Exploiting Degeneracy to Construct Good Ternary Quantum Error Correcting Code

Exploiting degeneracy to construct good ternary quantum error correcting code

Ritajit Majumdar∗ and Susmita Sur-Kolay† Advanced Computing & Microelectronics Unit, Indian Statistical Institute, India

Quantum error-correcting code for higher dimensional systems can, in general, be directly con- structed from the codes for qubit systems. What remains unknown is whether there exist eﬃcient code design techniques for higher dimensional systems. In this paper, we propose a 7-qutrit error- correcting code for the ternary quantum system and show that this design formulation has no equivalence in qubit systems. This code is optimum in the number of qutrits required to correct a single error while maintaining the CSS structure. This degenerate CSS code can (i) correct up to seven simultaneous phase errors and a single bit error, (ii) correct two simultaneous bit errors on pre-deﬁned pairs of qutrits on eighteen out of twenty-one possible pairs, and (iii) in terms of the cost of implementation, the depth of the circuit of this code is only two more than that of the ternary Steane code. Our proposed code shows that it is possible to design better codes explicitly for ternary quantum systems instead of simply carrying over codes from binary quantum systems. Keywords: Quantum Error Correction, Ternary Quantum System, Multi-valued Logic, Degenerate Code, CSS Code

I. INTRODUCTION [13, 14]. To the best of our knowledge, it is not known whether it is possible to explicitly construct more effi- The design of a large-scale, general-purpose quantum cient QECC for a particular higher-dimensional quantum computer is hindered by error. The interaction of qubits system, instead of using the construction technique from with the environment readily destroys the information some lower-dimensional system. content of the state. As a remedy, quantum error correction techniques have been proposed [1–5] which encodes the information of k qubits into n (> k) qubits to correct A ternary quantum system (or qutrit) is the simplest errors. However, the fidelity of a quantum circuit decays multi-valued system. The 7-qubit Steane code [2] can be exponentially with the depth of the circuit [6]. There- carried over to the ternary system, which is the optimal fore, quantum error-correcting code (QECC) with higher CSS QECC in the number of qutrits that corrects a single circuit depth not only reduces the speed of computation error [15]. In this paper we propose a 7-qutrit degenerate but also has the potential to incorporate further errors. CSS QECC which is (i) optimal in the number of qutrits The requirement for QECC construction is, therefore, to (for CSS type code), (ii) can correct up to seven simul- reduce the number of qubits required for encoding, as taneous phase errors, (iii) can correct a single bit error, well as reduce the depth of the QECC circuit. (iv) can be readily designed to correct two simultaneous Consider two linear classical codes C1 = [n, k1, d1] and bit errors on a predefined pair of qutrits (on 18 out of ⊥ 21 pairs), and (v) the depth of the circuit of our QECC C2 = [n, k2, d2] such that C2 ⊆ C1 and k2 < k1. The parity check matrices of C1 and C2 can be combined is only two more than that of the ternary Steane code. to construct an [[n, k1 − k2, min{d1, d2}]] QECC [2, 7], Its ability to correct multiple phase errors and two simul- called the CSS code. The parity check matrix of C1(C2) taneous bit errors on pre-defined qutrit pairs without a forms the stabilizers for bit error correction, while that significant increase in the depth of the circuit makes our of C2(C1) forms the stabilizers for phase error correction. proposed QECC a strong candidate for error correction Therefore, the stabilizers of a CSS code can be parti- in the ternary quantum system. We also show that our tioned into two sets such that the non-identity operators proposed construction has no equivalence in the qubit in the two partitions are different. CSS codes usually system in order to construct a linear degenerate 7-qubit arXiv:2008.00713v1 [quant-ph] 3 Aug 2020 have low depth circuit as compared to non-CSS codes QECC. Therefore, our QECC implies that there exist de- [8, 9]. For the rest of this paper, we shall concentrate on sign techniques of QECC for ternary quantum systems CSS codes only. which are more efficient than ternary QECCs which are Quantum systems are inherently multi-valued, and carried over from binary QECCs. multi-valued quantum computers can outperform their binary counterparts in certain cryptographic protocols [10], search algorithms [11], and machine learning [12]. The rest of the paper is organized as follows - Section QECC for multi-valued systems can be directly car- II gives a brief description of quantum error correction ried over from the QECC construction for qubit systems using stabilizers followed by the construction of the circuit of ternary Steane code in Section III. In Section IV we introduce our QECC and describe phase and bit error correction in detail. Section V compares the circuit cost ∗ [email protected] of our QECC to the ternary Steane code. We conclude † [email protected] in Section VI. 2

( II. PRELIMINARIES OF QUANTUM ERROR i = j when k 6= l, CORRECTION i 6= j when k = l.

A. Error model and degeneracy Proof. It is easy to verify that 2 XiZi = ω ZiXi, i ∈ {1, 2} (2) The error model considered in this paper From Eq. 2, the following can be derived

2 2 2 2 2 X X X X1Z2 = X1Z1Z1 = ω Z1X1Z1 = ω Z1(ω Z1X1) = ωZ2X1 E = δI3 + ηiZi + (µjXj + ξijYij) (1) 2 2 2 X2Z1 = X1X1Z1 = X1(ω Z1X1) = ω ω Z1X1X1 = ωZ1X2 i=1 j=1 i,j In summary, where I3 is the (3 × 3) identity matrix, XiZj = ωZjXi, i, j ∈ {1, 2}, i 6= j (3) i 2i Zi |ψi = α |0i + ω β |1i + ω γ |2i Now consider (Xi ⊗ Xj)(Zk ⊗ Zl) if either i = j, k 6= l Xi |ψi = α |0 + ii + β |1 + ii + γ |2 + ii or i 6= j, k = l, i, j, k, l ∈ {1, 2}. Thus, either i = k, j 6= l Yij = XiZj or i 6= k, j = l. From Eq. 2 and 3, commutation on one of the qutrits yields ω2, and on the other qutrit yresults for i, j ∈ {1, 2}, δ, η, µ, ξ ∈ C, spans the (3 × 3) operator in ω, and hence the product is 1. Therefore, for such a space [16]. Xi and Zi are termed as bit and phase errors scenario, [Xi ⊗ Xj,Zk ⊗ Zl] = 0. respectively. Conversely, consider (Xi ⊗ Xj)(Zk ⊗ Zl) where i = j For binary quantum systems, a set of mutually com- and k = l. Then from Eq. 2 and 3, if i = k (i 6= k), then muting operators S1,...,Sm ⊂ {I, ±σx, ±iσx, ±σy, commutation on both the qutrits produce value ω2 (ω), ⊗n ±iσy, ±σz, ±iσz} , where each σi is a Pauli operator and hence the product is ω (ω2). Therefore, for such a [17], is said to stabilize a quantum state |ψi if [4]: scenario, [Xi ⊗ Xj,Zk ⊗ Zl] 6= 0.

1. ∀i, Si |ψi = |ψi, 1 ≤ i ≤ m; III. TERNARY STEANE CODE AND ITS 2. ∀e ∈ E, ∃ j, Sj(e |ψi) = −(e |ψi) 1 ≤ j ≤ m; CIRCUIT 0 0 0 3. for e, e ∈ E, e 6= e , ∃ j, k Sj(e |ψi) 6= Sk(e |ψi), 1 ≤ j, k ≤ m; In this section, we carry over the 7 qubit QECC by Steane [2] to the ternary system and show its circuit im-

Weight of a stabilizer Si (wt(Si)) is defined as the plementation. The stabilizers for Steane code are number of non-identity operators in S . For an [[n, k, d]] i S1 = I ⊗ I ⊗ I ⊗ X ⊗ X ⊗ X ⊗ X QECC, let S be the set of stabilizers and O be the set S = I ⊗ X ⊗ X ⊗ I ⊗ I ⊗ X ⊗ X of all n-qubit operators. The distance d of the QECC is 2 defined as S3 = X ⊗ I ⊗ X ⊗ I ⊗ X ⊗ I ⊗ X S4 = I ⊗ I ⊗ I ⊗ Z ⊗ Z ⊗ Z ⊗ Z min{wt(i) | i ∈ O \ S and [i, Sj] = 0 ∀ Sj ∈ S} S5 = I ⊗ Z ⊗ Z ⊗ I ⊗ I ⊗ Z ⊗ Z d A distance d QECC can correct upto b 2 c errors. A S6 = Z ⊗ I ⊗ Z ⊗ I ⊗ Z ⊗ I ⊗ Z QECC is said to be degenerate if ∃ e, e0 ∈ E, e 6= e0, X and Z correspond to X and Z respectively in where E is the set of all correctable errors, such that for 1 1 ternary quantum systems [18]. Each Z operator is re- a codeword |ψi, e |ψi = e0 |ψi = |φi. For such a code, it alized by a single CNOT gate, and X = H CNOT H, is not necessary to distinguish between e and e0 as long where H is the Hadamard gate. The ternary equivalent as the error state |φi can be uniquely identified. of CNOT gate is the C + T [14] gate, where In accordance to the error model of Eq. 1, the operators X which form the stabilizers for ternary quantum system C + T : |x, (x + y)%3i hx, y| are x,y∈{0,1,2}

X1 |ji = |j + 1i mod 3 X2 |ji = |j + 2i mod 3 Chrestenson basis [19] is the equivalent of Hadamard j 2j Z1 |ji = ω |ji Z2 |ji = ω |ji basis in ternary quantum system. Two conjugate Chrestenson bases b1 and b2 are deﬁned as where j ∈ {0, 1, 2}, and ω3 = 1. It can be noted that 1 |+ii = √ (|0i + |1i + |2i) X2 = X1X1 Z2 = Z1Z1. 3 1 i 2i The ternary stabilizers for any [[n, k, d]]3 QECC will |−ii = √ (|0i + ω |1i + ω |2i) 3 be n-fold tensor products of {I,X1,X2,Z1,Z2} [15]. 1 2i i ||ii = √ (|0i + ω |1i + ω |2i) Lemma 1. [Xi ⊗ Xj,Zk ⊗ Zl] = 0 if and only if 3 3

seven qutrits as |ψi = α |0 i + β |1 i + γ |2 i, where FIG. 1. Circuit to correct a single bit error with ternary L L L L Steane code |0Li = |0000000i + |1020102i + |2010201i + |0102010i + |1122112i + |2112211i + |0201020i + |1221122i + |2211221i

|1Li = |1111111i + |2101210i + |0121012i + |1210121i + |2200220i + |0220022i + |1012101i + |2002200i + |0022002i

|2Li = |2222222i + |0212021i + |1202120i + |2021202i + |0011001i + |1001100i + |2120212i + |0110011i + |1100110i

This QECC is degenerate. For example, a single Z1 error on the first and the fifth qutrits generate the same error state. We shall discuss the impact of degeneracy in error correction in the following subsection. This encoding scheme satisfies the necessary and sufficient con- The Chrestenson gates Ch1 and Ch2 convert a qutrit ditions for error correction [20] which respectively states from computational basis to b1 and b2 respectively. that

h0L| σ |0Li = h1L| σ |1Li = h2L| σ |2Li 1 1 1  1 1 1  for any error σ ∈ E, and Ch = √1 1 ω ω2 Ch = √1 1 ω2 ω 1 3   2 3   1 ω2 ω 1 ω ω2 † hiL| σmσn |jLi = δijαmn

for any errors σm, σn ∈ E, where i, j ∈ {0, 1, 2}, αmn ∈ C One can verify that Ch1Ch2 = I and and δij is the Dirac-delta function.

A. Correction of phase errors Ch1X1Ch2 = Z1 Ch1X2Ch2 = Z2.

The stabilizers for correcting a single phase error are: The circuit for correcting a single bit error in a qutrit using the ternary Steane code is shown in Fig. 1. The S1 = X1 ⊗ I ⊗ X2 ⊗ I ⊗ X1 ⊗ I ⊗ X2 circuit for correcting bit errors follows from the stabilizers S2 = I ⊗ X1 ⊗ I ⊗ X2 ⊗ I ⊗ X1 ⊗ I S4,S5, and S6, and each Z operator is realized using a The stabilizer S1 and S2 operate on two disjoint sets of C+T gate. In the ﬁgure, q0 to q6 are data qutrits and the remaining are ancilla qutrits for syndrome measurement. qutrits. In other words, these two stabilizers partition the qutrits of the codeword in two sets, g1 = {q0, q2, q4, q6} To correct a single phase error, the qutrits must be and g2 = {q1, q3, q5}. The partitioning of the qutrits into converted to the Chrestenson basis. Therefore, the cir- two sets, and the action of S1 and S2 for phase errors cuit for correcting a single phase error is similar to Fig. 1, occurring on qutrits of each set, are depicted in Table I. except that there will be a single Ch1 gate at the beginning and a Ch2 gate at the end of the circuit for each qutrit. Therefore, the total number of gates required to TABLE I. Partition of the qutrits into probable error subsets correct a single error using the ternary Steane code is 38 Type of error S1 S2 Probable error qutrits 2 (24 C + T gates and 14 Chrestenson gates). The depth 1 ω 1 q0, q4 2 ω 1 q , q of the circuit, deﬁned as the maximum number of gates Z 2 6 3 1 1 ω2 q , q on any input to output path is 8, which is for q6. 1 5 4 1 ω q3 5 ω 1 q0, q4 2 6 ω 1 q2, q6 Z2 7 1 ω q1, q5 8 1 ω2 q IV. 7 QUTRIT DEGENERATE QECC 3

It is easy to see that ∃S/∈ {S1,S2} such that [S, Si] = In our proposed 7 qutrit QECC, the information of a 0, i ∈ {1, 2} and wt(S) < 3, which implies that the dis- single qutrit |ψi = α |0i + β |1i + γ |2i is encoded into tance of this code is < 3. For example, the operator 4

Z1 ⊗ I ⊗ Z1 ⊗ I ⊗ I ⊗ I ⊗ I commutes with both the sta- Theorem 1. The proposed QECC can correct upto seven bilizers, which implies that there exist phase errors on q0 simultaneous phase errors on the codeword. and q2 which this code fails to distinguish. However, we show that it is possible to correct the error even without Proof. The proof follows directly from Lemmata 2 and distinguishing them uniquely in certain cases. 3. Consider the error state |ψ¯i such that S |ψ¯i = ω2 |ψ¯i. 1 The circuit for correcting phase errors on the code- From Table I, we note that one cannot distinguish word, shown in Fig. 2, has a depth of 4 along q2, q3 and uniquely whether Z1 error occurred on q0, or Z2 error q6. occurred on q2. However, Z0 |0 i = |0000000i + ω |1020102i + ω2 |2010201i 1 L FIG. 2. Circuit for correcting phase errors with the proposed 2 + |0102010i + ω |1122112i + ω |2112211i QECC + |0201020i + ω |1221122i + ω2 |2211221i 2 = Z2 |0Li j th where, Zi implies the error Zi acting on the j phys- ical qutrit, i ∈ {1, 2}, j ∈ {0, 1,..., 6}. |1Li and |2Li also show similar behavior (which is obvious since this QECC satisfies the necessary condition for error correction). Therefore, it is not necessary to distinguish be- 0 2 tween the two errors Z1 , and Z2 . Rather, when the sta- 2 bilizer S1 gives ω eigenvalue, correcting any one of the two errors is sufficient to correct the error on the codeword. Similar other scenarios are observable in Table I. Thus, although this QECC cannot distinguish between certain phase errors, it can still correct them perfectly. We say that a stabilizer S triggers for some error E, if for that error E the eigenvalue of S is not unity. B. Correction of bit error Lemma 2. The proposed QECC can correct two simultaneous phase errors on two distinct qutrits qi and qj, if The stabilizers for correcting bit errors are as follows: these belong to two distinct sets defined above, i.e, qi ∈ g1 S3 = Z1 ⊗ Z2 ⊗ Z1 ⊗ Z2 ⊗ I ⊗ I ⊗ I and qj ∈ g2. S4 = I ⊗ I ⊗ I ⊗ Z1 ⊗ Z2 ⊗ Z1 ⊗ Z2 Proof. The stabilizers S and S operate on two disjoint 1 2 S = I ⊗ Z ⊗ Z ⊗ Z ⊗ Z ⊗ I ⊗ I sets of qutrits, so a single phase error cannot trigger both 5 1 2 1 2 of them. Both the stabilizers can trigger only when there S6 = I ⊗ I ⊗ Z1 ⊗ Z2 ⊗ Z1 ⊗ Z2 ⊗ I are phase errors on two qutrits q and q such that q ∈ g j i j i 1 Let us assume X indicate error X on the j-th qutrit, and q ∈ g . Each error can be individually detected and i i j 2 i ∈ {1, 2}, j ∈ {0, 1, ..., 6}. Table II shows the action of corrected according to Table I. Therefore, two simulta- the stabilizers for a single X error on different qutrits. neous phase errors can be corrected if they occur on two 1 Correction of X errors will be similar. The non-unity qutrits belonging to disjoint sets g and g . 2 1 2 values will be replaced by their complex conjugate (i.e. 2 Lemma 3. The proposed QECC can correct upto |gi| ω becomes ω and vice-versa). simultaneous phase errors on the qutrits belonging to the same disjoint set g , i ∈ {1, 2}. i TABLE II. Correcting a single bit error CHECK Proof. Consider 0 ≤ m ≤ |gi| phase errors occurring si- Error Type S3 S4 S5 S6 0 multaneously on individual qutrits belonging to the set X1 ω 1 1 1 1 gi, i ∈ {1, 2}. The action of these phase errors on the X1 ω 1 ω 1 2 codeword is X1 ω 1 ω ω 3 m X1 ω ω ω ω 4 O j q X1 1 ω ω ω Z |ψi = ω |ψi 5 X1 1 ω 1 ω i=1 6 X1 1 ω 1 1 j th where, Z ∈ {I,Z1,Z2} is the phase error on the j qutrit in gi, and q ∈ {0, 1, 2}. Therefore. multiple phase Note that the stabilizers S3,...,S6 are not unique. For errors on qutrits of gi behave like a single Z1 or Z2 er- example, S3 = Z2 ⊗ Z1 ⊗ Z2 ⊗ Z1 ⊗ I ⊗ I ⊗ I is valid ror acting on a qutrit of gi, which can be corrected as as well. The difference will be that some values of the depicted in Table I. Table II will be replaced by its complex conjugate. 5

The circuit for correcting a single bit error can be sim- and vice versa for Sj. Furthermore, the eigenvalues of ilarly obtained as for the phase error correction circuit Si and Sj together cannot be a syndrome for some other (Z1 and Z2 realized using a single or two C + T gates re- error. However, if Si operates on qi, then it must also spectively). Four stabilizers for bit error correction leads operate on qk in order to commute with the stabilizer S2 to four ancilla qutrits. The depth of the circuit for bit for phase error. Similarly Sj must again operate on qk error correction is 6 along q3. Therefore, the depth of the in order to commute with S2. Any other stabilizer which entire error correction circuit is 10, which is along q3. operates on qk, must also operate on qi or qj in order to commute with S2. Therefore, Si,Sj together form the Lemma 4. The set of stabilizers S3,...,S6 can correct syndrome for qk. If both Si and Sj show non-identity simultaneous bit errors occurring on q0 and q6. eigenvalues, then it is not possible to distinguish whether Proof. From Table II it is evident that if both q and q both qi and qj are erroneous, or only qk is erroneous. 0 6 Hence, two bit errors occurring simultaneously on two have bit errors, then both the stabilizers S3 and S4 will trigger. These are the only two stabilizers to trigger in qutrits of g2 cannot be corrected. this scenario, and there are no other errors for which only We now provide an algorithm (Algorithm 1) to gener- S3 and S4 trigger. Hence, if these two stabilizers trigger, ate the set of stabilizers S3,...,S6 in order to correct si- then it is possible to identify two simultaneous bit errors multaneous bit errors in a pair of qutrits which is allowed on q0 and q6. by Lemma 6. There are four stabilizers for correcting bit In the following, we show that it is possible to generate errors and seven qutrit positions for a 7-qutrit codeword. j th a set of stabilizers S ,...,S such that they can correct a Let Si denote the operator acting on the j qutrit for 3 6 th single bit error on any qutrit, as well as simultaneous bit the i stabilizer, 3 ≤ i ≤ 6, 0 ≤ j ≤ 6. Furthermore, let p denote the kth qutrit of the codeword, 0 ≤ k ≤ 6, errors on any pre-defined pair (qi, qj) of qutrits (except k for three pairs). We first state the sufficient condition and di denotes the number of non-identity elements at th for correcting two simultaneous bit errors and discuss the i qutrit position for the stabilizers S3,...,S6. For about the pairs for which simultaneous bit errors are not example, in the set of stabilizers shown at the beginning correctable. of Section IV B, d0 = 1, d1 = 2, d3 = 3 and so on. Without loss of generality, in accordance to Lemma 5, Lemma 5. Two simultaneous bit errors on qutrits qi 6= we shall consider Si = S3 and Sj = S4 when correcting qj can be corrected if there exist two stabilizers Si 6= Sj simultaneous bit errors on qi and qj, i < j. which trigger for bit errors on qi and qj respectively, and there does not exist any single bit error, or simultaneous Lemma 7. Algorithm 1 creates a valid set of stabiliz- bit errors on any other pair of qutrits, for which only Si ers which can correct simultaneous bit errors on the pre- and Sj trigger together. defined pairs of qutrits qi and qj excluding the ones by Lemma 6. Proof. Let Si and Sj be two distinct stabilizers such that Si triggers for a bit error on qi and Sj triggers for a bit Proof. We consider the proof in the following steps: error on qj. Therefore, if bit errors occur on both qi and q simultaneously, then the set of triggered stabilizers is (1) For each stabilizer Si, 3 ≤ i ≤ 6, two operators j placed in accordance with Lemma 1, operate on {Si,Sj}. Moreover, if there exist no other single error, or simultaneous bit errors on some different pair of qutrits qutrits of set g1 (g2). All these stabilizers S3,...,S6 commute with S1 and S2. Further, it is verified that for which the set of triggered stabilizers is {Si,Sj}, then this set of triggered stabilizers can be uniquely decoded each Si stabilizes the codeword. as simultaneous bit errors on qi and qj. (2) In Algorithm 1, S3 (S4) alone is chosen as a non- Lemma 5 is not a necessary condition for simultaneous identity operator on qutrit qi (qj), and only S3 (S4) bit error correction, because it may be possible to do so can trigger for a bit error on qi (qj). S3 and S4 even when multiple stabilizers trigger for one or both the together trigger when both qi and qj have simulta- errors. However, it is not possible to correct simultane- neous bit errors. ous bit errors on every pair of qutrits, as elaborated in (3) For unique detection of simultaneous bit errors, we Lemma 6. require that for any qutrit pair (qk,ql), such that qk, ql ∈ gi, i ∈ {1, 2}, other than (qi,qj), either Lemma 6. There are no stabilizers Si,Sj ∈ ⊗7 dk 6= dl, or if dk = dl, then the qutrits do not {I,Z1,Z2} for the proposed QECC design such that it can correct two simultaneous bit errors if they occur have non-identity operators for all the stabilizers at the same position. This condition is checked on two qutrits of g2. when putting the operators on the final stabilizer Proof. Let two bit errors occur simultaneously on qi and for both the sets g1 and g2. Therefore, even if such qj, where qi, qj ∈ g2. Let the third qutrit in g2 be qk. a scenario occurs upto stabilizer S5, the tie will be Thus, we need at least 2 stabilizers Si and Sj, Si,Sj ∈ broken in stabilizer S6, leading to a unique pattern ⊗7 {I,Z1,Z2} such that Si operates on qi and not on qj, of non-unity eigenvalues for each error. 6

our proposed QECC. The gate cost is in terms of C + T and Chrestenson gates. The time complexity of Algorithm 1 is O(1), since the number of qutrits and the number of stabilizers is fixed. TABLE III. Comparison of gate cost and depth of circuit Algorithm 1 To generate the set of stabilizers S3,...,S6 9-qutrit 6-qutrit Ternary Proposed to correct simultaneous bit errors QECC [14] AQECC [15] Steane QECC Input: Qutrits qi and qj on which simultaneous bit errors # qutrits 9 6 7 7 are to be corrected. Gate cost for Output: The set of stabilizers S3,...,S6. bit error 52 18 12 24 1: dh ← 0, 0 ≤ h ≤ 6. correction i j 2: S3,S4 ← Z1. Gate cost for 3: di ← di + 1, dj ← dj + 1. phase error 210 20 26 24 i 4: Sq ← I, 4 ≤ q ≤ 6. correction j 5: Sq ← I, 3 ≤ q ≤ 6, q 6= 4. Total gate cost 262 38 38 48 6: for all stabilizers S3,...,S6 do Depth of circuit 26 8 8 10 7: for i ∈ {1, 2} do 8: for all qutrits in set g do i Both (i) the reduction in speed of computation due to 9: qk, ql ← qutrits in g1 for which dk and dl have the minimum and the second minimum values, and operators have error correction, and (ii) the decay in fidelity depend on not been assigned for these qutrits on the stabilizer. Break the depth of the circuit [6]. Although the circuit cost of ties such that the operators at positions Pk and Pl of sta- our QECC is 10 more than that of the ternary Steane bilizers S1 and S2 respectively are not equal. If such a code, the increase in the depth of the circuit is only scenario is not available, break ties arbitrarily. If one po- 2. Therefore, our QECC is not expected to have any sition qk is already occupied by a non-identity operator, significant performance degradation in terms of (i) and choose only ql accordingly. (ii). Moreover, the ternary Steane code can correct at 10: if stabilizer is S6 then most a single bit and a single phase error. The 6-qutrit 11: if dk 6= dl then AQECC can correct multiple phase errors, but it cannot 12: continue correct a single bit error in all possible scenarios. There- 13: else fore our QECC surpasses both the 6-qutrit AQECC and 14: while both qk and ql has non-identity operators at the same positions for all prior stabilizers do the ternary Steane code in its ability to correct errors, 15: Discard ql. New ql ← qutrit having the minimum value without a significant increase in the depth of the circuit. of d apart from q or the previous q . l k l Theorem 3. It is possible to correct upto n simultane- 16: end while 17: end if ous phase errors for an n-qutrit QECC, n ≥ 3, having 18: end if stabilizers th th 19: Operator on k qutrit ← Zu, and operator on l qutrit n n ← Z , u, r ∈ {1, 2} such that Lemma 1 is satisfied, and O O r S1 = Xi; S2 = Xj Sk ⊗ Sl |ψi = |ψi. i i i=0 j=0 20: dk ← dk + 1, dl ← dl + 1. i is even j is odd 21: end for 22: end for Proof. The stabilizers S1 and S2 partition the qutrits into 23: end for two disjoint sets g1 and g2. The proof of Lemma 3 directly extends to any value of |gi|, i ∈ {1, 2}. Lemma 3, We show an example of the stabilizer construction us- together with Lemma 2, proves this theorem. ing Algorithm 1 in Appendix. Consider an n-qutrit QECC which can correct upto t ⊗7 Theorem 2. A set of stabilizers ∈ {I,Z1,Z2} can be errors. The number of stabilizers in an n-qutrit QECC is generated in O(1) time such that two simultaneous bit n − 1 [4]. Theorem 3 shows that for any n ≥ 3, only two errors on a given pair (qi, qj) of qutrits where both qi and stabilizers are sufficient to correct upto n simultaneous qj ∈/ g2, can be corrected. phase errors on the codeword. This opens up a rich field Proof. The proof follows directly from Algorithm 1 and of using n − 3 stabilizers to correct something more than just the t bit errors. For our QECC, where n = 7 and Lemmata 5, 6 and 7. t = 1, we could correct simultaneous bit errors on pre- defined qutrit pairs. V. COMPARISON OF CIRCUIT DEPTH AND Theorem 4. It is not possible to construct a linear de- SOME REMARKS generate CSS QECC for a qubit system that partitions the qubits into disjoint qubit sets. In Table III we compare the gate cost and the depth of the circuits of the 9-qutrit QECC [14], 6-qutrit approx- Proof. For an n-qubit linear CSS QECC, the codeword imate QECC (AQECC) [15], ternary Steane code, and |0i⊗n must be present in the superposition state of the 7 logical qubit. Any QECC which partitions the n qubit seven simultaneous phase errors. We have also shown codeword into disjoint sets gi must have the stabilizers Si, that the stabilizers can be generated to correct upto two 1 ≤ i ≤ n−1 operating on disjoint qubits. Moreover, for a simultaneous bit errors on a pre-defined pair of qutrits CSS code, the stabilizers can be partitioned into two sets, (except for three pairs). Our QECC is optimal in the ⊗n S1 and S2, such that stabilizers in S1 ∈ {I, σx} and number of qutrits required to correct a single error with ⊗n stabilizers in S2 ∈ {I, σz} . Therefore, the codewords the CSS structure. Moreover, the depth of the circuit are generated by operating the stabilizers in S1 multiple of our QECC is only two more than that of the ternary times on |0i⊗n. Steane code. We have also shown that this formulation can be extended to design any n ≥ 3-qutrit QECC, but Let qSi be the set of codewords which are generated by ⊗n ⊗n applying the stabilizer Si alone on |0i . Also |0i ∈ it is not possible to use this formulation technique to ⊗n ⊗n generate QECCs for qubit systems that can correct si- |0Li. Then (ΠiSi) |0i = |1i , such that Si ∈ S1. This implies that if such a set of stabilizers exist, then both multaneous phase errors. Therefore, our proposed code ⊗n ⊗n readily shows that there exist design techniques of QECC |0i and |1i ∈ |0Li. Hence, a QECC, in which the set of stabilizers operate on disjoint set of qubits, cannot for the ternary quantum system which are more efficient exist for qubit systems. than the ones which are a simple extension of binary quantum codes.

VI. CONCLUSION ACKNOWLEDGEMENT

In this paper, we have proposed a 7-qutrit degenerate All the ﬁgures have been generated using IBM qiskit CSS QECC that can correct a single bit error and upto simulator [21].

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1. Initially di = 0, ∀ i. For brevity, we represent d as operators in positions 0 and 2 must be Z2. a 7-tuple, all of whose entries are initialized to 0. S3 = Z2 Z1 Z2 − I − −

S4 = − I − − Z1 − − d = 0 0 0 0 0 0 0 S5 = − I − − I − −

S6 = − I − − I − −

The new tuple is d = 1 1 1 0 1 0 0 2. We shall start with putting Z1 in positions 1 and 4 of stabilizers S3 and S4 respectively. 6. Now if we look into S4 for g1, then here one position in g1 is already ﬁlled up. So we shall consider some other position only. From the d-tuple, we se- S3 = − Z1 − − − − − lect position 6. From Lemma 1, we note that the operators in q4 and q6 must be the same in order S4 = − − − − Z1 − − for it to commute with stabilizer S1. Therefore, the S5 = − − − − − − − operator at position 6 must also be Z1 (which im- S6 = − − − − − − − plies that when we look into the operators for g2 in S4, both the operators must be Z2).

S3 = Z2 Z1 Z2 − I − − 3. The tuple d is updated as S4 = − I − − Z1 − Z1 S5 = − I − − I − −

S6 = − I − − I − − d = 0 1 0 0 1 0 0 The new tuple is d = 1 1 1 0 1 0 1

7. If we now look into the set g2 for S3, position 1 4. Our requirement is that S3 and S4 alone will trigger is already occupied with a non-identity operator. for bit errors on q1 and q4 respectively. Therefore, Therefore, we need to select some other position for every other stabilizers, we put an identity in from g2, say position 3. From Lemma 1, and also positions 1 and 4. from the argument discussed in step 5, the operator at position 3 of S3 should be Z1.

S3 = Z2 Z1 Z2 Z1 I − − S3 = − Z1 − − I − − S4 = − I − − Z1 − Z1 S4 = − I − − Z1 − − S5 = − I − − I − − S5 = − I − − I − − S6 = − I − − I − − S6 = − I − − I − − The new tuple is d = 1 1 1 1 1 0 1

This step does not change d. 8. Working out in the above-mentioned way for each stabilizer, we have the ﬁnal set of stabilizers

S3 = Z2 Z1 Z2 Z1 III 5. The two disjoint sets are g1 = {q0, q2, q4, q6} and S = IIIZ Z Z Z g2 = {q1, q3, q5}. We shall first fill up the stabilizer 4 2 1 2 1 positions for qutrits in g1. We need to choose two S5 = Z1 IZ1 Z2 IZ2 I qutrit positions with minimum d values. Let us S6 = Z1 IIZ2 IZ2 Z1 select q0 and q2. From Lemma 1, we can either use Z1 on both, or Z2 on both. However, one can The new tuple is d = 3 1 2 4 1 3 2 verify that if all the four non-identity operators in a stabilizer are Z1, then such a stabilizer will not If a bit error occurs on q1, then only S3 has a non- stabilize the logical qubit. Take |1iL. A stabilizer identity operator at position 3 and hence only S3 will which has four or three Z1 operators, will produce a trigger. Similarly, for bit error on q4 only S4 will trigger. non-identity phase for the codeword |1111111i (and Finally, it can be verified that there exist no other single for others as well). Therefore, it is mandatory to qutrit or two qutrit bit error for which only S3 and S4 have 2 Z1 and 2 Z2 operators in each stabilizer. trigger together. Therefore, this set of stabilizers can Since we have already put one Z1 operator, the correct simultaneous bit errors on q1 and q4.