1

Throughout, assume that $ n,K,d $ are integers such that an $ ((n,K,d)) $ code exists

Recall the quantum Hamming bound: For a non-degenerate ((n,K,d)) qubit code we must have $$ K(\sum_{j=0}^{\lfloor d/2 \rfloor} 3^j \binom{n}{j} ) \leq 2^n $$

If all codes with parameters $ ((n,K,d)) $ are degenerate then must the parameters $ ((n,K,d)) $ violate the quantum Hamming bound?

In other words, are there any parameter $ ((n,K,d)) $ that satisfy the quantum Hamming bound but it is still the case that all $ ((n,K,d)) $ codes are degenerate?

Ian Gershon Teixeira
  • 3,722
  • 3
  • 21

1 Answers1

1

I think you've slightly mis-stated this: you need a $2^K$ instead of $K$.

So, it is true this bound applies rigorously to non-degenerate codes. That does not mean that all degenerate codes violate the bound. It just means that there could exist degenerate codes that violate it. In fact, most codes that you end up looking at typically do not violate the bound. For example, consider the Shor [[9,1,3]] code. This does not violate the bound.

You might want to check out this previous question: Violation of the Quantum Hamming bound

DaftWullie
  • 57,689
  • 3
  • 46
  • 124
  • I think my notation is unclear. By $ ((n,K,d)) $ I mean a (possibly nonadditive) code made by encoding a $ K $ dimensional logical space into the physical Hilbert space of $ n $ qubits. So an $ [[n,k,d]] $ stabilizer code is an $ ((n,2^k,d)) $ code since the logical space is size $ 2^k $. – Ian Gershon Teixeira Aug 08 '23 at 17:32
  • Oh, OK. Well, whether or not your notation was unclear, I missed that point! (Probably my fault) – DaftWullie Aug 15 '23 at 01:54