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Suppose we have a stabilizer code on n physical qubits, with stabilizer group S and assume fixed representatives of the logical operators L. Given this fixed set of generators for S and L, can we define a unique (up to certain stabilizer transformations) set of pure error generators (or destabilizers) which generate the group of pure errors in stim?

The commutation relations say that the pure error generator T_{i} anticommutes only with the stabilizer generator S_{i} and commutes with all other generators of S, L, and T.

Given S, L is there a possibility to get T from stim library?

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You want stim.Tableau.from_stabilizers. The z_outputs of the tableau are the stabilizers; the x_outputs are the corresponding destabilizers.

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  • $\begingroup$ Ah, Ok. Can it also provide logical operators consistent with stabilizers/destabilizers? $\endgroup$
    – reza
    Commented Jul 18 at 23:56
  • $\begingroup$ @reza Yes, it will also solve for the observables. $\endgroup$
    – Craig Gidney
    Commented Jul 18 at 23:57
  • $\begingroup$ @CraigGidney how will this work if the stabilizers are not independent?...e.g. if you give it all stabilizers of the toric code what would it return? $\endgroup$
    – unknown
    Commented Jul 20 at 15:19
  • $\begingroup$ @unknown First, it will force you to specify allow_redundant=True to warn you what's happening. Then it will simply go stabilizer by stabiilizer, in order, dropping any that are redundant given the presence of the previous ones in the order. Then it solves as normal. $\endgroup$
    – Craig Gidney
    Commented Jul 20 at 21:06
  • $\begingroup$ @CraigGidney, this is reasonable, but it does make the destabilizers dependent on the order the stabilizers are processed. I don't know if there's a "better" way that gives back a dependent set of destabilizers without favoring an order. For the toric code for example, the $L^2$ X (and $L^2$ Z) stabilizers are indistinguishable : you can drop any one of them to get an independent set ; you would think there's a way to get dependent destabilizers that are also indistinguishable. $\endgroup$
    – unknown
    Commented Jul 20 at 21:24

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