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Consider the $(u,r)$ plane at $h=0$ for $V=r{M}^2/2+u{M}^4/4+v{M}^6/6$, with fixed $v>0$. For $u>0$, the continuous boundary is $r=0$: above it the global minimum is zero, and below it there are two symmetry-related ordered pure thermodynamic phases. For $u<0$, the equilibrium boundary is the tricritical three-phase lineAbove this curve the disordered minimum wins; below it the two ordered minima win. On the curve,Its three zeros are exactly $M=0$ and $M=\pm \sqrt{-3u/(4v)}$, proving equality of three global minimum values. Spin inversion symmetry enforces the equality of the two ordered phases automatically, so tuning a single remaining coefficient can produce three-phase coexistence on a line in this symmetry plane. This does not contradict the generic phase-rule counting without symmetry restrictions.

The tricritical point is $(u,r)=(0,0)$, where the discontinuity vanishes, the first-order boundary meets the continuous boundary, and the positive sextic term becomes the leading stabilizer. The local metastability limits $r=0$ and $r=u^2/(4v)$, $u<0$, bracket the first-order curve but are not phase boundaries. In the full $(r,u,h)$ parameter space, the zero-field line is attached to tricritical wings; a nonzero field splits the equality of the two ordered phases.

Use inverse temperature ${\beta }_T=(k_{B}T{)}^{-1}$, reserving $\beta$ without a subscript for the critical exponent. Write $c=qJ$; each undirected bond occurs once, so the number of bonds is $Nq/2$. In the mean-field theory of the Ising model, write each Ising spin as ${\sigma }_i=m+\delta {\sigma }_i$ and neglect the product of fluctuations:The resulting independent-spin statistical Hamiltonian isThe constant corrects the double counting of interaction energy. The Ising spin sums now factorize and can all be evaluated:This is the approximate partition function at an assumed mean field; equilibrium fixes $m$ self-consistently. The corresponding Ising auxiliary mean-field free energy isIts equilibrium value gives the mean-field Helmholtz free energy in the imposed field. Away from a stationary point its parameter $m$ is an assumed field variable, not necessarily the actual mean Ising spin of that independent-spin distribution.

For the full small-$m$ expansion at fixed $h$, set $t_h=\tanh ({\beta }_{T}h)$ and $s_h={\mathrm{sech}}^2({\beta }_{T}h)$. Differentiating $\log \cosh z$ gives successive derivatives $t_h$, $s_h$, $-2t_h{s}_h$, and $2s_h(2-3s_h)$ at $z={\beta }_{T}h$. ThereforeAt zero field, spin inversion symmetry eliminates odd powers and this simplifies toThe quadratic coefficient changes sign and the quartic coefficient is positive atThus the zero-field mean-field prediction is a continuous, continuous phase transition: the stable zero spin magnetization develops two symmetry-related nonzero minima continuously below $T_c$.

Both requested routes give the same mean-field self-consistency equation. First, the one-spin expectation in the effective field isSecond, differentiating the Ising auxiliary mean-field free energy giveswhose stationary condition is precisely that mean-field self-consistency equation. Choose its stable, lowest-free-energy branch rather than every algebraic solution.

An equally useful mean-field approximation parametrizes the trial distribution by its actual mean Ising spin. Its probabilities are $(1\pm m)/2$, giving the Bragg-Williams free energy of the Ising modelThis energy-minus-entropy function has expansionIts stationary equation is $h=k_{B}T\mathrm{arctanh}m-cm$, again equivalent to the mean-field self-consistency equation. The two functions differ away from equilibrium, but agree on stationary branches: for $z={\beta }_T(h+cm)$ and $m=\tanh z$, the entropy bracket equals $mz-\log (2\cosh z)$. Their small-$m$ coefficients therefore need not agree at arbitrary temperature; at $T_c$ they have the same leading critical quartic coefficient $c/12$. This distinction prevents confusing the auxiliary-field expansion with the physical-magnetization variational expansion.

For the order-parameter critical exponent, expand the equation of state at $h=0$:On a nonzero stable branch,Hence $\beta =1/2$. The spontaneous magnetization is understood by selecting a branch with an infinitesimal field after the thermodynamic limit; a finite symmetric sample has zero exact zero-field mean Ising spin.

For the magnetic susceptibility per site, define $\chi =\partial m/\partial h$ with $h$ in energy units. Implicit differentiation givesAbove $T_c$, $m=0$ and $\chi =1/[k_B(T-T_c)]$. Below $T_c$, evaluated on a selected ordered branch, the expansion of $m^2$ gives $\chi \sim 1/[2k_B(T_c-T)]$. Thus the magnetic-susceptibility critical exponent is $\gamma =1$ on both sides, with different amplitudes. At $T=T_c$, the equation of state becomesso the critical-isotherm exponent is $\delta =3$. These are mean-field critical exponents, not a claim that neglecting fluctuations gives the exact Ising transition in every dimension.

Use dimensionless interaction couplings in the exponent of the Boltzmann weight. In the following formulas $K,L$ denote the physical interaction energies divided by $k_{B}T$ if that factor has not already been absorbed into the displayed Hamiltonian. This convention is required for temperature-independent numerical fixed-point coordinates.

Split the original square lattice into retained sites $E$ with even $x+y$ and eliminated sites $O$ with odd $x+y$, where coordinates are in units of $a$. Every nearest-neighbour bond joins $E$ to $O$, whereas a diagonal next-nearest-neighbour bond stays within one sublattice. Denote retained Ising spins by $s_e$ and eliminated Ising spins by ${\eta }_o$. For fixed retained Ising spins defineThe blocked Boltzmann weight in the checkerboard decimation of the square-lattice Ising model isUse the independent uniform product measure on Ising spins on $O$ to perform a cumulant expansion. Its one-spin mean vanishes, and $⟨{\eta }_o{\eta }_{o^{\prime }}{⟩}_0={\delta }_{o{o}^{\prime }}$. Consequently the first-order $K$ term, the first-order eliminated-sublattice $L$ term, and the mixed $KL$ term vanish. To the requested order,Here the truncation keeps $K^2$ and the linear term in $L$; equivalently count $L$ as order $K^2$. At higher orders the interaction family will generally not close.

Each eliminated site has four retained neighbours, soIts six distinct neighbour pairs comprise four corner pairs at distance $\sqrt{2}a$ and two opposite pairs at distance $2a$. A given retained diagonal pair shares two eliminated centres; a given opposite pair shares only one. Thus the generated pair couplings are $2K^2$ for diagonal pairs and $K^2$ for axial pairs separated by two original spacings. The original $L$ interaction directly contributes to each retained diagonal pair. No single-spin term survives spin inversion symmetry, and no three-spin or four-spin term is present in this second cumulant. This proves closure on exactly two nonconstant interactions at the retained order.

The primitive vectors of the retained square lattice are $a(1,1)$ and $a(1,-1)$. Rotate the axes by $45$ degrees and rescale lengths by $1/\sqrt{2}$; retained diagonal pairs become nearest neighbours, and the axial distance-$2a$ pairs become next-nearest neighbours. With $b=\sqrt{2}$, the leading checkerboard Ising decimation recursion is thereforeThe constant part of the weight is also determined: $4(K^2/2)(N/2)=N{K}^2$, so the original partition function equalsto the same order in the logarithm of the weights. That normalization changes free energy but not the retained-spin interaction operators. If physical energy couplings are kept instead, at fixed ${\beta }_T$ the relations read $K_{\mathrm{energy}}^{\prime }=L_{\mathrm{energy}}+2{\beta }_T{K}_{\mathrm{energy}}^2$ and $L_{\mathrm{energy}}^{\prime }={\beta }_T{K}_{\mathrm{energy}}^2$.

The fixed-point equations are $L_{\ast }=K_{\ast }^2$ and $K_{\ast }=3K_{\ast }^2$. Their only finite real solutions areThe origin is the high-temperature fixed point; its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, with both multipliers zero. The nontrivial candidate critical point is $(1/3,1/9)$. Linearize the discrete renormalization-group transformation there:The characteristic polynomial is ${\lambda }^2-(4/3)\lambda -2/3$, givingEigenvectors may be chosen as $(1,2/(3{\lambda }_{\pm }){)}^T$. The positive eigenvalue greater than one is the relevant direction of a fixed point; the other is irrelevant because its magnitude is less than one. Its negative sign only alternates the sign of the perturbation between steps. For a discrete map the magnitude relative to one decides relevance, not the sign rule used for continuous beta-function eigenvalues.

Let $t$ be the relevant thermal scaling coordinate, so $t^{\prime }={\lambda }_{+}t$ to first order. A correlation length measured in restored lattice units satisfies $\xi (t^{\prime })=\xi (t)/b$. If $\xi (t)\sim |t{|}^{-\nu }$, then ${\lambda }_{+}^{-\nu }=b^{-1}$. Hence the thermal exponent from a discrete renormalization map isThis is the value predicted by the stated low-order real-space renormalization group, whose nontrivial fixed point is evaluated beyond the strictly infinitesimal-coupling limit. It is an approximation, not an exact solution of the full square-lattice Ising model.

Use the positive exponent in the PDF. The TeX introduces a minus sign in the numerator while retaining the positive-sign normalizing constant, which would not define the stated probability distribution. The Curie–Weiss model assigns equal weights to $s$ and $-s$, since $({\sum }_j{s}_j{)}^2$ is unchanged by simultaneous reversal. This spin inversion symmetry pairs every configuration with one having the opposite value of $S_i$, so

Take the couplings in the Hamiltonian to include inverse temperature, so the Boltzmann factor is $e^{-H}$. If physical energies are used instead, first replace $J,L,D,K$ by their products with $1/(k_{B}T)$. Split the on-site term equally between its two bonds. For spin states $s,t\in \{1,0,-1\}$ the spin-chain transfer matrix has entriesUsing the specified coupling coordinates, in the order $(1,0,-1)$ this isFor example, $e^{D/2}=x/z$ and $e^{-J+L+D}=y/z$. This is the nearest-neighbour Blume–Emery–Griffiths model with an additive constant and the paper's sign convention for $D$.

Summing the periodic spin chain gives the partition function $Z_N=\mathrm{tr}{W}^N={\sum }_{a=1}^3{\lambda }_a^N$. For finite real couplings all entries of $W$ are strictly positive; the Perron–Frobenius theorem gives a unique positive dominant eigenvalue with ${\lambda }_1>|{\lambda }_a|$ for $a\ne 1$. Since $W$ is a symmetric matrix, all its eigenvalues are real. Thus in the thermodynamic limit,The positivity condition is stronger and more useful than mere ordering by signed value: subdominant eigenvalues can be negative. Also, the printed strict ordering between the other two is not guaranteed for all couplings. For instance, $J=L=D=0$ makes $W$ the positive constant matrix $e^{K}11^T$ with two equal zero eigenvalues. Degeneracy there does not affect the largest-eigenvalue limit; no explicit generic eigenvalues are needed.

For spin magnetization, introduce a dimensionless field $h$ through $-h{\sum }_i{\sigma }_i$, or insert $S=\mathrm{diag}(1,0,-1)$ into the trace. Thenwhere $v_1$ is a normalized eigenvector of the largest eigenvalue and $W_{st}(h)=e^{h(s+t)/2}{W}_{st}(0)$. Spin inversion symmetry gives $v_{1,+}=v_{1,-}$, since the positive eigenvector of the largest eigenvalue is unique. Thereforeat zero field, both at finite $N$ and in the finite-coupling thermodynamic limit. The finite-$N$ result follows directly by pairing each configuration with its spin-reversed partner. Eigenvalues at a single fixed field do not determine a general observable: a field derivative of the largest eigenvalue, or its eigenvector, is required. Positivity and the real analytic dependence on the couplings exclude a finite-temperature spontaneous symmetry breaking transition in this one-dimensional finite-range chain; singular zero-temperature coupling limits require separate treatment.

For even $N$, spin decimation on alternate sites sums the middle spin of each two-bond segment, hence the coarse spin-chain transfer matrix is $W^{\prime }=W^2$. DefineDirect multiplication gives $W^2=e^{2K}{z}^{-2}\left(\begin{array}{ccc}A& B& C\\ B& E& B\\ C& B& A\end{array}\right)$. Matching its entry ratios to the original parameterization yields the spin-1 chain decimation recursionThe remaining overall positive factor is absorbed into $K^{\prime }$. Keeping that factor preserves the free energy as well as normalized spin probabilities. Indeed $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$, exactly; on an odd ring an unmatched boundary segment needs separate handling rather than assuming a uniform two-site block decomposition.

For the isotropic zero-field classical Heisenberg model of a ferromagnet, the analytic vector Landau expansion uses the Cartesian components of the vector order parameter near $\mathbf{m}=0$. rotational symmetry makes its scalar terms functions of $\mathbf{m}\cdot \mathbf{m}$. Global spin inversion symmetry also requires $f(\mathbf{m})=f(-\mathbf{m})$. Through fourth order this givesA positive quartic coefficient stabilizes the unconstrained truncated free energy. The disordered phase has $b>0$ and its minimum at $m=0$; the aligned phase has $b<0$ and a minimum at nonzero $m$. At the continuous transition $b=0$. The attractive alignment interaction supplies an energetic tendency toward negative quadratic curvature; orientational entropy favors the disordered phase at high temperature, so the sign of $b$ is a free-energy question rather than a consequence of the pair energy alone.

The analyticity assumption matters: the variable printed as $m$ is a nonnegative magnitude, so reversing $\mathbf{m}$ does not send that scalar to $-m$. Terms proportional to $|\mathbf{m}|$ or $|\mathbf{m}{|}^3$ would be rotationally invariant but nonanalytic at the vector origin. Standard analytic Landau theory excludes them. Without that assumption, the stated pair symmetry alone would not exclude odd powers of the magnitude.^[ref]

At zero field, write the right-hand side of the self-consistency equation as $G_T(S)=g(4\beta JS)$, with $g(x)=2\sinh x/[1+2\cosh x]$. The origin is always a solution, and the slope there isThe critical crossing occurs when this equals one, givingThis is a continuous phase transition in the mean-field approximation. For $x>0$,so the positive branch is a strictly concave function, while $g$ is odd and bounded by one. For $T\ge T_c$, the slope at zero is at most one and strict concavity puts $G_T(S)$ below $S$ for every $S>0$, leaving only the unmagnetized solution. For $T<T_c$, it initially lies above $S$ but has $G_T(1)<1$; hence there is one positive intersection and, by spin inversion symmetry, one negative intersection.

The curvature of the Helmholtz free energy in the mean-field approximation is $4J[1-G_T^{\prime }(S)]$ per site. It is negative at $S=0$ below $T_c$ and positive at the nonzero intersections, identifying the ordered minima and the unstable zero solution.

The critical temperature and branches here belong to the mean-field approximation. In the exact finite system, spin inversion symmetry forces zero average magnetization at zero field; selecting a broken-symmetry phase requires the thermodynamic limit with a vanishing field of specified sign.

The small-argument Taylor series of the single-site response isLet $K=4\beta J$. For a small nonzero order parameter at zero field, the self-consistency equation becomesSince $K_c=3/2$ and $K=K_c/(1+t)$, where $t=T/T_c-1$, this givesConsequentlyThe positive branch is the expression printed in the question, while its negative partner is required by spin inversion symmetry. The square-root behavior gives the mean-field critical exponent $1/2$ for the spontaneous magnetization.

The coarse-grained order parameter is a real scalar field because the microscopic spin has one component. At $B=0$, spin inversion symmetry takes $m\mapsto -m$, so an analytic local Landau free energy contains only even powers. Near a continuous phase transition, the quadratic coefficient changes sign linearly, represented by $a(T-T_c)$ with $a>0$. A positive quartic coefficient stabilizes the density when the quadratic term becomes negative, giving the two minima $m=\pm \sqrt{a(T_c-T)/b}$ below $T_c$.

The ferromagnetic interaction favours neighbouring spins pointing alike, so spatial variation costs energy; its leading long-wavelength form is the positive gradient term $c({\partial }_{x}m{)}^2$. The source term $-Bm$ favours alignment with the applied magnetic field and explicitly breaks spin inversion symmetry at fixed nonzero $B$. Finally $f_0(T)$ is the spin-independent background free-energy density.

The quartic sign can also be checked from part (b): its expansion containswhose quartic coefficient is positive. This is consistent with the proposed Landau-Ginzburg theory, not a tricritical truncation requiring a stabilizing sixth-order term. The microscopic square lattice is two-dimensional; the written gradient describes a profile depending on one coordinate. For general profiles, replace it by $c[({\partial }_{x}m{)}^2+({\partial }_{y}m{)}^2]$.

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.

For the spin inversion symmetry parameterization $W=c\left(\begin{array}{ccc}1& x& y\\ x& z& x\\ y& x& 1\end{array}\right)$ with positive entries, spin decimation gives the same form with $x^{\prime }=x(1+y+z)/(1+x^2+y^2)$, $y^{\prime }=(x^2+2y)/(1+x^2+y^2)$ and $z^{\prime }=(z^2+2x^2)/(1+x^2+y^2)$. Multiply $W$ by itself and divide by its $(+,+)$ entry to prove these ratios. The leftover scalar is an additive free-energy coupling; dropping it leaves normalized expectations unchanged but loses the full free energy.

At zero field and with spin inversion symmetry in its boundary conditions, a finite Ising model pairs every configuration with its spin-reversed configuration. Their equal Boltzmann weights force the expected value of its spin magnetization $m=N^{-1}{\sum }_i{s}_i$ to vanish. Below a model's ordering critical temperature, an arbitrarily small positive field can instead select the positive spontaneous magnetization if size is taken to infinity first:The distinction is the order of limits and the stated symmetric boundary conditions; plus boundary conditions already break spin inversion symmetry. A zero-field symmetric mixture of the two ordered states can retain zero mean spin magnetization even in the thermodynamic limit. Also, $⟨|m|⟩>0$ in a finite system is not by itself proof of spontaneous symmetry breaking.^[ref]

A thermodynamic phase is a form of macroscopic equilibrium characterized by its structure, symmetries and thermodynamic response. Away from phase transitions, its free-energy density varies smoothly with the control parameters. Different phases can share a phase coexistence boundary. An ordered magnetic phase is distinguished from a disordered phase by its spontaneous order parameter; positive and negative ordered states are different pure thermodynamic phases related by spin inversion symmetry.