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Amplitude amplification increases the probability of obtaining an answer accepted by a specified Boolean function. It requires a coherent preparation $A$, its inverse, and a marked-state phase oracle that negates accepted components. Alternating that sign change with a reflection operator about the prepared state directs interference toward the accepted subspace. Brassard, Høyer, Mosca, and Tapp developed the general framework, extending Grover's algorithm beyond a uniform quantum superposition. ^[ref].

If a normalized quantum state has good amplitude $\sin \theta$, the product of the reflection in the bad axis and the reflection in the state line rotates its good-bad plane by $2\theta$. After $k$ iterations the good probability isso $O(1/\sin \theta )$ iterations raise a small good amplitude close to one. Success subsequently oscillates, making the stopping time part of the algorithm.

Compute the exact Grover search on four entries directly. Order the computational basis as $00,01,10,11$ and mark $11$. The Hadamard transform prepares amplitudes $(1/2,1/2,1/2,1/2)$. The marked-state phase oracle changes these to $(1/2,1/2,1/2,-1/2)$, whose arithmetic mean is $\overline{a}=1/4$. The Grover diffusion operator performs inversion about the mean:The three unmarked amplitudes become $1/2-1/2=0$, while the marked amplitude becomes $1/2-(-1/2)=1$. All three stages have squared norm one. Here $\sin \theta =1/2$, so $\theta =\pi /6$ and one iteration gives ${\sin }^2(3\theta )=1$, agreeing with the explicit calculation.

Grover's original search paper was submitted on 29 May 1996; its algorithm combines selective sign changes with inversion about the mean. The example exposes the beauty of that construction: a change initially invisible to computational-basis measurement becomes complete cancellation of every wrong answer. ^[ref].

The marked-state phase oracle is an assumed operation, whose implementation cost must be counted. Computational-basis measurement yields a candidate label, here $11$; using it to retrieve a record requires the relevant data access. Preparing a quantum superposition does not make all its entries available as classical outputs. The improvement concerns the number of oracle queries needed to obtain an accepted label.

A finite family of quantum states is antidistinguishable if a quantum measurement has labelled outcomes such that each outcome has zero probability for its corresponding state, while the outcomes form a complete quantum measurement. It can exclude a state without identifying which state was prepared. The PBR exclusion measurement for zero and plus supplies a four-state projective measurement example.

The central system of protons and neutrons in an atom. Its mass number $A$ is its baryon number in ordinary nuclear states; $Z$ is the proton number and $N$ the neutron number.

The nucleus contains almost all of an atom's mass and carries positive electric charge $Ze$, where $e$ is the elementary charge. The atomic number $Z$ identifies the chemical element; changing $N$ at fixed $Z$ gives another isotope. The mass number $A=Z+N$ counts nucleons, rather than specifying the exact nuclear mass: nuclear binding energy lowers the mass of a bound system relative to its separated constituents. The simplest nucleus is a single proton; other atomic nuclei can contain many protons and neutrons bound by the nuclear force.

A nuclear quantum state has properties such as spin and a spatial nuclear charge distribution. These are different questions from merely counting nucleons. A small-angle elastic scattering measurement can reveal a nuclear charge radius, while a wider range of momentum transfer tests the distribution's shape. The Hofstadter nuclear electron scattering experiment made this connection between scattering patterns and nuclear structure experimentally concrete.

Let the specimen rotate uniformly with angular velocity $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. In its rotating reference frame, the internal Hamiltonian containsHere $J_z$ is the internal angular momentum operator about the rotation axis, and ${\mathcal{H}}_0$ contains the internal interactions and any imposed magnetic field expressed in that frame. It does not include the separately treated rigid body's rotational kinetic energy. For a sector with magnetic moment $m_z=\gamma J_z$, the rotational term has exactly the form of a magnetic dipole potential energy:This Barnett equivalent magnetic field has tesla units, with the angular convention for the signed gyromagnetic ratio: $\Omega$ is in radians per second, not revolutions per second. Its sign matters. For an electronic contribution with $\gamma <0$, positive $\Omega$ favors positive $J_z$ and negative $m_z$. A sector with $\gamma >0$ has the opposite magnetic sign. Equal $\Omega$ need not correspond to equal Barnett fields for sectors having different gyromagnetic ratios.

A short derivation fixes the sign independently of any material model. Put $U(t)=\exp (-\mathrm{i}\Omega t{J}_z/\hslash )$ and transform a laboratory quantum state by $|{\psi }_{\mathrm{rot}}⟩=U^{†}|{\psi }_{\mathrm{lab}}⟩$. The Schrödinger equation givesThe transformed laboratory term is denoted by ${\mathcal{H}}_0$; a laboratory perturbation not stationary in the rotating frame can make it time dependent. Establishing a steady thermal response also requires exchange of internal angular momentum with the rotating body, for example through spin-lattice relaxation. A mere relabeling of coordinates cannot polarize isolated, unrelaxed spins.

The Barnett field is a useful description of spin-rotation coupling, not a claim that empty laboratory space acquires a uniform magnetic flux density $\Omega /\gamma$. In particular, inserting it into the Lorentz force on an unrelated test electric charge is unjustified. A nearby magnetometer detects the specimen's actual stray magnetic field and backgrounds, rather than directly detecting this internal Hamiltonian parameter.

Relative to a fixed orthonormal basis, the displayed linear map is an isometry because its basis images are orthonormal. Its adjoint obeys $C^{\ast }|k⟩|\ell ⟩={\delta }_{k\ell }|k⟩$. If vectors are described by coefficient functions, this adjoint maps $|f⟩|g⟩$ to $|fg⟩$, where the product is pointwise. Although $C$ copies every basis quantum state, it does not copy arbitrary superpositions. For example it sends $(|0⟩+|1⟩)/\sqrt{2}$ to $(|00⟩+|11⟩)/\sqrt{2}$, whereas two copies would also contain the cross terms $|01⟩$ and $|10⟩$.

Bell's theorem shows that no theory assigning outcomes through local hidden variables can reproduce every quantum correlation.

The force of the Bell theorem is that a disagreement about a local hidden-variable theory becomes a constraint on frequencies in a Bell test. Quantum entanglement alone does not guarantee a violation for every measurement in quantum mechanics: the quantum state, analyzer settings and treatment of missed events all matter. A violation of the CHSH inequality tests the conjunction of Bell local causality, measurement independence and the stated outcome and sampling assumptions; it remains compatible with quantum no-signalling.

An isolated qubit that acquires only a global phase has unchanged measurement probabilities. To reveal the Berry phase, use two coherent paths of a Mach-Zehnder interferometer, represented by a path qubit. In one arm the internal spin one-half qubit follows the loop; in the reference arm it stays in its starting energy eigenstate. Assume equal arm amplitudes, identical final internal states, cancelled relative dynamical phases, and calibrated fixed apparatus phases. In the model above, hold the reference Hamiltonian operator at its initial value for the same duration. Both arms then have energy eigenvalue $-\Delta /2$ and acquire the same dynamical phase $\Delta T/(2\hslash )$, so their relative dynamical phase vanishes in the ideal limit of the adiabatic theorem.

The joint quantum state before recombination then factors asThe Berry phase is now a relative quantum phase between the two paths. The calibrated plus output projects onto $|+⟩=(|\mathrm{ref}⟩+|\mathrm{loop}⟩)/\sqrt{2}$. The Born rule givesFor the Berry phase of a Bloch-sphere latitude, the plus-port probabilities at $\theta =0,60^{\circ },90^{\circ },180^{\circ }$ are respectively $1,1/2,0,1$. This ideal Mach-Zehnder interferometer measurement cannot distinguish $\gamma$ from $-\gamma$ or $\gamma +2\pi$ by itself; scanning a known reference phase can resolve the sign. An uncancelled relative dynamical phase would shift the fringe as well.

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ with constant $\Delta >0$. Its energy eigenvalues are $-\Delta /2$ and $+\Delta /2$, so its spectral gap is $\Delta$ everywhere. The spin points along $\mathbf{n}$; this explicit convention fixes the sign of the Berry phase.

Keep the polar angle $\theta$ fixed and increase the azimuth $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

At $\theta =60^{\circ }$, the solid angle is $\pi$ and the Berry phase is $-\pi /2$. At $\theta =90^{\circ }$, the solid angle is $2\pi$ and the Berry phase is $-\pi$. At $\theta =0$, the loop collapses and $\gamma =0$. At $\theta =\pi$, the same expression gives $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. Its winding is only a choice of global phase, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued observable.

An isotope of beryllium with 4 protons and 6 neutrons. Its nuclear spin concerns the coupled nuclear quantum state, not the electronic configuration.

In 1990, Wayne M. Itano, Daniel J. Heinzen, J. J. Bollinger and David J. Wineland reported inhibited transitions in about 5,000 ${}^9{\mathrm{Be}}^{+}$ ions held in a cylindrical Penning trap. Trapped magnesium ions, illuminated at 280 nm, kept the beryllium ions cold. A 320.7 MHz radio frequency field coupled two ground-state hyperfine structure sublevels during a 256 ms pulse adjusted for complete transfer. Up to 64 intervening 313 nm pulses addressed an optical transition from level 1: scattering distinguished it from level 2. Afterward, calibrated fluorescence counts over 100 ms measured the final quantum state population. This ensemble readout did not select ions that remained in the initial quantum state at every check; departure followed by return also contributes. The apparatus made inhibition by repeated quantum measurements experimentally accessible.^[ref]

A nuclear beta decay connects initial and final quantum states through vector and axial-vector parts of the weak charged current. At leading order in the lepton wavelength expansion, Fermi transitions carry lepton-pair spin $S=0$, while Gamow-Teller transitions carry $S=1$. If the lepton pair carries orbital angular momentum $l$, its total angular momentum is $J=l$ for $S=0$, or $J=|l-1|,\dots ,l+1$ for $S=1$. The nuclear states must obey $|J_i-J_f|\le J\le J_i+J_f$, and their parities satisfy ${\pi }_f/{\pi }_i=(-1{)}^l$ in this leading orbital classification. The least permitted $l$ labels the degree of forbiddenness. Real forbidden transitions involve several weak multipole operators; the elementary spin classification does not assert that only one operator or nuclear matrix element exists.

A bond-dissociation energy specifies the energy needed for a stated dissociation into separated fragments. An energy from the lowest vibrational state, an energy from the bottom of a potential energy well, and a finite-temperature molar dissociation enthalpy require different conventions; do not silently mix them in an electronegativity formula.

Which fragments, in which quantum states, and at which temperature? For a neutral gaseous molecule with a specified two-electron covalent bond, the usual thermochemical convention is homolysis,Each fragment receives one of the two bonding electrons; the dots denote unpaired electrons. The final fragments are isolated from each other, with their own relaxed molecular geometry and specified electronic quantum states. The energy cost includes this relaxation, rather than merely stretching the initial chemical bond while freezing everything else. For a neutral parent, heterolysis instead produces ${\mathrm{A}}^{+}(g)+{\mathrm{B}}^{-}(g)$, or the opposite assignment of electric charges; those are different final states and different energies. A value for gaseous neutral fragments cannot be transferred to ions, a chemical solution, or a solid without accounting for those changed conditions.^[ref]

The distinction is useful beyond terminology. Pauling electronegativity compares consistently defined bond-dissociation energies; photodissociation can reveal a specified ground-state threshold; a thermal chemical reaction balance requires compatible enthalpies. A threshold is also different from an activation energy: it compares initial and final quantum states, whereas a barrier can obstruct the path between them.

For a diatomic molecule, let $V(r)$ be the electronic potential energy at bond length $r$, with the atomic nuclei treated as fixed while finding the electronic energy. Choose zero at infinitely separated chemical atoms in the specified electronic quantum states. If the minimum occurs at $r_e$, the bond potential-well depth isThe bottom of this curve describes an equilibrium bond length, not an attainable motionless vibrational quantum state. Changing the zero of potential energy shifts both endpoints equally and leaves $D_e$ unchanged. A quantum harmonic oscillator approximates the curve near its minimum but rises without limit, so that local approximation alone cannot supply a dissociation asymptote. A Morse potential supplies both a minimum and a finite separated-fragment limit.^[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].

An isotope of boron with 5 protons and 5 neutrons. Its nuclear spin concerns the coupled nuclear quantum state, not the electronic configuration.

Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

A ket $|\psi ⟩$ denotes a vector representing a quantum state, and its bra $⟨\psi |$ is the linear functional taking $|\varphi ⟩$ to $⟨\psi |\varphi ⟩$. The map from a ket to its bra is conjugate-linear. In the physics convention, $⟨\psi |\varphi ⟩$ is conjugate-linear in $\psi$ and linear in $\varphi$. The adjoint operator is characterized by $⟨\psi |A\varphi ⟩=⟨A^{\ast }\psi |\varphi ⟩$. Products such as $|\psi ⟩⟨\varphi |$ are rank-one operators, while $⟨\varphi |\psi ⟩$ is a scalar inner product.

Real chemical bonds can stretch as molecular rotation increases. The resulting larger moment of inertia lowers the energy eigenvalues relative to a rigid rotor. Within one vibrational quantum state, a leading effective correction isBoth $B_v$ and $D_v$ are in frequency units here; $v$ labels the vibrational quantum state. For positive $D_v$, the adjacent-line spacing becomesso the departures grow with $J$. Molecular vibration also changes the effective rotational constant: even the lowest vibrational quantum state does not have motionless atomic nuclei.^[ref]

The quartic correction is a limited expansion, not a model to extrapolate to arbitrary $J$. It is small relative to the rigid contribution only while $(D_v/B_v)J(J+1)\ll 1$; eventually higher terms, changed internal motion, or chemical bond breaking matter. In the synthetic plot, the largest upper-state label is $J=5$, giving a stipulated correction ratio of $0.03$. No universal maximum valid $J$ follows without a particular molecule and precision target.

The quantum numbers also have limits of meaning. Here $J$ describes spin-free rotation. In a molecule with electronic spin, rotation alone is commonly labelled $N$, while $J$ includes electronic angular momentum; including nuclear spin introduces additional total-angular momentum labels and resolved splittings. An applied electric field can mix different field-free $J$ states through the Stark effect. For identical atomic nuclei, the Pauli exclusion principle can restrict combinations of rotational and nuclear-spin states. The simple $2J+1$ degeneracy and every-integer-$J$ ladder therefore describe the stated idealization, not all observed sublevels.^[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.

For a coherent state, write $a|\alpha ⟩=\alpha |\alpha ⟩$ and $|\alpha ⟩={\sum }_n{c}_n|n⟩$ in number states. The annihilation operator gives $c_{n+1}\sqrt{n+1}=\alpha c_n$, so $c_n=c_0{\alpha }^n/\sqrt{n!}$. Normalization yields $|c_0{|}^2=e^{-|\alpha {|}^2}$. Thus photon statistics follows a Poisson distribution with $\mu =|\alpha {|}^2$, and the normal ordering of every moment gives $⟨(a^{†}{)}^k{a}^k⟩={\mu }^k$. This factorization is the defining virtue of coherent states for optical calculations.^[ref]

As an original comparison, choose $\mu =2$. Then $p_0=e^{-2}$, $p_1=p_2=2e^{-2}$, $\mathrm{Var}N=2$, and $⟨N(N-1)⟩=4$. A two-photon Fock state also has mean two but has $\mathrm{Var}N=0$ and pair moment two. ConsequentlyIn particular, $|1⟩$ cannot supply a pair. Nevertheless, photon statistics alone does not determine a quantum state: averaging $|\alpha ⟩⟨\alpha |$ over a uniform quantum phase removes its off-diagonal entries while retaining exactly the same $p_n$ and these equal-time moments.