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For a qubit density matrix $\rho =(I+n\cdot \sigma )/2$ with nonzero Bloch vector, $\hat{n}=n/\Vert n\Vert$ and the multiplication rule for the Pauli matrices give these orthogonal rank-one spectral projectors, with eigenvalues $(1\pm \Vert n\Vert )/2$. Their Bloch vectors are opposite unit vectors $\pm \hat{n}$. At $n=0$ the state is maximally mixed and the direction is undefined; any orthonormal eigenbasis gives an antipodal pair, but no pair is distinguished by the state.

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]

For candidate density matrices ${\rho }_h$ with prior probabilities $p_h$, a projective measurement has outcome likelihood $p(m\mid h)=\mathrm{Tr}(P_m{\rho }_h)$. Bayes' theorem gives the displayed posterior whenever the denominator is positive. An outcome certifies one candidate only when every other candidate has zero likelihood there. If all candidates have the same outcome likelihood, observing it leaves their prior probabilities unchanged.

Suppose independent resource pairs have density matrices $v_j|{\Phi }^{+}⟩⟨{\Phi }^{+}|+(1-v_j)I_4/4$. An ideal complete Bell-basis measurement, followed by outcome-dependent Pauli gates, produces a chosen Bell state with white-noise visibility $v_1{v}_2$: either white-noise input makes the remote pair maximally mixed. If the reported label is reliable with probability $q$ and otherwise uniformly random, averaging the erroneous corrections makes that branch maximally mixed too, so $v_{\mathrm{out}}=v_1{v}_{2}q$. A second independent detector of the same quality during quantum teleportation adds another factor $q$ to the teleported Bloch vector.

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.

Exact quantum teleportation that preserves an arbitrary reference requires at least two classical bits from sender to receiver, even with preshared quantum entanglement. For the essential reduction, encode two classical bits in the four Bell states by Pauli gates on the sender's half of a Bell pair. Teleportation as an identity channel on a reference transfers this half while preserving the encoding. The receiver then has both halves and recovers both bits by a Bell-basis measurement, as in superdense coding.

The classical bit bound itself has a short no-communication theorem proof. For $N$ possible messages and $K$ possible classical transmissions, let ${\sigma }_{m,c}$ be the receiver's subnormalized conditional density matrix before decoding. The no-communication theorem gives ${\sum }_c{\sigma }_{m,c}=\tau$, independent of $m$, so $0\le {\sigma }_{m,c}\le \tau$. For each transmission $c$, let $\{E_{m|c}{\}}_m$ be the decoding POVM. Averaging over equiprobable messages,Perfect decoding of four messages requires $K\ge 4$, hence at least two classical bits. This applies to the preshared resources of the proposed quantum teleportation protocol as well as the Bell pair used for the encoding.

Coherence order classifies an operator component of NMR coherence by its response to rotation about the chosen total $z$ angular momentum. With dimensionless $F_z$ and rotation convention $U(\varphi )=e^{-i\varphi F_z}$, order $p$ meansThe order describes a phase transformation, rather than the amount of magnetization or the duration for which coherence persists.^[ref]

An original two-spin example shows why order zero is not synonymous with a diagonal population. Let $F_z=I_z+S_z$ for two spin one-half nuclei. For raising and lowering components with $[I_z,I_{+}]=I_{+}$ and $[S_z,S_{\pm }]=\pm S_{\pm }$,The first is zero-quantum coherence; the second is a double-quantum coherence component. In a product basis, the first connects the states with projections $(+1/2,-1/2)$ and $(-1/2,+1/2)$: different states with the same total projection. The second connects total projections−1 and+1. These operator components are not individually normalized density matrices.

At $\varphi =\pi /2$, the order-zero term remains unchanged while the order-two term changes sign. At $\varphi =\pi$, both remain unchanged. A phase comparison must therefore be chosen to discriminate the orders of interest; one phase change cannot distinguish every possible component.

Order zero also does not require zero time evolution. Under the Quantum Hamiltonian $H/\hslash ={\Omega }_I{I}_z+{\Omega }_S{S}_z$, the first component acquires $e^{-i({\Omega }_I-{\Omega }_S)t}$ when the two offsets differ. Total order and the separate offsets answer different questions.^[ref]

NMR phase cycling uses such phase responses to select contributions to an experiment. Selecting an order does not ensure that the spin system or pulse sequence actually generated it, nor that it is directly observable before further conversion.^[ref]

Choose the qubit computational basis $|0⟩,|1⟩$, also called the $Z$ basis, and the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, also called the $X$ basis. Compare a coherent preparation of $|+⟩$ with choosing $|0⟩$ or $|1⟩$ at random with equal probabilities and discarding that label:These are original ideal preparation models. Their $Z$-basis probabilities are both $(1/2,1/2)$, because their quantum state populations agree. Their $X$-basis probabilities are respectively $(1,0)$ and $(1/2,1/2)$. Indeed $⟨+|{\rho }_{+}|+⟩=1$, whereas $⟨+|I/2|+⟩=1/2$. The missing off-diagonal quantum coherence therefore changes a prediction that the common populations cannot determine.

For a relative quantum phase $\varphi$, the coherent family instead hasThe last formula follows by adding or subtracting the two probability amplitudes before taking the squared modulus. At $\varphi =\pi$, the state is $|-⟩$ and the $X$ outcomes switch to $(0,1)$, while the $Z$ outcomes remain $(1/2,1/2)$. In contrast, $-|+⟩$ changes both components by the same global phase and retains the $X$ outcomes $(1,0)$. In photon polarization, the identification $|0⟩=|H⟩$, $|1⟩=|V⟩$ makes $|+⟩$ and $|-⟩$ the two orthogonal diagonal linear polarizations; these labels describe measurement bases, not spatial axes for photon spin.

Even perfect statistics in one basis cannot certify a preparation. The above $Z$ data fit both density matrices. Conversely, distinct preparation histories can yield the same quantum state:The off-diagonal terms cancel in the right-hand ensemble. Every measurement on the qubit alone then has identical probabilities for those two histories. A retained classical label would allow conditioning on different members of the quantum state ensemble; it is extra information beyond the unlabelled density operator.

Correlated dephasing caused by the same fluctuating $z$ rotation on all qubits. Averaging $U_{\varphi }\rho U_{\varphi }^{†}$ over the unknown angle suppresses quantum coherence between sectors with different total $Z$ eigenvalues. Within one eigenspace, however, every state acquires the same global phase, so its density matrix is preserved. Such an eigenspace is a decoherence-free subspace for this noise model. The assumption concerns the shared rotation; independent local phase noise generally destroys this protection.

For a qubit, computational basis dephasing with complex overlap $c$ maps ${\rho }_{01}$ to $\overline{c}{\rho }_{01}$ and preserves the diagonal entries. Conjugation by the phase gate $P_{\phi }=\mathrm{diag}(1,e^{i\phi })$ maps the same entry to $e^{-i\phi }{\rho }_{01}$. Since these scalar factors commute, their quantum channels commute on every density matrix. More directly, an interaction $|j⟩|e⟩\mapsto |j⟩|e_j⟩$ and a diagonal phase gate give the same joint quantum state in either order on the specified initial environment state. This need not hold for an interaction that changes the quantum state populations.

The condensate fraction is the population $N_0$ of a macroscopically occupied one-particle state divided by the total particle number $N$. For an interacting system, use a macroscopic eigenvalue of the one-particle density matrix, rather than assume that the relevant mode is an undeformed single-particle trap ground state. The condensate fraction is not generally the fraction measured as superfluid by a flow or rotation response.

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 finite-dimensional quantum dynamical semigroup is Davies irreducible when no proper nonzero subspace can support a state for all future times. Equivalently, if a projection $P$ satisfies ${\Phi }_t(P\rho P)=P{\Phi }_t(P\rho P)P$ for every state $\rho$ and every $t\ge 0$, then $P=0$ or $P=I$. This is a property of invariant supports, rather than merely uniqueness of a stationary density matrix.

For a two-level system with only spontaneous decay, $L=\sqrt{\gamma }|0⟩⟨1|$ and zero Quantum Hamiltonian, the ground-state line is invariant. The stationary state is unique but pure, so the dynamics is reducible. Adding a positive excitation rate $L^{\prime }=\sqrt{\kappa }|1⟩⟨0|$ removes that invariant line; the stationary populations are $\gamma /(\gamma +\kappa )$ and $\kappa /(\gamma +\kappa )$ and the dynamics is irreducible. In finite dimensions irreducibility ensures a unique faithful stationary state, which has full rank. Related notions of quantum irreducibility must be distinguished.^[ref]

A decoherence time is a characteristic timescale for loss of specified off-diagonal coherence in a density matrix. In a simple qubit dephasing model, ${\rho }_{01}(t)={\rho }_{01}(0)e^{-t/T_2}$, so $T_2$ is the time for that coherence magnitude to fall by a factor $e$.

In Ramsey interferometry, two separated pulses convert accumulated phase into an interference signal; scanning the pulse separation measures the decay of its contrast. The inferred timescale depends on the basis, environment and pulse sequence. It differs from the energy-relaxation time, and nonexponential decay may require a stated threshold or fit convention rather than a unique decay constant.

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]

A dephasing channel suppresses off-diagonal entries of a density matrix in a preferred basis while preserving quantum state populations. Complete dephasing is identical to a projective measurement in that basis with its outcome discarded.

For the Bell correlations versus a classical mixture example, completely dephasing Alice's qubit in the computational basis maps ${\rho }_{\mathrm{Bell}}$ to ${\rho }_{\mathrm{mix}}$. It preserves their perfect $Z$ agreement and each reduced density matrix, yet destroys the $X$ correlation. This is why testing only populations can miss damage to a quantum memory: the stored quantum coherence may have disappeared while every preferred-basis measurement still looks correct. The joint Von Neumann entropy rises from zero to one bit, although each local entropy stays fixed.

Double-quantum coherence is a component of NMR coherence with coherence order $p=+2$ or $p=-2$. Under a common $z$ rotation $U(\varphi )=e^{-i\varphi F_z}$, an order-$p$ component transforms as $U{A}_p{U}^{†}=e^{-ip\varphi }{A}_p$. The term describes a phase response of the spin system.^[ref]

For two spin one-half nuclei with $F_z=I_z+S_z$, the component $I_{+}{S}_{+}$ has order+2: it connects the state with total projection−1 to that with total projection+1. Its adjoint $I_{-}{S}_{-}$ has order−2. A Hermitian density matrix can contain both components with conjugate coefficients.

An original constructed state makes the difference from populations explicit:Its density matrix has populations one half in each displayed basis state, plus the off-diagonal terms $|\uparrow \uparrow ⟩⟨\downarrow \downarrow |/2$ and its adjoint. A mixture of the two basis states with the same populations lacks those terms. This is an assumed mathematical preparation, rather than a claim about what a single pulse produces in a thermal sample.

Under the Quantum Hamiltonian $H/\hslash ={\Omega }_I{I}_z+{\Omega }_S{S}_z$, the positive-order off-diagonal component acquires phase $e^{-i({\Omega }_I+{\Omega }_S)t}$. A common phase rotation of $\pi /2$ changes its sign; a rotation of $\pi$ leaves it unchanged. The summed offsets and doubled phase response are consequences of the two projection changes.^[ref]

Conventional transverse detection reads appropriate single-quantum coherence. Experiments that select double-quantum coherence must generate it and subsequently convert it into an observable contribution. A phase-selection rule alone does not demonstrate that the necessary transfer occurred.^[ref]

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.

For an exact dimensionless Hamiltonian $H$ and trial Hamiltonian $H_0$, the free energies obeywhere $F=-\ln Z$, $F_0=-\ln Z_0$, and the expectation value uses the normalized trial equilibrium state. A physical Hamiltonian is multiplied by $\beta =1/(k_{B}T)$ in these formulas; multiplying the bound by $k_{B}T$ restores physical free energy units.

For classical configurations, the identity $e^{-F}=e^{-F_0}⟨e^{-(H-H_0)}{⟩}_0$ and the Jensen inequality for the convex exponential function give $\ln ⟨e^{-(H-H_0)}{⟩}_0\ge -⟨H-H_0{⟩}_0$. Taking minus the logarithm proves the bound.

The same bound holds for finite-dimensional, noncommuting quantum Hamiltonians, although the preceding exponential-factorization identity cannot generally be used. Define the density matrices $\rho =e^{-H}/Z$ and ${\rho }_0=e^{-H_0}/Z_0$. With natural logarithms, nonnegativity of quantum relative entropy givesRearrangement proves the quantum result without requiring $[H,H_0]=0$. Equality holds when ${\rho }_0=\rho$; in this finite-dimensional setting that means $H-H_0$ is a scalar multiple of the identity operator.^[ref]

A tractable trial family therefore supplies upper bounds that can be minimized over its parameters. This variational minimum need not reproduce exact correlations or phase transition locations.