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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.

A rank-one projective measurement associated with an orthonormal basis $\{|b_j⟩{\}}_j$. For a density matrix $\rho$, outcome $j$ has probability $⟨b_j|\rho |b_j⟩$ and leaves the measured system in the pure state $|b_j⟩$ whenever that probability is nonzero. It can be implemented by mapping this orthonormal basis to the computational basis with a unitary transformation and performing a computational-basis measurement.^[ref]

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]

A quantum system with two specified subsystems has a tensor product Hilbert space. A joint density operator determines subsystem $A$'s measurement statistics through its partial trace ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$. A joint pure state need not give a pure reduced state: the Bell state $(|00⟩+|11⟩)/\sqrt{2}$ gives ${\rho }_A=I/2$.

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 Bose--Marletto--Vedral experiment is a family of proposals to test gravitationally induced entanglement between initially independent masses. Sougato Bose and collaborators, and independently Chiara Marletto and Vlatko Vedral, described the idea in 2017. Each mass occupies two distinguishable spatial branches, encoding a qubit. Different branch separations give different Newtonian gravitational potential energies and hence different relative phases.^[ref]

For stationary branches at separations $r_{ab}$ held for duration $\tau$, the ideal model givesThe positive phase follows from negative Newtonian gravitational potential energy in $e^{-iV\tau /\hslash }$. A direct algebraic check separates local rephasing from genuine entanglement: the determinant of the two-by-two coefficient matrix vanishes exactly whenThus a large common phase alone is insufficient. For this ideal pure state, the concurrence is $|\sin (\Delta \varphi /2)|$, reaching one at an odd multiple of $\pi$.

The spin-based proposal uses two Stern-Gerlach interferometers: embedded spins label the spatial branches, the masses interact, and recombination transfers the branch information into measurable spin correlations. Measurements in different spin bases supply an entanglement witness. Unwanted electromagnetic interactions and quantum decoherence must be controlled for the gravitational interpretation.^[ref]

The mediator inference assumes initially unentangled systems, local coupling through a mediator, and no other entangling interaction. Under those assumptions, a mediator described only by commuting observables cannot generate the observed entanglement; at least two incompatible observables are required. This is an indirect test of nonclassical mediation, without selecting a complete high-energy theory of quantum gravity.^[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.

A chirped Gaussian wave packet adds a quadratic quantum phase to the zero-mean Gaussian wave packet already evaluated above. Here $\sigma >0$ and $\kappa$ is a real dimensionless parameter. Its probability density function $|{\psi }_{\sigma ,\kappa }(x){|}^2$ is unchanged for every $\kappa$, so $⟨x⟩=0$ and $(\Delta x{)}^2={\sigma }^2$. Nevertheless, applying the momentum operator givesThe sign of the quantum covariance follows directly, without assigning a classical trajectory to the wavefunction:Here $r$ is a normalized symmetrized quantum covariance, analogous to a correlation coefficient. Positive $\kappa$ gives positive $C_{xp}$, and reversing $\kappa$ reverses $C_{xp}$ while leaving both variances unchanged. In the negative-exponent Fourier transform convention used above, the momentum density isConsequently the quantum uncertainty product and the covariance matrix determinant tell different stories:Every member of this pure state family saturates the Robertson-Schrödinger uncertainty relation. Only $\kappa =0$ saturates the Heisenberg uncertainty relation for this fixed pair of position and momentum axes. Thus being a Gaussian wave packet alone does not imply a minimum $\Delta x\Delta p$.

For a concrete comparison, choose a reference length $L$, put $X=x/L$, $P=pL/\hslash$, and set $\sigma =L$. The choices $\kappa =0,+2,-2$ all have $(\Delta X{)}^2=1$, but give $(\Delta P{)}^2=1/4,5/4,5/4$ and $C_{XP}=0,+1,-1$, respectively. The two nonzero cases have $r=\pm 2/\sqrt{5}$ and product $\sqrt{5}/2$; all three covariance matrices have determinant $1/4$.

The phase space curves in the figure satisfy $z^T{\Sigma }^{-1}z=1$ in the dimensionless coordinates, where $z=(X,P{)}^T$. They display second moments, not observed simultaneous position and momentum pairs. Their common area is $\pi \sqrt{\det \Sigma }=\pi /2$, although their coordinate projections differ.

This distinction is useful for predicting focusing. Under free particle evolution with mass $m$, $x(t)=x(0)+tp(0)/m$, soA negative chirp initially narrows the position distribution; positive chirp initially broadens it. For $\kappa <0$, the minimum occurs at $t_{\ast }=-2m{\sigma }^2\kappa /[\hslash (1+{\kappa }^2)]>0$ and has variance ${\sigma }^2/(1+{\kappa }^2)$. At that time the quantum covariance vanishes, and the unchanged momentum spread gives $\Delta x(t_{\ast })\Delta p=\hslash /2$. The extra initial product was a usable position-momentum correlation, not extra measurement backaction.

For a Hermitian perturbation mode $X$ and its time derivative $\dot{X}$, subtract their expectations and use the symmetrized covariance to define variances ${\sigma }_X^2,{\sigma }_{\dot{X}}^2$. A commutator-to-fluctuation diagnostic isFor zero-mean modes this reproduces the usual expression with $⟨XX⟩$ and $⟨\dot{X}\dot{X}⟩$ in the denominator. A finite real Fourier mode or a suitably smeared field must be specified; unregulated coincident field products need not have finite variance.

The Robertson uncertainty relation gives $\mathcal{C}\le 2$ when the denominator is nonzero. Small $\mathcal{C}$ means the commutator is small relative to the marginal fluctuation scales, which can justify classical approximations for some symmetrized correlators. It does not alone show that the quantum state is mixed or that quantum decoherence has occurred. For example, a correlated Gaussian wave packet in a pure state can have a large product of marginal uncertainties while retaining quantum coherence. Which observables and accuracy are being approximated must therefore accompany a claim of effective classicality.

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.

For a pure state of three qubits, the squared concurrence obeys $C_{A:BC}^2\ge C_{AB}^2+C_{AC}^2$. In particular, if $A$ and $B$ form a maximally entangled pair, then $A$ has no entanglement with $C$.

For a pure state $a|0⟩|E_0⟩+b|1⟩|E_1⟩$ with normalized environment states, the upper-right device reduced density matrix entry is $a{b}^{\ast }⟨E_1|E_0⟩$. Distinguishability of the conditional environment states suppresses device coherence. States differing only by a global phase have unit-modulus overlap and do not cause quantum decoherence.

A nonzero bipartite pure state is a product state if and only if all its partial inner products on the first subsystem lie in a common one-dimensional vector subspace of the second. For a product $|a⟩\otimes |b⟩$, every contraction equals $⟨u|a⟩|b⟩$. Conversely, contract against each vector of an orthonormal basis to write $|\Psi ⟩={\sum }_i|i⟩\otimes |{\chi }_i⟩$. If $|{\chi }_i⟩=c_i|b⟩$ for a common nonzero $|b⟩$, then $|\Psi ⟩=({\sum }_i{c}_i|i⟩)\otimes |b⟩$. Zero conditional vectors are allowed. A nontrivial linear dependence relation requires multipliers not both zero; allowing both multipliers to vanish would give no test at all.

For a Bell pair $|{\varphi }_{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, Alice's basic measurement in the entrywise complex conjugate of any desired Bob orthonormal basis prepares each Bob pure state with probability $1/2$. Indeed, if $|u⟩={\sum }_{j=0}^1{u}_j|j⟩$, then $⟨\overline{u}|j⟩=u_j$, which proves the displayed contraction identity. The undisclosed-outcome density matrix stays $I/2$; quantum steering changes the conditional quantum state ensemble without violating the no-communication theorem.

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.

If a joint pure state is $\alpha |0⟩|E_0⟩+\beta |1⟩|E_1⟩$, the off-diagonal entry of the system's reduced density matrix is multiplied by the environmental overlap $⟨E_1|E_0⟩$. Independent repeated interactions multiply these overlaps.

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.