A Note on Quantum Hamming Bound

A Note on Quan tum Hamming Bound Salah A. Aly Departmen t of Computer Science, T exas A&M Univ ersity , College Station, TX 778 43-3112 , USA Email: salah@cs.tam u.edu Proving the quantum Ha mming b ound for dege nerate nonbinary stabilizer c o des has b een a n op en problem for a decade. In this note, I prov e this bound for double error-co rrecting degenerate stabilizer co des. Also, I co mpute the maximum length of single and double err or-co r recting MDS stabilizer codes o ver finite fields. 1 Bounds on Quan tum Co des Quantum stabilizer co des ar e a known class of quantum co des that ca n pr o tect qua nt um informa tion against noise and decoherence. Stabilizer co des ca n be constructed from self-o r thogonal or dual- containing c lassical codes, se e for example [3, 8, 11] and references therein. It is desirable to s tudy upper and low er bounds on the minim um distance o f class ical and quantum co des, so the computer search o n the code par ameters can b e minimized. It is a well-known fac t that Singleton and Hamming bo unds hold for classica l co des [10]. Also, upp er and low er b o unds on the achiev able minim um distance o f quantum stabilizer co des a re neede d. Perhaps the simplest upp er bound is the quantum Singleton b o und, also known as the Knill-Laflamme b ound. The binary version of the quantum Singleton bo und w as first proved b y Knill and La flamme in [12], see a lso [1 , 2], a nd la ter generalized by Ra ins using w eight enumerators in [16]. Theorem 1 (Qua n tum Singleto n Bound) . An (( n, K, d )) q stabilizer c o de with K > 1 satisfies K ≤ q n − 2 d +2 . (1) Co des which meet the quan tum Singleton b ound with eq uality are called qua ntum MDS co des. If we assume tha t K = q k , then this bo und can b e stated as k ≤ n − 2 d + 2 . In [1 1] It has b een shown that these codes cannot b e indefinitely long and sho wed tha t the maximal length of a q - ary quantum MDS codes is upp er b ounded b y 2 q 2 − 2. This could pr obably b e tightened to q 2 + 2. It would b e in teres ting to find quantum MDS co des of length greater than q 2 + 2 since it w o uld disprov e the MDS co njecture for classical co des [10]. A related op en questio n is re g arding the construction of co des with lengths b etw een q and q 2 − 1. A t the moment there a re no analytical metho ds for constr uc ting a quantum MDS code of arbitra ry length in this range (see [9] for some nu merical results). Another imp ortant b o und for quantum codes is the qua nt um Hamming b ound. The quantum Hamming bound states (see [6, 7]) that: 1 Theorem 2 (Qua n tum Hamming Bound) . A n y pur e (( n, K, d )) q stabilizer c o de satisfies ⌊ ( d − 1) / 2 ⌋ X i =0  n i  ( q 2 − 1) i ≤ q n /K. (2) The previous q uantum Hamming b ound holds only for nondegenerate (pure) quan tum c o des. How ever, the degenerate (impure) quantum co des are particular ly interesting cla ss of quantum co des b eca us e they can pack more quantum information. In addition, the er r ors of s mall weigh ts do not need ac tive er ror corre ction strategies . So far no deg e ne r ate quan tum code has b een found that beats this bound. Gottesman sho wed that impure single and double erro r-cor r ecting binary quantum code s cannot b eat the quantum Hamming bound [8 ]. It is pro ved in [11] that Hamming b ound holds for quan tum stabilizer codes with dista nc e d = 3 . In general, do es Hamming bound exist for an y distance d in (( n, K, d )) q stabilizer codes? This has b een a n o p en question for a decade. In this no te we pro ve Hamming bound for double er ror- correcting stabilizer co des with distance d = 5 and also give a sketc h to prov e it for genera l distance d . 2 Quan tum Hamming Bou nd Holds for Distance d = 5 There hav e been several appro aches to prove bounds on the qua nt um co de par ameters. In [1] Ashikhmin and Litsyn derived man y bo unds for quant um co des b y extending a nov el method originally intro duced b y Dels a rte [5 ] for classica l co des. Using this method they proved the binary versions of Theor ems 1,2. W e use this metho d to show that the Hamming b ound holds for all double error -corre c ting qua n tum codes. See [11] for a similar result for single error-co rrecting co des. But first w e ne e d Theo rem 3 and the K rawtc houk p olynomia l of degr ee j in the v ariable x , K j ( x ) = j X s =0 ( − 1) s ( q 2 − 1) j − s  x s  n − x j − s  . (3) Theorem 3. L et Q b e an (( n, K, d )) q stabilizer c o de of dimension K > 1 . Supp ose t hat S is a nonempty subset of { 0 , . . . , d − 1 } and N = { 0 , . . . , n } . L et f ( x ) = n X i =0 f i K i ( x ) (4) b e a p olynomial satisfying the c onditions i) f x > 0 for al l x in S , and f x ≥ 0 otherwise; ii) f ( x ) ≤ 0 for al l x in N \ S . Then K ≤ 1 q n max x ∈ S f ( x ) f x . (5) 2 Pr o of. See [11]. W e demo ns trate usefulness of the previo us The o rem by showing that quantum Hamming b o und holds for impure co des when d = 5. Lemma 4 (Quantum Hamming Bound) . An (( n, K , 5)) q stabilizer c o de with K > 1 satisfies K ≤ q n   n ( n − 1)( q 2 − 1) 2 / 2 + n ( q 2 − 1) + 1  . (6) Pr o of. Let f ( x ) = P n j =0 f j K j ( x ), wher e f x = ( P e j =0 K j ( x )) 2 , S = { 0 , 1 , . . . , 4 } a nd N= { 0,1,. . . ,n } . Calculating f ( x ) a nd f x gives us f 0 = (1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2) 2 f 1 = 1 4 ( n − 1) 2 ( n − 2) 2 ( q 2 − 1) 4 f 2 = ( 1 2 ( n − 3)( n − 2)( q 2 − 1) 2 − ( n − 2)( q 2 − 1)) 2 f 3 = (1 − 2( n − 3)( q 2 − 1) + 1 2 ( n − 4)( n − 3)( q 2 − 1) 2 ) 2 f 4 = (3 − 3( n − 4)( q 2 − 1) + 1 2 ( n − 5)( n − 4)( q 2 − 1) 2 ) 2 and, f (0) = q 2 n (1 + n ( q 2 − 1) + 1 2 ( n − 1) n ( q 2 − 1) 2 ) f (1) = q 2 n ( q 2 + 2( n − 1)( q 2 − 1) + ( n − 1)( q 2 − 2)( q 2 − 1)) f (2) = q 2 n (4 + 4( q 2 − 2) + ( q 2 − 2) 2 + 2( n − 2)( q 2 − 1)) f (3) = q 2 n (6 + 6( q 2 − 2)) f (4) = 6 q 2 n . Clearly f x > 0 for a ll x ∈ S . Also, f ( x ) ≤ 0 for all x ∈ N \ S since the binomial coefficients for the negative v alues ar e zero . The Hamming bo und is given by K ≤ q − n max s ∈ S f ( x ) f x (7) So, there ar e four differ e nt comparisons where f (0) /f 0 ≥ f ( x ) /f x , for x = 1 , 2 , 3 , 4 . W e find a low er b o und for n that holds for all v alues of q . F rom Lemmas 5 ,6,7, and 8, shown b elow, for n ≥ 7 it follows that max { f (0) /f 0 , f (1) / f 1 , f (2) / f 2 , f (3) / f 3 , f (4) / f 4 } = f (0) /f 0 (8) While the ab ov e method is a genera l metho d to pr ov e Hamming bo und for impure quantum co des, the num b er o f terms increa ses with a large minimum distance. It b ecomes difficult to find the tr ue b ound us ing this metho d. How ever, one can derive more conseq ue nc e s from Theore m 3; see, for instance, [1 , 2 , 13, 15]. 3 Lemma 5. The ine quality f (0) /f 0 ≥ f (1) /f 1 holds for n ≥ 6 and q ≥ 2 . Pr o of. Let f (0) / f 0 ≥ f (1) /f 1 then 1 1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2 ≥ 4 q 2 (( n − 1)( q 2 − 1) + 1) ( n − 1) 2 ( n − 2) 2 ( q 2 − 1) 4 ( n − 1) 2 ( n − 2) 2 ( q 2 − 1) 4 ≥ (1 + n ( q 2 − 1) + n ( n − 1) 2 ( q 2 − 1) 2 )(4 q 2 (( n − 1)( q 2 − 1) + 1)) in the left side ( n − 1) approximates to ( n − 2 ). Also , in the right side ( n − 2) and ( n − 1) approximate to ( n ). So, ( n − 2) 4 ( q 2 − 1) 4 ≥ 4(1 + n ( q 2 − 1) + n 2 2 ( q 2 − 1) 2 )( q 2 ( q 2 − 1)( n − 1) + 1) divide both sides by ( q 2 − 1) 2 ( q 2 − 1) 2 and appro ximate 1 q 2 − 1 ≤ 1, w e find that ( n − 2) 4 ≥ 8(1 + n + n 2 2 )( n − 1) by approximating b oth sides, the final res ult is ( n − 2) ≥ 4 or n ≥ 6 Lemma 6. The ine quality f (0) /f 0 ≥ f (2) /f 2 holds for n ≥ 7 and q ≥ 2 . Pr o of. Let q 2 n 1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2 ≥ q 2 n ( q 4 + 2( n − 2)( q 2 − 1)) ( − ( n − 2)( q 2 − 1) + ( n − 3)( n − 2)( q 2 − 1) 2 / 2) 2 by simplifying b oth sides ( − ( n − 2)( q 2 − 1) + ( n − 3)( n − 2 )( q 2 − 1) 2 / 2) 2 ≥ ( q 4 + 2( n − 2)( q 2 − 1))(1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2) Simplifying L.H.S, ( n − 2) to ( n − 3) then ( q 2 − 1) 4 (( n − 3) 2 / 2 − ( n − 2)) 2 ≥ ( q 4 + 2( n − 2)( q 2 − 1))(1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2) 4 by simplifying b oth sides (( n − 3) 2 / 2 − ( n − 2)) 2 ≥ ( q 2 ( q 2 − 1) 2 + 2( n − 2) q 2 − 1 )(1 + n + n ( n − 1) / 2) (( n − 3) 2 / 2 − ( n − 2)) 2 ≥ 2(2 n + 1 )( n 2 + 2 n + 2) ( n − 3) 2 (( n − 3) / 2 − 1) 2 ≥ 2(2 n + 1 )(( n + 1 ) 2 + 1) ( n − 5) 2 / 4 ≥ 2(2 n + 1) n ≥ 7 Lemma 7. The ine quality f (0) /f 0 ≥ f (3) /f 3 holds for n ≥ 7 and q ≥ 2 . Pr o of. Let q 2 n 1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2 ≥ 6 q 2 n ( q 2 − 1) (1 − 2( n − 3)( q 2 − 1) + ( n − 3)( n − 4)( q 2 − 1) 2 / 2) 2 by simplification (1 − 2( n − 3)( q 2 − 1) + ( n − 3)( n − 4)( q 2 − 1) 2 / 2) 2 ≥ 6( q 2 − 1)(1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2) (( n − 4) 2 ( q 2 − 1) 2 − 4( n − 4)( q 2 − 1) + 2) 2 4 ≥ 6( q 2 − 1)(1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2) by approximation to ( q 2 − 1) ( q 2 − 1) 4 4 (( n − 5) 4 ≥ 3( q 2 − 1) 3 (2 + 2 n + n 2 ) ( n − 5) 4 ≥ 4(2 + 2 n + n 2 ) ( n − 5) 2 ≥ 2 n ≥ 7 Lemma 8. The ine quality f (0) /f 0 ≥ f (4) /f 4 holds for n ≥ 7 and q ≥ 2 . Pr o of. Let q 2 n 1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2 ≥ 6 q 2 n (3 − 3( n − 4)( q 2 − 1) + ( n − 4)( n − 5)( q 2 − 1) 2 / 2) 2 divide b y q 2 n and simplifying (3 − 3( n − 4)( q 2 − 1) + ( n − 4)( n − 5)( q 2 − 1) 2 / 2) 2 ≥ 6(1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2) 5 then by a pproximating ( n − 4 ) to ( n − 5) in L.H.S and ( n − 1) to 4 n in R.H.S, we ca n find that ( − 3( n − 4)( q 2 − 1) + ( n − 5) 2 ( q 2 − 1) 2 / 2) 2 ≥ 6(1 + n ( q 2 − 1) + n 2 ( q 2 − 1) 2 / 2) (( n − 5) 2 ( q 2 − 1) + ( n − 5) 2 ( q 2 − 1) 2 / 2) 2 ≥ 6(1 + n ( q 2 − 1) + n 2 ( q 2 − 1) 2 / 2) ( q 2 − 1) 4 (( n − 5) 2 + ( n − 5) 2 / 2) 2 ≥ 6(1 + n ( q 2 − 1) + n 2 ( q 2 − 1) 2 / 2) dividing both sides by ( q 2 − 1) 4 and simplifying (9 / 4)( n − 5) 4 ≥ 6(1 + n ( q 2 − 1) + n 2 ( q 2 − 1) 2 / 2) ( q 2 − 1) 4 (9 / 4)( n − 5) 4 ≥ 6(1 + n + n 2 ) n ≥ 7 Since it is not known if the q ua nt um Hamming b ound holds for degenera te nonbinary quantum co des, it would b e interesting to find degenera te quantum co des that either meet or b eat the quantum Hamming b ound. This is obviously a challenging op en research problem. 3 Maximal L ength of MDS Co des In this section we deriv e some r esults on the maximal length of single and double err or-cor recting quantum MDS co des . These bounds hold for all additive quantum co des . 3.1 Maximal Length Single Error-cor recting MDS Co des Lemma 9. The maximal length of single err or c orr e ct ing additive quantum MDS c o des is given by q 2 + 1 . Pr o of. W e know that the quantu m Ha mming bound holds for K > 1 for d = 3, so K ≤ q n 1 + n ( q 2 − 1) (9) If the Hamming bo und is tigh ter tha n the Singleton b ound for a ny (( n, K , 3)) q quantum co de, then it mea ns that MDS co des canno t exist for that se t of n, K . This occur s when q n − 2 d +2 = q n − 4 ≥ q n 1 + n ( q 2 − 1) 1 + n ( q 2 − 1) ≥ q 4 n ≥ q 2 + 1 (10) Thu s there exist no single er r or cor recting quantum MDS co des for n > q 2 + 1. 6 3.2 Upp er Bound on the Maximal Length of Double Err or-correcting MDS Co des Lemma 10. The maximal length of double err or-c orr e cting quantum MDS c o des is upp er b ounde d by: n ≤ ( q 2 − 3) + p (( q 2 − 3) + 8( q 8 − 1)) 2( q 2 − 1) (11) Pr o of. It is known that the Hamming b ound for d = 5 is given b y: K ≤ q n 1 + n ( q 2 − 1) + n ( n − 1)( q 2 − 1) 2 / 2 (12) If the Hamming bo und is tigh ter tha n the Singleton b ound for a ny (( n, K , 5)) q quantum co de, then it means that MDS co des cannot exist for that set of co de parameters . By simple computation, we find that q n − 2 d +2 = q n − 8 ≥ q n 1 + n ( q 2 − 1) + n ( n − 1) 2 ( q 2 − 1) 2 n 2 ( q 2 − 1) 2 − n ( q 2 − 1)( q 2 − 3) − 2( q 8 − 1) ≥ 0 (13) So, the quadratic equa tion of n has t wo real s olutions. This inequalit y holds for n ≥ ( q 2 − 3) − p ( q 2 − 3) 2 + 8( q 8 − 1) 2( q 2 − 1) (14) n ≤ ( q 2 − 3) + p ( q 2 − 3) 2 + 8( q 8 − 1) 2( q 2 − 1) (15) Only the p ositive solution fo r n is v alid. So, the maxima l leng th of double erro r-cor r ecting MDS co de is upper b o unded by n ≤ ( q 2 − 3) + p ( q 2 − 3) 2 + 8( q 8 − 1) 2( q 2 − 1) (16) References [1] A. Ashikhmin and S. Litsyn. Upp er b ounds on the size of quantum co des. IEEE T r ans. Inform. The ory , 4 5 (4):1206 –1215 , 1 999. [2] A.E. Ashikhmin, A.M. Ba rg, E. Knill, and S.N. Lits y n. Q uantum error detection II: Bounds. IEEE T r ans. on Information The ory , 46(3):789– 800, 2 000. 7 [3] A.R. Calderbank, E.M. Rains, P .W. Shor, and N.J.A. Slo ane. Qua nt um err o r corr ection via co des over GF(4). IEEE T r ans. Inform. The ory , 44:1 369–1 387, 1998. [4] P . Delsarte. F our fundamental pa r ameters of a co de and their co m binatorial sig nificance. Information and Contr ol , 23(5):4 07-438, De c emb er 1973 . [5] P . Delsarte. Bounds for unrestricted co des by linear progra mming . Ph ilips R es. R ep orts , 27:272 –289 , 19 72. [6] K. F eng and Z. Ma. A finite Gilb e rt-Varshamov b ound for pure stabilizer quantum co des. IEEE T r ans. Inform. The ory , 50(12 ):3323– 3 325, 2004 . [7] D. Gottesma n. A clas s of quantum error- correcting codes saturating the quantum Hamming bo und. Phys. R ev. A , 54:1862 –186 8, 19 9 6. [8] D. Gottesman. Stabilizer co des a nd quan tum error co rrection. Caltech Ph. D. Thesis, eprint: quant-ph/9705052, 1 997. [9] M. Grass l, T. Beth, a nd M. R¨ otteler. On optimal q uantum co des. Internat. J. Quantu m Information , 2(1):757–775 , 2004 . [10] W. C. Huffman and V. Pless. F undamentals of Err or-Corr e cting Co des . Univ ersity Press , Cambridge, 2 003. [11] A. Ketk ar, A. Klapp eneck er, S. Kumar, and P .K. Sar vepalli. Nonbinary stabiliz e r co des ov er finite fields. IEEE T r ans. Inform. The o ry , 52(11):48 92 – 4914, 200 6. [12] E. K nill and R. Laflamme. A theo ry of quantum error –corr ecting co des. Physic al R eview A , 55(2):900 –911 , 1 9 97. [13] V.I. L evensh tein. Krawtc ho uk p olyno mials and univ er sal b ounds for co des and designs in Hamming space s. IEEE T r ans. Inform. The ory , 41(5 ):1303– 1321, 1995 . [14] J.H V an Lint. Introductio n to co ding theory . Thir d Edition, Springer-V erla g 1999 . [15] R.J. McE liece, E.R. Ro demic h, jr. H. Rumsey , and L.R. W e lch. New upp er b ounds on the rate of a co de via the Delsarte-MacWilliams inequalities. IEEE T r ans. Inform. The ory , 23(2):157, 1977. [16] E.M. Rains. Nonbinary quantum co des. IEEE T r ans. Inform. The o ry , 45:182 7–183 2, 19 99. 8 4 App endix An A pproac h (Sk etc h) to Pro v e Hamming Bound for Degen- erate Non binary S tabilizer Co des with Minim um Distance d One wa y to prov e the quantum Hamming b ound for impure no nbinary stabilizer co des with d ≤ ( n − k + 2) / 2 is to expand f ( x ) /f x in terms of Kr awtc houk p olynomia ls. Let f ( x ) = P n j =0 f j K j ( x ) and f x = ( P e i =0 K i ( x )) 2 . The Kr awtc houk p olynomial of deg r ee e in the v ar iables x and q is given by K e ( q , x ) = e X j =0 ( − 1) j ( q 2 − 1) e − j  x j   n − x e − j  (17) Theorem 1 1. L et Q b e an (( n, K, d )) q stabilizer c o de of dimension K > 1 . Supp ose that S is a nonempty subset of { 0 , 1 , ..., d − 1 } and N = { 0 , 1 , ..., n } . L et f ( x ) = n X i =0 f i K i ( x ) b e a p olynomial satisfying the c onditions: i) f x > 0 for al l x ∈ S , and f x ≥ 0 otherwise; ii) f ( x ) ≤ 0 for al l x ∈ N \ S . Then K ≤ 1 q n max x ∈ S f ( x ) f x . Notice that f ( x ) = P n i =0 f i K i ( x ) c an b e written as f i = q − 2 n P n x =0 f ( x ) K x ( i ). Lemma 12 (Sketc h) . L et Q b e an (( n, K, d )) q stabilizer c o de of dimension k ≥ 1 . Supp ose that S is a non-empty su bset of { 0,1,2, ....,2e } , wher e e = ⌊ d − 1 2 ⌋ . The Hamming b ound is given by K ≤ q − n max x ∈ S f ( x ) f x e quals to K ≤ q n P e i =0  n i  ( q 2 − 1) i (18) If and only if f (0) /f 0 is the maximum value for d ≥ 3 and n ≥ n 0 . Pr o of. In this pro of, we propo se f x satisfying Theor em 11. Let f x =  P e i =0 K i ( x )  2 and f ( x ) = P n j =0 f j K j ( q , x ). 9 f ( x ) f x = P n j =0 f j K j ( q , x ) f x (19) And our goal is to find max { f (0 ) / f 0 , f (1) / f 1 , ..., f ( d − 1 ) /f d − 1 } that ma y equal to f (0) /f 0 . Now, for x = 0, we find that f (0) f 0 = P n j =0 f j K j (0) f 0 = K 0 ( q , 0) + f 1 K 1 ( q , 0) f 0 + ...... + f n K n ( q , 0) f 0 (20) or f (0) f 0 = P n j =0  P e i =0 K i ( j )  2 K j (0)  P e i =0 K i (0)  2 and for an y other v alue of y ∈ { 1 , 2 , ..., d − 1 } , we find that f ( y ) f y = P n j =0 f j K j ( y ) f y = f 0 K 0 ( q , y ) f y + f 1 K 1 ( q , y ) f y + ...... + f n K n ( q , y ) f y (21) or f ( y ) f y = P n j =0  P e i =0 K i ( j )  2 K j ( y )  P e i =0 K i ( y )  2 F ro m 20 and 21, s imply we need to show that f (0) f 0 − f ( y ) f y ≥ 0 (22) f (0) f 0 − f ( y ) f y = P n j =0  P e i =0 K i ( j )  2 K j (0)  P e i =0 K i (0)  2 − P n j =0  P e i =0 K i ( j )  2 K j ( y )  P e i =0 K i ( y )  2 = n X j =0  e X i =0 K i ( j )  2  K j (0)  P e i =0 K i (0)  2 − K j ( y )  P e i =0 K i ( y )  2  ! = n X j =0  f j K j ( q , 0) f 0 − f j K j ( q , y ) f y  (23) 10 in the previous equatio n, f j > 0 and f y > 0, so, if w e pro ve that f j K j ( q , 0) f 0 − f j K j ( q , y ) f y ≥ 0 , (24) then the cla im holds . As s hown in [4 ], [14], we see k a constant v alue fo r the left side in 24, so, m ultiplying both sides by K e ( i ) K e ( i ) K i ( q , 0) f 0 − K e ( i ) K i ( q , y ) f y ≥ 0 (25) and tak e P n i =0 n X i =0  K e ( i ) K i ( q , 0) f 0 − K e ( i ) K i ( q , y ) f y  ≥ 0 P n i =0 K e ( i ) K i ( q , 0) f 0 − P n i =0 K e ( i ) K i ( q , y ) f y ≥ 0 (26) from [14], g iven tha t P n i =0 K e ( i ) K i ( q , j ) = q n δ ej , by substitution, q n δ e 0 f 0 − q n δ ey f y ≥ 0 δ e 0 f 0 − δ ey f y ≥ 0 (27) Now, δ e 0 = 1 , a nd δ ey = 1 or 0; and ob vious ly f y ≥ f 0 . So, if y = e = ⇒ δ e 0 /f 0 ≥ 0 , a nd similarly , δ ey = 1 = ⇒ f y − f 0 ≥ 0. 11