In quantum error correction, two codes are considered to be equivalent if they differ by a locally unitary operator and a permutation. I would imagine that such codes should have the same error-correcting capabilities. Maybe it is too basic, but I am failing to see this, and I am also failing to find it explained in literature.

To make the answer easier, I am looking for the following:

1) If an error set is correctable by a code, is it also correctable by an equivalent code? And why?

2) Is this equivalence notion a natural thing to do? And why?

If it helps, feel free to restrict yourselves on stabilizer codes.

UPDATE: Why do two equivalent codes have the same error correcting capabilities?For instance, do they have the same minimum distance?


2 Answers 2


Two equivalent quantum error-correcting codes have the same distance. They don't necessarily have the same set of correctable errors if it is an asymmetric set.

Consider the definition of distance: $$\langle \psi_i | E | \psi_j \rangle = C_{E} \delta_{ij}$$ for basis vectors $|\psi_i\rangle$ and errors $E$ with weight less than the distance $d$. Let $U$ be a local unitary and permutation mapping between two equivalent codes. Then a basis for the second code is $U |\psi_i\rangle$. We thus want to show that $$\langle \psi_i | U^\dagger E U |\psi_j \rangle = C'_{E} \delta_{ij}.$$

A local unitary is a tensor product of single-qubit operations, in which case $U^\dagger E U$ has the same weight as $E$. Similarly, if $U$ is a permutation, $U^\dagger E U$ also has the same weight as $E$. Letting $F=U^\dagger E U$, the result follows, with $C'_{E} = C_F$.

To be more explicit re question 2: Local unitaries plus permutations is exactly the set of unitaries that preserve weight for all $E$. If you care about distance, it is a completely natural thing to look at.

  • $\begingroup$ Thanks for answering! I understand that! What it puzzles me is the following: If one conjugates a stabilizer, it might not be a stabilizer anymore, unless the operator is Clifford. It will still be abelian and $-I\notin S$ but won't be a subset of the Pauli group. So in turn, it will not make sense to talk about the corresponding stabilizer code. That's why, to me, it seems more natural to talk (only) about local Clifford Equivalence. $\endgroup$
    – user 1987
    Commented Sep 29, 2016 at 18:35
  • $\begingroup$ One more thing: when the equivalence is concerned, is the local unitary applied to the stabilizer or to the code? It seems the last paragraph says is the former (I originally thought was the latter!) Thanks! $\endgroup$
    – user 1987
    Commented Sep 29, 2016 at 20:09
  • $\begingroup$ Yes, a stabilizer code can be equivalent to a non-stabilizer code; but in this case, the non-stabilizer code is just a stabilizer code in some non-standard basis (which might be different for each qubit). That's kind of a tautological statement, I know. $\endgroup$ Commented Sep 29, 2016 at 20:13
  • $\begingroup$ There is no real distinction between applying the LU to the code and to the stabilizer. Changing the code changes the stabilizer and vice-versa. $\endgroup$ Commented Sep 29, 2016 at 20:16

There's at least some trivial cases where the equivalence doesn't hold.

The error set {I, X} is corrected by classical error correcting codes (over the computational basis Z). But switch the computational basis from Z to X (which you can do with local transformations), and you aren't protected anymore. Classical codes only protect along one axis per qubit, and local unitary operations can push them off that axis.

(I'm not sure if that meets your criteria or not... classical ECCs also break in other ways in the presence of quantum operations. An I-vs-X between two Hs is an I-vs-Z, and so won't be corrected.)

  • $\begingroup$ Thanks, that helps with the first question. But I am now confused with my second question as to why this notion is used. Just as a side question, why do you call a code over Z a classical ECC? $\endgroup$
    – user 1987
    Commented Sep 29, 2016 at 2:18
  • $\begingroup$ @user1987 I probably have the terminology wrong. I just mean that classical codes protect against toggle errors but don't protect against phase errors. $\endgroup$ Commented Sep 29, 2016 at 2:24

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