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For spins $s_i\in \{-1,0,1\}$ with ferromagnetic nearest-neighbour exchange and no single-ion anisotropy, a homogeneous mean-field approximation givesThe self-consistency equation is $S=2\sinh h/(1+2\cosh h)$ with $h=\beta (zJS+B)$. Its small-$h$ expansion is $2h/3-h^3/9+O(h^5)$. Hence $k_B{T}_c=2zJ/3$ and, with $t=T/T_c-1$, the nonzero branches satisfy $S^2=(8/3)(-t)+O(t^2)$. The critical exponent is the mean-field value $1/2$, not an exact two-dimensional exponent. At zero source the finite-system equilibrium magnetization is zero by spin inversion symmetry; the broken branches describe the thermodynamic limit with a phase selected by a vanishing source.

The zero-field ferromagnetic Curie–Weiss model is an Ising model on $n$ spins with an equal interaction between every pair. For $s_i\in \{-1,1\}$ and $\beta >0$, its normalized probability mass function is $Z_{n,\beta }^{-1}\exp [\beta ({\sum }_i{s}_i{)}^2/(2n)]$. The partition function $Z_{n,\beta }$ sums these weights over all configurations. The square includes constant diagonal terms; these affect only normalization. Spin inversion symmetry makes every spin's expected value zero, while grouping configurations by their spin magnetization givesHere $m=-1+2k/n$, $0\le k\le n$.

The Edwards-Anderson order parameter measures squared local spin magnetization within a selected equilibrium state $\alpha$. Squaring prevents opposite local orientations from cancelling. The state selection and limits matter: the full finite, zero-field canonical ensemble has $⟨s_i⟩=0$ by spin inversion symmetry, so substituting that average would miss broken-state order.^[ref]

Finite-size rounding replaces a limiting discontinuity or divergence by a finite crossover or peak. For a spatial system near a continuous phase transition, the bulk description ceases to apply unchanged when the correlation length becomes comparable with the sample size. How the crossover width and location depend on size also depends on the model, geometry and boundary conditions; there is no universal $1/N$ rule for every phase transition.

An exact finite example is the Curie–Weiss model with $s_i=\pm 1$, $J>0$, and a conjugate field $h$ in energy units. Set $K=J/(k_{B}T)$ and $u=h/(k_{B}T)$. Its statistical Hamiltonian and finite partition function areThe binomial coefficient counts the microstates with $k$ positive Ising spins. The diagonal terms in the square add a configuration-independent constant to the energy; retaining them fixes the normalization used here. Divide each summand by $Z_N$ to obtain the probability mass function of $m_k=(2k-N)/N$. In particular,The variance expression makes the finite response nonnegative, while spin inversion symmetry gives $⟨m⟩=0$ exactly at $u=0$ for every plotted $N$.

Every curve is computed from the finite partition function, without selecting one minimum of the Bragg-Williams free energy of the Ising model. The field panel holds $K=1.5$ fixed and shows increasingly rapid switching between opposite spin magnetizations. The temperature panel uses $u=0$: its nonzero $⟨|m|⟩$ measures the typical magnitude, while the mean spin magnetization stays zero. The reference $k_{B}T/J=1$ marks the limiting critical temperature, not a singular point of any finite curve.

A short limiting argument explains the field panel. Combining the binomial coefficient with its leading entropy gives a Helmholtz free energy per spin whose stationarity equation is $m=\tanh (Km+u)$. For $K>1$ and $u=0$ the two ordered minima have $m=\pm m_{\ast }$, where $m_{\ast }=\tanh (K{m}_{\ast })$. For large $N$ well below the critical temperature, their relative Boltzmann weight is approximately $e^{2Nu{m}_{\ast }}$, giving $⟨m⟩\approx m_{\ast }\tanh (Nu{m}_{\ast })$. Thus this two-peak approximation has a field width of order $1/(N{m}_{\ast })$ in $u$. It is not valid at the critical temperature, where the peaks merge; the figure uses the exact finite sum throughout.

Take three Ising spins $s_i\in \{-1,1\}$, with an antiferromagnetic interaction of energy scale $J$ on each edge of a triangle. An individually satisfied bond has $s_i{s}_j=-1$. But multiplying all three bond products gives $(s_1{s}_2)(s_2{s}_3)(s_3{s}_1)=1$, whereas three satisfied bonds would give $-1$. At least one bond must therefore remain unsatisfied. This is an exactly finite example of geometrical frustration.

There are six ground states with one unsatisfied bond and ground-state energy $-J$. The two all-aligned microstates have energy $3J$. No interaction is switched off in a ground state: the system minimizes their sum by accepting one local disappointment. Allowing continuous three-component spins instead of Ising spins changes this optimization problem and permits the $120$-degree arrangement described under geometrical frustration.

Writing $\beta =1/(k_{B}T)$ for inverse temperature, the exact partition function and mean internal energy areAt $T>0$ these expressions are smooth: a finite triangle has no thermodynamic phase transition. In zero magnetic field, spin inversion symmetry also gives $⟨s_i⟩=0$ at every site, even as the Boltzmann distribution concentrates on the six frustrated ground states.

The Ising ring bond correlation is the neighboring two-point correlation function $c_1=⟨{\sigma }_i{\sigma }_{i+1}⟩$ in the four-spin Ising ring. All four edges are equivalent. At zero field each mean Ising spin is zero by spin inversion symmetry, so $c_1$ also equals the connected correlation function. Agreement gives bond product $+1$ and disagreement $-1$; thus $P({\sigma }_i={\sigma }_{i+1})=(1+c_1)/2$.

Differentiating the finite partition function with respect to $K=\beta J$ inserts the sum of four bond products. ThereforeHere $U$ is internal energy. At $K=(\ln 2)/4$, the class-weight calculation gives $c_1=(4-1)/17=3/17$ and $U=-12J/17$. The agreement probability for a selected bond is $10/17$, despite the exactly zero mean spin magnetization.

A second derivative gives the energy variance, and hence the heat capacity $C_H=dU/dT$ at fixed $J$ and $h=0$:To see the variance identity directly, a derivative of an expected value under weights $e^{KS}$ inserts $S-⟨S⟩$, where $S={\sum }_i{\sigma }_i{\sigma }_{i+1}=-H/J$. The factor $K^2$ follows from $dK/dT=-K/T$. At the same exact example, $\mathrm{Var}(H)/J^2=1216/289$. The heat capacity vanishes both as $K\to 0$ and as $K\to \infty$: at high temperature its dimensionless factor $K^2$ vanishes, while at low temperature excited microstates become exponentially rare.

An odd periodic nearest-neighbour Ising model with antiferromagnetic coupling has unavoidable geometrical frustration. For $k$ parallel bonds, $H=-NJ+2Jk$, while ${\prod }_j{S}_j{S}_{j+1}=1$ forces $N-k$ to be even. Thus odd $N$ requires at least one parallel bond. A ground state has exactly one, with $N$ possible locations and two global spin inversion symmetry choices, proving the displayed ground-state degeneracy and ground-state energy. The zero-temperature canonical ensemble distributes that defect uniformly. A directed arc of $n$ bonds contains it with probability $n/N$, giving two-point correlation function $(-1{)}^n(1-2n/N)$. The finite antiferromagnetic Ising-ring correlation gives the corresponding positive-temperature result.

Let $M=L/2$, so that the total number of Ising spins is $3M$. Write $K=\beta J$ and group each triangle's two base spins and its apex into one cell. The Ising chain with alternating triangular decorations has a triangle bond followed by an undecorated connecting bond. Summing an apex Ising spin first gives a decorated-bond Boltzmann weightThe undecorated bridge has Boltzmann weight $B_{t{s}^{\prime }}=e^{-Kt{s}^{\prime }}$. In the basis $+1,-1$,Each cell contributes $T=AB$, so periodic boundary conditions giveThe two matrices have common normalized eigenvectors $(1,1{)}^{\mathsf{T}}/\sqrt{2}$ and $(1,-1{)}^{\mathsf{T}}/\sqrt{2}$, expressing spin inversion symmetry. Their respective eigenvalues areConsequently the cell transfer matrix has eigenvaluesand the finite-size partition function isFor $K>0$, ${\lambda }_{-}\ge 0$ and ${\lambda }_{+}-{\lambda }_{-}=4+2e^{-2K}+2e^{2K}>0$, so only ${\lambda }_{+}$ survives in the thermodynamic limit. Divide the Helmholtz free energy by $3M$, the number of sites rather than the number of cells or base spins:As a counting check, $K=0$ gives ${\lambda }_{+}=8$, ${\lambda }_{-}=0$, hence $Z_L=8^M=2^{3M}$, exactly the number of independent Ising spin configurations. This model has separated decorated bonds and should not be confused with the corner-sharing sawtooth Ising chain, which has a different site count.

Each neighboring pair of Ising spins contributes at least $-J$ to the energy, with equality exactly when the two Ising spins are opposite. For even $N$, alternation closes consistently under the periodic boundary conditions. The ground states are thereforeHere $g_0$ is the ground-state degeneracy. Choosing $S_1$ fixes every other Ising spin, so these are all the ground states; translations do not give additional spin configurations. The two ground states are exchanged by spin inversion symmetry.

Write $K=\beta J$, where $\beta =1/(k_{B}T)$ is the inverse temperature, $T$ is temperature and $k_B$ is the Boltzmann constant. In the canonical ensemble, each spin configuration has probability $e^{-\beta H}/Z_N$. The nearest-neighbor Boltzmann factors multiply, so the spin-chain transfer matrix, in the order $+1,-1$, isMultiplying its entries around the ring and summing over every Ising spin yields the partition function:The trace implements the periodic boundary conditions; replacing it by an endpoint sum would describe a different partition function. The normalized eigenvectors and corresponding eigenvalues areThe negative eigenvalue is essential for the antiferromagnetic Ising model. Introduce their ratioThe exact finite-$N$ partition function and Helmholtz free energy areFor finite $K>0$, $|t|<1$, so $Z_N>0$ for both even and odd $N$. The last term retains the finite-ring effect of geometric frustration.

To insert an Ising spin into this trace, use the diagonal matrix $\mathsf{S}=\mathrm{diag}(1,-1)$. In the eigenbasis of $\mathsf{T}$, it exchanges $|+⟩$ and $|-⟩$, and hence has vanishing diagonal entries. The one-point correlation function is thereforeThe same conclusion follows directly from spin inversion symmetry: reversing every Ising spin preserves the energy and the Boltzmann factor while reversing $S_1$.

For the two-point correlation function, the two insertions divide the ring into paths of $n$ and $N-n$ bonds. ThusIn particular, the exact finite-ring antiferromagnetic Ising correlation isBecause the one-point correlation functions vanish, this is also the connected correlation function. It gives $C_0=1$ and $C_n=C_{N-n}$, as required by $S_1^2=1$ and exchanging the two paths around the ring. At $K=0$, distinct Ising spins are independent and $C_n=0$ for $1\le n\le N-1$; $C_0=1$ is understood separately if a power $0^0$ occurs in the displayed expression.

Let $k$ count parallel bonds. Then $N-k$ bonds are antiparallel and the energy isThe periodic boundary conditions imposeHence $N-k$ must be even. For odd $N$, $k$ is odd, so its minimum is one: one bond must be parallel. All other bonds alternate. Choosing that bond in $N$ ways and then choosing either value of $S_1$ determines every Ising spin, givingThis is an odd antiferromagnetic Ising ring: geometrical frustration forces one defect, whose location is unrestricted. The ground-state degeneracy includes both its possible locations and global spin inversion symmetry. The formula also includes the self-bond case $N=1$ under the stated Hamiltonian convention.

For $J>0$ each bond minimizes its energy when its two Ising spins are opposite, giving bond energy $-J$. An even ring allows every bond to do this simultaneously, including the closing bond required by the periodic boundary conditions. Its two ground states areThe two configurations are related by spin inversion symmetry. There is no geometrical frustration.

Set $K=\beta J$. The transfer matrix for the one-dimensional Ising model assigns a Boltzmann factor to each bond. In the Ising spin order $(+1,-1)$, it isMultiplying these bond factors and summing every Ising spin implements the canonical ensemble. The periodic boundary conditions close the product into a trace:The eigenvectors $|+⟩=(1,1{)}^T/\sqrt{2}$ and $|-⟩=(1,-1{)}^T/\sqrt{2}$ form an orthonormal basis, with eigenvaluesThe negative second eigenvalue encodes alternating correlations. Defining $q={\lambda }_{-}/{\lambda }_{+}=-\tanh K$, the exact partition function and total Helmholtz free energy areThe parity-dependent term must be retained for finite $N$; in particular it nearly cancels the leading term of $Z_N$ at low temperature when $N$ is odd.

Represent insertion of one Ising spin by $\Sigma =\mathrm{diag}(1,-1)$. In the eigenbasis of $T$, this matrix swaps $|+⟩$ and $|-⟩$. Its diagonal entries in that eigenbasis vanish, giving the exact one-point correlation functionEquivalently, configurations pair with their opposites under spin inversion symmetry, with the same Boltzmann factor and opposite $S_1$.

Two insertions separated by $n$ bonds give the two-point correlation functionThus the exact finite antiferromagnetic Ising-ring correlation isThe two numerator terms represent the two arcs between the sites. The formula has $C_N(0)=1$ and $C_N(n)=C_N(N-n)$, as required by the periodic boundary conditions. Since the one-point correlation function is zero, this is also the connected correlation function. At $K=0$, every Ising spin is independent, so $C_N(n)=0$ for distinct sites and $C_N(0)=1$; powers with exponent zero are understood as one.

Put $r=\tanh K$, so $r\to 1^{-}$ as the temperature tends to zero. For even $N$, the finite antiferromagnetic Ising-ring correlation becomesThe two perfectly alternating ground states give exactly the same product $S_1{S}_{n+1}=(-1{)}^n$; averaging them preserves its magnitude even though the one-point correlation function remains zero.

For odd $N$ the signs instead giveBoth numerator and denominator vanish. Set $r=1-\delta$ and expand at fixed $N,n$:HenceThere is a direct ground state interpretation. The mandatory parallel bond of an odd antiferromagnetic Ising ring is uniformly distributed among its $N$ bonds in the limiting canonical ensemble. Along the directed $n$-bond arc from site 1 to site $n+1$, the product is $(-1{)}^n$ if the defect lies outside that arc, and $-(-1{)}^n$ if it lies inside. The respective probabilities are $1-n/N$ and $n/N$, giving the displayed two-point correlation function. Global spin inversion symmetry doubles the number of ground states but leaves each product unchanged.

The reduction on an odd ring survives at separations comparable with $N$, despite the vanishing defect density $1/N$. For fixed $n$, however, taking $N\to \infty$ after the zero-temperature limit recovers $(-1{)}^n$ for either parity, consistently with the diverging correlation length. The linear envelope at finite odd $N$ comes from one freely located frustration defect; it is distinct from the exponential decay caused by thermal defects in the thermodynamic limit.

An order parameter distinguishes equilibrium thermodynamic phases. For a scalar magnetic system, a slowly varying coarse-grained scalar field $\varphi (x)$ represents local magnetization, and its uniform expectation $M=⟨\varphi ⟩$ is the order parameter. Under spin inversion symmetry, $\varphi \mapsto -\varphi$ and $h\mapsto -h$. The disordered phase has $M=0$ at $h=0$, while a pure ordered phase has $M\ne 0$. In a finite symmetric system the average still vanishes at $h=0$; spontaneous symmetry breaking means taking the thermodynamic limit before $h\to 0^{+}$ or $h\to 0^{-}$ to select a pure thermodynamic phase.

Landau-Ginzburg theory assigns a local free-energy functionalSpin inversion symmetry forbids odd powers at zero conjugate field; locality and slow variation motivate the gradient expansion. The gradient energy penalizes interfaces and determines long-wavelength response. Integrating $e^{-\mathcal{F}/(k_{B}T)}$ over fields includes fluctuations, whereas the Landau approximation replaces that integral by its stable stationary configuration. For a uniform phase this reduces to minimizing $V(M)$. If $f$ is the equilibrium free-energy density, then $M=-\partial f/\partial h$, so the order parameter is also the thermodynamic response to its conjugate field.

An order parameter distinguishes phases macroscopically. In an Ising model it is the local coarse-grained magnetization; the uniform mean is $M=V^{-1}\int \varphi (x)d^{D}x$. In a selected ordered phase $M\ne 0$, whereas the disordered phase has $M=0$ at zero conjugate field. The ordered signs are related by spin inversion symmetry. A finite zero-field system may average both signs to zero, so spontaneous symmetry breaking is defined using the thermodynamic limit before removing a symmetry-selecting field. For other systems the order parameter can be a vector, a density difference, or a tensor rather than a single magnetization.

Landau-Ginzburg theory makes the order parameter a spatial field and expands a local free energy consistent with the symmetries. For a short-range scalar system with spin inversion symmetry at $h=0$, a useful expression isThe linear source is the field conjugate to an order parameter. Odd terms at zero field are forbidden by spin inversion symmetry. The coefficients depend smoothly on control parameters away from singular fluctuation effects. With $u>0$ the quartic term suffices locally; if $u<0$, take $v>0$ for stability. The Landau approximation minimizes this functional; statistical field theory also integrates fluctuations about those minima.

An order parameter distinguishes phases by a macroscopic quantity that changes at the transition. For an Ising model ferromagnet it is the magnetization per Ising spin, $M=N^{-1}{\sum }_n⟨{\sigma }_n⟩$. At zero magnetic field the statistical Hamiltonian is invariant under spin inversion symmetry. The disordered phase preserves this symmetry, whereas a selected ordered phase has $M\ne 0$ and a spin-reversed partner with $-M$. At finite volume with symmetric boundary conditions the expected value is zero even below the transition; spontaneous magnetization means taking the thermodynamic limit before sending a symmetry-breaking field to zero.

Landau-Ginzburg theory introduces a coarse-grained local order parameter $\varphi (x)$ and a symmetry-constrained free-energy functional. For a short-range system with scalar spin inversion symmetry, a useful expansion isThe coefficients vary smoothly with microscopic control parameters at the level of the Landau approximation; normally $r=A(T-T_c)$ with $A>0$. The gradient term penalizes spatial variation. The magnetic field is conjugate to the order parameter, and odd powers at $h=0$ are excluded by spin inversion symmetry. With $u>0$, the sextic term may be omitted locally; with $u<0$, a stabilizing positive $v$ is essential. The equilibrium uniform order parameter minimizes the local potential $V(M)$, givingThis equation of state follows from minimization, rather than being imposed as an independent assumption. Vector or tensor order parameters and their allowed invariants describe other symmetry classes.

An order parameter distinguishes phases by an equilibrium expectation value. For an Ising model, the local field $M(x)$ is the coarse-grained magnetization, and the uniform value is $M=N^{-1}⟨{\sum }_r{\sigma }_r⟩$. The spin inversion symmetry maps $M$ to $-M$, so at zero field the Landau free energy contains even powers. Its disordered minimum is $M=0$; ordered pure thermodynamic phases have $M=\pm M_0$. Their sign is selected by taking the thermodynamic limit before sending $h$ to zero. A symmetric mixture can have zero average without being a single disordered phase. The term $-hM$ identifies $h$ as the field conjugate to an order parameter.

For a ferromagnet the scalar order parameter is the spontaneous spin magnetization. It distinguishes a symmetry-invariant phase with zero mean from one of two phases with a nonzero mean of opposite signs. At zero field, a finite system with spin inversion symmetry has exactly zero mean. The spontaneous value is defined by taking the thermodynamic limit first and then sending the symmetry-breaking conjugate field to zero from one side.

A Landau-Ginzburg theory promotes the coarse magnetization to a scalar field $M(x)$ and usesAt $h=0$, spin inversion symmetry permits only even local powers. The gradient term penalizes spatial variations. The equilibrium statistical weight is $e^{-\mathcal{F}/(k_{B}T)}$; the Landau approximation chooses a stable minimum of $\mathcal{F}$, whereas the full functional integral retains fluctuations around it. In a homogeneous phase this gives the equation of state $h=rM+u{M}^3+v{M}^5$. The field $h$ is conjugate to the order parameter, with $M=-\partial f_{\mathrm{eq}}/\partial h$.

In Landau-Ginzburg theory, a scalar order parameter $M$ distinguishes macroscopic phases: for an Ising model it is the magnetization per site, while near a liquid–gas endpoint it can be a shifted density. An ordered pure thermodynamic phase has a nonzero value even when the conjugate field is removed. For a symmetry-related pair, take the thermodynamic limit before $h\to 0^{\pm }$; a finite-volume symmetric ensemble instead has zero average.

Coarse-grain to a slowly varying real field and use a local, symmetry-constrained free-energy density. A useful convention isFor spin inversion symmetry, odd powers other than the externally imposed source are absent. Take $v>0$ when the quartic coefficient can be negative; for an ordinary transition $u>0$ already stabilizes the quartic truncation. The gradient energy penalizes spatial variations. In the Landau approximation, minimize this functional, usually first over constant fields. The corresponding statistical field theory also integrates fluctuations with weight $e^{-\mathcal{F}/(k_{B}T)}$. The order parameter identifies the phase; its spatial field describes both equilibrium order and its fluctuations.