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A CWS code can be defined in terms of stabilizer codes/stabilizer states and graph states. See What is an example of a non-additive code that is not a CWS code?

Is the $ ((11,2,3)) $ nonadditive quantum error-correcting code given in On the Structure of Additive Quantum Codes and the Existence of Nonadditive Codes a codeword stabilized code?

Certainly all the coefficients are $ \pm 1 $ up to a global scalar (in fact all the nonzero coefficients are $ +1 $) so that necessary condition is met.

Maybe there is some way of confirming this (or deriving a contradiction) by thinking about what graph the $ ((11,2,3)) $ code would have to correspond to?

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There exists a CWS code with parameters ((11,2,5)) with a graph specified in 1, p. 89.

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  • $\begingroup$ Do you know if this $ ((11,2,5)) $ CWS code is equivalent to a stabilizer code? $\endgroup$ Aug 8, 2023 at 18:11
  • $\begingroup$ Sorry, I do not know. In general, CWS codes can have better parameters than stabilizer codes, e.g., there is a ((9,12,3)) and a ((10,24,3)) code outperforming the stabilizer codes with the same number of qubits. $\endgroup$
    – Creeper
    Aug 25, 2023 at 8:07
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Theorem 6 of https://arxiv.org/abs/0803.3232 states "All $((n, 2, d))$ CWS codes are additive."

The $ ((11,2,3)) $ code described in the question is non-additive, which can be verified by observing that the weight enumerator coefficients $$ A = (1,0,0,0,\frac{110}{3},0,88,0,605,0,\frac{880}{3},0) \\ B = (1, 0, 0, \frac{55}{3}, \frac{110}{3}, 88, 88, 1210, 605, \frac{4400}{3}, \frac{880}{3}, 289 ) $$ are not all integers. Thus the code is not CWS.

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  • $\begingroup$ how long did it take you to compute this? the code is not permutationally invariant so I don't know of any shortcuts. $\endgroup$
    – unknown
    May 21 at 18:30
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    $\begingroup$ @unknown On my computer in Mathematica using sparse array for the Paulis it takes about 14.5 hours to compute the A weight enumerator for this $((11,2,3))$ code directly from the definition. I've never computed the B weight enumerator directly from the definition I just take A weight enumerator and apply quantum McWilliams identity arxiv.org/abs/quant-ph/9610040. If all computations are done numerically in Mathematica instead of symbolically then the time to find A weight enumerator reduces further to about half an hour. $\endgroup$ May 21 at 20:03

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