Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 303 / 3

If the singular free-energy density is set by one correlated region, then $f_s\sim {\xi }^{-d}$. With reduced temperature $t$, the definitions $\xi \sim |t{|}^{-\nu }$ and $f_s\sim |t{|}^{2-\alpha }$ therefore giveThis is the hyperscaling relation; it assumes that no dangerously irrelevant coupling introduces an additional scale.

For ${\mu }^2<0$ and $g>0$, minimizing the $O(N)$-invariant potential givesThe free energy has the full $O(N)$ symmetry, but choosing one point on this sphere leaves only the rotations $O(N-1)$ fixing that point. This is spontaneous symmetry breaking, $O(N)\to O(N-1)$. The $N-1$ tangent directions along the sphere cost no potential energy and are the massless Goldstone modes predicted by the Goldstone theorem.

Choose the vacuum $\varphi =(v+\sigma ,{\pi }_1,\dots ,{\pi }_{N-1})$. The quadratic free energy contains a positive mass term for the radial field $\sigma$ but no mass term for any transverse field ${\pi }_a$:Thus the momentum-space correlation function is $⟨{\pi }_a(k){\pi }_b(-k)⟩={\delta }_{ab}/k^2$. Since a massive correlator has denominator $k^2+{\xi }^{-2}$, these Goldstone modes have ${\xi }^{-1}=0$ and hence infinite correlation length.

The long-wavelength Goldstone fluctuation is proportional towhich diverges in the infrared for $d\le 2$. These fluctuations destroy finite-temperature long-range order with a broken continuous symmetry, as formalized by the Mermin-Wagner theorem. Therefore the lower critical dimension is $d_l=2$: conventional spontaneous $O(N)$ order is possible for $d>2$ but not for $d\le 2$.

The $O(2)$ rotation symmetry requires the quadratic dependence on $({\varphi }_2,{\varphi }_3)$ to be proportional to ${\varphi }_2^2+{\varphi }_3^2$, soThe quartic terms involving only that pair must be proportional to $({\varphi }_2^2+{\varphi }_3^2{)}^2$, and the terms coupling it to ${\varphi }_1$ must be proportional to ${\varphi }_1^2({\varphi }_2^2+{\varphi }_3^2)$. With the convention that the symmetric double sum counts off-diagonal terms twice, this giveswhile ${\mu }_1^2$, $g_{11}$, and the common mixed coupling are unrestricted. The stated ${\mathbb{Z}}_2$ symmetry imposes no further relation because every term is already even in ${\varphi }_1$.

Let $r_1={\mu }_1^2$, $r_2={\mu }_2^2={\mu }_3^2$, and ${\rho }^2={\varphi }_2^2+{\varphi }_3^2$. With every $g_{ij}=g>0$, the uniform potential isFor $r_1>0,r_2>0$, the minimum is the origin and ${\mathbb{Z}}_2\times O(2)$ is unbroken. If $r_1<\min (0,r_2)$, then ${\varphi }_1^2=-r_1/(4g)$ and $\rho =0$; the ${\mathbb{Z}}_2$ factor is broken, $O(2)$ remains, and there is no Goldstone mode. If $r_2<\min (0,r_1)$, then ${\varphi }_1=0$ and ${\rho }^2=-r_2/(4g)$; the first ${\mathbb{Z}}_2$ remains while $O(2)$ is broken to the reflection fixing the chosen direction, producing one Goldstone mode.

The positive coordinate half-axes are continuous transition lines. For $r_1=r_2<0$, the potential has an enhanced $O(3)$ symmetry and a sphere of minima; $O(3)\to O(2)$ gives two Goldstone modes on this line. Crossing the negative diagonal exchanges the two ordered phases and gives a first-order line at mean-field level.