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For a central potential, each angular-momentum channel has a scalar scattering coefficient $S_l(k)=e^{2i{\delta }_l(k)}$.

Each channel coefficient is a component of the scattering matrix in a basis adapted to rotational symmetry.

For a reflection-invariant localized potential, order incoming amplitudes from the left and right and outgoing amplitudes toward the left and right. Thenwhere $r$ and $t$ are the reflection and transmission amplitudes. Probability-current conservation makes $S$ unitary for a real potential.

The transmission probability is the ratio of transmitted to incident probability current. When the asymptotic wavenumbers agree, it is the squared modulus of the transmission amplitude.

A reflected wave is an outgoing component travelling back toward the side from which the incident wave arrived.

The spin coherent-state path integral contains a Berry phase equal, modulo gauge choices, to the spin $S$ times the solid angle swept on the Bloch sphere. Single-valued path-integral amplitudes require $2S\in \mathbb{Z}$.

A ferromagnetic spin chain is a one-dimensional array whose exchange interactions favor aligned neighbouring spins. For $J>0$, the nearest-neighbour Heisenberg model conventionhas fully aligned ground states. A collective small deviation from alignment produces a ferromagnetic magnon rather than a set of unrelated local spin flips.

For spin length $S$, lattice spacing $a$ and this exchange convention, the one-magnon energy is $E(k)=2JS[1-\cos (ka)]$, giving $E(k)\simeq JS{a}^2{k}^2$ at long wavelength. The gapless point reflects the freedom to rotate a uniform aligned state. This ground-state and excitation statement does not establish finite-temperature spontaneous order in an infinite one-dimensional short-range model. Finite length, anisotropy and coupling between chains can change what an experiment observes.

An antiferromagnetic spin wave is a small collective variation of the ordered spin pattern in an antiferromagnet. In a bipartite Néel state, neighbouring spins point oppositely; the slowly varying Néel order parameter describes their staggered orientation. A long-wavelength disturbance changes that orientation smoothly rather than independently flipping each site.

For an isotropic ordered antiferromagnet in the usual long-wavelength regime, the low-energy dispersion relation is linear, $\omega \simeq c|\mathbf{k}|$, with spin-wave velocity $c$. This differs from the quadratic low-wavenumber ferromagnetic magnon in a conventional isotropic ferromagnet. Anisotropy or an applied magnetic field can alter the branches or introduce gaps. The starting ordered state also needs justification: strong quantum fluctuations, especially in low dimension, can invalidate a naive linear spin-wave approximation. A computed divergence is then a warning about the approximation, not a negative physical magnetization.

Entanglement monogamy describes limits on sharing quantum entanglement among several partners. If two qubits form a pure Bell state, their joint state cannot also be correlated with a third system: the Bell pair already has a pure joint density operator, so the full state factorizes from the third system.

For partially entangled states, a quantitative sharing rule depends on the measure and subsystem dimensions. The Coffman--Kundu--Wootters inequality bounds the sum of squared pairwise concurrences for three qubits by the entanglement of one qubit with the other two together. Squaring and the chosen measure matter; one cannot assume an identical linear inequality for every entanglement measure. Classical correlations can be shared much more freely, distinguishing monogamy from a general ban on correlated triples.

Quantum binary hypothesis testing chooses a two-outcome measurement to distinguish two candidate density operators, trading the probabilities of the two kinds of error.

For ensemble $\{p_j,{\rho }_j\}$ with average $\bar{\rho }={\sum }_j{p}_j{\rho }_j$, the pretty good measurement has $M_j={\bar{\rho }}^{-1/2}{p}_j{\rho }_j{\bar{\rho }}^{-1/2}$ on the support of $\bar{\rho }$.

A bipartite quantum state is separable when its density operator is a convex combination ${\sum }_j{p}_j{\rho }_j^A\otimes {\rho }_j^B$ of product states; otherwise it has entanglement. Operationally, Alice and Bob can prepare a separable quantum state by sharing a classical random label $j$ and each preparing the corresponding local state.

Separable does not mean uncorrelated. The mixture in Bell correlations versus a classical mixture is separable with two terms of weight $1/2$, yet the parties' $Z$ records agree perfectly. A single product state would give independent local outcomes. Allowing a mixture of products captures shared classical randomness; excluding this possibility would incorrectly classify ordinary correlations as entanglement.

A quantum channel is a linear, completely positive, trace-preserving map taking input density operators to output density operators. Trace preservation keeps total probability one. Complete positivity requires positivity even when the map acts on one part of an entangled larger system, with the other part unchanged.

In finite dimensions, a Kraus representation writes $\mathcal{E}(\rho )={\sum }_a{K}_a\rho K_a^{†}$, with ${\sum }_a{K}_a^{†}{K}_a=I$. A unitary channel has one Kraus operator; a dephasing channel suppresses coherences in a specified basis. Different physical implementations can yield the same channel after unobserved degrees of freedom are traced out. A conditional measurement outcome alone is generally trace decreasing; only the sum over all outcomes gives a trace-preserving channel.

When $E-\mu \gg k_{B}T$, the additive $\mp 1$ is negligible and both quantum distributions reduce to the Maxwell--Boltzmann factor $e^{-(E-\mu )/(k_{B}T)}$.

The Fermi-Dirac distribution follows from Fermi-Dirac statistics and is $n(E)=[e^{\beta (E-\mu )}+1{]}^{-1}$.

The mean occupation of a fermionic one-particle state of energy $E$ isAt zero temperature this becomes the step function $1_{E<E_F}$.

The density of states $g(E)$ counts available single-particle states per energy interval: $g(E)dE$ is the number in a narrow interval of width $dE$. A stated convention may count the entire sample or divide by its volume and may include spin degeneracy separately.

It converts mode sums into integrals, such as $N=\int g(E)f(E)dE$ for an occupation function $f$. Available states and occupied states are therefore different quantities. For a quadratic free-particle dispersion, the bulk density of states scales as $E^{-1/2}$, a constant and $E^{1/2}$ in one, two and three dimensions, respectively, away from the appropriate threshold convention. These differences control thermal occupation and spectral features. Band extrema or saddle points can produce a Van Hove singularity, and an optical transition also needs matrix elements and joint initial/final phase space.

The Bose-Einstein distribution gives the mean occupation of a one-particle energy eigenstate by noninteracting bosons in thermal equilibrium:Here $\mu$ is the chemical potential and $T>0$ the temperature. It is a mean particle count in one state, so it can exceed one.

For that state, occupation numbers $n=0,1,2,\dots$ have weights $r^n$, with $r=e^{-\beta (E-\mu )}$. The one-state grand partition function is the geometric seriesConsequently $\overline{n}=r{\partial }_r\log {\mathcal{Z}}_E=r/(1-r)$, giving the distribution. Convergence requires $\mu <E$ for every included state, hence $\mu <E_0$ for a finite ideal gas with lowest energy $E_0$. ^[ref]

At $E-\mu =k_{B}T\log 2$, the mean occupation is one; at much larger $E-\mu$, it approaches $e^{-\beta (E-\mu )}$, the dilute classical limit. Approaching $\mu =E_0$ requires careful treatment of the ground-state occupation in Bose-Einstein condensation.

The Maxwell-Boltzmann distribution describes the random velocities and speeds of particles in a classical ideal gas at thermal equilibrium. Individual particles keep changing velocity through collisions, while the equilibrium probability distribution stays fixed. For one species of particle mass $m$ at positive temperature $T$, the characteristic scale iswhere $k_B$ is the Boltzmann constant. A signed velocity component follows a normal distribution; the nonnegative magnitude follows the Maxwell speed distribution. Their peaks differ because a speed measurement groups together all directions of motion.

Use the gas rest frame, three spatial dimensions, and nonrelativistic particles, with $k_{B}T\ll m{c}^2$ where $c$ is the speed of light. Assume a dilute gas whose interactions between encounters can be neglected in the ideal gas approximation and whose translational motion is classical. A sufficient condition for negligible quantum exchange effects is $n{\lambda }^3\ll 1$, where $n$ is the number density and $\lambda$ the thermal de Broglie wavelength. Diluteness alone does not guarantee thermal equilibrium; the gas must have equilibrated. These are the assumptions behind the classical Maxwell-Boltzmann velocity distribution. ^{[ref] [ref]}

The connection to Maxwell–Boltzmann statistics is the Boltzmann factor for translational kinetic energy, $E_{\mathrm{kin}}=m|\mathbf{v}{|}^2/2$. Normalizing $e^{-E_{\mathrm{kin}}/(k_{B}T)}$ over velocity gives a probability density; an occupation factor $e^{(\mu -E)/(k_{B}T)}$ instead specifies a mean occupation, with chemical potential $\mu$. Their normalizations serve different purposes. ^[ref]

This connects microscopic motion to measurements: a gas thermometer uses the ideal gas law, collisions and the mean free path govern transport, and measurements of spectral lines use thermal Doppler broadening to infer temperature. Each probes a different aspect of the same thermal motion.

In 1860, James Clerk Maxwell presented the component and speed laws in Part I, Proposition IV of Illustrations of the Dynamical Theory of Gases, pp. 22–23. His argument combined independence of the three velocity components with isotropy, and he explicitly compared the component law with the normal distribution of observational errors. Independence is an assumption of that argument; perpendicular coordinate axes alone do not establish it. ^[ref]

For independent frequencies $1$ and $2$,so the eigenspace at energy $k+3/2$ has one basis state $|n,m⟩$ for each nonnegative solution of $n+2m=k$.

For a quantum harmonic oscillator, creation and annihilation operators move between energy eigenstates in steps of one oscillator quantum. Here $|n⟩$ is a normalized number state with nonnegative integer $n$.

The bosonic ladder operators satisfy

Writing $N=A^{†}A$ gives $N|n⟩=n|n⟩$ and $H=\hslash \omega (N+\frac{1}{2})$. Thus applying $A^{†}$ raises the energy by $\hslash \omega$. The annihilation operator obeys $A|0⟩=0$; it does not turn the ground state into a state below the zero-point energy. The normalization factors make repeated application consistent with the commutator. ^[ref]

When two masses occupy spatial superpositions, their branch-dependent Newtonian gravitational potential energy produces relative phases. If those phases cannot be separated into one phase for each mass, the interaction turns an initial product state into an entangled state.