Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 81 / 4 / f / Solution

The spin-independent term cancels in the signed sum. The linearized stationary equation becomesThe even powers vanish because the equation is odd under $M\mapsto -M$. Equivalently, the mean-field action has curvatureThus the unpolarized stationary point becomes unstable whenAt low temperature, the derivative of the Fermi distribution samples a narrow window around $\mu$, so ${\chi }_0\simeq \nu (\mu )$ for a smooth density of states. Measuring energy from the Fermi level gives $U_c\nu (0)\simeq 1$, the Stoner criterion in the spin-summed convention fixed above. This identifies local instability of the paramagnetic state; the detailed transition order also depends on higher terms in the action.

In the isotropic ordered phase of itinerant ferromagnetism, continuous spin-rotation symmetry leaves a degenerate direction of magnetization. Its transverse fluctuations produce one gapless, quadratically dispersing mean-field itinerant ferromagnetic Goldstone mode, $E(\mathbf{q})\propto q^2$. The two broken spin generators form a canonical pair, as for the type-B Goldstone boson of the localized ferromagnet. The system also remains metallic, with exchange-split Fermi surfaces and low-energy spin-conserving particle-hole excitations. Longitudinal-amplitude fluctuations and the spin-flip continuum are distinct from the long-wavelength magnon; their energy and damping scales depend on the band structure. Spin anisotropy, if added, could gap the transverse mode, but it is absent from the stated Hamiltonian.