Logical Operators

A logical operator is an operation on the encoded qubit: it must transform the codewords the way $X$ and $Z$ transform a bare qubit, while leaving the codespace as a whole invariant. The stabilizer formalism pins down exactly which physical Pauli strings qualify.

The defining requirements

For a stabilizer code with group $\mathcal S$ encoding $k$ logical qubits, the logical Pauli operators $\bar X_i, \bar Z_i$ are elements of the normalizer $N(\mathcal S)$ that are not in $\mathcal S$, chosen to satisfy the qubit commutation relations:

Two conditions, two physical meanings:

• Commute with all stabilizers. This guarantees $\bar X, \bar Z$ map the codespace to itself — they act within the code, not out of it, and produce no syndrome.
• Anticommute with each other ($\{\bar X, \bar Z\} = 0$). This is what makes them a real qubit's operators: a logical $X$ and logical $Z$ must satisfy $\bar X \bar Z = -\bar Z \bar X$ just like the Pauli matrices do.

Defined only up to stabilizers

A logical operator is not unique. Multiplying $\bar X$ by any stabilizer $g \in \mathcal S$ gives another valid representative $g\bar X$, since $g$ acts as identity on codewords. We exploit this freedom to find a low-weight representative; the minimum weight over all representatives of any nontrivial logical operator is the code distance $d$.

Example: the bit-flip code

For the three-qubit bit-flip code with $\mathcal S = \langle Z_1Z_2, Z_2Z_3\rangle$:

Check: $\bar Z = Z_1$ commutes with both $Z$-type stabilizers (all $Z$'s commute) and $\bar X = X_1X_2X_3$ commutes with $Z_1Z_2$ and $Z_2Z_3$ because each overlap is on an even number of qubits. But $\bar X$ and $\bar Z$ overlap on qubit 1 alone, so they anticommute — exactly the qubit algebra we need.

Try it

Verify, in the symplectic formalism, that for the bit-flip code $\bar X = X_1X_2X_3$ and $\bar Z = Z_1$ each commute with both stabilizer generators and anticommute with each other. The grader checks your computed commutation counts.

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