Checkpoint: Stabilizer Workout

Time to put the whole module together. This checkpoint analyzes the five-qubit perfect code $[[5,1,3]]$ — the smallest code that corrects an arbitrary single-qubit error — using only the binary symplectic machinery you have built across the module.

The code under study

The five-qubit code is defined by four cyclic stabilizer generators, each a cyclic shift of the string $X Z Z X I$ :

\begin{aligned} g_1 &= X_0\, Z_1\, Z_2\, X_3\, I_4, \\ g_2 &= I_0\, X_1\, Z_2\, Z_3\, X_4, \\ g_3 &= X_0\, I_1\, X_2\, Z_3\, Z_4, \\ g_4 &= Z_0\, X_1\, I_2\, X_3\, Z_4. \end{aligned}

It is called perfect because it saturates the quantum Hamming bound: with $n=5$ , the $1 + 3n = 16$ single-qubit error operators (identity plus $X,Y,Z$ on each of 5 qubits) map onto exactly $2^{n-k} = 2^4 = 16$ distinct syndromes, leaving no waste.

What you will verify

This is a synthesis exercise touching every lesson in the module:

It is a valid stabilizer code (the abelian check). The four generators pairwise commute, computed with the symplectic inner product from the Pauli-group and stabilizer-group lessons.
It encodes one logical qubit ( $k = 1$ ). Stack the generators into the $4 \times 10$ check matrix and take its $\mathbb{F}_2$ rank: $\operatorname{rank} = 4$ , so $k = n - \operatorname{rank} = 1$ .
Its logical operators are correct. $\bar X = X^{\otimes 5}$ and $\bar Z = Z^{\otimes 5}$ each commute with all four stabilizers (so they preserve the codespace) yet anticommute with each other (so they act as a genuine logical $X$ / $Z$ pair).
Its distance is $d = 3$ . Every nontrivial logical operator has weight at least $3$ , which is why the code corrects any single-qubit error: $\lfloor (3-1)/2 \rfloor = 1$ .

Putting it together

Encode the four generators as $(x \mid z)$ vectors, then compute all four facts: pairwise commutation, the $\mathbb{F}_2$ rank (hence $k$ ), the logical-operator (anti)commutation relations, and the distance. Return them as an object { k, abelian, logicalOK, distance }. If your reasoning across the module is sound, you will find $k = 1$ , abelian = true, logicalOK = true, and $d = 3$ — the full signature of the $[[5,1,3]]$ perfect code.

Run your code to see the quantum state.