# Does every code have transversal Pauli group?

What is an example of a code for which the Pauli group cannot be implemented transversally?

Mostly interested in an $$[[n,1,d]]$$ code where $$d \geq 3$$.

Edit: By transversal I mean that there exists $$n$$ single qubit operators $$g_1 \dots g_n$$ such that the physical operator $$g_1 \otimes \dots \otimes g_n$$ implements logical $$X$$ (or $$Z$$ or $$Y$$ as the case may be). For example, in this sense every stabilizer code has a transversal implementation of the Pauli group (indeed in that case the $$g_i$$ can all be picked to be Paulis).

• A clarification of "transversally" would help here. The destabilizers of the code are a subgroup of the pauli group; they are generated by a pauli strings but they don't preserve the codespace. Jul 20 at 18:13
• ...or maybe by "Pauli group" you mean the subgroup generated by the $2k$ logicals? Jul 20 at 21:45

Remember that the definition of the code space (which determines the distance of the code) and the logical operators within the code space are effectively separate, and the choice of logical operators is largely arbitrary. We can use this to construct examples. For instance, let's take the usual Steane 7-qubit code. Let $$R_Z(\theta)$$ be the logical $$Z$$ rotation relative to the standard construction. So, $$R_Z(\pi/8)$$ is the $$T$$ gate (up to some global phases).
Now, if $$|0_L\rangle$$ and $$|1_L\rangle$$ are the usual definitions of the logical states, we can choose to work in a new basis with $$|0_L'\rangle=R_Z(\pi/16)|0_L\rangle,\qquad |1_L'\rangle=R_Z(\pi/16)|1_L\rangle$$ This means that the logical $$X$$ operator is $$X_L'=R_Z(\pi/16)X_LR_Z(-\pi/16)=R_Z(\pi/8)X_L.$$ Since the Steane code doesn't have transversal $$T$$, we know that this operation cannot be implemented transversally. So, it's a very arbitrary construction, but the redefined code does not have transversal logical $$X$$.