Ji-Hao Fan(樊繼豪), Pei-Wen Xia(夏沛文), Di-Kang Dai(戴迪康), and Yi-Xiao Chen(陳一驍)

School of Cyber Science and Engineering,Nanjing University of Science and Technology,Nanjing 210094,China

Keywords: asymmetric quantum channel,entanglement fidelity,entanglement-assisted quantum error correction code,quantum memory channel

1.Introduction

Quantum error correction (QEC) is crucial to realization of quantum communications and the building of quantum computers.[1,2]The stabilizer code formalism[3]is one of the most successful schemes in quantum information theory for realizing QEC.Stabilizer codes can be constructed by classical error correction codes with certain dual-containing conditions.Such constraint is not easy to be satisfied in many situations.[4]The entanglement-assisted (EA) quantum error correction codes(EAQECCs)[5-8]generalize the standard stabilizer codes, and EAQECCs can be constructed by any classical linear code without the dual-containing constraint.The price is that some ebits need to be preshared between the sender(Alice)and the receiver(Bob)before the transmission.In general, the preshared ebits need to be carefully preserved and thus they are assumed to exist without error.However,maintaining a number of ebits in perfect condition is usually difficult in practical applications.Such a phenomenon is crucial to the entanglement-assisted quantum communication formalism since imperfect ebits may degrade the performance of EAQECCs largely.[9]In Refs.[9,10], performance of EAQECCs with imperfect ebits was evaluated in entanglement fidelity.In Ref.[11], the entanglement fidelity of entanglement-assisted concatenated quantum codes was computed based on the concatenation scheme.

In most quantum channels, the probabilities of different types of quantum noise usually exhibit a large asymmetry.It was shown that the phase-flip errors (Zerrors) happen much more frequently than the bit-flip (Xerrors).[12]Asymmetric quantum codes (AQCs) which have a more biased error correction towardsZerrors thanXerrors,were constructed to better cope the significant asymmetry in quantum channels.[13-17]Moreover, in fault-tolerant quantum computation, asymmetric errors have been further explored to help improve the fault-tolerant thresholds.[18-22]What’s more, the standard discretization error model[23]supposed that quantum errors are independent of each other.However, correlated errors are more practical in reality.[24-26]So far, the asymmetry and memory effect on the performance of EAQECCs with noisy ebtis are still unknown.

In this work,we study the performance of EAQECCs with noisy ebits over asymmetric quantum channels and quantum memory channels by computing the entanglement fidelity of several EAQECCs.For aQE=[[n,k,d;c]]EAQECC,we consider that asymmetric and correlated noises exist not only in thenphysical qubits but also in thecebits.We compute the entanglement fidelity of the[[3,1,3;2]]entanglement-assisted repetition code and Bowen’s[[3,1,3;2]]EAQECC over asymmetric quantum channels.We know that the [[3,1,3;2]] repetition code is also an asymmetric quantum code,[13]which is assumed to be more powerful than the standard stabilizer codes.Our computation results suggest that the performance of the two EAQECCs is quite diverse and is related not only to the channel asymmetry but also to the ratio(denoted byω)of the error probabilities of ebits and qubits.Specifically, if the asymmetry of quantum channels is sufficiently large and the error probability of ebits is much smaller than that of qubits,then the [[3,1,3;2]] repetition code will perform much better than Bowen’s[[3,1,3;2]]EAQECC and the[[5,1,3]]stabilizer code.

In quantum memory channels,the numerical results suggest that the entanglement fidelity of the two EAQECCs is lowered down in most range of depolarizing probabilities.Moreover,the performance of the two EAQECCs is related not only to the channel memoryμbut also to the ratioω.We show that the two EAQECCs can outperform the[[5,1,3]]stabilizer code over quantum memory channels as the ratioωbecomes smaller and smaller.For the two EAQECCs, we show that Bowen’s[[3,1,3;2]]EAQECC can outperform the[[3,1,3;2]]repetition code over quantum memory channels if the ratioωis relatively large,e.g.,ω ≥0.5.However,if the ratio is small,e.g.,ω=0.01, the performance of the two EAQECCs is diverse and is effected largely by the channel memory.In such a case, the numerical results suggest that the[[3,1,3;2]]repetition code can outperform Bowen’s[[3,1,3;2]]EAQECC if the channel memoryμis small,e.g.,μ ≤0.1.However,Bowen’s[[3,1,3;2]] EAQECC can beat the [[3,1,3;2]] repetition code if the channel memoryμis large,e.g.,μ=0.5.

The rest of this article is given as follows.In Section 2,we present the needed background about quantum codes.In Section 3, we compute and plot the entanglement fidelity of several EAQECCs over asymmetric quantum channels.In Section 4, we compute and plot the entanglement fidelity of several EAQECCs over Markovian quantum memory channels.The conclusion is given in Section 5.

2.Preliminaries

In this section, we present some background knowledge about quantum channels and entanglement-assisted quantum codes.

We use Pauli channels as the basic channel models to perform the computation of entanglement fidelity.Denote a two-dimensional Hilbert space by C2.For a qubit|x〉 in C2,we suppose|x〉=a|0〉+b|1〉,whereaandbare two complex numbers anda2+b2=1.Define the Pauli operators

The quantum stabilizer codeQ= [[n,k]] is a 2kdimensional subspace of the Hilbert space C2n,andQis specified by a stabilizer groupSwith stabilizer generatorsSi(1≤i ≤n-k), i.e.,S=〈S1,...,Sn-k〉.Denote the parity check matrix ofQbyH=(HX|HZ).ThenHcan be obtained from Eq.(3)accordingly.Qhas minimum distancedif and only if it can detect any error of weight less than or equal tod-1,but cannot detect some error of weightd.

DefineQE= [[n,k,d;c]] as an entanglement-assisted quantum stabilizer code(EAQSC)which can detect any qubit error of weight less than or equal tod-1.The number of ebits isc.Consider the Pauli operators given by

Combining Eqs.(3) and (6), we can derive the parity check matrix ofQE.We denote the parity check matrix ofQEbyHE= (HX|HZ).Moreover, we denote the set of syndrome representatives that correspond to the parity check matrix ofHE=(HX|HZ) byTE.ThenTEis a set of correctable error operators ofQEand|TE|=2n-k+c.

The Pauli channel on a single qubit can be written as

whereφis a single qubit,0≤p0,px,py,pz ≤1,andp0+px+py+pz=1.Letp=1-p0be the error probability of a Pauli channel.Ifpx=py=pz=p/3,then the standard depolarizing channel is given by

Moreover, quantum channels usually have memories in practice.[24]The standard discretization error model in Ref.[12] supposes that quantum errors are independent of each other.However, correlated errors are more practical in reality.Similarly to classical information theory, an important family of quantum memory channels is specified by the channels with Markovian correlated errors.Thus we use a Markovian correlated depolarizing quantum channel model(see Refs.[17,24]) to present the behavior of quantum channels with memory.For ann-qubits quantum system,we denote

wherep0=1-p,p1,2,3=p/3,pl|i=(1-μ)pl+μδ(i,l)for 0≤i,l ≤3 and 1≤j ≤n,andμ∈[0,1]is the degree of memory.

3.Fidelity of entanglement-assisted concatenated quantum codes over asymmetric quantum channels

Entanglement fidelity of quantum codes is a reasonable measure of how well the qubits and entanglement of a subsystem of a larger quantum system are preserved.In practice,quantum codes with very high entanglement fidelity are preferred.In Ref.[11],entanglement fidelity of several EAQSCs over depolorizing channels was computed.In this section,we compute and compare the entanglement fidelity of several EACQCs over asymmetric quantum channels.

Let H be a Hilbert space of dimensionn.Let?be a quantum operator acting on a mixed stateρ=∑i piρi=TrHR|α〉〈α|in terms of a purification|α〉∈H?HR, where HRis a reference system.The output state?(ρ) also lies in the Hilbert space H.Then the entanglement fidelity is defined as

For aQE=[[n,k,d;c]]EAQSC,denote the encoding and decoding operations ofQEbyUandU?,respectively.DenoteTandDby the encoding and decoding channels,respectively.Let{Mj}be the syndrome measurement operations and let{Dj}be the correction operations.Let the channelNbe the encoded quantum information transmitted.During the decodingU?, denote the basis of then-k-cancillas andcebits by{|l〉}and{|s}, respectively.Let|Φ+〉A(chǔ)Bbe the Bell state shared between Alice and Bob.We define

whereTEis a set of syndrome representatives andSEis the stabilizer group.

For EAQSCs over asymmetric quantum channels, we have the following result about the computation of entanglement fidelity.

Theorem 1 LetQEbe an an EAQSC code, with a stabilizer groupSEand a set of syndrome representativesTE.Forqwx,wzandrvx,vzdefined in Eqs.(18) and (19), denoteΛ={qwx,wzrvx,vz}by the probability distribution of elements inTE×SE.The entanglement fidelity ofQEover asymmetric depolarizing channelsNA?NBis the weight enumerator ofΛ.

Proof According to Eq.(15)and Lemma 1,the entanglement fidelity ofQE=[[n,k,d;c]]is given by

whereλwx,wz,vx,vzis the number of error operators inTE?SEof weight distribution{wx,wz,vx,vz}.

In most quantum systems, there exists a large asymmetry between the occurrence probabilities ofXerrors,Yerrors,andZerrors.[13,14,16]Furthermore, it was shown that the occurrence probability ofZerrors is much larger than that ofXerrors.Since theYoperator can be formulated byY=iXZ,aYerror can be seen as a combination of anXerror and aZerror.Therefore we letpy=px.Moreover, we suppose that Alice and Bob transmit quantum information through two independent quantum channelsNAandNB.[9]Suppose that bothNAandNBare depolarizing channels.Letpx,py,andpzbe the occurrence probabilities ofX,Y, andZerrors inNA, respectively.Denotepa=px+py+pz=2px+pz.Letrx,ry, andrzbe the occurrence probabilities ofX,Y,andZerrors inNB,respectively.Denotera=rx+ry+rz=2rx+rz.We have

where 0≤wx+wz ≤n,and denote by

where 0≤vx+vz ≤c.Defineω=ra/paas the ratio of the error probabilities of ebits and qubits.In asymmetric quantum channels,we takeη=pz/px=rz/rxas the channel asymmetry.It is reasonable to set the same channel asymmetry for the qubit and ebit channels.There may exist quantum channels that qubtis and ebits suffer from different channel asymmetry.But in this paper we do not consider that case.Therefore we haveω=ra/pa=rx/px=rz/pz.

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LetQE1=[[3,1,3;2]]be an entanglement-assisted quantum repetition code.The parity check matrix ofQE1=[[3,1,3;2]](see Ref.[9])is given by

According to Eqs.(3) and (4), the stabilizer group ofQE1is given as follows:

LetQE2=[[3,1,3;2]] be an EAQSC given by Bowen in Ref.[27].The parity check matrix ofQE2= [[3,1,3;2]] is given by

According to Eqs.(3) and (4), the stabilizer group ofQE2is given as follows:

and the syndrome representives ofQE2is given as

In asymmetric Pauli channels,we rewriteSE2andTE1as

LetQE= [[5,1,3]] be the quantum stabilizer code.[3,9]The stabilizer group ofQEis given by

Since theQE= [[5,1,3]] stabilizer code is equivalent to Bowen’sQE2=[[3,1,3;2]] EAQSC, the syndrome representive ofQEis the same as that of Bowen’s code.

According to Theorem 1,the entanglement fidelity ofQE1over asymmetric quantum Pauli channels is given by

The entanglement fidelity ofQE2over asymmetric quantum Pauli channels is given by

In Fig.1, we plot the entanglement fidelity of several EAQSCs over asymmetric quantum channels with different asymmetryη, and we choose different ratiosω.For all the three codes in Fig.1, it is shown that the entanglement fidelity of each code gets improved as the channel asymmetryηgrows.In particular, the effect of channel asymmetry is more obvious to theQE1=[[3,1,3;2]] repetition code.For example, the entanglement fidelity ofQE1withη= 100 is much higher than that ofQE2withη=10 andη=1.Moreover, the three codes perform diversely under different ratiosω.In particular, theQE1=[[3,1,3;2]] repetition code performs much better than Bowen’sQE2=[[3,1,3;2]] EAQSC and the[[5,1,3]]stabilizer code whenω=0.01.It should be noted that the curve of Bowen’sQE2=[[3,1,3;2]]EAQSC coincides with that of the [[5,1,3]] stabilizer code whenω=1 in Fig.1.The reason is that the stabilizer group of Bowen’sQE2=[[3,1,3;2]]EAQSC is equivalent to that of the[[5,1,3]]stabilizer code.Whenω= 1, the entanglement fidelity of Bowen’sQE2= [[3,1,3;2]] EAQSC is equal to that of the[[5,1,3]]stabilizer code.

Fig.1.Entanglement fidelity of EAQSCs and the[[5,1,3]]stabilizer code over asymmetric quantum channels.The channel asymmetry is denoted by η=pz/px=rz/rx.The ratio of the error probabilities of ebits and qubits is denoted by ω=ra/pa=rx/px=rz/pz.We use the error probability pz and the entanglement fidelity as the X and Y axes,respectively.

4.Fidelity of entanglement-assisted quantum codes over quantum memory channels

In practical quantum channels, errors in qubits are usually not absolutely independent of each other.In this section,we compute and compare the entanglement fidelity of several entanglement-assisted quantum codes over Markovian quantum memory channels.We not only consider the memory in qubits but also consider the memory in ebits.Since qubits and ebits are usually stored in different space or in different time,we assume that errors in qubits are independent of those in ebits.Therefore Alice and Bob transmit quantum information through two independent quantum channelsMAandMB.[9]Suppose thatMAandMBare two independent Markovian depolarizing channels.The conditional probability ofMAsatisfies

wherep0=1-p,p1,2,3=p/3,pl|i=(1-μA)pl+μAδ(i,l)for 0≤i,l ≤3,andμA∈[0,1]is the degree of memory of channelMA.The conditional probabilities ofMBsatisfy

wheret0= 1-t,t1,2,3=t/3,tl|i= (1-μB)tl+μBδ(i,l)for 0≤i,l ≤3,andμB∈[0,1]is the degree of memory of channelMB.

LetSEandTEbe the stabilizer group and the set of syndrome representatives of an EAQSC, respectively.According to Lemma 1 and Theorem 1, the entanglement fidelity of EAQSCs over Markovian depolarizing channels is the weight enumerator of the corresponding probability distribution of elements inTE×SE.We not only consider the memory in the qubits but also consider that in the ebits.For simplicity,we let the degree of channelMAbe the same as the degree of channelMB, i.e., we letμ=μA=μB.Moreover, from Eqs.(27) and(28),we have

Then the entanglement fidelity of theQE1= [[3,1,3;2]]entanglement-assisted repetition code is given by

The entanglement fidelity of Bowen’sQE2=[[3,1,3;2]]EAQSC is given by

For comparison with standard quantum stabilizer codes,we also compute the entanglement fidelity of the[[5,1,3]]stabilizer codes as follows:

In Fig.2, we plot the entanglement fidelity of several EAQSCs over Markovian quantum memory channels.From Fig.2, we know that the channel memory lowers down the performance of EAQSCs in most range of the depolarizing probability.Moreover, theQE1=[[3,1,3;2]] repetition code and Bowen’sQE2=[[3,1,3;2]] EAQSC can outperform the[[5,1,3]] stabilizer code provided that the error probability of ebits is sufficiently smaller than that of qubits.Whenω=1, the [[5,1,3]] stabilizer code performs better than the two EAQSCs over quantum memory channels;whileω=0.5,Bowen’sQE2=[[3,1,3;2]]EAQSC can outperform theQE1=[[3,1,3;2]]repetition code and the[[5,1,3]]stabilizer code.As the ratioωis smaller, e.g.,ω= 0.1,0.01, the performance of the two EAQSCs becomes much better than the [[5,1,3]]stabilizer code over quantum memory channels.Specifically,whenω=0.01, the performance of the two EAQSCs is diverse and is effected largely by the channel memory.In such a case, theQE1=[[3,1,3;2]] repetition code can outperform Bowen’sQE2=[[3,1,3;2]]EAQSC if the channel memoryμis small,e.g.,μ=0,0.1.However,Bowen’sQE2=[[3,1,3;2]]EAQSC can beat theQE1=[[3,1,3;2]] repetition code if the channel memoryμis large, e.g.,μ= 0.5.It should be noted that the curve of Bowen’sQE2= [[3,1,3;2]] EAQSC does not coincides with that of the [[5,1,3]] stabilizer code whenω=1 (μ/=0) in Fig.2.Here the situation is different from that in asymmetric quantum channels.For Bowen’sQE2= [[3,1,3;2]] EAQSC over Markovian quantum memory channels, we separate the three qubits and the two ebits.The two parts are independent of each other.However, for the[[5,1,3]]stabilizer code over Markovian quantum memory channels,the five qubits are considered as a whole.

Fig.2.Entanglement fidelity of EAQSCs and the[[5,1,3]]stabilizer code over Markovian quantum memory channels.The channel memory is denoted byμ.The ratio of the error probabilities of ebits and qubits is denoted by ω=ra/pa=rx/px=rz/pz.We use the error probability p and the entanglement fidelity as the X axis and the Y axis,respectively.

5.Conclusion

In summary,we have computed the entanglement fidelity of two entanglement-assisted quantum stabilizer codes over asymmetric quantum channels and quantum memory channels.Moreover, we have not only considered asymmetric errors and correlated errors in qubits but also in ebits.In asymmetric quantum channels, we have shown that the performance of the two EAQSCs is quite diverse and is related not only to the channel asymmetry but also to the ratio of the error probabilities of ebits and qubits.We have shown that the[[3,1,3;2]]repetition code performs much better than Bowen’s[[3,1,3;2]]EAQECCs and the[[5,1,3]]stabilize code if the asymmetry is sufficiently large and the error probability of ebits is much smaller than that of qubits.In Markovian quantum memory channels, we have shown that the entanglement fidelity of the two EAQSCs is lowered down by the channel memory.We have shown that the performance of the two EAQSCs is not only related to the channel memoryμbut also to the ratio of the error probabilities of ebits and qubits.We have shown that the two EAQSCs can outperform the [[5,1,3]] stabilizer code over quantum memory channels if the ratio is small.We have shown that Bowen’s[[3,1,3;2]] EAQSC can outperform the [[3,1,3;2]] repetition code over quantum memory channels if the ratio of the error probabilities of ebits and qubits is relatively large.However,if the ratio is small,we have shown that the performance of the two EAQSCs is diverse and is effected greatly by the channel memory.We have shown that the [[3,1,3;2]] repetition code can outperform Bowen’s[[3,1,3;2]]EAQSC if the channel memory is small.However, Bowen’s [[3,1,3;2]] EAQSC can beat the[[3,1,3;2]]repetition code if the channel memory becomes larger.How to compute the entanglement fidelity of other EAQSCs over asymmetric quantum channels and quantum memory channels needs further research.However, the exact computation is extremely difficult as the code length becomes longer.One possible method is to approximate the computation of the entanglement fidelity.The other method is to use the entanglement-assisted concatenated quantum codes scheme.However,the exact computation of the entanglement fidelity of entanglement-assisted concatenated quantum codes is not a direct generalization and needs further research in the future work.Moreover,how to use entanglement-assisted quantum codes in quantum key distribution[28,29]to improve the entanglement fidelity is one important future work which needs further studied.

Acknowledgments

Project supported by the National Key R&D Program of China (Grant No.2022YFB3103802), the National Natural Science Foundation of China (Grant Nos.62371240 and 61802175),and the Fundamental Research Funds for the Central Universities(Grant No.30923011014).