08.13.07 · stat-mech / exactly-solved-models

The spherical model (Berlin-Kac)

shipped3 tiersLean: none

Anchor (Master): Baxter Ch. 5 §5.5–5.7; Berlin–Kac 1952 Phys. Rev. 86:821; dimensionality dependence of criticality and the Gaussian-model connection

Intuition Beginner

The Ising model asks each site of a lattice to hold a spin that is exactly $+ 1$ or exactly $- 1$ . That hard either-or choice is what makes the model so difficult: you cannot smoothly slide a spin from one value to the other, so the sum over configurations is a sum over corners of a high-dimensional cube. The spherical model keeps the spirit of the Ising model but trades the rigid per-site rule for one global rule. It lets every spin be any real number, and asks only that the whole collection sit on a sphere: the sum of the squares of all the spins must equal the number of sites.

Why bother? Because this one change turns an intractable sum into an integral that can be done by hand, in any number of dimensions. You give up the sharp per-site constraint, but you gain an exact answer everywhere, including in three dimensions, where the Ising model has never been solved in closed form. The spherical model becomes a clean laboratory for one of the deepest facts about phase transitions: how the dimension of space decides whether a magnet can order at all.

The single sphere keeps the spins honest on average. No spin can run off to infinity, because the others would have to shrink to compensate, and the total length is fixed. So the model still has the competition between energy, which wants neighbouring spins aligned, and entropy, which wants them spread out, that drives every phase transition.

Visual Beginner

Picture two ways of choosing many spins. In the Ising picture, each spin is a switch that is either up or down, and the allowed states are the corners of a cube. In the spherical picture, the spins are coordinates of a single point that is free to move, as long as that point stays on the surface of a high-dimensional ball of a fixed radius.

The picture captures the trade. The cube's corners are a finite, awkward set; the sphere's surface is smooth, and integrals over a sphere can be evaluated exactly. The radius of the sphere grows with the number of sites, so on average each spin still has size about one, matching the Ising scale, but no individual spin is pinned.

Worked example Beginner

Take the smallest interesting case: two sites with a single bond between them, at zero magnetic field. The spherical rule says the two spins $s_{1}$ and $s_{2}$ must satisfy $s_{1}^{2} + s_{2}^{2} = 2$ , so the pair sits on a circle of radius about $1.41$ . The energy rewards alignment: it is lower when $s_{1}$ and $s_{2}$ have the same sign and similar size.

Step 1. Parametrise the circle by an angle, writing $s_{1} = 2 cos θ$ and $s_{2} = 2 sin θ$ . Every allowed configuration is one value of $θ$ .

Step 2. The bond energy is proportional to $- s_{1} s_{2} = - 2 cos θ sin θ = - sin (2 θ)$ . This is most negative when $2 θ = 90$ degrees, that is $θ = 45$ degrees, giving $s_{1} = s_{2} = 1$ . The two spins want to be equal and both about one.

Step 3. At high temperature the system does not care about energy and spreads evenly over the circle, so the average of $s_{1} s_{2}$ is near zero: no alignment. At low temperature it concentrates near $θ = 45$ degrees (and the opposite point $θ = 225$ degrees), where $s_{1} s_{2}$ is largest: strong alignment.

What this tells us: even in this tiny case the spherical constraint keeps the spins finite while letting them vary smoothly, and the same competition between energy and spread that governs real magnets is already visible. The full model does exactly this calculation for a whole lattice at once, and the smooth circle becomes a smooth sphere that an integral can handle.

Check your understanding Beginner

Formal definition Intermediate+

Let $Λ$ be a $d$ -dimensional hypercubic lattice of $N$ sites with periodic boundary conditions. To each site $i$ assign a real spin $s_{i} \in R$ . The energy of a configuration $s = (s_{1}, \dots, s_{N})$ in a uniform field $H$ is $$ \mathcal{E}(\mathbf{s}) = -J \sum_{\langle i j \rangle} s_i s_j - H \sum_i s_i, $$ the sum running over nearest-neighbour bonds. The spherical model of Berlin and Kac is the classical statistical-mechanical system with this energy, subject to the single spherical constraint $$ \sum_{i=1}^{N} s_i^2 = N. $$ Writing $K = β J$ for the dimensionless coupling and $h = β H$ for the scaled field (the Baxter convention $K = J / k_{B} T$ ), the partition function is the constrained integral $$ Z_N = \int_{\mathbb{R}^N} \Big(\prod_i ds_i\Big), \delta!\Big(\sum_i s_i^2 - N\Big), \exp\Big(K\sum_{\langle ij\rangle} s_i s_j + h \sum_i s_i\Big). $$

The constraint is imposed by a Dirac delta, which is then written as a Laplace contour integral over a Lagrange multiplier $μ$ : $$ \delta!\Big(\sum_i s_i^2 - N\Big) = \frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} d\mu; \exp\Big(\mu\Big(\sum_i s_i^2 - N\Big)\Big), $$ with the contour to the right of all singularities of the spin integral. The multiplier $μ$ is the single auxiliary variable that carries the entire solution; the saddle-point value $μ = μ_{s}$ in the thermodynamic limit plays the role of an inverse susceptibility, vanishing in distance from the band edge as the critical point is approached.

The model is isotropic in spin space: there is no preferred direction, and the spherical constraint is the mean-field-exact stand-in for the Ising hard constraint. It is distinct from the mean spherical model of Lewis and Wannier, in which the strict constraint $\sum_{i} s_{i}^{2} = N$ is replaced by the thermodynamically equivalent requirement $⟨ \sum_{i} s_{i}^{2} ⟩ = N$ on the average; the two agree in the thermodynamic limit away from the critical point, and the strict version is the one solved exactly below.

Counterexamples to common slips

The spherical constraint is global, not local. Imposing $s_{i}^{2} = 1$ at every site recovers the Ising model and loses exact solvability. Only the single sphere $\sum_{i} s_{i}^{2} = N$ keeps the integral Gaussian.
The Lagrange multiplier $μ$ is not a free parameter; it is fixed by the constraint through the saddle-point equation. Treating $μ$ as adjustable, rather than determined by the spherical condition, gives the unconstrained Gaussian model, which has no ordered phase and a different free energy.
The model has a transition only for $d > 2$ . Asserting a finite-temperature transition in one or two dimensions contradicts the convergence of the lattice Green function at the band edge, examined below.

Key theorem with proof Intermediate+

Theorem (exact free energy of the spherical model; Berlin–Kac 1952). On the $d$ -dimensional hypercubic lattice, the spherical model at zero field has free energy per site, in the thermodynamic limit, $$ -\beta f = \frac{1}{2}\ln\frac{\pi}{K} + \frac{1}{2}\Big(2\mu_s - 1\Big) - \frac{1}{2},W_d(\mu_s), $$ where $$ W_d(\mu) = \int_{[-\pi,\pi]^d} \frac{d^d k}{(2\pi)^d}, \ln!\Big(\mu - \frac{1}{d}\sum_{j=1}^d \cos k_j\Big), $$ and the saddle-point value $μ_{s} \geq 1$ is the unique solution of the constraint equation $$ \frac{1}{2K} = g_d(\mu_s), \qquad g_d(\mu) = \int_{[-\pi,\pi]^d}\frac{d^d k}{(2\pi)^d},\frac{1}{2\big(\mu - \tfrac{1}{d}\sum_j \cos k_j\big)}, $$ as long as $1/ (2 K) \leq g_{d} (1)$ ; for smaller $1/ (2 K)$ the saddle sticks at $μ_{s} = 1$ and the system is in the ordered phase.

Proof. Begin with the constrained partition function at $h = 0$ and replace the delta-constraint by its Laplace representation, so that $$ Z_N = \frac{1}{2\pi i}\int_{a - i\infty}^{a+i\infty} d\mu; e^{-N\mu} \int_{\mathbb{R}^N}\Big(\prod_i ds_i\Big)\exp\Big(\mu\sum_i s_i^2 + K\sum_{\langle ij\rangle} s_i s_j\Big). $$ The exponent in the inner integral is a quadratic form $s^{T} A s$ with $A = μ 1 - \frac{1}{2} K B$ , where $B$ is the adjacency matrix of the lattice. Translation invariance diagonalises $B$ by the discrete Fourier transform: its eigenvalues are $2 \sum_{j} cos k_{j}$ indexed by wavevectors $k$ in the Brillouin zone. The Gaussian integral converges when $A$ is positive definite, that is when $μ$ exceeds the largest eigenvalue contribution, and evaluates to $$ \int_{\mathbb{R}^N}\Big(\prod_i ds_i\Big),e^{-\mathbf{s}^{\mathsf T}\mathsf{A}\mathbf{s}} = \frac{\pi^{N/2}}{(\det \mathsf{A})^{1/2}}. $$ Taking the logarithm and using $ln det A = \sum_{k} ln λ_{k} (A)$ converts the determinant into a sum over the Brillouin zone, which becomes the integral $W_{d}$ in the limit $N \to \infty$ after rescaling the coupling so that the band has unit half-width. Writing $A$ 's eigenvalues with the normalisation $μ - \frac{1}{d} \sum_{j} cos k_{j}$ absorbs $K$ into an overall prefactor.

The $μ$ -integral is then $Z_{N} = \frac{1}{2 π i} \int d μ e^{N φ (μ)}$ with $$ \varphi(\mu) = -K\mu + \tfrac{1}{2}\ln\frac{\pi}{K} - \tfrac{1}{2},W_d(\mu) + \text{const}, $$ the linear term $- K μ$ coming from the Laplace factor and supplying the spherical constraint, while $W_{d}$ carries the Gaussian determinant.

For large $N$ the integral is dominated by the saddle where $φ^{'} (μ) = 0$ . Differentiating $W_{d}$ under the integral sign, $$ W_d'(\mu) = \int_{[-\pi,\pi]^d} \frac{d^d k}{(2\pi)^d},\frac{1}{\mu - \tfrac1d\sum_j\cos k_j} = 2,g_d(\mu), $$ so the saddle condition $φ^{'} (μ_{s}) = 0$ balances the constraint term against the determinant term and, with the multiplier normalised conjugate to $\sum_{i} s_{i}^{2} = N$ , takes the standard form $$ \frac{1}{2K} = g_d(\mu_s), $$ the constraint equation that fixes $μ_{s}$ . Equivalently, $g_{d} (μ_{s})$ is the per-site mean square spin from the Gaussian propagator, so the constraint $⟨ s_{i}^{2} ⟩ = 1$ selects the same saddle. Because $g_{d} (μ)$ is positive, decreasing, and finite at the band edge $μ = 1$ precisely when $d > 2$ , the equation has a solution $μ_{s} > 1$ for $1/ (2 K) < g_{d} (1)$ and no solution below; in the latter case the saddle is pinned at the edge $μ_{s} = 1$ and the surplus weight condenses into the zero-wavevector mode, the ordered phase. Substituting $μ_{s}$ into $φ$ produces the stated free energy. $□$

Bridge. This computation builds toward the dimension-resolved theory of critical behaviour, and it is exactly the place where the lattice Green function $g_{d} (μ)$ decides everything. The foundational reason the spherical model is solvable is that its single global constraint keeps the configuration integral Gaussian, so the only hard step is the one-dimensional saddle-point integral over the multiplier $μ$ ; this is exactly the mechanism that appears again in 08.13.02 (six-vertex / Bethe ansatz), where a single auxiliary spectral parameter likewise reduces an $N$ -body sum to a one-variable consistency condition. The central insight is that the constraint equation $1/ (2 K) = g_{d} (μ_{s})$ identifies the susceptibility with the lattice Green function evaluated at the saddle, and the convergence or divergence of that integral at the band edge is what generalises the sharp Ising transition into a dimension-dependent statement. Putting these together, the spherical model becomes the cleanest exactly-solved bridge from the mean-field picture of 08.02.01 to the dimensional-crossover physics organised by the critical exponents of 08.05.01.

Exercises Intermediate+

Exercise 3 (medium, symbolic).

Show that near the band edge, $μ \to 1^{+}$ , the singular part of the lattice Green function behaves as $g_{d} (μ) - g_{d} (1) \sim - c_{d} (μ - 1)^{(d - 2) /2}$ for $2 < d < 4$ , and identify the exponent of $(μ - 1)$ .

Hint

Near $k = 0$ the denominator behaves as $μ - 1 + \frac{1}{2 d} k^{2}$ . Replace the sum by a small- $k$ radial integral $\int k^{d - 1} d k / (μ - 1 + k^{2})$ and count powers.

Answer

Expand the denominator near the zone centre: $μ - \frac{1}{d} \sum_{j} cos k_{j} \approx (μ - 1) + \frac{1}{2 d} k^{2}$ . The singular contribution is $$ \int_0^{\Lambda} \frac{k^{d-1},dk}{(\mu - 1) + \tfrac{1}{2d}k^2}. $$ Substituting $k = 2 d (μ - 1) u$ extracts a factor $(μ - 1)^{(d - 2) /2}$ from the measure relative to the denominator, times a convergent $u$ -integral for $2 < d < 4$ . So $g_{d} (μ) - g_{d} (1) \sim - c_{d} (μ - 1)^{(d - 2) /2}$ , with exponent $(d - 2) /2$ . Rubric: full credit for the exponent $(d - 2) /2$ and the power-counting argument.

Exercise 6 (medium, short-answer).

Explain why the mean spherical model (with the constraint imposed only on the average $⟨ \sum_{i} s_{i}^{2} ⟩ = N$ ) gives the same free energy as the strict spherical model in the thermodynamic limit.

Hint

The strict constraint is imposed by a saddle point in $μ$ ; the mean constraint introduces $μ$ as a true thermodynamic parameter fixed by the same equation $1/ (2 K) = g_{d} (μ)$ .

Answer

In the strict model the multiplier $μ$ is integrated and evaluated at its saddle $μ_{s}$ , fluctuations about which are suppressed by $N$ . In the mean model $μ$ is a Lagrange multiplier (an external field conjugate to $\sum_{i} s_{i}^{2}$ ) chosen so that $⟨ \sum_{i} s_{i}^{2} ⟩ = N$ , which is precisely $- \partial (ln Z) / \partial μ = N$ , the same equation $1/ (2 K) = g_{d} (μ)$ . Both procedures select the identical $μ_{s}$ and identical free energy; they differ only in the $O (ln N / N)$ correction terms and in the behaviour of constraint fluctuations, which do not affect the bulk free energy. Rubric: full credit for identifying the shared saddle equation and the vanishing of the difference in the bulk limit.

Exercise 7 (hard, short-answer).

Derive the spherical-model correlation-length exponent $ν$ and the specific-heat exponent $α$ for $2 < d < 4$ , and verify the hyperscaling relation $2 - α = ν d$ .

Hint

The singular free energy comes from the $μ$ -dependence near the saddle. Use $μ_{s} - 1 \sim t^{2/ (d - 2)}$ with $t = T - T_{c}$ and integrate the singular part of $W_{d}$ twice in temperature.

Answer

The correlation length is set by the gap: $ξ^{- 2} \sim μ_{s} - 1 \sim t^{2/ (d - 2)}$ , so $ξ \sim t^{- 1/ (d - 2)}$ and $ν = 1/ (d - 2)$ . The singular free energy density scales as $f_{sing} \sim (μ_{s} - 1)^{d /2} \sim t^{d / (d - 2)}$ , read off from the band-edge integral $\int k^{d - 1} d k ln (μ - 1 + k^{2})$ after the substitution $k = μ - 1 u$ . The specific heat is two temperature derivatives, $C \sim t^{d / (d - 2) - 2} = t^{(4 - d) / (d - 2)}$ , defining $α$ through $C \sim t^{- α}$ : $$ \alpha = \frac{d-4}{d-2}, \qquad 2 < d < 4. $$ For $2 < d < 4$ this is negative, so the specific heat vanishes at $T_{c}$ as a cusp rather than diverging — the hallmark of the spherical model. Hyperscaling holds directly: $2 - α = 2 - (d - 4) / (d - 2) = (2 (d - 2) - (d - 4)) / (d - 2) = d / (d - 2) = ν d$ . At $d = 3$ : $α = - 1$ , $ν = 1$ . Rubric: full credit for $α = (d - 4) / (d - 2)$ , $ν = 1/ (d - 2)$ , and the hyperscaling check.

Exercise 8 (hard, short-answer).

Show that the spherical model coincides with the $n \to \infty$ limit of the $n$ -vector model (Stanley 1968), and explain why this identifies the spherical-model exponents with those of the $O (n \to \infty)$ universality class.

Hint

In the $n$ -vector model each site carries an $n$ -component unit vector $S_{i}$ with $∣ S_{i} ∣^{2} = n$ . Summing over the $n$ components and using the central-limit concentration of $\sum_{a} (S_{i}^{a})^{2}$ recovers a single scalar with a global constraint.

Answer

In the $n$ -vector model the partition function integrates over $S_{i} \in R^{n}$ with $∣ S_{i} ∣^{2} = n$ at each site. As $n \to \infty$ , each Cartesian component $S_{i}^{a}$ becomes a weakly-constrained scalar whose per-site constraint relaxes, by the law of large numbers over the $n$ components, into a single collective constraint on $\sum_{i, a} (S_{i}^{a})^{2}$ . The saddle-point analysis of the $O (n)$ model in the large- $n$ limit produces exactly the constraint equation $1/ (2 K) = g_{d} (μ_{s})$ of the spherical model, with $μ_{s}$ the large- $n$ self-energy. Stanley (1968) made this identification precise. It places the spherical-model exponents $γ = 2/ (d - 2)$ , $ν = 1/ (d - 2)$ , $η = 0$ in the $O (n \to \infty)$ universality class, the exactly-tractable endpoint of the family $n = 1$ (Ising), $n = 2$ (XY), $n = 3$ (Heisenberg). Rubric: full credit for the component-counting argument and the universality-class identification.

Advanced results Master

Theorem (critical dimension via the lattice Green function; Berlin–Kac 1952). Let $g_{d} (μ) = \int_{[- π, π]^{d}} \frac{d ^{d} k}{( 2 π ) ^{d}} [2 (μ - \frac{1}{d} \sum_{j} cos k_{j})]^{- 1}$ be the constraint integral. Then $g_{d} (1)$ is finite if and only if $d > 2$ . Consequently the spherical model has a finite-temperature ordered phase precisely for $d > 2$ , with critical coupling $K_{c}$ determined by $1/ (2 K_{c}) = g_{d} (1)$ ; for $d \leq 2$ no $K$ drives the saddle to the band edge and the model is disordered at every positive temperature.

The integrand near the band edge $k = 0$ behaves as $1/ k^{2}$ , so the small- $k$ integral $\int k^{d - 1} d k / k^{2} = \int k^{d - 3} d k$ converges at the origin exactly when $d - 3 > - 1$ , that is $d > 2$ . The divergence at $d = 2$ is logarithmic, the same marginal behaviour that forbids long-range order in two dimensions for continuous symmetries, here arising from the spherical constraint rather than from Goldstone modes. The lower critical dimension is therefore $d_{ℓ} = 2$ , in agreement with the Mermin–Wagner threshold of the $O (n)$ models the spherical model represents.

Theorem (continuously varying exponents in $2 < d < 4$ ). For $2 < d < 4$ the spherical-model critical exponents are $$ \gamma = \frac{2}{d-2}, \quad \nu = \frac{1}{d-2}, \quad \eta = 0, \quad \beta = \frac{1}{2}, \quad \alpha = \frac{d-4}{d-2}, \quad \delta = \frac{d+2}{d-2}, $$ all rational functions of $d$ , satisfying the scaling relations $γ = ν (2 - η)$ , $α + 2 β + γ = 2$ , and the hyperscaling relation $2 - α = ν d$ .

The exponents are pinned by the single non-analytic feature of the saddle, the band-edge singularity $g_{d} (μ) - g_{d} (1) \sim - (μ - 1)^{(d - 2) /2}$ . Because $η = 0$ exactly, the spherical model has no anomalous dimension: its two-point function decays with the Gaussian power $∣ x ∣^{- (d - 2)}$ at criticality, the same as free-field theory, while the thermodynamic exponents still depend on $d$ through the constraint. This combination — Gaussian correlations with non-Gaussian thermodynamics — is the signature of the $O (n \to \infty)$ class and the reason the spherical model is the natural exactly-solved companion to the $ε$ -expansion around $d = 4$ organised by the Wilson–Fisher fixed point of 08.04.02.

Theorem (spherical-model free energy is non-analytic at $K_{c}$ ). The free energy per site $f (K)$ of the spherical model in $d > 2$ is analytic in $K$ except at $K = K_{c}$ , where its $(⌈ d /2 ⌉ + 1)$ -th derivative first diverges or develops a cusp, with the precise order set by the exponent $α = (d - 4) / (d - 2)$ . For $d > 4$ the singularity is the mean-field cusp ( $α = 0$ , specific-heat jump); for $2 < d < 4$ it sharpens to a power-law divergence as $d ↓ 2$ .

The non-analyticity is genuine and arises in the thermodynamic limit alone: at finite $N$ the saddle-point integral is a smooth function of $K$ , and only as $N \to \infty$ does the saddle reach and stick at the band edge, producing the sharp transition. The spherical model is thus a fully explicit instance of the general principle that phase transitions are non-analyticities created by the infinite-volume limit, computable here without approximation in every dimension.

Synthesis. The spherical model is the cleanest exactly-solved demonstration that the dimension of space is the foundational reason a phase transition exists or fails to exist. The central insight is that the entire solution is controlled by one scalar saddle equation $1/ (2 K) = g_{d} (μ_{s})$ , and the convergence of the lattice Green function $g_{d}$ at the band edge is exactly what generalises the sharp Ising transition into the statement that there is order for $d > 2$ and none below. Putting these together, the lower critical dimension $d_{ℓ} = 2$ , the upper critical dimension $d_{c} = 4$ , and the rational exponents $γ = 2/ (d - 2)$ , $ν = 1/ (d - 2)$ , $η = 0$ emerge from the single band-edge singularity, and this is exactly the dimensional-crossover structure that the Wilson–Fisher $ε$ -expansion of 08.04.02 reproduces perturbatively near $d = 4$ . The spherical model identifies the $n \to \infty$ vector model with a Gaussian theory carrying a global constraint, so the model is dual to a free field in its correlations while remaining non-mean-field in its thermodynamics; the bridge is the constraint multiplier $μ$ , whose deviation from the band edge is the inverse susceptibility and whose saddle value organises every critical exponent. It builds toward the modern large- $n$ field theory that underlies the $1/ n$ expansion, where the same auxiliary-field constraint reappears as a Hubbard–Stratonovich variable.

Full proof set Master

Proposition (the constraint integral $g_{d} (1)$ converges iff $d > 2$ ). With $g_{d} (μ) = \int_{[- π, π]^{d}} \frac{d ^{d} k}{( 2 π ) ^{d}} [2 (μ - \frac{1}{d} \sum_{j} cos k_{j})]^{- 1}$ , the value $g_{d} (1) = lim_{μ \to 1^{+}} g_{d} (μ)$ is finite for $d > 2$ and divergent for $d \leq 2$ .

Proof. The only possible divergence of $g_{d} (1)$ is at the point where the denominator vanishes, namely $k = 0$ , since $\sum_{j} cos k_{j} = d$ only there in the closed Brillouin zone. Split the integral into a small ball $B_{ϵ}$ around the origin and its complement. On the complement the integrand is bounded and the region has finite measure, so that contribution is finite. On $B_{ϵ}$ expand $\frac{1}{d} \sum_{j} cos k_{j} = 1 - \frac{1}{2 d} ∣ k ∣^{2} + O (∣ k ∣^{4})$ , giving the integrand $\sim [2 \cdot \frac{1}{2 d} ∣ k ∣^{2}]^{- 1} = d /∣ k ∣^{2}$ up to bounded corrections. In $d$ -dimensional spherical coordinates $d^{d} k = Ω_{d - 1} k^{d - 1} d k$ with $k = ∣ k ∣$ , so the singular contribution is proportional to $$ \int_0^\epsilon \frac{k^{d-1}}{k^2},dk = \int_0^\epsilon k^{d-3},dk, $$ which converges at $k = 0$ if and only if $d - 3 > - 1$ , that is $d > 2$ . At $d = 2$ the integral is $\int_{0}^{ϵ} k^{- 1} d k$ , logarithmically divergent; for $d = 1$ it is $\int_{0}^{ϵ} k^{- 2} d k$ , power divergent. Hence $g_{d} (1)$ is finite precisely when $d > 2$ . $□$

Proposition (existence and uniqueness of the saddle $μ_{s}$ ). For $K$ with $1/ (2 K) \leq g_{d} (1)$ and $d > 2$ , the constraint equation $1/ (2 K) = g_{d} (μ)$ has a unique solution $μ_{s} \geq 1$ , and $μ_{s} = 1$ exactly at and below $K_{c}$ defined by $1/ (2 K_{c}) = g_{d} (1)$ .

Proof. On $μ \in [1, \infty)$ the function $g_{d}$ is positive, continuous, and strictly decreasing, since $\partial_{μ} g_{d} (μ) = - \int \frac{d ^{d} k}{( 2 π ) ^{d}} [2 (μ - \frac{1}{d} \sum_{j} cos k_{j})^{2}]^{- 1} < 0$ . As $μ \to \infty$ , $g_{d} (μ) \to 0$ ; at $μ = 1$ it attains its maximum $g_{d} (1)$ , finite by the previous proposition. A strictly decreasing continuous function from $g_{d} (1)$ down to $0$ takes each value in $(0, g_{d} (1)]$ exactly once, so for every $1/ (2 K) \in (0, g_{d} (1)]$ there is a unique $μ_{s} \geq 1$ . When $1/ (2 K) > g_{d} (1)$ — equivalently $K < K_{c}$ — no $μ > 1$ satisfies the equation and the saddle is pinned at the boundary $μ_{s} = 1$ ; the unsatisfied portion of the constraint is absorbed by a macroscopic occupation of the zero mode, the spontaneous magnetisation of the ordered phase. $□$

Proposition (the exponent $γ = 2/ (d - 2)$ ). For $2 < d < 4$ , the zero-field susceptibility diverges as $χ \sim (T - T_{c})^{- γ}$ with $γ = 2/ (d - 2)$ .

Proof. Near the band edge write $μ = 1 + ϕ$ with $ϕ \to 0^{+}$ . From the band-edge expansion the singular part of $g_{d}$ is $g_{d} (1) - g_{d} (μ) \sim c_{d} ϕ^{(d - 2) /2}$ for $2 < d < 4$ , obtained by the substitution $k = ϕ u$ in $\int_{0}^{ϵ} k^{d - 1} d k / (ϕ + \frac{1}{2 d} k^{2})$ , which factors out $ϕ^{(d - 2) /2}$ times a convergent integral. The constraint equation gives $$ g_d(1) - \frac{1}{2K} = g_d(1) - g_d(\mu_s) \sim c_d,\phi^{(d-2)/2}. $$ Linearising $1/ (2 K)$ in temperature about $K_{c}$ , $g_{d} (1) - 1/ (2 K) = 1/ (2 K_{c}) - 1/ (2 K) \propto (T - T_{c})$ , so $ϕ = μ_{s} - 1 \sim (T - T_{c})^{2/ (d - 2)}$ . The susceptibility is the inverse gap, $χ = ⟨(\sum_{i} s_{i})^{2} ⟩ / N \sim 1/ (μ_{s} - 1) = ϕ^{- 1}$ , by the $k = 0$ value of the Gaussian propagator $[2 (μ_{s} - \frac{1}{d} \sum_{j} cos k_{j})]^{- 1}$ at $k = 0$ . Therefore $χ \sim (T - T_{c})^{- 2/ (d - 2)}$ and $γ = 2/ (d - 2)$ . $□$

Proposition ( $η = 0$ exactly). The spherical-model critical correlation function decays as $⟨ s_{0} s_{x} ⟩_{c} \sim ∣ x ∣^{- (d - 2)}$ , so the anomalous dimension vanishes.

Proof. At criticality $μ_{s} = 1$ , the two-point function is the lattice propagator $$ \langle s_0 s_x\rangle_c = \int_{[-\pi,\pi]^d}\frac{d^dk}{(2\pi)^d},\frac{e^{i\mathbf k\cdot \mathbf x}}{2(1 - \tfrac1d\sum_j\cos k_j)}. $$ For large $∣ x ∣$ the integral is dominated by small $k$ , where the denominator is $\sim \frac{1}{d} ∣ k ∣^{2}$ , reducing the propagator to the continuum massless form $\int d^{d} k e^{i k \cdot x} /∣ k ∣^{2} \propto ∣ x ∣^{- (d - 2)}$ . The standard definition $⟨ s_{0} s_{x} ⟩_{c} \sim ∣ x ∣^{- (d - 2 + η)}$ then forces $η = 0$ . The decay exponent is the free-field one, so the spherical model has Gaussian correlations even though its thermodynamic exponents depend on $d$ . $□$

Connections Master

Partition function 08.01.01. The spherical model is solved entirely at the level of its partition function: the constrained configuration integral is reduced, by the Laplace representation of the delta-constraint and a Gaussian evaluation, to a one-dimensional saddle-point integral over the Lagrange multiplier. The whole content of the model is in how this partition function fails to be analytic as $N \to \infty$ , making it the most explicit worked instance of the partition-function machinery developed there.
Transfer matrix 08.03.02. Where the transfer-matrix method diagonalises a row-to-row operator to obtain the free energy of lattice models, the spherical model reaches the same free energy by a complementary route — Fourier-diagonalising the quadratic form and integrating out a single constraint variable. The Brillouin-zone eigenvalues $2 \sum_{j} cos k_{j}$ that appear in the spherical-model determinant are exactly the transfer-matrix band, so the two methods compute the same spectral data by different means.
Critical exponents 08.05.01. The spherical model supplies a complete, exactly-computed set of critical exponents $γ, ν, η, α, β, δ$ as rational functions of the dimension $d$ , satisfying every scaling and hyperscaling relation. It is the canonical exactly-solved test of the exponent framework and the cleanest realisation of $d$ -dependent (non-mean-field) exponents that still respect $η = 0$ , sharpening the universality discussion that the critical-exponents unit frames.
Wilson–Fisher fixed point 08.04.02. The spherical model is the $n \to \infty$ endpoint of the $O (n)$ family whose finite- $n$ critical behaviour the Wilson–Fisher $ε$ -expansion computes perturbatively near $d = 4$ . The spherical exponents are the exact large- $n$ limit against which the $ε$ -expansion and the $1/ n$ expansion are checked, and the constraint multiplier $μ$ is the lattice precursor of the Hubbard–Stratonovich auxiliary field of the large- $n$ field theory.

Historical & philosophical context Master

Theodore Berlin and Mark Kac introduced the spherical model in a 1952 paper in the Physical Review, "The spherical model of a ferromagnet" ^{[Berlin-Kac 1952]}. Their stated motivation was to construct a model close enough to the Ising model to share its qualitative physics, yet tractable enough to solve exactly in three dimensions, where Onsager's transfer-matrix solution of the two-dimensional Ising model had no known extension. Kac had recognised that the obstruction to solving the Ising model was the per-site constraint $s_{i} = \pm 1$ , and that replacing it with a single global constraint $\sum_{i} s_{i}^{2} = N$ — geometrically, confining the configuration to one high-dimensional sphere — would render the partition function Gaussian. Berlin carried out the saddle-point analysis. The resulting solution gave, for the first time, an exact account of how critical behaviour depends on spatial dimension: a transition for $d > 2$ , none for $d \leq 2$ , and continuously varying exponents between the lower critical dimension $d = 2$ and the upper critical dimension $d = 4$ .

The model acquired a deeper meaning in 1968, when H. Eugene Stanley proved that it is the $n \to \infty$ limit of the $n$ -vector model ^{[Stanley 1968]}, placing the spherical exponents in a genuine universality class rather than treating them as an artefact of an artificial constraint. This identification made the spherical model the exactly-solvable anchor of the large- $n$ expansion that Wilson, Fisher, and others were developing in parallel as part of the renormalisation-group programme. G. S. Joyce's 1972 review in the Domb–Green series collected the dimension-dependent exponents and the lattice-Green-function technology in the form used since ^{[Joyce 1972]}. Baxter's Chapter 5 presents the model as the simplest exactly-solved system exhibiting genuine dimension-dependence of criticality, a foil to the two-dimensional vertex and Ising models that occupy the rest of his book and that cannot, by their fixed dimensionality, display the crossover the spherical model makes explicit.

Bibliography Master

@article{BerlinKac1952,
  author  = {Berlin, T. H. and Kac, M.},
  title   = {The Spherical Model of a Ferromagnet},
  journal = {Physical Review},
  volume  = {86},
  number  = {6},
  year    = {1952},
  pages   = {821--835}
}

@article{Stanley1968,
  author  = {Stanley, H. E.},
  title   = {Spherical Model as the Limit of Infinite Spin Dimensionality},
  journal = {Physical Review},
  volume  = {176},
  number  = {2},
  year    = {1968},
  pages   = {718--722}
}

@incollection{Joyce1972,
  author    = {Joyce, G. S.},
  title     = {Critical Properties of the Spherical Model},
  booktitle = {Phase Transitions and Critical Phenomena, Vol. 2},
  editor    = {Domb, C. and Green, M. S.},
  publisher = {Academic Press},
  address   = {London},
  year      = {1972},
  pages     = {375--442}
}

@book{Baxter1982,
  author    = {Baxter, R. J.},
  title     = {Exactly Solved Models in Statistical Mechanics},
  publisher = {Academic Press},
  address   = {London},
  year      = {1982}
}

@article{LewisWannier1952,
  author  = {Lewis, H. W. and Wannier, G. H.},
  title   = {Spherical Model of a Ferromagnet},
  journal = {Physical Review},
  volume  = {88},
  number  = {3},
  year    = {1952},
  pages   = {682--683}
}

@article{Watson1939,
  author  = {Watson, G. N.},
  title   = {Three Triple Integrals},
  journal = {Quarterly Journal of Mathematics},
  volume  = {os-10},
  number  = {1},
  year    = {1939},
  pages   = {266--276}
}

Prerequisites

08.01.01
08.03.02
08.05.01

Tier anchors

beginner: Baxter Ch. 5 §5.1; the spherical constraint as a relaxation of the Ising $\pm 1$ constraint
intermediate: Baxter Ch. 5 §5.3–5.5; saddle-point / Lagrange-multiplier solution of the spherical model in $d$ dimensions
master: Baxter Ch. 5 §5.5–5.7; Berlin–Kac 1952 Phys. Rev. 86:821; dimensionality dependence of criticality and the Gaussian-model connection

References

raw/pdfs/kardar-1-3-laws-entropy-potentials.pdf · Kardar statistical physics notes (saddle-point method, Gaussian integrals, critical exponents)
quantum-well · statistical physics index (mean-field models, critical dimension)
Baxter — Exactly Solved Models in Statistical Mechanics · Ch. 5 (the spherical model)
Berlin — Kac — The spherical model of a ferromagnet · Phys. Rev. 86:821 (1952), originator of the spherical model and its exact solution
Stanley — Spherical model as the limit of infinite spin dimensionality · Phys. Rev. 176:718 (1968), identification with the $n \to \infty$ vector model
Joyce — Critical properties of the spherical model · in Domb–Green eds., Phase Transitions and Critical Phenomena Vol. 2 (Academic Press, 1972), Ch. 10, spherical-model exponents in general $d$

Estimated time

beginner: 16m
intermediate: 38m
master: 70m