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Say we have a $n$-qubit stabilizer code $Q$ with stabilizer group $S$. If unitary $\bar{K}$ is a logical operation that preserves the codespace, what conclusion can we draw for $\bar{K}$ and $S$?

If $\bar{K}$ is Clifford, I believe it's straightforward to have $\bar{K}s\bar{K}\in S, \forall s\in S$. In another words, $\bar{K}$ should map the stabilizer group $S$ to itself. However, this seems doesn't hold in the general case. The $[[15, 1, 3]]$ Reed-Muller code admits $\bar{T} = T^{\otimes 15}$. $[[15, 1, 3]]$ codes have several weight-8 X stabilizer generators $s_x$, but apparently $T^{\otimes 15} s_x T^{\otimes 15} $ isn't a stabilizer.

Another equivalent way to ask the question can be: Given arbitrary $n$-qubit operation $U$, how can I judge if $U$ is logical operation for $Q$ or not by checking its relation with the stabilizer group $S$? For example, if $UsU\in S, \forall s\in S$, can we derive that $U$ is logical operation?

Yunzhe
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Often logical operations have this property, because it makes things simple, but it's in no way required.

A fault tolerant logical operation can flip the signs of stabilizers, as long as it's known which ones flip. It can even transform the stabilizers, as long as it's known how they transform. There's no rule against a logical operation also doing code switching. The only rule is it must tolerate faults.

Craig Gidney
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To my understanding, there isn't a straightforward definition for the group of all unitaries that preserve the code in terms of the stabilizer group. Of course, if $UsU^\dagger = S$ for all $s \in S$, $U$ is a logical operation, but this is not a necessary condition. A broader sufficient condition is that $U^\dagger s U$ acts as a logical identity for all $s \in S$, which I believe should also be a necessary condition.

Rohan Mehta
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