Past exam of the theoretical physics I course of the University of Cambridge / 2014 / Paper 1 / 6 / c / Solution

Set $r={\varphi }_0$ and $A_k=a+c{k}^2+k^4$. The trial density is $A_k{r}^2+r^4/2$. At fixed $k$, its minimum is $r=0$ if $A_k\ge 0$ and $r^2=-A_k$ otherwise, with density $-A_k^2/2$. Minimizing $A_k$ usesFor $c\ge 0$, $k=0$ minimizes the quadratic coefficient. For $c<0$, the minima are at $k=\pm \sqrt{-c/2}$ and have $A_k=a-c^2/4$. ThusEverywhere else the minimum is $\varphi =0$; at the transition boundary the ordered amplitude tends to zero. The zero-field minimum is respectively $-a^2/2$ or $-(a-c^2/4{)}^2/2$ in the two ordered regions. For $c<0$ the helical solution is lower than a uniform ordered extremum whenever the latter exists, and lower than the disordered state whenever its amplitude is nonzero. The assumed positive-amplitude ansatz must allow its zero-amplitude boundary to describe disorder.

The phase $\delta$ remains arbitrary because the functional has global U(1) symmetry. At nonzero $k$, a translation shifts that phase; separate translation and phase operations retain a combined unbroken subgroup. The two signs of $k$ also select opposite handedness. The meeting point of the uniform, modulated and disordered mean-field regions is the Lifshitz point $(a,c)=(0,0)$.