Two quantum codes are said to be equivalent if they are related by a non-entangling gate.

Over the range of all parameters $((n,K,d))$ with $d> 1$ we know there are at least countably many non-equivalent quantum codes. For example, there exists countable families of codes such as https://errorcorrectionzoo.org/c/diagonal_clifford .

On the other hand, the integer programming bounds (https://arxiv.org/pdf/quant-ph/9608006.pdf extended to non-additive codes) indicate that they may be an uncountable number of codes. For example, a code with parameters $((4,2,2))$ must have the following $A$-quantum weight enumerator: $$ A = (1, 0, t, 4 - 2 t, 3 + t) $$ where $0 \leq t \leq 2$. Of course, this doesn't mean that for each choice of $t$ there is a quantum code with this weight enumerator but it at least suggests an uncountable number of quantum codes is possible.

In summary, I am wondering about the cardinality of the set of all $((n,K,d>1))$ quantum codes up to equivalence.

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    $\begingroup$ Just pick a pair of qubits and conjugate the code's stabilizers by $\text{CZ}^t$ on those qubits for any $t \in (0, 1)$. Sure it can reduce the distance by 1, and the check circuit is going to start by undoing this silly partial CZ and end by reinstating it, but the result's still a thing that can correct errors. It's still a code. A continuous family of codes. You need a more flexible notion of equivalence to prevent these kinds of shenanigans. $\endgroup$ Apr 24, 2023 at 13:57
  • $\begingroup$ Why does this produce an uncountable collection of codes? Wouldn't this procedure produce a countable collection (indeed finite)? $\endgroup$ Apr 24, 2023 at 21:08
  • $\begingroup$ Because there are uncountably many values of t between 0 and 1. $\endgroup$ Apr 24, 2023 at 21:09
  • $\begingroup$ @CraigGidney Interesting so you are applying the entangling gate $ diag(1,1,1,e^{2 \pi i t}) $ to the code space, for some $ t \in (0,1) $ . It seems intuitive to me that your approach would almost never work because the resulting code would almost always have distance $ 1 $ and the OP has asked specifically if there uncountably many code with $ d > 1 $. Do you have a code in mind such that for uncountably many values of $ t $ it is the case that $ diag(1,1,1,e^{2 \pi i t}) $ applied to the codespace still has $ d>1 $? $\endgroup$ Apr 25, 2023 at 1:38
  • $\begingroup$ To be more precise, I would conjecture that for any code it is the case that applying $ diag(1,1,1,e^{2 \pi i t}) $ to the code space yields a $ d>1 $ code for only finitely many values of $ t \in(0,1) $. This conjecture may be naive, so feel free to provide a counterexample! $\endgroup$ Apr 25, 2023 at 1:45

1 Answer 1


The space of codes is continuously infinite, because you can take any stabilizer code and conjugate its stabilizer generators by a local unitary to produce another code. For example, conjugate two qubits of a distance 25 surface code by $\text{CZ}^t$. For all intents and purposes the result is the same code, but it isn't a stabilizer code and it's not equal up to non-entangling gates.


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