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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 scattering interpretation connects an incoming preparation to the relative frequency of outgoing directions and final atomic states. Max Born's 1926 paper treats a distant incoming electron and a chemical atom, expands the outgoing wavefunction into final atomic states, and imposes conservation of energy. Its translated correction footnote specifies the square of the coefficient; the modern complex modulus notation below is ours.^[ref]

In his 11 December 1954 lecture, Max Born recalled the difficulty of interpreting bound electrons alone. He instead described an incident electron wave generating an outgoing spherical wave, with direction-dependent squared amplitude determining relative scattering probabilities. Distinct final atomic states describe excitation with corresponding electron energy loss. This is a retrospective account of an interpretation, rather than the report of a new apparatus.^[ref]

The distinction from the modern Born rule matters. The historical question was which collision outcome occurs: a direction of motion and a final atomic state. The general projective measurement formula applies to arbitrary observables, including degenerate outcomes and mixed states; a positive operator-valued measure describes still more general detector statistics. Those formulations should not be read back as the explicit contents of the short 1926 scattering paper. Nor does the scattering interpretation itself supply a rule for the conditional post-measurement state.

Actual experimental settings are described in the Davisson–Germer experiment and the Franck-Hertz experiment. The former scans collected electric current as an electron beam scatters from prepared nickel; changing direction, accelerating voltage and surface preparation probes electron diffraction. The latter uses acceleration through mercury vapor and a retarding collector to infer discrete energy losses from current minima, with vapor density and retarding settings as controls. Angular resolution and surface backgrounds limit the first inference; initial energy spread and collision positions limit the second. These linked experiments supply the preparation, apparatus and readout; the examples below are original calculations, not Born's measured data.

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 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]

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.

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.

When two density operators are commuting elements, choose a common orthonormal eigenbasis and write $\rho =\mathrm{diag}(p_i)$, $\sigma =\mathrm{diag}(q_i)$. Their quantum fidelity becomes the classical overlap of the two probability distributions, while their trace distance becomes total variation distance:This follows by computing the positive square root of an operator entry by entry in the common eigenbasis. For the qubit mixed states $\rho =\mathrm{diag}(0.9,0.1)$ and $\sigma =\mathrm{diag}(0.6,0.4)$,The two contributions to quantum fidelity are about $0.734847$ and $0.2$. In contrast, the chance that two independent computational basis measurements, one on each preparation, print the same label is $0.9(0.6)+0.1(0.4)=0.58$. That particular event has a specified probability; neither $F$ nor $F_{\mathrm{sq}}$ is that probability.

Here trace distance $0.3$ means an equal-prior quantum state discrimination success of $0.65$ by the Holevo–Helstrom theorem. Measuring in the common computational basis and guessing $\rho$ on outcome zero and $\sigma$ on outcome one achieves $[0.9+0.4]/2=0.65$. Also $\sqrt{1-F^2}\approx 0.3551>D$: the pure-state trace distance and fidelity identity is an equality for pure states, rather than a formula for arbitrary mixed states.

For all finite-dimensional density operators, including pure states and mixed states, the unsquared quantum fidelity and trace distance obeyIn the squared quantum fidelity convention this is $1-\sqrt{F_{\mathrm{sq}}}\le D\le \sqrt{1-F_{\mathrm{sq}}}$. The bounds translate a high quantum fidelity into a small operationally meaningful trace distance. If $F\ge 1-\epsilon$ with $0\le \epsilon \le 1$, then $D\le \sqrt{2\epsilon -{\epsilon }^2}$; if $D\le \delta$, then $F\ge 1-\delta$. The upper bound is attained by the pure-state trace distance and fidelity identity. For the fidelity of commuting mixed states example, the bounds read $0.0652\lesssim 0.3\lesssim 0.3551$. ^[ref]

The infinite square well is an ideal confinement model with potential energy $V(x)=0$ for $0<x<L$ and infinite walls outside. The walls impose Dirichlet boundary conditions $\varphi (0)=\varphi (L)=0$ and a vanishing exterior wavefunction. Inside, the time-independent Schrödinger equation isThe first wall removes the cosine term; the second requires $\sin (kL)=0$, hence $k=n\pi /L$. The $n=0$ solution is identically zero and cannot satisfy wavefunction normalization. Each energy eigenstate has $n-1$ interior nodes of a standing wave, at $x=jL/n$ for $j=1,\dots ,n-1$. More curvature costs more kinetic energy, explaining the $n^2$ growth. The positive ground-state energy is a confinement effect: a constant zero-energy wavefunction cannot satisfy both walls and remain normalized.

For an electron with mass $m_e=9.1093837139\times 10^{-31}\mathrm{kg}$ in a box of length $L=1\mathrm{nm}$, use $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$. These constants are from the 2022 CODATA adjustment. ^[ref]. Substitution givesDoubling $L$ quarters every energy eigenvalue; doubling $m$ halves them. This scaling makes the particle in a box useful for estimating confinement energies before attempting a more realistic calculation. A finite square well permits exterior tails for bound states and has only finitely many bound energy eigenstates. For an electron, the scalar box calculation assumes fixed spin and no spin-dependent interactions.

For the ground state, the probability of finding the particle in the middle half of the box is fully computable:Thus half the available length contains about $81.8\%$ of the probability. The figure contrasts the signed wavefunctions with their nonnegative probability densities: a negative lobe is an amplitude with opposite quantum phase, not negative probability.

An equal quantum superposition of the first two energy eigenstates gives a simple example of moving density:Its density period is $T=h/(E_2-E_1)\approx 3.66608\mathrm{fs}$ for the electron box above. At $t=0$ the cross term enhances the left side; at $T/4$ it vanishes; at $T/2$ it enhances the right side. More quantitatively,which takes the values $0.924413$, $0.5$, and $0.075587$ at those times. Orthogonality keeps the total probability one throughout. An energy measurement still returns $E_1$ or $E_2$ with equal probabilities: changing position density arises from relative quantum phase, not changing populations. A mixed state describing a statistical mixture with those populations has the constant probability density $({\varphi }_1^2+{\varphi }_2^2)/2$; it lacks this matter-wave interference term.

The maximally mixed density operator on an $N$-dimensional Hilbert space assigns equal weights to every vector of any orthonormal basis. It is invariant under every unitary operator. Its purity is $\mathrm{Tr}({\rho }_{\ast }^2)=1/N$, the minimum possible because the nonnegative eigenvalues sum to one and obey the Cauchy-Schwarz inequality. For $N>1$ it is a mixed state; for $N=1$ it is the sole pure state. A rank-$k$ projector has probability $k/N$ in this state.

A POVM $\{E_i\}$ gives outcome probability distributions $p_i=\mathrm{Tr}(\rho E_i)$ and $q_i=\mathrm{Tr}(\sigma E_i)$ by the Born rule. Their classical overlap, the Bhattacharyya coefficient, is $B(p,q)={\sum }_i\sqrt{p_i{q}_i}$. Fuchs and Caves showed that minimizing this overlap over all POVMs gives the unsquared quantum fidelity:The same POVM is applied to either possible input quantum state. Each measurement in quantum mechanics has $B(p,q)\ge F$; some measurement attains equality, including when the density operators have zero eigenvalues. Thus quantum fidelity quantifies the smallest classical overlap that observation can reveal. For fidelity of commuting mixed states, measurement in a common eigenbasis attains the minimum. This operational interpretation compares statistics without assigning a probability that two mixed states are equal. ^{[ref] [ref]}

A mixed state is a density operator that is not a rank-one orthogonal projection. In finite dimension it is characterized by $\mathrm{Tr}({\rho }^2)<1$, so its purity of a density operator distinguishes it from a pure state.

Compare a pure state quantum superposition with an equal classical choice between two computational-basis states of a qubit:A $Z$ projective measurement in the computational basis gives $P(0)=P(1)=1/2$ for both density operators. An $X$ projective measurement in the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ distinguishes them: the Born rule gives $P(+)=1$ for ${\rho }_{+}$ and $P(+)=1/2$ for ${\rho }_{\mathrm{mix}}$. These comparisons use separate copies of each preparation; measuring $Z$ first changes the state tested by a subsequent $X$ measurement.

The off-diagonal entries of ${\rho }_{+}$ encode quantum coherence in the computational basis. Such quantum coherence depends on the basis: ${\rho }_{+}$ is diagonal in the Hadamard basis but remains a pure state. Mixedness is independent of the basis, because a unitary change of basis preserves the eigenvalues and the purity of a density operator.

Even the quantum state ensemble is not unique:The off-diagonal entries of the two Hadamard basis projectors cancel. Thus the density operator retains precisely what measurements on the system can reveal, while an accessible preparation label carries additional information. A mixed state can also be the reduced density matrix of an entangled state, even when the whole system is in a pure state; classical ignorance about a selected component is not its only physical origin. ^[ref]

A density matrix has unit trace. The Pauli matrices are traceless, so $\mathrm{Tr}\hat{\rho }=C$ and hence $C=1$. Using $\mathrm{Tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$ in the expectation value givesConsequently the required Bloch vector and density matrix areThe two eigenvalues are $3/4$ and $1/4$, both positive, so this is a valid density operator. Its purity of a density operator isTherefore it is a mixed state. Equivalently, its spectral decomposition is $\rho =(3/4)|+x⟩⟨+x|+(1/4)|-x⟩⟨-x|$, an ensemble of the two distinct spin one-half along an axis states, rather than a single pure state.

Fix the qubit polarization convention by $\rho =(I+\vec{P}\cdot \vec{\sigma })/2$, where the Pauli matrices are ordered in the basis $(|1⟩,|2⟩)$ and ${\sigma }_z=|1⟩⟨1|-|2⟩⟨2|$. Thus the ground state is the north pole of the Bloch sphere. In a rotating frame and within the rotating-wave approximation, choose the drive phase and detuning sign so thatFor example, with this ground-state convention one may take $\Delta ={\omega }_L-{\omega }_0$ and choose the transverse coupling positive. The Bloch vector precesses around the effective field $(\Omega ,0,\Delta )$ at angular frequency ${\Omega }_R=\sqrt{{\Omega }^2+{\Delta }^2}$, tracing a circle at fixed projection on that axis. Its length is conserved under coherent unitary time evolution. On resonance the axis is transverse and the Rabi oscillation can fully invert an initially ground-state atom. Away from resonance the axis tilts, andThere is no relaxation towards an equilibrium Bloch vector in this ideal model. An inhomogeneous ensemble can nevertheless lose its averaged qubit polarization through dephasing even without spontaneous emission; the circle description assumes a common drive and detuning.

For $|a{|}^2+|b{|}^2=1$, the stated quantum state givesThe diagonal elements are the two state populations. The off-diagonal elements are quantum coherences: they retain the relative quantum phase and determine interference under a subsequent pulse. This is a pure state, even if both populations are nonzero: ${\rho }^2=\rho$, $\mathrm{Tr}{\rho }^2=1$, and $|\vec{P}|=1$. A mixed state averages projectors over preparation probabilities and generally has $\mathrm{Tr}{\rho }^2<1$. Equal populations alone do not distinguish an equal coherent superposition from an unpolarized mixture.

For use in all three cases, multiplication of the gates gives the Hadamard-phase-Hadamard qubit rotationThe global phase cancels in ${\rho }^{\prime }=U\rho U^{†}$. Thus the Bloch vector map is

The final unlabelled Ramsey spectroscopy comparison applies to the three cases as follows. In an atomic clock, an initial coherent pulse prepares a superposition, a free interval $T$ accumulates a relative quantum phase proportional to detuning times $T$, and a final coherent pulse converts that phase to a population. Here $\alpha$ plays the role of that accumulated phase, with sign and fringe offset set by pulse conventions. In (a), the first Hadamard gate creates the needed superposition and the final gate reveals full-contrast Ramsey interferometry. In (b), the first Hadamard gate instead puts the state in $|1⟩$, so the intervening phase gate has no relative quantum phase to imprint: the output is independent of $\alpha$. In (c), the unpolarized density matrix remains unchanged under every gate, so the ensemble has no Ramsey interferometry contrast. A coherent unitary gate alone cannot initialize a maximally mixed ensemble into a pure clock state.

The Ramsey interferometry analogy concerns preparation, phase accumulation, and readout; a Hadamard gate is not literally a transverse $\pi /2$ rotation on every input. In this basis $H=R_y(\pi /2){\sigma }_z$, and its Bloch sphere action is a $\pi$ rotation about $(\hat{x}+\hat{z})/\sqrt{2}$, up to a global phase. The phase conventions of physical Ramsey spectroscopy pulses must be specified to compare their outputs for arbitrary initial states.

Use the ordered basis $(|1⟩,|2⟩)$ and ${\sigma }_z=\mathrm{diag}(1,-1)$, withA pure quantum superposition $|+x⟩=(|1⟩+|2⟩)/\sqrt{2}$ has $\mathbf{R}=(1,0,0)$ on the Bloch sphere. In contrast, a mixed state $\rho =p|1⟩⟨1|+(1-p)|2⟩⟨2|$ has $\mathbf{R}=(0,0,2p-1)$, inside the sphere for $0<p<1$. At equal populations its vector is zero. Equal populations alone therefore do not distinguish a coherent superposition from a mixture; their off-diagonal density matrix elements differ.

Choose the drive phase so the rotating-frame Hamiltonian is $H_{\mathrm{rot}}=\hslash (\Omega {\sigma }_x+\delta {\sigma }_z)/2$, where $\delta =\omega -{\omega }_0$ as specified. The Rabi vector is the angular-velocity vectorHere $\Omega$ is the on-resonance population oscillation angular frequency, not the unhalved coefficient of a cosine coupling. From $\mathbf{R}(0)=\hat{z}$, rotation about $\mathbf{W}$ at $W=\sqrt{{\Omega }^2+{\delta }^2}$ givesFor $\Omega \gg |\delta |$, the rotation axis is nearly transverse, the trajectory nearly a great circle, and transfer is almost complete. For $\Omega =\delta >0$, the axis bisects $+x$ and $+z$, the orbit lies in $R_x+R_z=1$, and $p_2=\frac{1}{2}{\sin }^2(\sqrt{2}\Omega t/2)$ never exceeds $1/2$. Each member remains a pure state in this coherent evolution.

For the Ramsey free interval, let atom $j$ have detuning ${\delta }_j=\delta +{\eta }_j$. Instantaneous ideal pulses give the Ramsey ensemble coherence factor $\chi (T)=⟨e^{i{\eta }_{j}T}⟩$. Averaging the free-evolution vector obtained in step (ii) givesDifferent atoms fan out around the equator, shortening the ensemble Bloch vector even though every individual's vector remains unit length. A broad smooth detuning distribution usually makes the average spiral towards the centre and damps the Ramsey fringes towards $1/2$. The dephasing time is of order $1/\Delta {\omega }_0\sim 5/\delta$. inhomogeneous ensemble dephasing is not necessarily irreversible decoherence of individual atoms; an appropriate spin echo can rephase static frequency offsets.

The given spread does not specify a probability distribution or whether it denotes an RMS width. An exact envelope is therefore not determined. The figure illustrates, rather than infers, a Gaussian distribution with RMS ${\sigma }_{\delta }=0.2\delta$, for which $\chi =e^{-{\sigma }_{\delta }^2{T}^2/2}$. A uniform distribution would instead give a sinc envelope and side lobes. Both have the same qualitative initial loss of contrast on the quoted inverse-spread scale.

In the rubidium-87 ground-state hyperfine structure, $J=1/2$ and $I=3/2$, so $F=1,2$. Taking $H_{\mathrm{hf}}=A\mathbf{I}\cdot \mathbf{J}/{\hslash }^2$ givesThe zero-field splitting is ${\Delta }_{\mathrm{hf}}=2A\simeq h\times 6.8\mathrm{GHz}$. Ignoring the small nuclear magnetic term, the weak-field Landé g-factors are $g_{F=1}\simeq -1/2$ and $g_{F=2}\simeq +1/2$. The diagram labels all three $F=1$ and five $F=2$ substates, with linear shifts $g_F{\mu }_B{m}_{F}B$.

The pair $|1⟩=|F=1,m_F=0⟩$ and $|2⟩=|F=2,m_F=0⟩$ is connected by a magnetic dipole transition with $\Delta F=1$, $\Delta m_F=0$, driven with its oscillating magnetic component along the quantization axis. Both first-order shifts vanish, giving the rubidium-87 clock transition first-order insensitivity near zero field. Its insensitivity does not imply absence of quadratic shifts.

For these two $m_F=0$ levels the Breit-Rabi formula gives the quadratic Zeeman shift of a clock transition:where the last expression neglects the nuclear term and sets $g_J\simeq 2$. HenceThis uses ${\mu }_B/h\simeq 1.40MHz/G$ and the supplied $A/h\simeq 3.4\mathrm{GHz}$. With physical moments defined by ${\mu }_i=-\partial E_i/\partial B$, the differential magnetic moment is ${\mu }_2-{\mu }_1\simeq -8.2\times 10^{-3}{\mu }_B$. If effective magnetic moment denotes the positive energy slope instead, that slope difference is positive. The magnitude and physical frequency sensitivity are the same. Retaining the full atomic factors gives the familiar quadratic clock-frequency coefficient about $575Hz/{\mathrm{G}}^2$; the simplified supplied parameters give about $576Hz/{\mathrm{G}}^2$.^[ref]

Use a far-detuned optical dipole trap, such as a red-detuned focused-beam trap, to confine both clock states. Their first-order magnetic moments vanish; their residual quadratic magnetic slopes are opposite, so a simple static magnetic-field minimum does not confine both in the same way. An optical potential acts through the electronic polarizability and can trap both ground hyperfine states. It is not perfectly state independent: differential AC Stark shifts, intensity noise and residual magnetic gradients still need control during Ramsey evolution.

A basic measurement is specified by an orthonormal basis $\{|b_j⟩{\}}_{j=0}^{d-1}$, with outcome $j$ associated with the rank-one orthogonal projection $P_j=|b_j⟩⟨b_j|$. For a normalized pure state $|\psi ⟩$, the Born rule givesThe probabilities sum to one because ${\sum }_j{P}_j=I$. Conditional on an outcome with $p(j)>0$, the post-measurement state iswhich represents $|b_j⟩$ up to a global phase. An outcome of zero probability has no conditional state to normalize. For a density matrix $\rho$, the corresponding expressions are $p(j)=\mathrm{Tr}(P_j\rho )$ and $P_j\rho P_j/p(j)$. If the outcome is ignored, the nonselective projective measurement instead produces ${\sum }_j{P}_j\rho P_j$, generally a mixed state.^[ref]

The displayed computational basis expansion is already a Schmidt decomposition, with coefficients $\sqrt{1-p}$ and $\sqrt{p}$. For $0<p<1$ both are nonzero, so the Schmidt rank is two. A pure product state has Schmidt rank one. The state is therefore entangled throughout the specified open interval.

Equivalently, the reduced density matrix isThe reduced state is a mixed state, which is impossible for one factor of a pure product state. At the excluded endpoints the Schmidt rank drops to one, giving $|00⟩$ or $|11⟩$.

Take the trine state ensembleAll three states are normalized. Since $1+\omega +{\omega }^2=0$, their average density matrix isGeometrically, their unit Bloch vectors form an equilateral triangle on the equator and sum to zero. Orthogonality is unnecessary for an ensemble decomposition of a mixed state: three distinct, nonorthogonal pure states can give the same maximally mixed average as two orthogonal states.

A bipartite pure state is entangled exactly when it is not a product state: there are no normalized quantum states $|u{⟩}_A,|v{⟩}_B$ such thatEquivalently, its Schmidt rank exceeds one, or either reduced density matrix is a mixed state. This definition concerns quantum entanglement across the specified division into systems $A$ and $B$.

Let $a_{ij}$ be the coefficient of $|ij⟩$. The partial trace contracts the repeated index of the discarded qubit:For the first qubit, each diagonal entry is $1/3+1/6=1/2$ and the upper off-diagonal entry is $1/3-1/6=1/6$. For the second qubit, the diagonal entries are $2/3$ and $1/3$, while the upper off-diagonal entry is $-1/\sqrt{18}+1/\sqrt{18}=0$. Thus the two reduced density matrices, each in computational basis order $0,1$, areBoth have eigenvalues $2/3,1/3$ and trace one, so both are mixed states even though the joint quantum state is pure. Their different eigenvectors are visible in the useful rewriteAbsorbing the minus sign into one local basis vector makes this a Schmidt decomposition with nonnegative coefficients. It also checks both partial traces and exhibits the entanglement responsible for the mixed local descriptions.