Plaquette and Vertex Stabilizers

Each tile of the surface-code lattice carries a stabilizer: a multi-qubit Pauli operator whose value the code measures every round. The two colours of tile correspond to the two species of check.

The two check types

Label the data qubits around an interior tile $1, 2, 3, 4$. The two stabilizer types are

A $Z$-type plaquette measures the joint parity of its data qubits in the computational ($Z$) basis; it detects bit-flip ($X$) errors. An $X$-type vertex measures parity in the conjugate ($X$) basis; it detects phase-flip ($Z$) errors. Together the two families catch any single-qubit Pauli error, because every Pauli is a product of $X$ and $Z$.

The protected code space is the simultaneous $+1$ eigenspace of all the stabilizers: a state $|\psi\rangle$ is a valid codeword exactly when $A_v|\psi\rangle = +|\psi\rangle$ and $B_p|\psi\rangle = +|\psi\rangle$ for every tile.

Why the checks commute

We can only measure all these operators in one round without disturbing each other if they commute. A $Z$-type and an $X$-type tile always share an even number of data qubits — in the surface code, either zero or two. On each shared qubit the local operators are $X$ and $Z$, which anticommute ($XZ = -ZX$); two anticommutations cancel, so the full operators commute:

so two shared qubits give two sign flips that cancel and $[A_v, B_p] = 0$. (Tiles that share no qubits commute trivially; sharing exactly one qubit would instead give a single sign flip and anticommutation, which is precisely why the lattice is arranged so overlaps are always even.)

Because they commute, the stabilizers can be assigned definite values simultaneously, and measuring them tells us about errors without revealing — or collapsing — the logical state.

A two-qubit example you can build

The smallest non-trivial stabilizer state lets us see the idea directly. Consider a single $X$-type check $X_0 X_1$ on two data qubits. Its $+1$ eigenstate with the data also stabilized by $Z_0 Z_1$ is the Bell state

Check it: $X_0X_1|\Phi^+\rangle = \tfrac{1}{\sqrt2}(|11\rangle + |00\rangle) = +|\Phi^+\rangle$, and $Z_0Z_1|\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle) = +|\Phi^+\rangle$. The state is a simultaneous $+1$ eigenstate of both an $X$-type and a $Z$-type check — a one-line toy version of a codeword.

Try it

Prepare the state $|\Phi^+\rangle$ on two qubits with a Hadamard followed by a CNOT. The grader checks the resulting state vector against the $+1$ eigenstate of $X_0X_1$ and $Z_0Z_1$.

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