Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 303 / 1

Write $t=T-T_c$. Analyticity gives ${\alpha }_2=at+O(t^2)$ with $a>0$ and ${\alpha }_4(T_c)=u>0$. For $t<0$, minimization gives $m^2=-{\alpha }_2/{\alpha }_4\sim -at/u$, so $\beta =1/2$. At the minima,Two temperature derivatives give a finite heat-capacity jump, hence $\alpha =0$. This is the ordinary mean-field transition of Landau theory.

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At the tricritical point, let ${\alpha }_2=at+O(t^2)$, ${\alpha }_4=bt+O(t^2)$ and ${\alpha }_6(T_c)=u>0$. The nonzero stationarity equation isThe middle term is subleading, so $m^4\sim -at/u$ and $\beta =1/4$. Substitution gives $F_{\min }\asymp -|t{|}^{3/2}$, so the heat capacity behaves as $|t{|}^{-1/2}$ and $\alpha =1/2$.

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Fourier expansion in volume $V$ diagonalizes the quadratic Landau-Ginzburg theory:whereEach independent real mode contributes a Gaussian integral proportional to $(T/K{)}^{1/2}$. Taking $-T\log Z$, dividing by $V$, and replacing the sum by an integral gives, up to field-independent conventions,

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For one mode put $f=-\frac{T}{2}\log (CT/K)$. Since $c=-T{\partial }_T^{2}f$,ThusAt $(T_c,0)$, $K_T=({\mu }^2{)}^{\prime }(T_c)\ne 0$ because ${\mu }^2$ has a simple zero, so $A(T_c,0)\ne 0$.

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Near the ordinary critical point, $K\sim at+\gamma k^2$. Rescaling $k=t^{1/2}q$ gives the Gaussian fluctuation correction near a critical pointThus ${\alpha }_{\mathrm{fl}}=2-d/2$. It becomes as important as the mean-field value $\alpha =0$ at $d=4$, so the ordinary upper critical dimension is $d_c=4$.

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The quadratic kernel still has $K\sim at+\gamma k^2$, so ${\alpha }_{\mathrm{fl}}=2-d/2$. The tricritical mean-field exponent is $\alpha =1/2$. Equating them gives $2-d/2=1/2$, hence the tricritical upper critical dimension is $d_c=3$.

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