Minus sign in logical operators

Question

Logical operators are defined as $C(S)\backslash S$, where $S$ is a stabilizer group, but I am confused about this definition. For instance, in the 3-qubit repetition code, the stabilizer group is $\langle ZZI, IZZ\rangle$, and logical $Z$ operators are defined to be $ZII$ (or $IZI$ or $IIZ$) usually. The logical states are $|0_L\rangle=|000\rangle$ and $|1_L\rangle=|111\rangle$. However, $-ZII$ (or $-IZI$ or $-IIZ$) also satisfy the definition of the logical operators, but the action of this logical operator with minus sign does not have the expected action to the logical states.

If we change the definition of the logical states to $|0_L\rangle=|111\rangle$ and $|1_L\rangle=|000\rangle$, $-ZII$ works well, but this is not my question. My question is that, it's obvious that $ZII$ and $-ZII$ should not be the logical $Z$ operator at the same time, but I have never seen explanation about such kind of implicit definition in the definition of logical operators $C(S)\backslash S$. Is it true that we should define either $+$ type logical operator or $-$ type logical operator as a logical operator? Is the reason people only use the $+$ type logical operator almost all the time because of its simplicity and is using $-$ type logical operator also totally fine?

Answer 1

The logical Pauli operators are commonly defined as belonging to the quotient group $C(S)/S$, which is different from $C(S)\backslash S$. Notably, the logical identity operator $\overline{I}$ appears in $C(S)/S$ as the equivalence class of any stabilizer (including $III$), but does not appear in $C(S)\backslash S$. The centralizer $C$ is here understood over the Pauli operators and not the whole possible unitaries.

Because they are equal up to a multiplication by an element of $S$, $ZII$, $IZI$ and $IIZ$ belongs to the same equivalence class and are representated by the same element in $C(S)/S$. Let's denote it $\overline{Z}$.

$-III \in C(S)$ but $-III \not \in S$ by definition, so it is a logical operator different from the identity. It cannot be $\overline{X}$ or $\overline{Z}$ since it wouldn't anticommutes with anything else. It is a representative of its own $\overline{(-I)} = -\overline{I}$.

The logical operators form a group, so there is an element $\overline{Z^{\prime}}$ of $C(S)/S$ such that $\overline{Z^{\prime}}$ = $-\overline{I}\cdot\overline{Z}$. One representative of $\overline{Z^{\prime}}$ could be $-ZII$, $-IZI$ or $-IIZ$.

When defining your stabilizer code, you are free to choose a basis for $C(S)/S$. To match what happen in the non-logical setting, it is convenient to choose a basis of the form:

$$ C(S)/S = \langle \overline{I}, -\overline{I}, i\overline{I}, -i\overline{I}, \overline{X_1}, \overline{Z_1},\dots,\overline{X_k}, \overline{Z_k}\rangle $$

where the $(\overline{X_i}, \overline{Z_i})$ anticommutes pairwise and otherwise commutes. You need to choose one pair for each of the $k$ logical qubits.

Whether you choose $\overline{Z_1}$ to be $\overline{Z}$ or $\overline{Z^{\prime}}$ is a matter of taste. People like defining $|0\rangle_L$ such that $|0\rangle_L = |000\rangle = \overline{Z_1}|0\rangle_L$ but it is indeed just a convention and you are free to choose the other option.