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The Berezinskii–Kosterlitz–Thouless transition in the two-dimensional XY model separates a low-temperature phase with algebraically decaying correlations from a high-temperature phase with unbound vortices and exponentially decaying correlations. It does not produce conventional long-range order.

How can something resist a twist when its orientation never settles down? The distinction is between long-range order, a nonzero limiting correlation function, and phase stiffness, a free energy cost for a slow spatial twist. In a uniform two-dimensional XY model at positive temperature, smooth spin waves can erase the first while leaving the second finite. Vortex-antivortex binding then decides whether that resistance survives on arbitrarily large scales.

Vadim Lvovich Berezinskii developed an earlier description of two-dimensional order. John Michael Kosterlitz and D. J. Thouless independently developed the defect mechanism; Kosterlitz's 2016 lecture explicitly acknowledges their initial ignorance of Berezinskii's work. The central step was to include rare vortices alongside smooth spin waves.^[ref]

The Mermin-Wagner theorem concerns continuous symmetry and sufficiently short-range interactions in the thermodynamic limit. It does not say that every two-dimensional system is disordered in every sense, nor that every continuous symmetry gives a Berezinskii–Kosterlitz–Thouless transition. Here the order parameter has a circular phase, permitting integer winding numbers. Finite-scale vortex renormalization explains why a finite sample can look coherent even when its infinite counterpart lacks long-range order.

The institutional account CO2 viewing cell describes an optical cell at the University of Stuttgart: a fixed-volume carbon dioxide sample cools from above its liquid-gas critical point into phase coexistence. The apparatus supplies a view through the cell and displayed temperature and pressure readings. Near roughly $31{}^{\circ }\mathrm{C}$, the account reports critical opalescence: enhanced scattering of light makes reflected light appear whitish and transmitted light dark. This readout reveals growing density thermal fluctuations; it is not a direct count of molecules or a measurement of an infinite correlation length. ^[ref]

Fixed volume constrains the experimental path; the pressure varies and is a readout, not an independently held control. The account explicitly identifies the clip as part of a longer cooldown and says its endpoints are not steady states. To test an equilibrium interpretation more closely, one would hold temperature until the readout settles, compare cooling and warming, and check optical background away from the liquid-gas critical point; these are proposed controls, not measurements reported on that page. The account does not provide cell dimensions, filling number density, calibration uncertainties or such control runs. The sample is finite: its striking appearance motivates the thermodynamic limit as a model of macroscopic sharpness, rather than demonstrating a literal equilibrium singularity.

Fix the spatial mean of a scalar order parameter to $M$ and sum the Boltzmann weights of all configurations with that mean. The resulting constrained partition function $Z_M$ defines this free-energy density, with an additive normalization fixed consistently. For an applied conjugate field $h$, $Z(h)=\int dM{e}^{-{\beta }_{T}V[F_c(M)-hM]}$, up to the chosen measure normalization. The thermodynamic limit gives $f(h)={\inf }_M[F_c(M)-hM]$. A homogeneous Landau free energy is an approximation to this constrained potential; phase-separated configurations can convexify it in the full thermodynamic limit.

An energy gap is a positive separation between a ground-state sector and its lowest excitations, with the sector, momentum and thermodynamic limit specified. A local positive curvature of an auxiliary field is not alone a proof of a dynamical gap: the kinetic terms and conjugate variables determine the excitation poles.

Two related properties of an extensive quantity must be distinguished. Additivity says that a quantity $X$ for independent subsystems A and B obeysThe subsystems may be in different states. Extensivity expressed through a homogeneous function says that the equilibrium fundamental relation obeyswith composition held fixed. This is degree-one homogeneity along a replication ray; it does not say that internal energy is a linear function of arbitrary independent changes of entropy, volume and amount of substance.

Additivity gives integer replication by repeated addition. Subdividing a uniform bulk sample into $b$ equivalent parts gives rational scaling $a/b$; continuity then extends it to positive real factors. This argument requires negligible boundary contributions and a consistently prepared state. Conversely, a degree-one homogeneous function need not be additive between different rays: $f(x,y)=\sqrt{x^2+y^2}$ has $f(\lambda x,\lambda y)=\lambda f(x,y)$, but $f(1,0)+f(0,1)=2\ne \sqrt{2}=f(1,1)$. Thus a scaling identity alone cannot justify summing arbitrary states.

For finite samples, the surface correction to thermodynamic free energy can matter. Under geometric enlargement with volume multiplied by $\lambda$, surface area grows as ${\lambda }^{2/3}$, so a surface energy term is not a bulk extensive quantity. Replicating separated droplets retains every original surface, whereas merging them changes the total surface area. Likewise, appreciable mutual gravitational potential energy makes interacting subsystems fail the independent-copy premise. The thermodynamic limit explains when bulk scaling becomes accurate; it does not remove these qualifications for every small or strongly interacting system.

The Fermi energy is the highest occupied one-particle energy in the ground state of a noninteracting Fermi gas at zero temperature, measured here relative to the lowest one-particle energy. The Pauli exclusion principle allows only one fermion in each complete one-particle quantum state, including its spin, so adding more particles fills progressively higher energy eigenstates. A cold gas can therefore have substantial kinetic energy even when its temperature is zero. In the thermodynamic limit, with the same energy reference, $E_F=\mu (T=0)$ where $\mu$ is the chemical potential.^[ref]

For a uniform, three-dimensional gas of noninteracting electrons, with spin one-half, equal populations of the two spin states, no applied magnetic field, and nonrelativistic particles of mass $m=m_e$,Here $n$ is the total number density, $V$ the volume, $p_F$ the Fermi momentum, $\hslash =h/(2\pi )$ the reduced Planck constant, and $k_B$ the Boltzmann constant. The electron Fermi temperature $T_F$ expresses the same energy scale in kelvin; it is not the actual gas temperature. Strong degeneracy of a degenerate electron gas means $T\ll T_F$. The momentum formula counts states and survives changing the energy–momentum relation, whereas $E_F=p_F^2/(2m)$ requires $p_F\ll mc$.^[ref]

The reference matters: the displayed $E_F$ excludes rest energy and sets the zero at $p=0$. Including rest energy shifts both the one-particle energy and the zero-temperature chemical potential by $m{c}^2$, leaving occupations unchanged. More generally, adding a constant $C$ to every one-particle energy also adds $C$ to $\mu$; the Fermi-Dirac distribution depends only on their difference. In a finite quantum system the chemical potential at zero temperature can lie between the last occupied and first empty energy eigenstates; its equality with the filling edge is a continuum statement.

At positive temperature, $E_F$ remains the zero-temperature scale associated with the specified number density. The Fermi level $\mu (T)$ must instead satisfy the particle-number conditionwhere $g(E)$ is the extensive density of states, including both spin states. Smearing the Fermi-Dirac distribution while arbitrarily holding $\mu =E_F$ generally changes $N$. This distinction matters when interpreting electron heat capacity and any experiment that infers temperature from an occupation profile.^[ref]

The Fermi sea is the zero-temperature many-fermion quantum state with every one-particle state below the Fermi energy occupied and every state above it empty. For a uniform isotropic Fermi gas, the occupied region in momentum space is a solid sphere; its boundary is the Fermi surface. These are momenta, not positions: filling the sphere does not put the particles in a ball in ordinary space. Opposite momentum contributions cancel, so the gas at rest has zero total momentum even though its total kinetic energy is positive.

The filled states are forced upward in energy by the Pauli exclusion principle. This is different from the zero-point energy of one particle confined by a boundary. At fixed number density, the Fermi energy remains positive as the box grows to the thermodynamic limit, while the lowest box confinement energy tends to zero. Low-energy excitations move particles from occupied states near the Fermi surface into nearby empty states; deep states have no nearby available destination. This explains why only a small fraction of electrons contribute to the low-temperature electron heat capacity.^[ref]

The exact zero-field two-point correlation function $C_n=⟨S_1{S}_{n+1}⟩$ in the finite canonical ensemble of an antiferromagnetic Ising ring. The spin-chain transfer matrix has eigenvalues ${\lambda }_{+}=2\cosh (\beta J)$ and ${\lambda }_{-}=-2\sinh (\beta J)$. In their eigenbasis, an Ising spin insertion exchanges the two eigenvectors. The trace with two insertions is therefore ${\lambda }_{-}^n{\lambda }_{+}^{N-n}+{\lambda }_{+}^n{\lambda }_{-}^{N-n}$, while the partition function is ${\lambda }_{+}^N+{\lambda }_{-}^N$, proving the formula. Its two terms retain both paths around the ring. It has $C_0=1$ and $C_n=C_{N-n}$. At fixed $n$ and finite positive inverse temperature its thermodynamic limit is $(-\tanh \beta J{)}^n$.

At fixed finite $L$, the Helmholtz free energy of a periodic Potts chain receives its first thermal correction from a pair of domain walls. The two-domain-wall degeneracy of a Potts ring divided by the ground state degeneracy factor supplies the coefficient. Expanding the exact canonical partition function proves the formula. The expansion is not uniform in $L$: taking the thermodynamic limit first gives $F/L=-J-{\beta }^{-1}\log [1+(Q-1)e^{-\beta J}]$, whose leading correction is of order $e^{-\beta J}$.

For a polarized spin chain with dispersion relation $ϵ(k)=2S[J(1+u)-J\cos k-D\sin k]$, the first zero mode appears at the maximum over the allowed ring momenta. In the thermodynamic limit this threshold is $\sqrt{1+(D/J{)}^2}-1$, since the maximum of $J\cos k+D\sin k$ is $\sqrt{J^2+D^2}$. At finite size the minimizing wave vector need not be available, so a positive continuum threshold need not give a positive finite-ring threshold. For $L=4$ and $0<D<J$, the latter is zero.

An experiment of size $L$ stops the renormalization-group flow at a finite ${\ell }_{\max }$. In the leading scaled model take the original initial conditions $Y_0=0.06$ and $X_0=-0.10,-0.06,-0.03$. The invariant is respectively $0.0064,0,-0.0027$. The first approaches $X=-0.08$, hence $K=(2.08)/\pi$; the second follows the critical separatrix; the third eventually leaves the small-vortex fugacity regime. These are chosen model trajectories, with no fitted experimental parameters.

On the critical separatrix, $X=-Y$ implies $Y^{\prime }=-Y^2$, henceEven a critical finite sample therefore has a phase stiffness above its asymptotic jump value, with a slow logarithmic correction in $L$. Slightly above the transition, a sample with $L\ll \xi$ can resemble the algebraic phase before the flow reaches unbound vortices. A trap additionally varies the local density and phase stiffness; neither its whole-cloud coherence nor a finite observation window automatically measures the homogeneous thermodynamic limit.

For $H=\int [a{\varphi }^2/2+\kappa |\nabla \varphi {|}^2/2+\gamma ({\nabla }^2\varphi {)}^2/2+B{\varphi }^2|\nabla \varphi {|}^2]$ and an even positive trial kernel $J(q)$, define $S_0={\sum }_q^{+}1/J(q)$ and $S_2={\sum }_q^{+}{q}^2/J(q)$ using a positive-wavevector sum for a real field. Isserlis theorem gives $⟨H_4{⟩}_0=4B{S}_0{S}_2/V$. The Feynman-Bogoliubov inequality therefore has stationary kernelsThus both the mass and gradient coefficient shift. In the thermodynamic limit, their self-consistency equations areA coarse-graining ultraviolet cutoff $\Lambda$ is necessary for the first integral in three dimensions: its large-$k$ radial integrand tends to a constant. Physical solutions require $J(q)>0$.

For $J>0$ and $S_i^z=\pm 1/2$, a parallel bond has energy $-J/2$ while an antiparallel bond has $+J/2$. A wall therefore costs J, rather than the $2J$ of a differently normalized $-J\sum s_i{s}_{i+1}$ model. At zero field its density is $[1-\tanh (\beta J/2)]/2$, positive at every nonzero temperature. Walls occur in pairs on a periodic ring, but that global parity restriction does not remove their nonzero density in the thermodynamic limit. The low-temperature correlation length is approximately $a{e}^{\beta J}/2$.

Define the instantaneous spin magnetization $m=\frac{1}{4}{\sum }_i{\sigma }_i$. Global reversal sends $m$ to $-m$ without changing the zero-field energy. Pairing every microstate with its reversed partner proves $⟨m⟩=0$ at every positive temperature, even when almost all probability lies on the two fully aligned patterns. A vanishing mean does not imply weak alignment: the limiting probability distribution is half at $m=+1$ and half at $m=-1$.

The Ising ring magnetization fluctuations measure this distinction through variance. In the $d=0$ class, both patterns have $m^2=1$. Among the $d=2$ patterns, eight have $m^2=1/4$ and four have $m^2=0$; the two $d=4$ patterns also have $m^2=0$. ConsequentlyIt equals $6/17$ at $K=(\ln 2)/4$, approaches $1/4$ for four independent fair Ising spins at $K\to 0$, and approaches $1$ as $K\to \infty$. Unlike the energy variance, the spin magnetization variance remains large in the low-temperature symmetric mixture.

The same thermal fluctuation controls the field response. Allow $h$ to vary before setting it to zero; differentiating the Boltzmann distribution inserts $\beta {\sum }_i{\sigma }_i=4\beta m$, givingThis is a spin-normalized magnetic susceptibility with respect to the energy-valued parameter $h$; conversion to physical magnetization and magnetic field requires the magnetic moment and sample volume factors.

For finite $N$, $Z$ is a finite positive sum of exponential functions. Its free energy $-k_{B}T\log Z$ is a real analytic function for $T>0$, so the smooth heat capacity peak here is not a critical phase transition. Spontaneous magnetization concerns the order of limits: first a thermodynamic limit, then removal of a symmetry-breaking field. The four-spin Ising ring alone establishes neither that limit nor any two-dimensional Ising model result.^[ref]

Differentiate $f_s=|t{|}^{2-\alpha }{g}_{f,\pm }(h/|t{|}^{\Delta })$ with respect to its conjugate field. A finite nonzero one-sided ordered-branch amplitude gives $\beta =2-\alpha -\Delta$. To approach a finite critical isotherm as $t\to 0$ at fixed small $h$, the large-argument function must behave as $g_m(x)\sim C\mathrm{sgn}(x)|x{|}^{1/\delta }$. Cancelling the remaining thermal power yields $\beta =\Delta /\delta$. An analytic background, branch selection and thermodynamic limit must be distinguished from the singular free energy being differentiated.

If $\Omega (E,V,N;\delta E)$ counts states in a narrow experimental energy window, the Boltzmann entropy isChanging the macroscopic resolution $\delta E$ by a fixed factor changes $S$ only by an additive quantity of order $k_B$, whereas $S$ itself is of order $N{k}_B$ in the thermodynamic limit. The dependence on the precise energy-window width is therefore negligible for macroscopic thermodynamics.

The one-body density matrixuses a creation operator and an annihilation operator to describe first-order spatial quantum coherence. Its diagonal is particle-number density and its trace is the expectation value of the particle number operator, $\mathrm{Tr}{G}^{(1)}=⟨N⟩$. Dividing by $⟨N⟩$ instead gives a unit-trace convention, so the normalization matters when interpreting its eigenvalues.

For bosons, an eigenvalue that remains proportional to particle number in the thermodynamic limit identifies Bose-Einstein condensation. This criterion extends occupation of a single-particle mode to interacting systems.^[ref]

In a homogeneous system of volume $V$, an occupied plane-wave mode contributesA constant long-distance limit therefore corresponds to a zero-momentum condensate; a moving condensate can retain a spatial phase. Finite local density or short-distance quantum coherence alone does not establish off-diagonal long-range order.

For $N\ge 2$, the open-chain Ising model coupling $[J_N{]}_{ij}=J{\rho }^{|i-j|}$, $0<\rho <1$, has a tridiagonal inverse. Its endpoint diagonal entries are $1/[J(1-{\rho }^2)]$, its interior diagonal entries are $(1+{\rho }^2)/[J(1-{\rho }^2)]$, and adjacent off-diagonal entries are $-\rho /[J(1-{\rho }^2)]$. Direct matrix multiplication verifies the inverse because the geometric sequences cancel in every interior row. The displayed quadratic form follows with $D=\rho /[J(1-{\rho }^2)]$, $A=(1-\rho )/[J(1+\rho )]$ and $B=\rho /[J(1+\rho )]$. For $\rho =e^{-\kappa }$, $D=(2J\sinh \kappa {)}^{-1}$ and $A=\tanh (\kappa /2)/J$. Omitting the endpoint term requires an infinite chain, a correctly periodized coupling, or a bulk thermodynamic limit.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.

For fixed $H_f>0$, the momentum sum has a finite thermodynamic limit obtained by $L^{-1}{\sum }_k\to (2\pi {)}^{-1}{\int }_{-\pi }^{\pi }dk$. At zero field the uniform mode already has no oscillator vacuum, as explained in part d. Even after removing that mode, the nonzero modes behave asThus assigning a finite ordered moment by simply setting $h=0$ in the infinite momentum sum is invalid. The divergence indicates failure of this expansion about a fixed planar orientation; it is not a negative physical magnetization. At finite size an exact zero-field state chosen to have definite total $S^z$ has zero planar one-point magnetization by the continuous rotation symmetry. Establishing its full correlations requires a treatment beyond the divergent ordered-state approximation.

Let $\beta =1/(k_{B}T)$ and use the two spin values $+1,-1$ as row and column labels. Define the spin-chain transfer matrixIt is a real symmetric matrix because $E_0(s,t)=E_0(t,s)$; every entry is strictly positive for finite bond energies at $T>0$. The Boltzmann factor of a spin configuration factorizes into the product $W_{{\sigma }_1{\sigma }_2}\cdots W_{{\sigma }_L{\sigma }_1}$. Summing the intermediate spin labels is exactly matrix multiplication, and the final closure imposes periodic boundary conditions. ThusWriting $W=\left(\begin{array}{cc}a& b\\ b& d\end{array}\right)$, its eigenvalues areA positive-entry matrix need not have two positive eigenvalues; ${\lambda }_{-}$ can be negative. The Perron–Frobenius theorem ensures ${\lambda }_{+}>|{\lambda }_{-}|$ here. Therefore the Helmholtz free energy is $F_L=-k_{B}T\ln Z_L$, and its value per spin in the thermodynamic limit is $-k_{B}T\ln {\lambda }_{+}$.