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For an $n$-qubit quantum code with projector $P$, the quantum code distance isHere the weight counts nonidentity tensor factors. Thus every error supported on fewer than $d$ qubits is detectable, by the Pauli expansion of a quantum error. A scalar action on the code, including a nonidentity stabilizer, is harmless and does not reduce this distance. For a stabilizer code, the equivalent definition minimizes weight among Pauli operators which commute with every stabilizer generator but act nontrivially on the logical information. The Knill--Laflamme condition then guarantees correction of all errors of weight at most $t=\lfloor (d-1)/2\rfloor$, since each pairwise product has weight at most $2t<d$.

Suppose the code encodes $k$ logical qubits and is nondegenerate for all errors through weight $t$. There aredistinct Pauli errors modulo phase: choose the $j$ affected positions and choose $X,Y$ or $Z$ at each. Each error is unitary, so its image of the code has dimension $2^k$. Nondegeneracy makes these images mutually orthogonal, equivalently the combined vectors $E_a|i_L⟩$ are orthonormal. They all lie in the physical Hilbert space of dimension $2^n$. Counting dimensions proves the quantum Hamming boundThis dimension argument requires nondegeneracy. For a degenerate code, different errors can share an image, as in the Shor code, and this counting proof cannot be applied unchanged.

For a binary classical code of $M$ words, the Hamming bound is $M{\sum }_{j=0}^t\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le 2^n$: disjoint Hamming balls around the words must fit into all binary strings. The quantum factor $3^j$ is needed because at each erroneous qubit there are three linearly independent nonidentity Pauli operators, not just a classical bit flip. A phase flip is invisible to a computational-basis classical record but changes a quantum superposition. A $Y$ error combines bit and phase effects, and the identity with $X,Y,Z$ spans every one-qubit operator. Quantum correction must protect amplitudes and coherences, and consequently count whole orthogonal logical subspaces rather than just classical words.

For parameters $[[5,1,3]]$, $t=1$, andThus a nondegenerate such code saturates the bound: the sixteen two-dimensional error images exhaust the physical space. It is a perfect quantum error-correcting code for single-qubit errors. It also has the smallest possible length for nondegenerate single-error protection of one logical qubit: for $1\le n\le 4$, the required inequality $1+3n\le 2^{n-1}$ fails.

Here is an explicit realization and its five-qubit stabilizer syndrome table. Write tensor products without multiplication symbols and useEach pair has an even number of anticommuting single-site factors, so the four stabilizer generators commute. Their binary Pauli labels are independent: no nonempty product has identity label. For example, the $X$-support at site three fixes the exponent of $g_3$ in such a product to zero, site five then fixes that of $g_2$, site one fixes that of $g_1$, and site two fixes that of $g_4$. Thus their group has sixteen elements and excludes $-I$. The common $+1$ eigenspace has projector $P={\prod }_{a=1}^4(I+g_a)/16$ and dimension $\mathrm{Tr}P=32/16=2$, since all nonidentity Pauli strings have zero trace.

Record a syndrome bit as one when an error anticommutes with the corresponding generator, in order $g_1,g_2,g_3,g_4$. Direct local commutation givesThese are all fifteen nonzero four-bit error syndromes; the identity has $0000$. Distinct errors in this list have a product with nonzero syndrome, so it anticommutes with some $g_a$. Inserting $g_{a}P=P$ on both sides gives $P{E}_b^{†}{E}_{c}P=-P{E}_b^{†}{E}_{c}P=0$. The diagonal product is the identity. Hence the Knill--Laflamme condition is exactly $C_{bc}={\delta }_{bc}$, proving nondegenerate correction. Measure the four commuting checks, apply the unique indicated Pauli error again, and every logical state is recovered. The Pauli expansion of a quantum error extends this recovery to an arbitrary one-qubit noise channel, not just the fifteen discrete errors.

The table also proves that every Pauli operator of weight one or two has nonzero syndrome: a weight-two operator is the product of distinct single-site errors, whose syndromes differ. Therefore the distance is at least three. The operators $\overline{X}=X^{\otimes 5}$ and $\overline{Z}=Z^{\otimes 5}$ commute with all four generators and anticommute with each other. They are nontrivial logical Pauli operators. Multiplication of $\overline{X}$ by $g_1$ gives $-IYYIX$, a weight-three operator with the same logical action; it cannot be scalar on the code because it anticommutes with the invertible logical operator $\overline{Z}$. This proves $d=3$ exactly.

The code consequently detects every two-qubit error and corrects two known-location erasures by the quantum erasure correction criterion, but it cannot correct all two-qubit errors of unknown location. A nontrivial weight-three logical operator can be split into a weight-one and a weight-two error, whose pairwise product violates the correction condition. It also saturates the quantum Singleton bound, since $n-k=4=2(d-1)$. These properties distinguish single-error perfectness from unrestricted protection against larger errors.

Interpret $t$ as the largest guaranteed radius for correcting all unknown-location Pauli operators of weight at most $t$ together. Products of two such errors have weight at most $2t$, so the Knill--Laflamme condition holds when $2t<d$. Conversely, let $F$ be a minimum-weight Pauli operator with nonscalar compression $PFP$. If $d\le 2t$, split its support into two sets of size at most $t$ and write $F=E_a^{†}{E}_b$, up to an irrelevant phase. These two admissible errors violate the Knill--Laflamme condition. Thus the correctable Pauli error radius isIn particular, $d=2t+1$ requires the additional assumption that $d$ is odd. Distance is not uniquely determined by $t$ alone. Nor does the weight of a particular correctable error characterize this radius: particular higher-weight errors can also be correctable.

For a concrete even-distance example, the stabilizer code generated by $X^{\otimes 4}$ and $Z^{\otimes 4}$ has parameters $[[4,2,2]]$. Every single-qubit nonidentity Pauli operator anticommutes with a generator, so it has zero compression. But $Z_1{Z}_2$ commutes with both generators and acts nonscalarly on the code: it is unitary there and its code trace is zero. Hence the distance is two and its guaranteed radius is zero. This shows why an unconditional odd-distance formula would be incorrect.

Ignoring overall phases, an error of weight $j$ has $\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ choices of support and three choices, $X,Y,Z$, at each affected qubit. The identity is the weight-zero term, soFor a nondegenerate quantum code, distinct correctable Pauli errors have orthogonal error-image subspaces. Each has dimension $2^k$, since a Pauli operator is unitary. All $N(t)$ subspaces lie in the physical Hilbert space of dimension $2^n$. Counting dimensions proves the quantum Hamming boundDegeneracy would allow distinct errors to share an error-image subspace, so this particular counting argument would then fail.

A nondegenerate quantum code is perfect at correction radius $t$ if its distinct correctable Pauli error images exhaust the physical Hilbert space. Each image has dimension $2^k$, and their orthogonal direct sum is the whole $2^n$-dimensional space precisely when the quantum Hamming bound is an equality. This perfectness concerns unknown-location errors of weight at most $t$; it does not imply correction of every higher-weight error.