Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 1

The equilibrium magnetization is a global minimum of the Landau free energy. Its stationary values solve the polynomial equationand a local minimum must satisfyOne compares the value of $f$ at every such local minimum and chooses the smallest. Since $a_6>0$, the polynomial tends to positive infinity as $|m|$ tends to infinity, so a global minimum exists.

Put $t=T-T_c$, so $a_4\sim t$. At zero field the stationary equation factors asFor $t>0$, $m=0$ is the unique minimum. For $t<0$, it is unstable and the two minima areThe order parameter therefore tends continuously to zero, and its order-parameter critical exponent is $\boxed{\beta =1/2}$.

At either ordered minimum the singular free-energy density iswhereas it is zero for $t>0$. Two temperature derivatives give a singular heat capacity proportional to $|t|$ below the transition and zero above it, so the heat-capacity critical exponent is $\boxed{\alpha =-1}$.

The inverse magnetic susceptibility at a stable minimum is the curvature ${\chi }^{-1}={\partial }^{2}f/\partial m^2$. Below $T_c$,and hence ${\chi }_{-}\sim |t{|}^{-2}$ and $\boxed{{\gamma }_{-}=2}$. Above $T_c$, however, the curvature at $m=0$ vanishes for every $t>0$. Indeed, at small field $B=4a_4{m}^3+O(m^5)$, so $m\sim (B/(4a_4){)}^{1/3}$ and the linear susceptibility is already infinite away from the critical point. Consequently the usual magnetic-susceptibility critical exponent ${\gamma }_{+}$ is not defined for this exceptional free energy; assigning it a finite value would incorrectly assume a quadratic term.

At $T=T_c$, the equation of state is $B=6a_6{m}^5$, so $m\sim |B{|}^{1/5}$ and the critical-isotherm exponent is $\boxed{\delta =5}$. Thus the transition is continuous, although its missing quadratic term makes the high-temperature linear response singular throughout that phase.

For every $T<T_c$, the two zero-field minima $m_{+}$ and $m_{-}$ coexist. A positive field selects $m_{+}$ and a negative field selects $m_{-}$, so crossing $B=0$ makes the equilibrium magnetization jump between nonzero values. Hence $B=0$, $T<T_c$, is a line of first-order phase transitions, ending at the continuous critical point $(T_c,0)$.

Write ${\mathbf{\sigma }}_i=\mathbf{m}+\delta {\mathbf{\sigma }}_i$ and neglect products of two fluctuations. ThenEach site has $q$ neighbours and each bond is counted once, so the mean-field approximation givesChoose the $x$ axis along $\mathbf{m}$. For the four-state clock model, the single-site partition function isConsequentlyand therefore

Differentiating the mean-field free energy and imposing stationarity givesThe hyperbolic-function identity $\sinh x/(1+\cosh x)=\tanh (x/2)$ turns this into the self-consistency equation

The Taylor series at $x=0$ isSubstituting $x=Jqm/T$ givesThe quartic coefficient is positive, while the quadratic coefficient changes sign atThis is therefore a continuous mean-field phase transition.

For finite $p=4$, the clock model has a discrete symmetry. Domain walls have finite energy per unit boundary area, so thermal disorder destroys long-range order in one dimension but a finite-temperature ordered phase can exist in two dimensions. Its lower critical dimension is therefore $\boxed{d_{\mathrm{l}}=1}$.

As $p\to \infty$, the permitted angles become continuous and the model becomes the XY model with $O(2)$ symmetry. The Mermin-Wagner theorem forbids spontaneous long-range order at positive temperature in two dimensions, so the lower critical dimension for conventional symmetry breaking is $\boxed{d_{\mathrm{l}}=2}$. The two-dimensional model can nevertheless undergo a Berezinskii–Kosterlitz–Thouless transition between algebraic and exponential correlation decay.