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For spin one-half, choose the normalized spin eigenstates $|+z⟩,|-z⟩$ with $S_z|\pm z⟩=\pm (\hslash /2)|\pm z⟩$. The spin one-half along an axis pure state in the $xz$ plane isIts norm is one because the two eigenstates are orthogonal and ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$. Applying the Born rule to a $z$ projective measurement givesConsequently the expectation value, second moment, and variance areAt $\theta =\pi /3$, the outcome probabilities are $3/4$ and $1/4$, the mean is $\hslash /4$, and the variance is $3{\hslash }^2/16$. Every individual outcome remains $\pm \hslash /2$. At $\theta =\pi /2$, the mean is zero and the variance is maximal, ${\hslash }^2/4$. At either pole, one outcome has probability one and the variance vanishes.

The diagonal mixed state ${\rho }_{\mathrm{diag}}=p_{+}|+z⟩⟨+z|+p_{-}|-z⟩⟨-z|$ has the same $S_z$ probabilities, expectation value, and variance as $|{\psi }_{\theta }⟩$. It is nevertheless a different quantum state for $0<\theta <\pi$: the displayed pure state has $⟨S_x⟩=\hslash \sin \theta /2$, while ${\rho }_{\mathrm{diag}}$ has $⟨S_x⟩=0$. A single component's mean does not reveal all relative quantum phases or establish purity of a density operator.

For a spin one-half system, an ideal projective measurement along a unit vector $\mathbf{n}$ measures ${\sigma }_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{\sigma }$. Its eigenvalues are $\pm 1$, corresponding to physical spin angular momentum components $\pm \hslash /2$, and its orthogonal projections are ${\Pi }_{\pm }=(I\pm {\sigma }_{\mathbf{n}})/2$. For density matrix $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, the Born rule and the Pauli matrix multiplication law give the displayed probabilities. Thus a spin measurement along one direction tests only that component of the Bloch vector; measurements along three independent directions determine the entire density matrix.

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].

A readout in which one logical quantum state is coupled to an optical cycling transition and scatters many photons, while another is dark. An ideal bright/dark record implements a projective measurement in the logical basis. Neglecting that record removes their relative quantum coherence. Missed photons, off-resonant excitation and pumping out of the logical subspace qualify the ideal model.

The ideal projective measurement of a spin-one-half component along a unit axis $\mathbf{n}$. The two outcomes have Born rule probabilities $⟨\psi |P_{\pm }|\psi ⟩$. For an input with Bloch vector $\mathbf{r}$, they are $(1\pm \mathbf{n}\cdot \mathbf{r})/2$. This model distinguishes a mathematical spin measurement from engineering a literal free-electron beam splitter.

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.

The time-evolution operator maps an initial quantum state to its state at a later time. For a time-independent Hamiltonian operator $H$, it is the unitary operator $U(t)=e^{-iHt/\hslash }$.

More generally, write $|\psi (t)⟩=U(t,t_0)|\psi (t_0)⟩$. The time-evolution operator obeysFor a constant self-adjoint Hamiltonian operator, $U(t,t_0)=\exp [-iH(t-t_0)/\hslash ]$. With zero as the initial time, this gives the group law $U(t+s)=U(t)U(s)$. Successive evolution is reversible; a measurement followed by conditioning on an outcome is a separate operation described by projective measurement.

For a time-dependent Hamiltonian operator, the defining equation is $i\hslash {\partial }_{t}U(t,t_0)=H(t)U(t,t_0)$. In finite dimension, a continuous Hermitian $H(t)$ still gives unitary time evolution, sinceThe solution is a time-ordered exponential, with later factors acting on the left. One may replace it by $\exp [-i{\int }_{t_0}^{t}H(s)ds/\hslash ]$ when $[H(s),H(s^{\prime })]=0$ at all times; independently changing pulse axes generally requires time ordering.^[ref]

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 $\sqrt{p}|000⟩+\sqrt{1-p}|111⟩$, 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.

After a computational-basis projective measurement, an observer who lacks the result assigns a diagonal density matrix. If the observer's preparation prior has mean $q=\mathbb{E}|\alpha {|}^2$, the assignment is the displayed matrix. No outcome record and no stated prior do not logically force $q=1/2$; that value follows from a symmetric or maximum-entropy ignorance convention. An observer who knows the preparation instead uses $q=|\alpha {|}^2$.

For Kraus operators satisfying ${\sum }_i{A}_i^{†}{A}_i=I$, the map $V|\varphi ⟩={\sum }_i{A}_i|\varphi ⟩\otimes |i⟩$ is an isometry. Implement it by a unitary operator on the system and a quantum ancilla, initially in an extra unused state $|0⟩$. Projecting the output quantum ancilla onto $|i⟩$ gives joint unnormalized state $A_i\rho A_i^{†}\otimes |i⟩⟨i|$. The outcome probability is $\mathrm{Tr}(A_i\rho A_i^{†})$, and the conditional system state is $A_i\rho A_i^{†}$ divided by this probability. The unused level can be merged into the first orthogonal projection without changing any outcome on the prepared state, giving a complete projective measurement with exactly the original outcome labels.

// Target: quantum-theory.bigb

The variational characterization of trace distance turns a matrix distance into the best possible binary experiment. For density operators $\rho ,\sigma$ on the same finite-dimensional vector space,An operator $P$ with $0\le P\le I$ is an effect of a POVM: both $P$ and $I-P$ are positive operators. The Born rule assigns the first outcome probabilities $\mathrm{Tr}(P\rho )$ and $\mathrm{Tr}(P\sigma )$. Their signed gap is $\mathrm{Tr}(P\Delta )$, where $\Delta =\rho -\sigma$. Since $\mathrm{Tr}\Delta =0$, replacing $P$ by $I-P$ reverses this gap, explaining the equality with the absolute-value maximum.

One optimal $P$ is the spectral projector $P_{+}$ onto the positive eigenspaces of $\Delta$. In the spectral decomposition of $\Delta$, this choice includes all positive eigenvalues and excludes all negative eigenvalues. Their positive sum equals the trace distance by the spectral-parts formula for trace distance. Any zero eigenspace can be assigned to either outcome without changing the gap. This is the effect selected by the equal-prior Holevo–Helstrom theorem. ^[ref]

The same optimum has a classical description. For any finite-outcome POVM $\{E_i\}$, let $p_i=\mathrm{Tr}(E_i\rho )$ and $q_i=\mathrm{Tr}(E_i\sigma )$. ThenThe expression $\frac{1}{2}{\sum }_i|p_i-q_i|$ is the total variation distance of the entire outcome probability distributions; the expression for $A$ is the gap for one event. Grouping outcomes in $A$ gives the single effect $P_A={\sum }_{i\in A}{E}_i$. Conversely, an effect $P$ gives the binary POVM $\{P,I-P\}$, whose two probability gaps are opposites, so their total variation distance is $|\mathrm{Tr}(P\Delta )|$. Thus the optimum over all POVMs is already attained by one binary projective measurement. ^[ref]

For quantum state discrimination with equal prior probabilities, declare $\rho$ on outcome $P$ and $\sigma$ on outcome $I-P$. Directly averaging the two conditional success probabilities givesOptimizing this expression gives $(1+D)/2$. For the diagonal example in trace distance, $P=|0⟩⟨0|$ gives a gap $0.8-0.3=0.5$. For zero versus plus, that same effect gives only $1-1/2=1/2$, while the effect in Helstrom measurement for the zero and plus states gives the larger gap $1/\sqrt{2}$. The trace-distance-preserving binary measurement depends on the pair of preparations; a fixed measurement basis does not generally attain the trace distance for every pair.

Wave function collapse describes the conditional change of a quantum state associated with a measurement outcome. For an ideal projective measurement with projector $P_a$, the outcome probability is $p_a=\mathrm{Tr}(P_a\rho )$ and the conditional state is $P_a\rho P_a/p_a$ when $p_a>0$, according to the Lüders rule.

If the outcome is ignored, the state becomes ${\sum }_a{P}_a\rho P_a$, which is generally a mixture rather than any one conditional state. These are different information conditions. General measurement instruments can produce other state changes with the same outcome probabilities, so the ideal projection rule is a specified measurement model. Its physical interpretation is part of the quantum measurement problem.