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For a bipartite pure state, entanglement entropy is the Von Neumann entropy of either reduced density matrix. When Alice makes a local measurement with pure conditional branches, Bob's reductions average to his original reduction. Concavity of Von Neumann entropy therefore gives ${\sum }_r{p}_{r}S({\rho }_{B,r})\le S({\rho }_B)$. When Bob measures, apply concavity to Alice's reductions. In each pure branch the two reduced entropies coincide by the Schmidt decomposition. Refining measurement outcomes to individual Kraus operators and iterating proves the inequality for arbitrary finite-round LOCC. Discarded local systems can be measured and their outcomes refined as well, so discarding does not evade the bound.

Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

The Born rule connects a quantum state to measurement probabilities. Let $\{|j⟩\}$ be a complete orthonormal basis, finite or countably infinite, for a Hilbert space, and letAn ideal measurement distinguishing these basis vectors returns label $j$ with probabilityThis assignment is a physical measurement rule. Its compatibility with total probability one follows from orthonormality and normalization:The probabilities are the squared moduli, not the amplitudes themselves or the squares $a_j^2$. ^[ref].

If the basis is an eigenbasis of an observable, the reported numerical value is its eigenvalue. Several vectors can share one eigenvalue; the probability of that value is then the sum over the corresponding basis labels. More generally, a projective measurement with orthogonal projections $P_r$ assignsfor a pure state and a general density matrix $\rho$, respectively. This also covers mixed states and degenerate eigenspaces. The Born rule determines outcome statistics; the ideal conditional state change is specified separately by the Lüders rule. See projective measurement for that formulation and its assumptions. ^[ref].

For a worked example, take the normalized qubit quantum stateThe squared moduli add to $3/4+1/4=1$. A computational-basis measurement, equivalently a measurement of the $Z$ Pauli matrix, hasThese probabilities do not depend on the relative quantum phase $\varphi$. Now choose the Hadamard basiswhose vectors are eigenstates of the $X$ Pauli matrix, also used by the Pauli X gate. The new probability amplitudes and probabilities areBoth probabilities are nonnegative and their sum is one. The relative quantum phase becomes visible because the two probability amplitudes add before their squared modulus is taken:

A Hadamard gate followed by a computational-basis measurement realizes these $X$ statistics, since $H|\pm ⟩=|0⟩,|1⟩$, respectively. Compare the two bases using fresh, identically prepared copies: a preceding measurement may change the state. The quantum superposition section shows why replacing the preparation by the mixed state $\mathrm{diag}(3/4,1/4)$ keeps the $Z$ statistics but gives $p_{+}=p_{-}=1/2$ for every $\varphi$.

The usefulness of the Born rule is this conversion from linear algebra to experimental predictions: calculate overlaps in the basis actually measured, then take squared moduli. Its appealing geometry is that changing basis redistributes probability while preserving total squared norm. The same rule relates an energy eigenstate expansion to an energy measurement and a position wavefunction to a probability density.

Born's 1926 paper on quantum scattering interpreted the outgoing-wave coefficients in terms of the probabilities of collision outcomes; a corrective footnote specified proportionality to the square of the amplitude. This is a historical scattering formulation, rather than the general density matrix formula above. ^[ref]; ^[ref].

The canonical ensemble is the probability distribution for a system in thermal equilibrium with a heat bath at fixed temperature $T$, with fixed particle number and external parameters. It allows energy exchange, so individual energy measurements fluctuate. With Boltzmann constant $k_B$ and inverse temperature $\beta =1/(k_{B}T)$,The heat bath is idealized as large enough that exchanges barely change its temperature; coupling must permit equilibration while contributing negligible interaction energy. A finite heat bath approximates this description only when its temperature change is negligible over the relevant exchanges. Strong coupling or a small heat bath can require a different description.

For one two-level system at positive temperature, the Boltzmann weights are $1,e^{-\beta \Delta }$, henceAt $\beta \Delta =\ln 3$, $Z_1=4/3$ and $p=1/4$. This is a probabilistic thermal mixed state. For quantum levels, its density matrix is $\rho =(1-p)|0⟩⟨0|+p|1⟩⟨1|$. The pure state $\sqrt{1-p}|0⟩+\sqrt{p}|1⟩$ has the same energy measurement probabilities, but also has quantum coherence between the energy eigenstates. Matching those probabilities does not make the states equivalent.

For two pure states of overlap $0<c<1$, a quantum channel attempts to turn $M$ copies into $N>M$ copies. Let $e_i$ be the failure probability of the all-copies projector test on input $i$. The pure-state trace distance and fidelity identity, trace-distance contraction under quantum channels, pure-target upper bound on trace distance and triangle inequality give the displayed strictly positive gap $\Delta$. Hence ${\max }_i{e}_i\ge {\Delta }^2/4$, and for positive priors ${\pi }_0+{\pi }_1=1$, weighted Cauchy-Schwarz inequality gives ${\pi }_0{e}_0+{\pi }_1{e}_1\ge {\pi }_0{\pi }_1{\Delta }^2$. No positive state-by-state bound holds: preparing one candidate perfectly makes its own conditional failure probability zero.

A convex roof extension extends a function $\mu$ on pure states to arbitrary density operators by minimizing its ensemble average over all pure state decompositions. It agrees with $\mu$ on pure states: every positive-weight state in a decomposition of a rank-one density operator has the same one-dimensional support. Concatenating nearly minimizing ensembles proves that $F$ is a convex function. Moreover, every convex function that is no larger than $\mu$ on pure states is no larger than $F$, by applying convexity to each decomposition. Infima need not be attained; an arbitrarily small additive error suffices when $F$ is finite.

Let distinct nonorthogonal pure states have overlap magnitude $0<c<1$, and let one quantum channel take $n$ identical inputs to $m$ outputs. Let ${ϵ}_u,{ϵ}_v$ be the respective probabilities that a test for the ideal $m$-copy pure state fails. Trace distance contraction, its triangle inequality and the pure-target upper bound on trace distance give $\sqrt{1-c^{2m}}\le \sqrt{{ϵ}_u}+\sqrt{1-c^{2n}}+\sqrt{{ϵ}_v}$. Squaring with the Cauchy-Schwarz inequality gives the displayed strategy-independent positive bound. It establishes both a worst-case bound and an equal-prior average bound. Averaging rotated pairs establishes a positive Haar-average bound for unknown qubit inputs, without assuming independent clone errors.

A finite-dimensional density operator $\rho$ represents a quantum state on a Hilbert space. It is Hermitian, positive semidefinite, and has trace one:The density matrix is its matrix in a chosen orthonormal basis. In dimension $N$ it has $N^2-1$ real parameters: a Hermitian matrix has $N^2$, and fixing its trace removes one; positivity restricts the allowed values.

A quantum state ensemble of normalized pure states $|{\psi }_i⟩$ with positive probabilities $p_i$ givesZero-weight terms can be omitted. Conversely, the spectral theorem gives such an ensemble for every density operator, using its orthonormal eigenvectors and nonnegative eigenvalues, with zero eigenvalues omitted. A pure state has $\rho =|\psi ⟩⟨\psi |$; a mixed state needs at least two nonzero eigenvalues. ^[ref]

For a projective measurement with orthogonal projections ${\Pi }_a$, the Born rule isThis is the average of the component pure state probabilities. Positivity makes each outcome probability nonnegative, and ${\sum }_a{\Pi }_a=I$ makes them sum to one. The same density operator therefore predicts every measurement on the system without requiring a unique preparation history. ^[ref]

For a closed system, Schrödinger picture evolution under a unitary operator iswhen ${\rho }_H={\rho }_S(0)$ denotes the time-independent Heisenberg picture state and $U(0)=I$. In the Schrödinger picture the density operator evolves; in the Heisenberg picture the observables evolve instead, with identical measurement predictions. This unitary operator preserves the eigenvalues and hence the purity of a density operator. ^[ref]

A historical source for the probabilistic formulation of quantum mechanics is von Neumann's 1927 paper Wahrscheinlichkeitstheoretischer Aufbau der Quantenmechanik. ^[ref]

For real $-1\le \alpha \le 1$, this two-qubit density operator has equal $00$ and $11$ populations and off-diagonal coherences $\alpha /2$. It is a pure state exactly when $\alpha =\pm 1$, and a separable quantum state exactly when $\alpha =0$. Its partial transpose has eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$, directly witnessing entanglement for every nonzero admissible coherence.

The energy eigenspace at $E$ is the vector subspacewhere $H$ is the Hamiltonian operator and $I$ is the identity operator. It is nonzero exactly when $E$ is an energy eigenvalue. Its dimension is the multiplicity of that energy eigenvalue; degenerate energy levels have multiplicity greater than one. The ground-state subspace is ${\mathcal{E}}_{E_0}$, and its dimension is the ground-state degeneracy. A vector in an energy eigenspace need not be one of a specially chosen orthonormal basis of energy eigenstates.

Linearity makes the useful point immediate: if $H{\varphi }_1=E{\varphi }_1$ and $H{\varphi }_2=E{\varphi }_2$, then $H(a{\varphi }_1+b{\varphi }_2)=E(a{\varphi }_1+b{\varphi }_2)$. Every normalized nonzero combination stays in the same energy eigenspace. For a time-independent Hamiltonian operator, its evolution is multiplication by the common global phase $e^{-iEt/\hslash }$, so it is a stationary state. A quantum superposition within a degenerate energy eigenspace can change spatial shape when its coefficients are chosen differently, but its density does not subsequently beat under this unchanged Hamiltonian operator.

Across distinct energy eigenspaces, relative quantum phases evolve. For orthonormal eigenfunctions ${\varphi }_1,{\varphi }_2$ of the Hamiltonian operator with energy eigenvalues $E_1\ne E_2$, an equal-weight quantum superposition hasand position probability densityThe interference term can change local probabilities even though the Born rule assigns the same time-independent probability $1/2$ to each energy measurement outcome. Orthogonality makes its integral zero, preserving wavefunction normalization. This pure state is not a stationary state; some particular measurement distributions may nevertheless remain constant. An incoherent mixed state with the same energy populations has no such interference term.

For the pictured infinite square well, $\theta =3E_{1}t/\hslash$ and the interference repeats after $T=2\pi \hslash /(3E_1)$. At $t=0$, direct integration gives $⟨x⟩=L/2-16L/(9{\pi }^2)$; halfway through the period it is $L/2+16L/(9{\pi }^2)$. This is a moving expectation value, rather than a trajectory of an individual particle. The distinction between an energy eigenspace and a combination of different energy eigenspaces is what makes the different behavior possible.

Entanglement distillation uses local operations and classical communication on many noisy entangled states to produce fewer high-fidelity Bell pairs. It is also called entanglement purification. The protocol may fail or abort; naming a purification step does not guarantee successful output for arbitrary inputs. If the output is an exact pure state of Bell pairs, every purification of a density operator factors between those pairs and the environment, so measuring each pair in the computational basis gives a shared uniform bit independent of an eavesdropper. Approximate privacy follows when the entire retained output is sufficiently close to that pure target. A separable quantum state cannot be distilled into a Bell pair by LOCC, because each local branch preserves separability.

If a density operator $\sigma$ on an $m\ge 2$ dimensional Hilbert space has overlap $t=⟨\psi |\sigma |\psi ⟩$ with a fixed pure state, then $S(\sigma )\le h(t)+(1-t){\log }_2(m-1)$. Dephase in a basis containing $|\psi ⟩$ and apply the entropy bound with one prescribed probability. The abstract state bound is attained by $\sigma =t|\psi ⟩⟨\psi |+(1-t)(I-|\psi ⟩⟨\psi |)/(m-1)$.

A mixture of two orthogonal pure qubit states has Von Neumann entropy equal to the binary entropy of its weights. Its entropy difference from the first pure state is exactly $h_2(u)$ and its trace distance from that state is $u$. The elementary bound $-(1-u)\ln (1-u)\le u$ gives $h_2(u)\le u{\log }_2(e/u)$; entropy also never exceeds one bit. For small $u$, the leading term is $u{\log }_2(1/u)$.

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.