# For stabilizer codes, is a certain logical operation unique？

Suppose we have a $$[[n, 1]]$$ stabilizer code $$Q$$ and a single-qubit unitary $$U$$. We define the logical counterpart of $$U$$ as $$\bar{U}$$. My question is: Is there just one $$\bar{U}$$ up to stabilizers of $$Q$$?

I am asking this because I have seen the definition of transversal gate to be like $$\bar{U}=U^{\otimes n}$$, while the right side seems to be unique if $$U$$ is given.

• What does logical counterpart of $U$ mean? Mar 22 at 6:34
• @FDGod I mean it's one of the logical $U$. I was not sure about the uniqueness of the logical $U$ so I tried to be cautious about my word-use. Mar 22 at 7:46

The answer turns out to be NO as denoted in this answer. There seems to be infinite number of operations that preserves the codespace of $$Q$$ and acts as if $$\bar{U}$$ for any given single-qubit unitary $$U$$.
As an example, we can define $$Q$$ to be a two-qubit trivial code with stabilizer $$IZ$$ and logical operator $$X_L = XI, Z_L=ZI$$. If $$U=H$$ is a Hadamard gate, then any operations with the following form $$H\otimes |0\rangle\langle0| + A\otimes |1\rangle\langle1|$$ is a logical $$\bar{H}$$, where $$A$$ is an arbitrary single-qubit unitary.