The four Pauli rows
XZZX_, _XZZX, X_XZZ, ZX_XZ
generate an abelian rank-four non-CSS stabilizer on five qubits. Its codespace encodes one logical qubit with symplectic basis Xbar=XXXXX, Zbar=ZZZZZ, and its exact distance is three.
Write an n-qubit Pauli as (x|z) with one binary X indicator and one binary Z indicator per site. Thus X is (1|0), Z is (0|1), Y is (1|1), and identity is (0|0). Two rows commute exactly when
<a,b>_s = x_a dot z_b + z_a dot x_b = 0 mod 2.
This convention discards Pauli phase because stabilizer commutation, rank, syndrome, and support depend only on the phase-free row. The constructor uses the same [x|z] order as qeclib.pauli.Pauli.symplectic().
For Pauli strings, the symplectic product counts positions with unlike nonidentity Paulis—X versus Z, X versus Y, or Y versus Z—modulo two. For example, XZZX_ and _XZZX disagree in Pauli type at two overlapping positions. Their local minus signs multiply to plus.
Running this count for all six generator pairs gives zero. The result is also checked by CheckMatrix.validate(), which rejects any nonzero symplectic product.
GF(2) row reduction of the 4 x 10 symplectic matrix has pivot columns 0,1,2,3. The four generators are independent. A rank-r stabilizer on n qubits has codespace dimension 2^(n-r), hence
k = 5 - 4 = 1.
The omitted fifth cyclic translate is therefore not needed as an independent constraint: its phase-free symplectic row is the XOR of the four printed rows. Adding it as if independent would incorrectly predict a zero-logical state.
A CSS stabilizer group is generated by elements whose binary rows lie wholly in the X sector or wholly in the Z sector. Changing generators only chooses a different basis of the same row space, so the decisive test is the row-space intersection with those two sectors.
Enumerate the fifteen nonzero linear combinations of the four rows. Every one has a nonzero X part and a nonzero Z part. Both pure-sector intersections are therefore trivial. Their dimensions sum to zero, far below stabilizer rank four, so a CSS basis does not exist for this exact Pauli subspace.
For Xbar=XXXXX, each stabilizer has two Z components, so the symplectic product is zero. For Zbar=ZZZZZ, each stabilizer has two X components, again giving zero. Neither vector lies in the stabilizer row space. Finally, the five X/Z overlaps between Xbar and Zbar give symplectic product one. They are a canonical logical pair.
The representatives are intentionally uniform, not minimum weight. Multiply by the first stabilizer and discard global phase:
Xbar * XZZX_ = _YY_X,
Zbar * XZZX_ = Y__YZ.
Stabilizer multiplication preserves logical action, and both products have weight three.
The single-site syndrome table contains no zero row, so no weight-one Pauli is in the normalizer outside the stabilizer. Suppose a weight-two Pauli E F were in the normalizer, with E and F supported on different sites. Syndrome is linear, hence
s(E F)=s(E)+s(F)=0, so s(E)=s(F).
The complete table assigns distinct syndromes to all single-site Paulis, a contradiction. Thus no weight-two normalizer exists and d>=3. The weight-three logical _YY_X gives d<=3. Together they prove d=3; the generic exact Pauli enumerator independently returns the same result.
Qubit rotation, generator multiplication, local Clifford conjugation, or a different logical pair can produce an equivalent code. The Atlas constructor fixes one row order and one logical convention so syndrome bits, tours, tests, and reproduction all mean the same thing. Rains' uniqueness result is up to equivalence (primary source); it does not erase this executable convention.