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A ferromagnetic spin chain favors aligned neighbouring spins. Its conventional long-wavelength magnons have quadratic dispersion.

An antiferromagnetic spin wave is a long-wavelength fluctuation of staggered magnetic order. In an isotropic ordered antiferromagnet its low-energy dispersion is linear.

Entanglement monogamy limits how strongly one subsystem can be entangled with several independent partners.

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 between operator algebras.

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 $g(E)$ is defined so that $g(E)dE$ counts one-particle states with energies in an interval of width $dE$. A sum over closely spaced states then becomes an integral weighted by $g(E)$.

The mean occupation of a bosonic one-particle state of energy $E$ is

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 $n+2m=k$.

The bosonic ladder operators satisfy

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.

If $U|\psi ⟩=e^{2\pi ij/N}|\psi ⟩$, controlled powers of $U$ applied to a uniform $N$-state control register produceThese states are the Fourier basis. Applying the inverse quantum Fourier transform and measuring returns $j$ with probability one.

For the cyclic shift $S|j⟩=|j+1\mathrm{mod}N⟩$ and the conventionthe Fourier states obeyConsequently $S={\mathrm{QFT}}_{N}D{\mathrm{QFT}}_N^{-1}$, where $D|k⟩=e^{-2\pi ik/N}|k⟩$.

For a function of period $r$ dividing $N$, prepare a uniform superposition of inputs, evaluate the function into a second register, and measure that register. The first register becomes a periodic coset state. Applying the quantum Fourier transform and measuring gives

Let $r$ divide $N$ andThe geometric sum in the quantum Fourier transform vanishes unless $c$ is a multiple of $N/r$. Thus measurement after the transform returns the $r$ valueswith equal probabilities $1/r$.

The root-of-unity filter givesThus ${\mathrm{QFT}}_N^4=I$, while the square is an involution whose eigenvalues lie in $\{1,-1\}$.