Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 42 / 1 / v / Solution

A tricritical point joins continuous and first-order transition loci and requires tuning an additional temperature-independent control. In the scalar sextic potentialit occurs at $r=u=h=0$: both the quadratic and quartic terms vanish. At $h=0,u>0$, the transition is continuous at $r=0$; at $h=0,u<0$, it is first-order at $r=3u^2/(16v)$. The jump tends to zero as $u\uparrow 0$, so these loci meet at the tricritical point.

For the three-control phase diagram with coordinates $(u,r,h)$, the sheet $h=0$ below these transition loci has coexistence of the two ordered phases. For $u>0$ its boundary is the ordinary continuous critical line $r=h=0$. For $u<0$ the boundary $r=3u^2/(16v),h=0$ is a tricritical three-phase line, where $m=0$ and $m=\pm \sqrt{-3u/(4v)}$ all coexist. Two symmetry-related tricritical wings extend from that line into $h>0$ and $h<0$, describing first-order coexistence between small- and large-magnitude phases of the same sign. Each wing ends along an ordinary wing critical edge.

The critical-edge conditions $V^{\prime }=V^{\prime \prime }=V^{(3)}=0$ giveHere $V^{(4)}(m_c)=-12u>0$, so these are ordinary quartic endpoints. The two wing critical edges and the zero-field ordinary critical line are three continuous critical lines meeting at the tricritical point; the three-phase first-order line also ends there. The “3D” phase diagram refers to three independent controls; it is distinct from the spatial dimension $D$ of the field theory.

The figure uses actual sextic-potential coexistence surfaces rather than drawing arbitrary wings. For two coexisting nonnegative minima $a\le b$, put $s=a+b,q=ab$. The tricritical wing coexistence factorizationholds when $u=(2v/3)(-2s^2+3q)$, $r=(v/3)(s^4-s^{2}q+3q^2)$ and $h=(v/3)s^{3}q$. It proves that both minima are globally stable. The limits $q=0$ and $q=s^2/4$ produce the three-phase line and critical edge, respectively; reflection gives the other wing.

At $u=h=0$, $m_0=(-r/v{)}^{1/4}$ on the ordered side and $V(m_0)=-(-r{)}^{3/2}/(3\sqrt{v})$. At the critical isotherm $h=v{m}^5$. Thus the tricritical mean-field critical exponents areThe source's extra control is required to tune $u$ to zero; simply changing temperature in a generic quartic system does not produce tricriticality. The tricritical upper critical dimension is three, as established in Question 3.