ check matrix $H=[I\ A|B^{\prime}\ O]$, there exists a linear transformation that maps $H$ to $[I\ O|B^{\prime}\ O]$ using $O\left((2^{m}+k)\cdot\frac{n-k}{m}\right)$ CNOT gates for some $m$ smaller than $n-k$.

Proof.

For simplicity, we first assume that $n-k$ is divisible by some integer $m$. Partition $[I~{}A]$ into $(n-k)/m$ blocks

[TABLE]

where $A_{i}$ is an $m\times k$ binary matrix. Beginning with $i=1$, we perform appropriate column operations so that the columns of $I_{m\times m}$ in the $i$th block will go over all nonzero $m$-bit vectors and transform to a matrix $D_{m\times m}$ with all ones in the upper triangular part. For example, if $m=3$,

[TABLE]

This requires $2^{m}-m-1$ operations per block. During this process, if there is any column in $A_{i}$ that is identical to a generated column, we can eliminate that column and this takes at most $k$ operations (since $A_{i}$ has $k$ columns). Next, we transform $D_{m\times m}$ back to $I_{m\times m}$, and this needs $m-1$ operations. For example, consider $A_{i}=\left[\begin{smallmatrix}1&1&1&1\cr 1&1&1&1\cr 1&0&0&1\cr\end{smallmatrix}\right]$:

[TABLE]

So it takes at most $((2^{m}-m-1)+k+(m-1))<(2^{m}+k)$ gates to eliminate one block. By repeating the above process for the $\left\lceil{\frac{n-k}{m}}\right\rceil$ blocks, it requires at most $(2^{m}+k)\left\lceil{\frac{n-k}{m}}\right\rceil=O((2^{m}+k)\cdot\frac{n-k}{m})$ gates to eliminate $A$. ∎

Similarly we propose the following decomposition algorithm for certain phase and CZ circuits.

Lemma 2**.**

For an $(n-k)\times 2n$ check matrix $H=[I\ O|B^{\prime}\ O]$ with symmetric $B^{\prime}$, there exists a linear transformation that maps $H$ to $[I\ O|O\ O]$ using $O\left((2^{m}+n-k)\cdot\frac{n-k}{m}\right)$ gates.

Proof.

By Rule 4), it needs at most $n-k=O(n-k)$ phase gates to eliminate non-zero diagonal entries in $B^{\prime}$. After that, CZ gates can further reduce $H$ to $[I\ O|O\ O]$ by Rule 2). To decompose an efficient CZ circuit, after phase gates, we first reduce $B^{\prime}$ to

[TABLE]

To obtain $O_{m\times m}$, every block needs fewer than $m^{2}$ CZ gates. To eliminate the * parts, every block needs fewer than ${(n-k)}$ CZ gates by the same technique as in Lemma 1 (though additional $2^{m}$ CNOT gates are required). Thus eliminating $B^{\prime}$ needs a gate complexity

[TABLE]

∎

By the two lemmas and (4), we have the following theorems.

Theorem 3**.**

For an $(n-k)\times 2n$ check matrix $H=[I\ A|B\ C]$ and some $m$ smaller than $n-k$, there exists a linear transformation that maps $H$ to $[I\ O|B^{\prime}\ C]$ with $B^{\prime}=B+CA^{T}$ using $O\left((2^{m}+k)\cdot\frac{n-k}{m}\right)$ CNOT gates, and there exists a linear transformation that maps $H$ to $[I\ A|B^{\prime\prime}\ O]$ with $B^{\prime\prime}=B+AC^{T}$ using $O\left((2^{m}+k)\cdot\frac{n-k}{m}\right)$ gates, where $B^{\prime}$ and $B^{\prime\prime}$ are both symmetric.

Theorem 4**.**

An $[[n,k]]$ QSC has an encoding complexity $O(n(n-k)/\log n)$.

Proof.

Without loss of generality, let $H=[I~{}A|B~{}C]_{(n-k)\times 2n}$ be the check matrix of a target stabilizer code. We add $k$ seed generators for the purpose of elimination as in [4], [19], and their corresponding binary matrix can be represented by $[O~{}I|C^{T}O]_{k\times 2n}$. The procedure of reduction is as follows.

[TABLE]

Note that Hadamard gates on the last $k$ qubits are not required since the last $k$ rows correspond to the seed generators [6]. The overall complexity is

[TABLE]

Similar to [8], taking $m=\left\lfloor{\alpha\log_{2}n}\right\rfloor$ for some $\alpha<1$ and we have $O((2^{m}+n)\cdot\frac{n-k}{m})=O(n\cdot\frac{n-k}{\log n})$ since ${2^{m}\leq n^{\alpha}=o(n)}$. ∎

The lemmas and theorems in this section are based on a trick: Originally it costs $O(n)$ gates to reduce each of the $(n-k)$ rows, but if we group every $m$ rows as a block, then we only have $O((n-k)/m)$ blocks and every block needs at most $O(2^{m}+n)=O(n)$ gates.

Theorem 5**.**

An $[[n,k;c]]$ EAQSC has an encoding complexity $O(n(n-k+c)/\log n)$.

Proof.

It suffices to show that the required number of gates is proportional to $(n-k+c)\times 2n$ to reduce (3) to (1). By (4) and Rules 1) to 4) above, we start to reduce (3) to (1):

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We have shown that, to reduce (3) to (1), the required number of gates is proportional to $(n-k+c)\times 2n$. By a deduction similar to the proof of Theorem 4 (cf. the mentioned trick prior to this theorem), the overall complexity is $O(n(n-k+c)/\log n)$. Notice that, when $c$ is large, a process like (7) is required to reduce $M_{11}$ to $I$ efficiently. ∎

IV Hardness of decoding EAQSCs

We first describe the decoding procedure of an $[[n,k;c]]$ EAQSC with a standard-form check matrix $H$ in (5). Suppose that a Pauli error $E\in{\cal G}_{n+c}$ occurs during the transmission of a codeword. The error syndrome vector is defined by $y=\varphi(E)\Lambda H^{T}\in{\mathbb{Z}}_{2}^{(s+2c)}$ [15]. Since Bob’s original qubits are error-free by assumption, $E=E^{\prime}\otimes I^{\otimes c}\in{\cal G}_{n+c}$ for some error $E^{\prime}\in{\cal G}_{n}$. Thus we have $y=\varphi(E^{\prime})\Lambda H^{\prime T}$ where $H^{\prime}$ is the corresponding simplified check matrix as in (3). Given $y$, the receiver (Bob) has to find a proper error vector $e\in{\mathbb{Z}}_{2}^{2n}$ such that $e\Lambda H^{\prime T}=y$. Then a correction operator ${\hat{E}}\in\varphi^{-1}(e)\otimes I^{\otimes c}$ is applied.

Two types of decoders are considered regarding to the degeneracy of EAQSCs. The quantum maximum likelihood decoder finds a minimum-weight error with syndrome $y$. If degeneracy is considered, the (optimal) degenerate quantum maximum likelihood decoder finds the most probable coset of degenerate errors with syndrome $y$ [13, 15]. For example, consider the independent $X$–$Z$ channel model, where a qubit suffers an $X$ error with probability $p$ and independently it suffers a $Z$ error with probability $p$ for $p\in[0,0.5)$. The coset probability of an $e\in{\mathbb{Z}}_{2}^{2n}$ with respect to a check matrix $H^{\prime}$ is defined as

[TABLE]

where $\operatorname{Row}(H^{\prime})\subset{\mathbb{Z}}_{2}^{2n}$ is the row space of $H^{\prime}$, $\mathrm{wt}(u)$ is the Hamming weight of $u\in{\mathbb{Z}}_{2}^{2n}$, and $P(u)=p^{\mathrm{wt}(u)}(1-p)^{2n-\mathrm{wt}(u)}$ is the probability that an error in $\varphi^{-1}(u)$ occurs.

Previously the quantum maximal-likelihood decoding (QMLD) of general QSCs is known to be NP-hard for the independent $X$–$Z$ channel [13] and depolarizing channel [15]. Moreover, the degenerate quantum maximum likelihood decoding (DQMLD) of QSCs is shown to be #P-complete [16]. We would like to generalize these results to EAQSCs with respect to the independent $X$–$Z$ channel. \[8pt] EAQSC Maximum Likelihood Decoding (EMLD)** \[2pt] Input: A full row-rank $(s+2c)\times 2n$ binary matrix $H^{\prime}$ satisfying the requirements in (2) and (3), and a binary vector $y\in{\mathbb{Z}}_{2}^{s+2c}$.\[1pt] Output: A binary vector $e\in{\mathbb{Z}}_{2}^{2n}$ satisfying $e\Lambda H^{\prime T}=y$ and minimizing $\mathrm{wt}(e)$. \[-8pt] \[8pt] Degenerate EAQSC Maximum Likelihood Decoding (DEMLD) \**[2pt] Input: A full row-rank $(s+2c)\times 2n$ binary matrix $H^{\prime}$ satisfying the requirements in (2) and (3), a binary vector $y\in{\mathbb{Z}}_{2}^{s+2c}$, and a real number $p\in[0,0.5)$.\[1pt] Output: A binary vector $e\in{\mathbb{Z}}_{2}^{2n}$ satisfying $e\Lambda H^{\prime T}=y$ and maximizing the coset probability $P(e+\operatorname{Row}(H^{\prime}))$ in (9). \[-2pt]

Corollary 6**.**

EMLD is NP-complete.

Proof.

A general EAQSC decoder supports the case $c=0$. Thus the QMLD in [13] trivially reduces to EMLD in polynomial time. By the main theorem of [13], EMLD is NP-complete. ∎

Corollary 7**.**

DEMLD is #P-complete.

Proof.

Consider $c=0$ and $k=1$. Then DQMLD defined in [16] reduces to DEMLD in polynomial time. By Theorem 2 of [16], DEMLD is #P-complete. ∎

It is straightforward to extend the results to the case of depolarizing channel by redefining the $P(u)=P(x|z)$ in (9) and in DEMLD as $P(u)=(p/3)^{\mathrm{gw}(u)}(1-p)^{n-\mathrm{gw}(u)}$, where $0\leq p<3/4$ and $\mathrm{gw}(x|z)\triangleq\mathrm{wt}(x\vee z)$ is the generalized weight, where $x\vee z$ is the bitwise OR of $x$ and $z$. By Theorem 5 of [15] and the conclusion remark in [16], we have the following remark:

Remark 8**.**

Corollaries 6 and 7 also hold for the depolarizing channel with respect to generalized weight.

V Conclusion

We have improved the encoding complexity to $O(n(n-k+c)/\log n)$ for an $[[n,k;c]]$ EAQSC. This also suggests that entanglement benefits an EAQSC at the cost of additional encoding complexity. On the other hand, we showed that decoding general EAQSC is #P-complete if degeneracy is considered and NP-complete if not.

CYL was financially supported from the Young Scholar Fellowship Program by Ministry of Science and Technology (MOST) in Taiwan, under Grant MOST107-2636-E-009-005.