Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 43 / 4

Assume short-range ferromagnetic exchange $J>0$ and zero external field. In the one-dimensional Ising model, a domain wall costs only $2J$, independent of the lengths of the domains it separates. There are order-$N$ positions for one wall, so its positional entropy offsets any finite energy cost at nonzero temperature as the system grows. Domain walls consequently have a positive density at every $T>0$, destroying a uniform spontaneous spin magnetization. Periodic boundaries require an even number of walls, but that constraint does not remove their nonzero thermodynamic-limit density.

The One-dimensional Ising correlation function makes this argument explicit. A bond is broken with probability $p_w=e^{-2{\beta }_{\mathrm{th}}J}/(1+e^{-2{\beta }_{\mathrm{th}}J})$, and the infinite-chain correlation isThe dominant eigenvalue of the transfer matrix for the one-dimensional Ising model is $2\cosh {\beta }_{\mathrm{th}}J$, so the free energy per spin $-k_{B}T\log [2\cosh {\beta }_{\mathrm{th}}J]$ is analytic for $T>0$. There is no finite-temperature transition in one dimension; only the zero-temperature limit orders.

In two dimensions a reversed cluster is enclosed by a Peierls contour. A contour of length $\ell$ costs $2J\ell$, while the number of such contours surrounding a fixed site grows at most as a polynomial in $\ell$ times $3^{\ell }$: after the initial edge, a nonbacktracking square-lattice walk has at most three choices per step. With plus boundary conditions, flipping the interior of a specified contour yields the Boltzmann suppression $e^{-2{\beta }_{\mathrm{th}}J\ell }$. ThereforeFor sufficiently low temperature this bound is below $1/2$, giving a positive spontaneous magnetization. This is the energy-entropy mechanism of the Peierls argument.

At sufficiently high temperature, the graphical expansion bounds correlations by sums of connecting paths weighted by $(\tanh {\beta }_{\mathrm{th}}J{)}^{\ell }$. The same nonbacktracking count converges geometrically when $3\tanh {\beta }_{\mathrm{th}}J<1$, giving exponential decay and a disordered phase. Since an ordered low-temperature phase and a disordered high-temperature phase both occur, the two-dimensional model has a finite-temperature phase transition. The argument establishes its existence and the dimensional distinction, not the exact value of its critical temperature.

Write $S=H/T$ and take $n\ge 2$. The unit-length constraint prevents an arbitrary multiplicative field rescaling, so the running coupling is found by renormalizing the coefficient of $|\nabla \sigma {|}^2$. We compute that coefficient in a slowly rotating planar background, which is enough because $O(n)$ symmetry makes the two-derivative stiffness isotropic on the sphere.

Let $n_0=(\cos \theta ,\sin \theta ,0,\dots )$, $e_1=(-\sin \theta ,\cos \theta ,0,\dots )$ and $e_a$ for $2\le a\le n-1$ the remaining constant orthonormal tangent vectors. Parameterize the fast tangent fluctuations byThis satisfies ${\sigma }^2=1$ exactly. Put ${\Omega }_{\mu }={\partial }_{\mu }\theta$, and choose it locally constant and small compared with the shell momenta. Since ${\partial }_{\mu }{n}_0=e_1{\Omega }_{\mu }$ and ${\partial }_{\mu }{e}_1=-n_0{\Omega }_{\mu }$, direct differentiation givesAt quadratic order in the fast fields, the cross term is $2{\Omega }_{\mu }{\partial }_{\mu }{\pi }_1$, whose integral is zero for constant ${\Omega }_{\mu }$. The term $(\partial q{)}^2$ is quartic in the fast fields. Remaining cross terms are cubic and have zero Gaussian expectation; their effects on the two-derivative stiffness start at higher loop order. Field-measure corrections independent of the background contribute normalization terms and do not alter the leading gradient coefficient in this calculation.

The quadratic fast action is $(2g{)}^{-1}\int |\nabla \pi {|}^2$. In the shell $\Lambda /b<|p|<\Lambda$ its connected correlation function is $⟨{\pi }_a(p){\pi }_b(-p)⟩=g{\delta }_{ab}/p^2$ with the momentum delta function understood. Thus $⟨{\pi }_a(x{)}^2⟩=g{I}_D$, wherewith the limit $I_2(b)=(2\pi {)}^{-1}\log b$. Here $S_D=2{\pi }^{D/2}/\Gamma (D/2)$ is the angular measure of the unit $(D-1)$-sphere. Averaging the background-gradient coefficient gives the one-loop stiffness of a spherical nonlinear sigma model:The coefficient is $n-2$, not $n-1$: the fluctuation tangent to the rotation plane does not reduce the gradient cost, because its contribution cancels between $q^2$ and ${\pi }_1^2$. All other $n-2$ transverse components do reduce it.

After integrating out the shell, put $x=b{x}^{\prime }$ to restore the cutoff. The constrained field keeps unit norm, so $g^{\prime }=b^{-ϵ}{g}_{\mathrm{eff}}$, or at this orderFor $b=e^{d\ell }$, $I_D=(S_D/(2\pi {)}^D){\Lambda }^{ϵ}d\ell +O(d{\ell }^2)$. Therefore the renormalization-group beta function isThe displayed terms are the requested one-loop flow. They are a weak-coupling expansion; the discrete $n=1$ target has no tangent modes and must instead be treated by domain walls, as in the first part.

Set $A_D=S_D{\Lambda }^{ϵ}/(2\pi {)}^D>0$. The one-loop flow is $\mathcal{B}(g)=-ϵg+(n-2)A_D{g}^2$. For $ϵ>0$ and $n>2$, its physical fixed points areThe zero point attracts small positive couplings, whereas the nonzero point is the separatrix between the low-temperature weak-coupling flow and the strong-coupling regime. This is a perturbative description for small $ϵ$; extrapolation of its nonzero root to large $ϵ$ is not controlled.

At $ϵ=0$ and $n>2$, the only finite one-loop fixed point is $g=0$, a double root. Positive $g$ flows to strong coupling. For $n=2$ the loop coefficient vanishes: at $ϵ>0$ only $g=0$ is fixed, while at $ϵ=0$ every $g$ is fixed in this perturbative spin-wave description. The compact two-component model also admits vortices; their Berezinskii–Kosterlitz–Thouless transition is absent from a calculation retaining only the smooth tangent modes. Thus a zero beta function here does not exclude that transition. The formula should not be interpreted as a continuum result for the discrete $n=1$ Ising target.

For a fixed point $g_f$, put $g=g_f+\delta g$. Linearizing the renormalization-group flow givesBecause $g$ is proportional to temperature at fixed microscopic stiffness, this derivative is the thermal eigenvalue. The thermal eigenvalues of the spherical nonlinear sigma model areThe first value describes the stable zero-temperature ordered fixed point. At the nonzero critical point the thermal perturbation grows as $e^{ϵ\ell }$, so the stopping scale gives $\xi \propto |g-g_{\ast }{|}^{-1/ϵ}$ and $\nu =1/ϵ$ to leading order in the expansion. Higher-loop terms change this to $y_T=ϵ+O({ϵ}^2)$.

At $ϵ=0$ the linear eigenvalue is zero, so its sign alone does not decide the flow. For $n>2$ the first nonlinear term is positive. Integrating it givesThe flow reaches order-one coupling at a length logarithm of order $2\pi /[(n-2)g(0)]$, explaining the exponentially large correlation length in a two-dimensional spherical sigma model rather than a finite-temperature power-law critical point in this smooth-field regime. For $n=2$, $ϵ=0$, the linear eigenvalue is zero everywhere on the perturbative line and vortex effects require an additional coupling. For $n=2$, $ϵ>0$, the sole zero-coupling fixed point has $y_T=-ϵ$.