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A stabilizer code is the common $+1$ eigenspace of a stabilizer group $S$, an abelian subgroup of the Pauli group excluding $-I$. Choose $r$ independent stabilizer generators $g_1,\dots ,g_r$ on $n$ physical qubits. They are commuting Hermitian operators, each squaring to $I$. The stabilizer subspace and its orthogonal projector areThe stabilizer-projector formula gives $\dim \mathcal{C}=2^{n-r}$, so $k=n-r$ logical qubits remain. Independent here means that no nonempty product of the chosen stabilizer generators is $I$. Excluding $-I$ is necessary: a vector cannot simultaneously be fixed by $I$ and $-I$ unless it is zero.

For the three-qubit bit-flip repetition code, take $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$, with $Z_j$ the Pauli Z gate on site $j$ and identities elsewhere. The stabilizer group is $S=\{I,Z_1{Z}_2,Z_2{Z}_3,Z_1{Z}_3\}$. Both stabilizer generators have eigenvalue $+1$ exactly on the span of $|000⟩$ and $|111⟩$, giving $2^{3-2}=2$ dimensions for one logical qubit. A projective measurement of these checks distinguishes four orthogonal two-dimensional sectors rather than eight individual computational basis states.

A Pauli error that anticommutes with a stabilizer generator changes that check's sign, producing an error syndrome. A Pauli operator commuting with every check can still act on the logical qubits; the centralizer of a stabilizer group describes these possibilities. Thus the number of checks determines the encoded dimension, while the quantum code distance determines protection against arbitrary local errors. The five-qubit error correcting code, for instance, has four stabilizer generators and quantum code distance three; the three-qubit bit-flip repetition code has two checks but quantum code distance one because a single phase flip is unflagged. ^[ref].

Consider $q$ identical length-$n$ arms joined to one hub, with XY exchange interactions of strength $J_0/\sqrt{q}$ from the hub to each first site and strength $J_k$ between sites $k,k+1$ on each arm. Let $|s_0⟩$ be the hub excitation and $|s_k⟩=q^{-1/2}{\sum }_{a=1}^q|a,k⟩$. This orthonormal symmetric single-excitation subspace is invariant, and its restricted Hamiltonian operator iswith endpoint terms omitted. At the first link the $q$ hub contributions add to $J_0$; all other links act identically on the arms. Thus the linear isometry $F|k⟩=|s_k⟩$ intertwines $H$ with the path Hamiltonian operator $H_T={\sum }_{k=0}^{n-1}{J}_k(|k⟩⟨k+1|+|k+1⟩⟨k|)$: $HF=F{H}_T$. Expanding the matrix exponential gives $e^{-iHt}F=F{e}^{-i{H}_{T}t}$. Perfect quantum state transfer from zero to $n$ on the path therefore produces an equal excitation superposition at the $q$ arm tips. Since all other qubits are zero, the tip state is a pure W state, up to the transfer's common phase.

If two pure quantum states overlap by $(1/\sqrt{2}{)}^{1/n}$, their $n$-copy states have overlap $1/\sqrt{2}$. Each block spans an effective qubit, with orthonormal basis $e_0={\Psi }_1$ and $e_1=\sqrt{2}{\Psi }_2-{\Psi }_1$. Applying the two-block PBR exclusion measurement for zero and plus uses $2n$ independently prepared copies. A common single-copy ontic overlap of positive mass remains positive under this finite tensor power.

Measure the commuting stabilizer generators $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$ in that order. Use error syndrome bit $0$ for eigenvalue $+1$ and bit $1$ for eigenvalue $-1$. For $X_j$ the Pauli X gate on physical qubit $j$, every row holds for arbitrary normalized $\alpha ,\beta$. This ordered parity-check convention follows ^[ref]:

Each pair of computational basis words within a row has the same two parities. Consequently the projective measurement reveals the row without distinguishing its two amplitudes. For example, $|{\psi }_L⟩=(\sqrt{3}/2)|000⟩+(i/2)|111⟩$ becomes $(\sqrt{3}/2)|010⟩+(i/2)|101⟩$ under $X_2$. Both checks give $-1$ with probability one, and applying $X_2$ restores both coefficients exactly, including the relative phase $i$.

One implementation uses two quantum ancillas initialized to $|00⟩$. Two controlled-NOT gates from data qubits 1 and 2 to the first quantum ancilla write their parity; two more from data qubits 2 and 3 to the second quantum ancilla write the other parity. Measuring only the quantum ancillas in the computational basis gives $(s_1,s_2)$. Measuring the individual data qubits instead would reveal the encoded computational basis value and destroy the unknown quantum superposition. The existing coherent syndrome extraction for the three-qubit repetition code uses the alternative ordered checks $(Z_2{Z}_3,Z_1{Z}_3)$, so its bit labels differ from this table.

The recovery also handles coherent errors. Put $E_0=I,E_1=X_1,E_2=X_2,E_3=X_3$. Their images of the code are mutually orthogonal, giving $P{E}_a^{†}{E}_{b}P={\delta }_{ab}P$, the Knill--Laflamme condition for this family. For any linear combination $E={\sum }_{a=0}^3{c}_a{E}_a$, its component in sector $s(E_a)$ is $c_a{E}_a|{\psi }_L⟩$. Measuring that error syndrome has probability $|c_a{|}^2/{\sum }_b|c_b{|}^2$ on the normalized erroneous state, and recovery $E_a$ returns $|{\psi }_L⟩$. More generally one recovery quantum channel corrects every noise quantum channel whose Kraus operators belong to this span. The assumption concerns the whole jointly correctable family; it does not extend to the span of arbitrary single-site Pauli errors. ^[ref].

A fully computed coherent example is the unitary operator $U=\exp (-i\theta X_2)=\cos \theta I-i\sin \theta X_2$. At $\theta =\pi /6$ it givesThe error syndromes $00$ and $11$ occur with probabilities $3/4$ and $1/4$, independently of $\alpha ,\beta$. Doing nothing for $00$ and applying $X_2$ for $11$ restores the same logical qubit in both branches; the scalar $-i$ in the second branch is an irrelevant global phase. This is an example of linearity of coherent quantum error correction, rather than an assumption that the original error was a classical random bit flip.

A path graph state on three qubits simulates two successive gates $U(\theta )=H\mathrm{diag}(1,e^{-i\theta })=J(-\theta )$. Measure vertex one in the equatorial qubit measurement basis $|v_r(\alpha )⟩=(|0⟩+(-1{)}^r{e}^{i\alpha }|1⟩)/\sqrt{2}$, then vertex two at angle $(-1{)}^r\beta$, obtaining bits $r,s$. Two one-bit teleportations and $U(\theta )X=e^{-i\theta }ZU(-\theta )$ leave vertex three in $X^s{Z}^{r}U(\beta )U(\alpha )|+⟩$, up to global phase. A computational output bit $z$ is corrected to $z\oplus s$; the $Z$ factor affects only phase. Adaptive sign choice is required for general angles, although the final bit correction is purely classical.

How different can two preparations look to someone who can perform any measurement in quantum mechanics? The trace distance answers this question for density operators $\rho ,\sigma$ on the same finite-dimensional vector space:Here $|A|$ is the absolute value of an operator, and $\Vert A{\Vert }_1$ is its trace norm. Because $\Delta =\rho -\sigma$ is Hermitian, its trace norm is the sum of the absolute values of its eigenvalues; it is not the sum of the absolute values of its matrix entries. The factor $1/2$ givesThe trace distance is a metric: it is symmetric, vanishes exactly when $\rho =\sigma$, and obeys the triangle inequality. Identical preparations have trace distance zero; preparations with orthogonal occupied eigenspaces have trace distance one and can be identified perfectly in a single measurement in quantum mechanics.

Operationally, the trace distance is the largest total variation distance between the outcome probability distributions obtainable by a POVM. Equivalently, the variational characterization of trace distance says it is the largest possible probability gap for one event in a binary measurement in quantum mechanics. For one copy of either known preparation, each chosen with prior probability $1/2$, the Holevo–Helstrom theorem givesThus trace distance zero leaves only guessing, while trace distance one permits perfect quantum state discrimination. The optimum ranges over measurements; different measurement bases can expose different amounts of distinguishability. ^[ref]

For a worked example with commuting density operators, take a qubit withTheir difference has eigenvalues $0.5,-0.5$, so $D=(|0.5|+|-0.5|)/2=0.5$. Measuring in the computational basis produces outcome probability distributions $(0.8,0.2)$ and $(0.3,0.7)$, whose total variation distance is also $0.5$. Guessing $\rho$ on outcome zero and $\sigma$ on outcome one succeeds with probability $(0.8+0.7)/2=0.75$. A projective measurement in the Hadamard basis instead produces $(0.5,0.5)$ for both preparations and gives no discrimination advantage.

The trace-distance contraction under quantum channels ensures that a common quantum channel, including the partial trace, cannot increase the trace distance. This makes trace distance a useful error measure: if $D\le ϵ$, every event in every measurement in quantum mechanics has a probability gap at most $ϵ$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.

For qubit density operators, the trace distance has a particularly simple geometry. Write their Bloch vectors as $\mathbf{r},\mathbf{s}$ in the Bloch ball:where $I$ is the identity operator and $\mathbf{\sigma }=({\sigma }_x,{\sigma }_y,{\sigma }_z)$ denotes the Pauli matrices. This Bloch vector representation includes both pure states on the Bloch sphere and mixed states in the interior of the Bloch ball. ^[ref]

The Pauli matrix multiplication law gives $[(\mathbf{r}-\mathbf{s})\cdot \mathbf{\sigma }{]}^2=|\mathbf{r}-\mathbf{s}{|}^{2}I$. Hence the two eigenvalues of $\Delta$ are $\pm |\mathbf{r}-\mathbf{s}|/2$, and their absolute sum givesThe trace distance is half the straight-line Euclidean distance between the Bloch vectors, not the angular or surface distance along the Bloch sphere. Opposite Bloch vectors have trace distance one; the centre $I/2$ is at trace distance $1/2$ from every pure state.

For the pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, the Bloch vectors are $\mathbf{r}=(0,0,1)$ and $\mathbf{s}=(1,0,0)$. Their straight-line separation is $\sqrt{2}$, soThese Bloch vectors are perpendicular, but the pure states are not orthogonal: $|⟨0|+⟩|=1/\sqrt{2}$. The Helstrom measurement for the zero and plus states measures along the difference direction $\mathbf{n}=(\mathbf{r}-\mathbf{s})/\sqrt{2}=(-1,0,1)/\sqrt{2}$, rather than along either input Bloch vector.

For the pure qubit $\alpha |0⟩+\beta |1⟩$, subtracting the output of dephasing channel leaves a Hermitian matrix with zero diagonal and off-diagonal modulus $|\alpha \beta ||1-c|$. Its two eigenvalues are that number and its negative, giving the displayed trace distance. For real $c=1-2p$, this is $2p|\alpha \beta |$.

Choose three pure qubit states whose unit Bloch vectors are coplanar, sum to zero and have pairwise angle $120$ degrees. The three effects $E_j=\frac{2}{3}|{\varphi }_j⟩⟨{\varphi }_j|$ sum to the identity, giving a POVM. If an input is the state orthogonal to ${\varphi }_1$ or to ${\varphi }_2$, outcome one excludes the first input and outcome two excludes the second. Each conclusive outcome has probability one half on its corresponding input; the third outcome is inconclusive with probability one half on either input. This gives unambiguous quantum state discrimination, rather than identification on every trial.

For $\sqrt{p}|000⟩+\sqrt{1-p}|111⟩$, a projective measurement of the number of ones in one party's $n$ local qubits gives a common type class of size $d_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The conditional pure state is $d_k^{-1/2}{\sum }_{|x|=k}|x,x,x⟩$. It can be converted exactly and deterministically to a uniform common-label state of any dimension $m\le d_k$: one party measures an $m$-element subset $S$ with Kraus operator ${\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)}$, and all parties relabel their common retained labels after receiving $S$. Choosing $m=2^f$ gives $f$ independent GHZ states. Stirling's formula and the weak law of large numbers give $f/n\to h_2(p)$ in probability. Measuring every local bit separately would instead destroy the coherence and leave a product state.

Take an isolated two-level system, or a spin one-half qubit, with Pauli matrices ${\sigma }_x,{\sigma }_z$. Let ${\sigma }_z|0⟩=|0⟩$ and ${\sigma }_z|1⟩=-|1⟩$. The initial and final Hamiltonian operators areThe energy gaps are ${\Delta }_i,{\Delta }_f$. Prepare the initial ground state $|0⟩$, with energy eigenvalue $-{\Delta }_i/2$. A final orthonormal eigenbasis isDirect matrix multiplication gives $H_f|g_f⟩=-{\Delta }_f|g_f⟩/2$ and $H_f|e_f⟩=+{\Delta }_f|e_f⟩/2$. The unchanged initial quantum state is $|0⟩=c|g_f⟩-s|e_f⟩$. Its new probability amplitudes are the inner products $c,-s$; the Born rule gives $p_g=c^2,p_e=s^2$. The angle between the real amplitude vectors is $\theta /2$, whereas the Hamiltonian axis on the Bloch sphere rotates through $\theta$.

At $\theta =\pi /3$, $c=\sqrt{3}/2$ and $s=1/2$, so the final energy measurement gives $-{\Delta }_f/2$ with probability $3/4$ and $+{\Delta }_f/2$ with probability $1/4$. At $\theta =0$ the initial ground state remains the final ground state; at $\theta =\pi$ it is the final excited energy eigenstate. These statements are about the new energy labels, not an instantaneous physical rotation of the quantum state.

Let $c,s\ge 0$ and $c^2+s^2=1$. Starting from the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, Alice applies the measurement in quantum mechanics with Kraus operators $M_0,M_1$ displayed above. Completeness follows from $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. Each outcome has probability $1/2$. The normalized outputs are $c|00⟩+s|11⟩$ and $s|00⟩+c|11⟩$. Communicating the outcome allows both parties to apply a Pauli X gate in the second branch, making both outputs equal to the first. Thus this LOCC protocol deterministically converts one Bell pair into any two-qubit pure state with these Schmidt coefficients, after appropriate local unitary operations. It includes the product state endpoint without postselection.

A two-qubit gate is a unitary operator on two qubit lines, extended by the identity on the others. It may be entangling, as a CNOT gate is, or a product of local operations. In a Hamiltonian simulation model allowing arbitrary two-qubit gates, the exponential of a Hermitian term supported on two lines counts as one such gate. Compiling it into a fixed discrete universal quantum gate set introduces a separate approximation cost.

The rotation-invariant average of pure qubit states is uniform area measure on the Bloch sphere, induced by Haar measure on unitary transformations. For a uniformly distributed unit Bloch vector $r$, rotational symmetry gives $\mathbb{E}{r}_i=0$ and $\mathbb{E}{r}_i{r}_j={\delta }_{ij}/3$. Uniform polar angle without the factor $\sin \theta$ is a different distribution.

The $n$-qubit W state is the Dicke state with one excitation:For three qubits it is $(|001⟩+|010⟩+|100⟩)/\sqrt{3}$. Its reduced density matrix on any one qubit is $\mathrm{diag}((n-1)/n,1/n)$, from the Dicke state recursion. For $n>1$ both diagonal entries are nonzero, so that qubit is entangled with the rest: a pure product state would have a pure marginal. A symmetric-arm reduction of an exchange Hamiltonian supplies a dynamical preparation from one initial excitation.