= Solution
For the isotropic zero-field <classical Heisenberg model> of a <ferromagnet>, the <analytic vector Landau expansion> uses the Cartesian components of the <vector order parameter> near $\mathbf m=0$. <rotational symmetry> makes its scalar terms functions of $\mathbf m\cdot\mathbf m$. Global <spin inversion symmetry> also requires $f(\mathbf m)=f(-\mathbf m)$. Through fourth order this gives
$$
\boxed{a=c=0,\qquad f=bm^2+dm^4,\qquad d>0.}
$$
A positive quartic coefficient stabilizes the unconstrained truncated <free energy>. The disordered phase has $b>0$ and its minimum at $m=0$; the aligned phase has $b<0$ and a minimum at nonzero $m$. At the continuous transition $b=0$. The attractive alignment interaction supplies an energetic tendency toward negative quadratic curvature; orientational <entropy> favors the disordered phase at high <temperature>, so the sign of $b$ is a free-energy question rather than a consequence of the pair <energy> alone.
The analyticity assumption matters: the variable printed as $m$ is a nonnegative magnitude, so reversing $\mathbf m$ does not send that scalar to $-m$. Terms proportional to $|\mathbf m|$ or $|\mathbf m|^3$ would be rotationally invariant but nonanalytic at the <vector> origin. Standard analytic <Landau theory> excludes them. Without that assumption, the stated pair <symmetry> alone would not exclude odd powers of the magnitude.<Statistical Field Theory by David Tong>[§1.2, small-order-parameter expansion; §4.2, rotationally invariant vector free energy]{ref}
Back to article page
