Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 209 (Lattice Models)

For a finite graph $\Lambda =(V,E)$ with free boundary conditions, write ${\sigma }_x=(\cos {\theta }_x,\sin {\theta }_x)\in S^1$. The ferromagnetic O(2) model isThe Ginibre inequality says, in particular, that for $a,b\in {\mathbb{Z}}_{\ge 0}^V$,

For the proof, take two independent replicas $\theta ,{\theta }^{\prime }$ and write the covariance as one half of the expectation ofSet $u=(\theta +{\theta }^{\prime })/2$ and $v=(\theta -{\theta }^{\prime })/2$. Product-to-sum identities turn each difference into $-2\sin (a\cdot u)\sin (a\cdot v)$, while every replicated interaction becomesExpand every exponential in a power series and then every cosine power into Fourier modes. Integration over each angle kills all unmatched modes. Because the couplings, field, and entries of $a,b$ are nonnegative, every surviving paired coefficient in the covariance is nonnegative. Their sum is therefore nonnegative, proving the inequality. The same replica expansion proves the usual product version.

Differentiate the finite-volume magnetization:Every summand is nonnegative by the Ginibre inequality with $a=1_x$ and $b=1_y$. Hence the magnetization is nondecreasing for $h\ge 0$.

At zero field the finite-volume law is invariant under the global rotation ${\theta }_z\mapsto {\theta }_z+\alpha$. Averaging $\cos (2{\theta }_x)$ over $\alpha$ gives zero. Reflection ${\theta }_z\mapsto -{\theta }_z$ likewise gives $⟨\sin {\theta }_y⟩=0$. Thusfor every finite $\Lambda$ containing $x,y$, so its infinite-volume limit exists and equals zero.

Let the square contain the Euclidean ball of radius $R$, set $r_x=\max (1,\Vert x\Vert )$, and define the logarithmic cutoffThen ${\phi }_0=1$ and $\phi =0$ on the boundary. On an edge at radius comparable to $r$, the mean value theorem gives $|{\phi }_x-{\phi }_y|\le C/(r\log R)$. There are $O(r)$ edges in the annulus of radius $r$, hence the discrete Dirichlet energy satisfiesThis logarithmic cutoff is the discrete manifestation of recurrence in two dimensions.

Let $\Delta$ be the torus graph Laplacian and let $v:{\Lambda }_L\to {\mathbb{R}}^n$ have zero spatial mean. Gaussian domination states, in a standard normalization, thatDifferentiating twice at zero gives the infrared bound: for every nonzero torus momentum $k$,up to the harmless normalization convention used for the Fourier transform and Hamiltonian.

The spin-length sum rule and Fourier inversion giveThe zero mode contains the square of the field-directed magnetization, while the infrared bound controls all nonzero modes. After $L\to \infty$ and then $h\downarrow 0$,Near zero, $\epsilon (k)\asymp |k{|}^2$, so the integral is finite exactly when $d\ge 3$. Choose ${\beta }_0$ larger than the resulting finite constant. Then for $\beta >{\beta }_0$ the right side is positive, and $m(\beta )\ge c>0$.

If $h\downarrow 0$ before $L\to \infty$, every finite torus retains global $O(n)$ symmetry and its magnetization is zero. The reversed iterated limit is therefore zero. The order of limits is what permits spontaneous magnetization.

The field leaves rotations fixing its unit direction $e$ as symmetries. Consequently the expectation vector is parallel to $e$:For any $u\in {\mathbb{R}}^n$, linearity therefore gives

Let $\vartheta$ be either permitted torus reflection and let ${\Lambda }_{+}$ be one reflected half. Under the product measure ${\mu }^{\otimes {\Lambda }_L}$, variables in the two open halves are independent and corresponding variables have the same law. For any $F$ depending on ${\Lambda }_{+}$,when the reflection has no fixed sites. If it fixes a layer of sites, condition on that layer; the same factorization gives a conditional square, whose expectation is nonnegative. This proves reflection positivity through sites. For an edge reflection the parity assumption makes the two halves pair exactly, so the first factorization applies there as well.

In the one-dimensional zero-field Ising model, the bond variables ${\tau }_j={\sigma }_j{\sigma }_{j+1}$ are independent andSince ${\sigma }_0{\sigma }_x={\prod }_{j=0}^{x-1}{\tau }_j$ for $x>0$,

In a finite interval, the effect at the origin of a fixed boundary spin a distance $N$ away is bounded by its two-point correlation, $(\tanh \beta {)}^N$. With two boundaries,Thus the plus and minus boundary limits coincide with the unique spin-flip-symmetric one-dimensional Gibbs state, and $⟨{\sigma }_0{⟩}_{\beta ,0}^{\mathbb{Z},\pm }=0$ for every finite $\beta$.

The product ${\sigma }_0{\sigma }_1\cdots {\sigma }_{2022}$ contains $2023$ spins and changes sign under the global spin flip $\sigma \mapsto -\sigma$. By part ii, the infinite-volume plus state is spin-flip symmetric. Therefore

Order the sites as $x_1<x_2<x_3<x_4$. Expressing spins through independent bond variables gives the One-dimensional Ising correlation functionIf the minimum pairwise separation tends to infinity, both displayed gaps tend to infinity. Since $\tanh \beta <1$, the expectation tends to zero.

Under plus boundary conditions, if ${\sigma }_0=-1$, the negative cluster containing the origin is surrounded by a Peierls contour $\gamma$. Flipping all spins inside $\gamma$ is injective after $\gamma$ is specified and increases the Boltzmann weight by $e^{2\beta |\gamma |}$. ThereforeThe number of length-$m$ contours surrounding the origin is at most $Cm{3}^m$, so the sum tends to zero as $\beta \to \infty$, uniformly in $L$. For sufficiently large $\beta$ it is below $1/4$, and then

Ferromagnetic monotonicity makes the plus-boundary expectation decrease as the finite volume grows, soexists. The uniform lower bound from part i passes to the limit, proving positive spontaneous magnetization at sufficiently large $\beta$.

The three-dimensional Peierls argument replaces planar contours by closed dual plaquette surfaces surrounding the negative cluster of the origin. A surface of area $m$ costs $e^{-2\beta m}$, and the number of connected surfaces of area $m$ containing a fixed nearby plaquette is at most $C^m$. HenceThis is below $1/2$ for sufficiently large $\beta$, so $⟨{\sigma }_0{⟩}_{\beta ,0}^{{\mathbb{Z}}^3,+}>0$.

The on-site potential of the Phi-four lattice model isIn the random-walk representation of a lattice-field covariance, $⟨{\varphi }_x{\varphi }_y⟩$ is a Green function for a nearest-neighbour walk with a nonnegative environment-dependent killing rate bounded below by a positive constant depending on $\nu$. Dropping the quartic contribution can only decrease that killing, soThe massive lattice Green function has the killed-walk expansionwith normalization adjusted to the chosen Laplacian. Reaching $y$ requires at least $|x-y{|}_1$ steps, while the geometric survival factor is strictly below one. Summing the tail gives constants $C,c>0$ such that