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If $|b⟩$ lies in the kernel of $A$, then $A|b⟩=0$ has no normalized quantum state. A request to produce that normalized vector with positive probability is consequently undefined. For example, $A=\mathrm{diag}(0,1/2)$ on one qubit has distinct dyadic eigenvalues, but annihilates $|0⟩$. The corrected filtering assertion requires $A|b⟩\ne 0$; it cannot be repaired by assigning a positive success probability to this input.

A Slater determinant extends two-particle antisymmetrization to $N$ fermions. Here $x_i$ contains position and spin or other internal variables. Distinct orthonormal one-particle states give the normalized wavefunction displayed in the title. Repeating a one-particle state repeats a column of the determinant, so the wavefunction vanishes. Exchanging particle slots exchanges rows and reverses its sign. A general fermionic quantum state can be a quantum superposition of Slater determinants; it need not be one Slater determinant.

For an illustrative $n=2\to 1s$ emission calculation, retain only the first-order shifts above and neglect the quadratic shift of the hydrogen atom ground state. The unperturbed Lyman-alpha line frequency is ${\nu }_0=(E_2-E_1)/h$. The possible offsets areThe outer pair originates in the $p_0$ parts of $|{\psi }_{\pm }⟩$ and has transition electric dipole moments along the applied electric field, conventionally called $\pi$ components. The central pair originates in $p_{\pm 1}$ and has transverse rotating transition electric dipole moments, conventionally called ${\sigma }^{\pm }$ components. Because electromagnetic radiation is transverse, the $\pi$ components disappear when viewed along the electric field. Viewed from the side, the $\pi$ and $\sigma$ components have polarization of an electromagnetic wave respectively parallel and perpendicular to the applied electric field. Line intensities additionally require the initial quantum state populations and the collection geometry.

For a self-adjoint operator $A$, its projection-valued spectral measure of a normal operator assigns each Borel set $S\subseteq \mathbb{R}$ an orthogonal projection $E_A(S)$. For a normalized quantum state $\psi$, the probability of a measurement result in $S$ is $⟨\psi ,E_A(S)\psi ⟩$. In finite dimensions, $E_A(S)$ is the sum of the orthogonal projections onto eigenspaces whose eigenvalues lie in $S$. This generalizes an individual eigenvector probability to degenerate eigenvalues and continuous spectra. The functional-calculus notation $1_S(A)$ uses the indicator function of the set.

The laboratory expectation value of the axial electric quadrupole moment in the stretched nuclear quantum state $|I,M=I⟩$; $M\hslash$ is the laboratory $z$ component of total angular momentum. It differs from the intrinsic nuclear quadrupole moment $Q_0$. In an axial nuclear rotational band with $K=0$, projection onto definite angular momentum givesHere $K\hslash$ denotes the component of total angular momentum along the intrinsic symmetry axis. Consequently a positive $Q_0$ for a prolate spheroid corresponds to a negative $Q_s(2)$ in this model. The sign of the laboratory moment alone must be interpreted together with the band assignment.

The $I=0$ result has a short symmetry explanation. A spin-zero state has no preferred spatial direction. Its second charge moments satisfy $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩$, so $⟨2z^2-x^2-y^2⟩=0$. Equivalently, averaging an intrinsic axis uniformly over directions gives $⟨P_2(\cos \theta )⟩=0$, where $P_2(u)=(3u^2-1)/2$ is a Legendre polynomial. This is a statement about a quantum expectation value, not a classical picture of a spinning object sweeping through orientations.

For an independent check of the $I=2$ sign, the $K=0$, $M=2$ rotor orientation probability is proportional to $|Y_2^2{|}^2\propto {\sin }^4\theta$, using a spherical harmonic. With $u=\cos \theta$,A zero static laboratory moment does not make every quadrupole observable zero. With electric multipole operator ${\mathcal{M}}_{2\mu }=\int {\rho }_{\mathrm{ch}}{r}^2{Y}_2^{\mu }{d}^{3}r$, define the reduced transition strength using a reduced matrix element:In the axial $K=0$ rotor, angular projection gives $B(E2;0^{+}\to 2^{+})=5Q_0^2/(16\pi )$ and $B(E2;2^{+}\to 0^{+})=Q_0^2/(16\pi )$. The factor of five is the ratio of the numbers of magnetic substates. Thus an electric quadrupole transition can reveal collective deformation even when $Q_s(0)=0$. These strengths determine $|Q_0|$ within the model; their squares alone cannot distinguish its sign.

In 1925, Wolfgang Pauli connected two puzzles: alkali metal doublets and their anomalous Zeeman effect, and the capacities of closed electron shells. His paper assigned the extra two-valued property to the optically active electron, rather than an angular momentum of the closed core. Its Paschen-Back effect classification used four quantum numbers; these historical labels should not be identified literally with modern $n,\ell ,m_{\ell },m_s$. The exclusion rule forbade matching all four. This connected resolved one-electron states with closed-group populations. An additional test was the missing triplet counterpart of the equivalent two-$s$-electron ground state in alkaline earth metals. Pauli explicitly claimed counts of allowed states, not their energies or ordering.^[ref]

The modern question is especially concrete: if a Quantum Hamiltonian has a lowest one-particle state, why cannot every electron occupy it? Minimizing energy answers which of the allowed quantum states is preferred; the Pauli exclusion principle determines which quantum states are allowed in the first place. Spectral lines then test energy differences between allowed multiparticle quantum states. A line count is not itself a direct count of occupied spin orbitals: transitions also depend on selection rules and the prepared quantum state populations. The observed anomalous Zeeman effect supplied constraints for the historical classification; the two-particle antisymmetrization below supplies the modern mathematical explanation of exclusion.

For spin one-half, choose the normalized spin eigenstates $|+z⟩,|-z⟩$ with $S_z|\pm z⟩=\pm (\hslash /2)|\pm z⟩$. The spin one-half along an axis pure state in the $xz$ plane isIts norm is one because the two eigenstates are orthogonal and ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$. Applying the Born rule to a $z$ projective measurement givesConsequently the expectation value, second moment, and variance areAt $\theta =\pi /3$, the outcome probabilities are $3/4$ and $1/4$, the mean is $\hslash /4$, and the variance is $3{\hslash }^2/16$. Every individual outcome remains $\pm \hslash /2$. At $\theta =\pi /2$, the mean is zero and the variance is maximal, ${\hslash }^2/4$. At either pole, one outcome has probability one and the variance vanishes.

The diagonal mixed state ${\rho }_{\mathrm{diag}}=p_{+}|+z⟩⟨+z|+p_{-}|-z⟩⟨-z|$ has the same $S_z$ probabilities, expectation value, and variance as $|{\psi }_{\theta }⟩$. It is nevertheless a different quantum state for $0<\theta <\pi$: the displayed pure state has $⟨S_x⟩=\hslash \sin \theta /2$, while ${\rho }_{\mathrm{diag}}$ has $⟨S_x⟩=0$. A single component's mean does not reveal all relative quantum phases or establish purity of a density operator.

Spin isomers of hydrogen are two forms of molecular hydrogen, ${}^1{\mathrm{H}}_2$, distinguished by the combined nuclear spin of their two protons. They have the same chemical formula and electronic ground state, but different allowed molecular rotation states. Parahydrogen has total nuclear spin zero and even rotational quantum number $J$; orthohydrogen has total nuclear spin one and odd $J$. The difference makes heat capacity a probe of fermionic exchange symmetry, and makes cooling history matter for a tank of molecular hydrogen.

Assume the electronic ground state $X{}^1{\Sigma }_g^{+}$ and the vibrational ground state of a free molecule. The electronic factor is symmetric under exchanging the two protons, and the vibrational factor depends only on their separation, so it is also symmetric. These assumptions concern the exchange of protons; the electronic wavefunction separately obeys fermionic exchange symmetry under exchange of electrons. Neglect electronic excitation, vibrational excitation, external fields, and small interactions involving nuclear spin when calculating rotational energy eigenvalues.

The two protons are identical particles with spin $1/2$. Their complete wavefunction must change sign under their particle exchange operator. Interchanging the two ends reverses the internuclear direction $\hat{\mathbf{r}}$, and the spherical harmonic rotational factor satisfiesThe antisymmetric spin-one-half singlet state therefore requires symmetric, even-$J$ rotation; the symmetric spin-one-half triplet state requires antisymmetric, odd-$J$ rotation. This is exchange symmetry of two identical spin-one-half fermions applied inside a diatomic molecule, not an extra rotational force. A linear rigid rotor may omit spin-dependent splittings while still requiring this restriction on its allowed quantum states.

When spatial and spin factors separate, a spin orbital combines a spatial orbital ${\varphi }_a$ with a normalized spin state ${\chi }_{\sigma }$. For an electron, choose spin one-half along an axis with $\sigma =\uparrow ,\downarrow$. The two spin orbitals $u_{a\uparrow }$ and $u_{a\downarrow }$ are orthogonal even though their spatial orbital is identical, since their inner product contains $⟨{\chi }_{\uparrow }|{\chi }_{\downarrow }⟩=0$. When spatial and spin variables do not separate, the complete one-particle state is instead a multicomponent wavefunction $u(\mathbf{r},s)$; exclusion still applies to that complete quantum state.

A spin rotation acts on a quantum state through the unitary representation generated by its angular momentum operators. With $\hslash =1$, rotation by angle $\theta$ about axis $\hat{\mathbf{n}}$ is $e^{-i\theta \hat{\mathbf{n}}\cdot \mathbf{S}}$. Simultaneously rotating every site is a global symmetry of an isotropic Heisenberg model.

For a qubit, use the dimensionless Pauli matrices as observables $A={\sigma }_x$, $B={\sigma }_z$. Their Pauli matrix multiplication law givesThese observables have eigenvalues $\pm 1$; the physical spin components are $S_j=(\hslash /2){\sigma }_j$, where $\hslash$ is the reduced Planck constant, so their standard deviations acquire a factor $\hslash /2$.

In the computational basis, prepare $|+y⟩=(|0⟩+i|1⟩)/\sqrt{2}$, an eigenstate of ${\sigma }_y$ with eigenvalue $+1$. Direct multiplication givesBoth Pauli measurements return $+1$ and $-1$ with equal probabilities, and the Robertson uncertainty principle is saturated: $1\cdot 1=1$.

Instead prepare $|+z⟩=|0⟩$. Now the expectation values are $⟨{\sigma }_x⟩=⟨{\sigma }_y⟩=0$, $⟨{\sigma }_z⟩=1$, so $\Delta {\sigma }_x=1$, $\Delta {\sigma }_z=0$. The Pauli measurement of ${\sigma }_z$ is certain, and both sides of the bound vanish. The commutator is still nonzero as an operator: noncommuting observables do not guarantee a positive lower bound independent of the quantum state. This example is useful for distinguishing the algebra of the observables from the expectation values in a particular preparation.

Spontaneous emission is radiative decay of an excited matter quantum state by emission of a photon, without an applied resonant drive. In a simple Markov decay model the excited-level population obeys ${\dot{P}}_e=-\Gamma P_e$. During driven control, the expected emission count is governed by $\Gamma \int P_e(t)dt$; suppressing the lossy level's population is therefore the purpose of dark-state adiabatic transfer.

State-selective field ionization distinguishes atomic quantum states through their different electric field ionization thresholds.

A readout in which one logical quantum state is coupled to an optical cycling transition and scatters many photons, while another is dark. An ideal bright/dark record implements a projective measurement in the logical basis. Neglecting that record removes their relative quantum coherence. Missed photons, off-resonant excitation and pumping out of the logical subspace qualify the ideal model.

For an equally populated ideal pair of angular momentum eigenstates, the magnetic moment component takes two values, $\pm {\mu }_B$, and the displacements at fixed velocity are $\pm d(v)$. Compare a specified classical alternative: every chemical atom carries a permanent magnetic moment of fixed magnitude ${\mu }_B$, with independently isotropic initial orientations and no dissipative alignment. If $u=\cos \theta$, the sphere's area element is $2\pi du$, so $u$ has a uniform distribution on $[-1,1]$. With precession preserving this component in the ideal geometry, the displacement $d(v)u$ fills the interval $[-d(v),d(v)]$ continuously. A pair of bands therefore tests allowed components rather than merely proving that silver is magnetic. Other classical preparation assumptions would be different models and must be stated.

To include finite instrumental width, choose a normalized Gaussian distributionwith standard deviation $\sigma =0.035\mathrm{mm}$, representing a combined field-off spatial width. At fixed velocity the model probability density functions areEach has unit integral and zero mean; the different shapes, rather than a different average displacement, distinguish them. The discrete model has variance ${\sigma }^2+d(v{)}^2$, whereas the classical model has variance ${\sigma }^2+d(v{)}^2/3$. Their common mean also explains why the raw average landing position is insufficient to identify the internal quantum state.

Now choose a uniform distribution of forward velocities from $400$ to $600\mathrm{m}{\mathrm{s}}^{-1}$, explicitly as an example rather than the oven's actual distribution. Average either conditional probability density function over that interval. The positive-channel displacement ranges from $0.1776$ to $0.3995\mathrm{mm}$, since slower chemical atoms spend longer accelerating and drifting. The two components broaden into bands even though their internal magnetic moment values remain discrete. The corresponding classical profile continues to populate the center. The displayed strip darkness is proportional to predicted deposited number density, with equal total arrivals; it is not a photograph or a calibrated optical response of a developed film.

Increasing the spatial width until it is comparable with $d$ can hide the two maxima. At fixed velocity, differentiating the two-state probability density function twice gives $p_{\mathrm{two}}^{\prime \prime }(0)={\varphi }_{\sigma }(d)(d^2-{\sigma }^2)/{\sigma }^4$: the central minimum becomes a central maximum when $\sigma >d$. Conversely, a broad distribution of velocities need not create a new internal component: it spreads each existing component through the $v^{-2}$ dependence. Further instrumental limits include collisions, aperture clipping of strongly deflected trajectories, asymmetric pole geometry, and nonlinear film development. Field-off widths and a mapped magnetic field gradient are needed to separate such effects. For a modern counting experiment, channel calibration and efficiency corrections belong to Stern-Gerlach expectation estimation; spatial separation alone does not supply those statistics.

Let the atomic beam travel along $x$, with velocity $v>0$, and let $z$ measure displacement along the dominant magnetic field gradient. For a fixed effective magnetic moment component ${\mu }_z$, the magnetic dipole potential energy is $U\simeq -{\mu }_z{B}_z$. Taking its negative partial derivative givesA spatially uniform magnetic field can produce torque and Larmor precession, but cannot produce this translational separation. In a quantum state following a local internal energy eigenvalue $E_n(B)$, the more general expression is $F_z=-{\partial }_z{E}_n$, with effective magnetic moment $-\partial E_n/\partial B$. Treating that moment as constant is an approximation with a physical regime, not an extra measured fact. Assume the local magnetic field direction stays nearly fixed across the sampled beam, and that internal states follow it without transitions, as in the adiabatic theorem. The scalar gradient model describes the beam's neighborhood; it omits the full magnetic field's transverse components and fringe regions.

Assume a constant $G$ over magnet length $\ell$, followed by a field-free drift $D$. Neglect transverse motion at entry and changes in longitudinal velocity. Newton's second law gives transverse acceleration $a_z={\mu }_{z}G/M$ for whole-atom mass $M$. The magnet transit time is $t_m=\ell /v$, and the drift time is $t_d=D/v$. The magnet supplies both a displacement and an exit transverse velocity:The $\ell D$ contribution is motion with the acquired transverse velocity, not continued acceleration outside the magnet. For the plate-at-exit geometry, set $D=0$. For a nonzero entrance transverse velocity $w$, add $w(\ell +D)/v$. These terms show why initial collimation and flight geometry matter when inferring a magnetic moment from a deposit.

Choose hypothetical parameters $M=1.79\times 10^{-25}\mathrm{kg}$, $G=300\mathrm{T}{\mathrm{m}}^{-1}$, $\ell =0.035\mathrm{m}$, $D=0.100\mathrm{m}$, and $v=500\mathrm{m}{\mathrm{s}}^{-1}$. Approximate the two effective magnetic moments by ${\mu }_z=\pm {\mu }_B$, where the Bohr magneton is ${\mu }_B\simeq 9.274\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$. Then $|a_z|\simeq 1.55\times 10^4\mathrm{m}{\mathrm{s}}^{-2}$, $t_m=70\mu \mathrm{s}$ and $t_d=200\mu \mathrm{s}$. The positive displacement is about $0.0381\mathrm{mm}$ inside the magnet plus $0.2176\mathrm{mm}$ during drift: $d=0.2557\mathrm{mm}$, with full separation $2d=0.5114\mathrm{mm}$. These chosen values illustrate the model and are not reconstructed 1922 measurements.

The exit transverse velocity is only about $1.09\mathrm{m}{\mathrm{s}}^{-1}$, small compared with the chosen forward velocity; this checks the small-deflection assumption. Halving $G$ halves $d$, doubling $v$ divides $d$ by four, and setting $D=0$ removes most of this example's separation. A real pole has fringe magnetic fields and a position-dependent gradient; these require a more complete trajectory calculation.

Stimulated emission is emission induced by incident light, in contrast with the unprompted decay in spontaneous emission. It supplies the amplification mechanism of a laser. ^[ref].

For an emitter with upper and lower energy values $E_2>E_1$, the resonant photon has frequency $\nu$ satisfyingwhere $h$ is the Planck constant. In the elementary photon picture, an incident photon stimulates an emitter in the upper quantum state to emit an additional photon into the same optical normal mode, with the same frequency, propagation direction and photon polarization. The emitter supplies the extra energy; the stimulating photon is not consumed. This builds the existing field and its optical coherence. An emitter in the lower quantum state can instead undergo photon absorption, so stimulated emission alone does not establish net optical gain.

For example, a $2.0\mathrm{eV}$ gap corresponds to $\nu \approx 4.8360\times 10^{14}\mathrm{Hz}$ and a vacuum wavelength of about $620\mathrm{nm}$. One ideal stimulated emission event converts one incoming photon into two outgoing photons in the selected optical normal mode, adding $2.0\mathrm{eV}$ of light energy while lowering the emitter's energy by that amount.

Take an invented diatomic molecule with a Morse potential, setting the separated-atom energy to zero:Here $a$ is an inverse length, $r_e$ the equilibrium bond length, and ${\omega }_e=a\sqrt{2D_e/\mu }$, with reduced mass $\mu$. These stipulated parameters describe a simple model, not a measured chemical substance. For the one-dimensional Morse oscillator, energy eigenvalues above the minimum areConsequently,The harmonic estimate $A/2=0.1000\mathrm{eV}$ is slightly larger than this model's exact zero-point energy. The physical bound-level restriction matters: the quadratic expression is not an infinite ladder of vibrational quantum states.

For a separate thermal illustration using this same stipulated $D_0$, assume an ideal gas, fully active molecular rotation of the parent, negligible thermal molecular vibration, and negligible electronic excitation for all species. Each monatomic product contributes $\delta h=5RT/2$; the diatomic molecule contributes $7RT/2$. Thus the net increment is $3RT/2$, with molar gas constant $R$. At $T=300\mathrm{K}$,This last result follows from the stated thermal model, not a universal correction for every chemical bond. Frozen rotations, accessible excited states, different fragment composition, or a condensed phase change the thermal increments.

Real atoms have multiple internal quantum states. Optical pumping and spatially varying electromagnetic polarization can remove kinetic energy by additional mechanisms, producing sub-Doppler cooling below the ideal Doppler cooling limit. In polarization-gradient cooling, state-dependent energy shifts from the Stark effect and delayed optical pumping make atoms repeatedly climb potential energy hills. This mechanism needs the internal-state structure omitted by the independent-beam two-level system above. ^[ref].

William D. Phillips's team tested unexpectedly cold sodium molasses with time-of-flight atom detection. A probe above the cloud initially saw no atoms; deliberately heating them verified that the detector worked. Moving the probe below the cloud produced arrival signals consistent with temperatures below the Doppler cooling limit. Independent release/recapture checks used sharply apertured beams and included gravity; uncertain cloud and probe geometry still limited the inferred temperature. These controls tested the thermometer as well as the cooling theory. ^[ref].

The sudden approximation treats a rapid, finite change of a Hamiltonian operator as leaving the quantum state approximately unchanged during the switch. The energy measurement can nevertheless change immediately: the same quantum state has different probability amplitudes in the new eigenbasis. Compute their inner products with the new energy eigenstates, then apply the Born rule. The two-level sudden quench below makes this distinction quantitative without assuming the change is small.

An ideal finite step has exact quantum state continuity at a finite Hamiltonian step; a finite-duration implementation has a small evolution error under the bounded-model conditions below. An impulsive quantum kick is different: its Hamiltonian operator grows as its duration shrinks, retaining a finite integrated action and a finite change of quantum state. A sudden approximation therefore concerns short evolution during switching, not conservation of energy during externally driven change. The subsequent unitary time evolution can be substantial.^[ref]