Past exam of the theoretical physics I course of the University of Cambridge / 2022 / 3

The mean-field approximation replaces the fluctuating environment of each microscopic variable by a uniform average field. For spins, write $S_i=S+\delta S_i$ and drop products $\delta S_i\delta S_j$ between different sites. The resulting single-site problem can be solved independently, and its average spin must reproduce the assumed order parameter $S$: this is the self-consistency equation.

It is reliable when correlated thermal fluctuations about the mean are small, for example with many interacting neighbours, long-range interactions or high spatial dimension. Near a thermodynamic critical point the quantitative test is the Ginzburg criterion. For short-range scalar ordering the upper critical dimension is four, with logarithmic qualifications at four; a two-dimensional nearest-neighbour lattice is below this and its asymptotic critical region of a phase transition is not described accurately by simple mean-field theory.^[ref]

The square lattice has coordination number of a lattice $z=4$. The mean-field decoupling isAccounting for the factor $1/2$ in the directed-neighbour sum gives the Hamiltonian in the mean-field approximation:The constant term is necessary to correct the double counting of the interaction. Let $h=4JS+B$. The one-site partition function isso $Z_{\mathrm{MF}}=e^{-2\beta JN{S}^2}{Z}_1^N$. The Helmholtz free energy is thereforeThe equilibrium mean-field approximation is obtained by minimizing this expression over the allowed $S$. The $s=0$ state supplies the extra $1$ in $Z_1$; omitting it would instead describe the two-state Ising model. This is the Blume-Capel mean-field transition at zero single-ion anisotropy.

The self-consistency equation requires that the mean spin computed using the single-site Boltzmann distribution equals the spin $S$ used to form the effective field. With $h=4JS+B$,ThusEquivalently, differentiating part (b) gives $N^{-1}\partial F_{\mathrm{MF}}/\partial S=4J[S-⟨s{⟩}_h]$, so stationarity of the Helmholtz free energy in the mean-field approximation is exactly this condition. Among multiple stationary points one must select the stable minimum.

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 $c[({\partial }_{x}m{)}^2+({\partial }_{y}m{)}^2]$.

For a homogeneous saddle-point approximation define $r=a(T-T_c)>0$. With the cubic source coupling, minimizing the Landau free energy givesIts curvature $2r+6b{m}^2$ is positive, so there is a unique local minimum and global minimum. The solution near $B=0$ hasbecause substituting this leading term makes the cubic self-interaction of order $B^9$. Therefore the cubic source coupling and third-order susceptibility giveThus the first nonzero response diverges like $(T-T_c{)}^{-1}$ from above. The factor $3!$ is essential: the coefficient of $B^3$ is not itself the third derivative. This is susceptibility to the specified cubic source variable, rather than the usual third-order response to a linearly coupled field.