Remarks on Bounds for Quantum codes

We present some results that show that bounds from classical coding theory still work in many cases of quantum coding theory. 1 Nondegenerate Quantum Codes Let Q be an ((n,K, d)) nondegenerate quantum code with projection operator P . Let Ai and Bi be the quantum enumerators of Q, [1],[2] Ai = ∑ wt(E)=i Tr(EP )Tr(EP ), Bi = ∑ wt(E)=i Tr(EPEP ), where E is an error operator. (See [1],[2] for definition of error operators and their weights.) It is shown in [1],[2] that At = 1 2n n ∑ i=0 BiPt(i, n), where Pt(i, n) = ∑i j=0(−1) 3 (i j )(n−j t−j ) is the Krawtchouk polynomial. Replace Ai by Ai/K 2 and Bi by Bi/K. Then we get At = 1 2nK n ∑ i=0 BiPt(i, n) and A0 = B0 = 1. (1) Now we can use well known techniques to get upper bounds for the quantity 2K (see for example [3]).We describe this technique here for completeness. 1 Let f(x) = ∑n i=0 aix i be a polynomial . Consider its expansion by Krawtchouk polynomials f(x) = n ∑ i=0 fiPi(x, n). (2) Then from (1) and (2) we have n ∑ i=0 Bif(i) = 1 2nK n ∑ i=0 Aifi. Suppose that f0 > 0, fi ≥ 0, if i = 1, . . . , n, andf(0) > 0, f(i) ≤ 0, if i = d, . . . , n. (3) Then taking into account that for nondegenerate codes Ai = Bi = 0 for i = 1, . . . , d− 1, we have 2K = f(0) + ∑n i=1 f(i)Bi f0 + ∑n i=1 fiAi ≤ f(0) f0 . (4) From classical coding theory we know many useful polynomials that satisfy conditions (3). Here we list some of them. Let f(x) = 4q−d+1 ∏n j=d(1− x j ) [4]. Then fi = (n−k d−1 ) ( n d−1 ) ≥ 0. So we get a quantum Singleton bound [5][6]: K ≤ 2n−2d+2. Note that fi = 0 for i ≥ n − d + 2 and hence if a quantum code meets the Singleton bound it should be nondegenerate up to n− d+2 (a quantum code is called nondegenerate up to t if Ai = 0 for i = 1, . . . , t−1 [6].)(In [4] this proposition is proven for all quantum codes.) Let d = 2e+ 1 and fi = ( Pe(i− 1, n − 1) ∑e i=0 3 i (n i ) )2 . Then [4] f(i) = 0 for i = d, . . . , n and f(0) = 4/ ∑e i=0 3 i (n i ) , and from (4) we get the quantum analog of the Hamming bound: K ≤ 2 ∑e i=0 3 i (n i ) . 2 Analogously, from [3] it is straightforwardly follows that K ≤ L(d)/2, where L(x) = { Lk(x), if dk(n− 1) + 1 < x < dk−1(n− 2) + 1 4L k (x), if dk(n− 2) + 1 < x < dk(n− 1) + 1. Lk(x) = k−1 ∑