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Rotational Raman spectroscopy observes a frequency change of scattered light associated with molecular rotation. The coupling uses the induced electric dipole moment, ${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$, where $\mathbf{\alpha }$ is the electric polarizability tensor. For a linear molecule, the useful anisotropy is ${\alpha }_{\parallel }-{\alpha }_{\perp }$: the response differs for an electric field parallel or perpendicular to the molecular axis. A permanent electric dipole moment is not required.^[ref]

For the spin-free linear rigid rotor in one internal quantum state, the anisotropic tensor depends quadratically on the molecular direction. Its angular part has even parity and degree two in spherical harmonics. The triangle inequality for coupled angular momenta permits $|\Delta J|\le 2$, and even parity removes odd $\Delta J$, leaving $\Delta J=0,\pm 2$. The $\Delta J=0$ contribution is unshifted light; shifted rotational lines have $\Delta J=\pm 2$. Consequently the rigid-model Stokes shifts, in ordinary frequency units, areSuccessive allowed shifts are separated by $4B$, rather than the $2B$ absorption spacing of pure rotational electric-dipole selection rules. In spectroscopic wavenumber units replace $B$ by $\tilde{B}=B/c$ and $\Delta \nu$ by $\Delta \tilde{\nu }=\Delta \nu /c$, using the speed of light in $\mathrm{cm}{\mathrm{s}}^{-1}$ for $\mathrm{c}{\mathrm{m}}^{-1}$.

A homonuclear diatomic molecule such as molecular hydrogen has no permanent electric dipole moment in its symmetric internal state, so it lacks the ordinary electric-dipole microwave ladder. Its anisotropic electric polarizability can nevertheless give rotational Raman spectroscopy lines. Nuclear-spin symmetry can change the populations of even and odd $J$ levels, so the spin-free population example above must not be reused unchanged for molecular hydrogen. The method of coupling, as well as the energy eigenvalues, determines what the experiment can see.

For the spin-free linear rigid rotor in thermal equilibrium at temperature $T>0$, assume all integer $J\ge 0$ are allowed and ignore nuclear-spin restrictions. Each $J$ level has $2J+1$ quantum states sharing one energy eigenvalue, an instance of degenerate energy levels, so the Boltzmann distribution givesHere $Q_{\mathrm{rot}}$ is a rotational partition function, $h$ the Planck constant, $k_{\mathrm{B}}$ the Boltzmann constant, and $B$ a frequency in $\mathrm{Hz}$. Thus $hB/(k_{\mathrm{B}}T)$ is dimensionless. The $2J+1$ factor counts magnetic quantum numbers; it does not mean that a whole level is one quantum state. For the stipulated $B=5\mathrm{GHz}$, the most populated levels are $J=2$ at $3\mathrm{K}$ and $J=7$ at $30\mathrm{K}$.

Population alone is insufficient to predict intensity. In weak, unsaturated absorption, the integrated absorption coefficient includes the transition frequency, the squared electric dipole moment matrix element, and the difference of lower and upper substate populations because stimulated emission returns energy to the field. A pulsed FTMW spectroscopy signal additionally depends on excitation, quantum coherence, cavity tuning, and receiver response.^[ref]

For an isotropic sample with polarization of an electromagnetic wave along $z$, write $d_z=d\cos \theta$. Normalized spherical harmonics giveThe last equality follows from ${\sum }_{M=-J}^J{M}^2=J(J+1)(2J+1)/3$. Combining this sum with the per-substate Boltzmann distribution and subtracting stimulated emission gives relative integrated absorption areasThe common factor includes $d^2/3$ and cancels upon normalization. This formula describes integrated weak absorption in a thermal, spin-free linear rigid rotor, not arbitrary peak heights. The example's strongest integrated lines originate at $J=3$ and $J=13$ for $3$ and $30\mathrm{K}$, respectively: they differ from the most populated levels. A missing line can reflect population, coupling, coverage, or sensitivity; it does not prove that the corresponding quantum state is absent.

Ground-state electronic angular momentum $J=1/2$ couples to nuclear spin $I=3/2$, giving $F=1,2$. For contact hyperfine interaction $A\mathbf{I}\cdot \mathbf{J}/{\hslash }^2$, their zero-field energies are $-5A/4$ and $3A/4$. The splitting is $2A$. Electronic weak-field contributions give $g_{F=1}\simeq -1/2$ and $g_{F=2}\simeq +1/2$; the small nuclear contribution refines these numbers.

Compare two entrance channels feeding the same isolated resonance and the same selected exit channel, each at its own resonance energy. The exit partial decay width and the resonant denominator cancel. Different entrance momenta, spin weights and entrance widths remain. The formula assumes the same resonance, the stated unpolarized spin average and negligible nonresonant interference; a ratio at one shared laboratory energy would generally be different.

An Ising model on a chain of corner-sharing triangles, with one base spin and one apex spin per cell. If the apex interacts only with the two base corners, summing each apex independently gives a two-by-two transfer matrix on base spins. For symmetric Ising spins and energy $Js{s}^{\prime }+\bar{J}\bar{s}(s+s^{\prime })$, its entries are $2e^{-\beta Js{s}^{\prime }}\cosh [\beta \bar{J}(s+s^{\prime })]$.

A scalar particle has zero intrinsic spin. A scalar-field excitation provides its relativistic description. Spin zero alone does not force its decay matrix element to be independent of internal invariant masses; resonance dynamics can still produce a nonuniform three-body distribution.

Through-bond interaction between nuclear spins that splits NMR resonances. In the weak-coupling limit, $n$ equivalent spin-$\frac{1}{2}$ neighbours produce $n+1$ lines with binomial coefficient intensities. Equivalent nuclei do not split each other.

The Schrödinger equation governs the evolution of a quantum state between instances of measurement in quantum mechanics. For a particle of mass $m>0$ moving in a prescribed real potential energy $V(\mathbf{r},t)$, its position wavefunction satisfiesHere $h$ is the Planck constant, ${\nabla }^2$ is the Laplace operator, and the Hamiltonian operator combines kinetic energy with potential energy. Specify an initial wavefunction, an operator domain and any boundary conditions: the differential expression alone does not completely specify a physical model. The Born rule then turns the evolving probability amplitude into position probabilities through $\rho =|\psi {|}^2$ and integration over a region.

This scalar Schrödinger operator describes nonrelativistic quantum mechanics, with relevant momenta much smaller than $mc$, where $c$ is the speed of light. It omits the constant rest energy $m{c}^2$, assumes fixed particle number, and suppresses active spin dynamics and coupling to a magnetic vector potential. A free electron with a fixed, uncoupled spin state can still use this scalar model; the Pauli equation supplies the additional spin components and magnetic coupling when needed. A prescribed real $V$ describes a closed evolution model or an approximation in which the source of that potential energy is treated as external. Dissipation, particle creation and a general measurement in quantum mechanics need further modelling.

For a time-independent self-adjoint Hamiltonian operator, the spectral theorem defines unitary time evolution byInitial states in the operator domain satisfy the differential equation in the Hilbert space sense; the unitary operator also evolves all other states in that Hilbert space. Time-dependent Hamiltonian operators need suitable regularity and domain hypotheses; their evolution is not generally the exponential of $-itH(t)/\hslash$.

The local reason for conservation is especially tangible. Multiply the Schrödinger equation by the complex conjugate ${\psi }^{\ast }$, subtract its conjugate multiplied by $\psi$, and use the fact that the real potential energy terms cancel. For smooth wavefunctions this gives the probability continuity equationThe divergence theorem identifies the change in probability inside a region with minus the outward probability current. Decay at infinity, or appropriate probability-conserving boundary conditions, preserves wavefunction normalization. Conservation of total probability does not require a stationary probability density.

Its usefulness comes from being a linear partial differential equation: quantum superpositions of solutions are solutions, while different quantum phases can produce interference. For a fixed Hamiltonian operator, the time-independent Schrödinger equation finds the energy eigenstates; each then acquires just its own time-dependent quantum phase. One eigenvalue problem therefore organizes both allowed energies and the evolution of arbitrary quantum superpositions, with a continuous spectrum included when required. This is the attractive bridge between a spatial wave problem and a clock: energy differences determine the rate at which relative quantum phases change. It makes atomic spectral lines, quantum tunnelling and quantum scattering calculable from a specified model.

For an explicit example without confinement, take a free particle on the whole line, $V=0$, initially in a Gaussian wave packet with position standard deviation ${\sigma }_0$ and mean momentum $\hslash k_0$:Unlike a single plane wave, this wavefunction satisfies wavefunction normalization. A Fourier transform writes it as a quantum superposition of plane waves. Their dispersion relation is $\omega (k)=\hslash k^2/(2m)$, so each Fourier component gains the factor $e^{-i\hslash k^{2}t/(2m)}$. Completing the square in the resulting Gaussian integral giveswith the square root chosen continuously from $t=0$. Its position probability density and standard deviation areThe centre moves with the group velocity ${\omega }^{\prime }(k_0)=v_0$ and the wave packet spreads because different wavenumbers have different group velocities. This is unitary wave dispersion, without a random force or loss of probability. Independently, the initial momentum standard deviation is $\Delta p=\hslash /(2{\sigma }_0)$ and the initial position-momentum covariance vanishes. Free evolution gives $x(t)=x(0)+p(0)t/m$ in the Heisenberg picture, henceThis independently checks the spreading law. The Gaussian integral of $\rho$ is one at every time; the decreasing peak is exactly offset by increasing width.

As hypothetical preparation parameters, choose a free electron with ${\sigma }_0=1\mathrm{nm}$ and $k_0=2{\mathrm{nm}}^{-1}$. Using $\hslash \simeq 1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$ and $m\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ givesThe rounded electron mass uses the 2022 adjustment.^[ref] At $t=t_s$, the centre is $4\mathrm{nm}$ and ${\sigma }_t=\sqrt{2}\mathrm{nm}$; at $t=2t_s$, they are $8\mathrm{nm}$ and $\sqrt{5}\mathrm{nm}$. The mean kinetic energy staysIt includes both mean motion and the momentum spread. The mean speed is about $7.72\times 10^{-4}c$ and the momentum spread corresponds to $1.93\times 10^{-4}c$, making the nonrelativistic model accurate for the appreciably occupied part of this ideal Gaussian distribution. Its mathematically unbounded tails are an idealization.

The complex width also explains where the probability goes. Write $\psi =\sqrt{\rho }{e}^{i\theta }$, where $\theta$ is the dimensionless quantum phase. Differentiating the exact solution givesRelative to the moving centre, the probability current $j_{\mathrm{rel}}=j-v_0\rho$ points outward on both sides for $t>0$. This is a flux statement, not an additional assumption about individual particle trajectories. At $t=t_s$ and $x-v_{0}t=\pm {\sigma }_0$, the relative flow speeds $j_{\mathrm{rel}}/\rho$ are $\pm {\sigma }_0/(2t_s)$, about $\pm 2.8942\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$. The probability within a moving interval of half-width ${\sigma }_0$ falls from about $0.6827$ initially to $0.5205$ at $t_s$, exactly because of this outward probability current.

Historically, the 1926 Quantisierung als Eigenwertproblem (Erste Mitteilung) by Erwin Schrödinger began with the stationary hydrogen atom problem. It distinguishes discrete negative energies from a positive continuous spectrum; it is not yet the general evolution equation displayed above.^[ref] The opening of quantum mechanics discusses its original vibrating-string analogy.

In the 12 December 1933 lecture The fundamental idea of wave mechanics by Erwin Schrödinger, Erwin Schrödinger explains the move from independent rays to waves through diffraction, then applies that analogy to atomic physics. His closing pages explicitly acknowledge the difficulty of reconciling wave language with observed particle tracks. This retrospective physical argument motivates the model; it does not derive the modern Born rule from classical mechanics.^[ref]

The dates separate proposal from measurement: the de Broglie wavelength belongs to the 1924 thesis, the Schrödinger equation to 1926, and the Davisson–Germer experiment's electron diffraction beams to 1927.^[ref] That linked experiment describes the prepared nickel crystal, evacuated electron gun, angular collector scans, retarding-potential selection and heating controls. Its measured collector electric current tests wave-dependent scattering probabilities; it neither measures a wavefunction directly nor establishes this equation uniquely.

A free radical monoanion intermediate in redox state between a quinone and a catecholate; an isolated ligand has spin $S=1/2$. Its spin may couple to a metal spin in a complex.

Using the orbital order after 28 supplied in the simple shell model, proton counts 32 and 38 fill $2p_{3/2}$ and $1f_{5/2}$, so the odd 39th proton occupies $2p_{1/2}$. With paired valence neutrons this gives a $1/2^{-}$ yttrium-91 ground-state prediction. The daughter has paired 40 protons and one neutron beyond the 50-neutron closure in $2d_{5/2}$, giving $5/2^{+}$. The inferred ground-state beta transition is therefore first-forbidden unique. This is a stated paired-shell model deduction; an endpoint diagram alone does not determine the spin assignment or an evaluated isotope mass.

Single-ion anisotropy is an orientation-dependent local spin energy, commonly $D(S_j^z{)}^2$. It differs from a linear Zeeman magnetic field term. For a classical planar spin, positive $D$ penalizes the out-of-plane component; for a spin-half quantum degree of freedom this particular square is constant.

A finite molecular spin system exhibiting slow reversal of its magnetic moment on an experimental timescale. Blocking is timescale-dependent and does not require bulk long-range magnetic order.

An isolated spin-degenerate orbital coupled to a particle reservoir has Hamiltonian $H=ϵ(n_{\uparrow }+n_{\downarrow })+U{n}_{\uparrow }{n}_{\downarrow }$. The four occupation microstates have weights $1,z,z,q{z}^2$, where $z=e^{\beta (\mu -ϵ)}$ and $q=e^{-\beta U}$. Thus $\Xi =1+2z+q{z}^2$ and $\bar{n}=(2z+2q{z}^2)/\Xi$. The factor two counts distinct spin states. At $U=0$ the grand partition function factorizes into two independent fermionic orbitals; positive U separates the two electron-addition thresholds.

In the simplest nuclear shell model of an odd-mass number nucleus, paired nucleons contribute zero total angular momentum and magnetic dipole moment. The last unpaired nucleon supplies the nuclear spin $j$ and the stretched-state moment $\mu =g_{j}j{\mu }_N$, where $g_j$ is the Landé g-factor and ${\mu }_N$ the nuclear magneton.

Using addition of angular momentum with $s=1/2$ givesFor $j=\ell +1/2$, this reduces to $\mu /{\mu }_N=g_{\ell }\ell +g_s/2$. For $j=\ell -1/2$, it givesThe free-nucleon orbital factors are $g_{\ell }=1$ for a proton and $0$ for a neutron. Residual nuclear interactions and configuration mixing can shift measured moments away from this ideal prediction. A single hole in an otherwise closed $j$ shell has the same stretched-state moment as a particle in that shell: removing the $m_j=-j$ particle leaves total $J_z=j$ and moment $g_{j}j{\mu }_N$.

A single-particle wavefunction used as a mode of a many-particle description. An orbital can include spatial and spin variables; a spatial orbital specifies the spatial part. Orthonormal orbitals define creation and annihilation operators. Occupation of one normalized orbital by every boson gives a Hartree product state, while antisymmetrized occupation by fermions gives a Slater determinant.

A helium atom electronic state with total spin zero. Its antisymmetric two-spin function requires a symmetric spatial wavefunction. Its approximate spectroscopic family is distinct from the triplet family under spin-conserving electric dipole transitions.

An ordinary scalar wavefunction assigns one complex amplitude to each physical configuration. In polar coordinates without an additional twisted boundary condition, the points $(r,\varphi )$ and $(r,\varphi +2\pi )$ are identical, so the angular factor must be periodic. An angular eigenfunction $e^{im\varphi }$ therefore requires integer $m$. This scalar orbital condition does not rule out half-integer intrinsic spin.

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.

A spatial orbital is a normalized spatial wavefunction used in a one-particle description. An atomic orbital is the corresponding notion for a chemical atom; a molecular orbital belongs to a molecule. A spatial orbital is not a trajectory and does not by itself specify spin.

A spin-boson model couples a two-state quantum system to a collection of quantum harmonic oscillators. The oscillator modes provide environmental degrees of freedom that can exchange energy with the spin and change its reduced density matrix. One transverse-coupling convention isThe choice of coupling axis matters: here the Pauli matrix ${\sigma }_x$ flips the energy-basis spin states. Coupling only through ${\sigma }_z$ has a different effect, since it commutes with the free spin Hamiltonian operator.