= Solution
Work at zero <temperature> and fixed total <carrier density>. Write $n_\uparrow=n/2+\delta$ and $n_\downarrow=n/2-\delta$. Let $g_s(E_F)$ be the <single-spin density of states> per unit volume in the unpolarised state. Since $d\mathcal E_s/dn_s=E_{Fs}$ and $dE_{Fs}/dn_s=1/g_s$, the linear kinetic-energy shifts cancel and the quadratic cost is
$$
\Delta\mathcal E_{\rm kin}=\frac{\delta^2}{g_s(E_F)}+O(\delta^4).
$$
The repulsive interaction instead decreases under <spin polarization>:
$$
\mathcal E_{\rm int}=U\left(\frac{n^2}{4}-\delta^2\right),\qquad
\Delta\mathcal E=\left[\frac1{g_s(E_F)}-U\right]\delta^2+O(\delta^4).
$$
Hence the <Stoner criterion for an opposite-spin contact repulsion> is
$$
\boxed{U_c=\frac1{g_s(E_F)}=\frac2{g_{\rm tot}(E_F)},\qquad U>U_c\text{ destabilises the unpolarised state}.}
$$
The two forms differ only in whether the stated <density of states> counts one spin or both. For the three-dimensional <free electron model>, $g_{\rm tot}(E_F)=3n/(2E_F)$, so
$$
\boxed{U_c=\frac{4E_F}{3n}.}
$$
Equivalently, for fractional polarisation $\zeta=(n_\uparrow-n_\downarrow)/n$, the kinetic-energy density is $\tfrac3{10}nE_F[(1+\zeta)^{5/3}+(1-\zeta)^{5/3}]$; its quadratic coefficient is $nE_F/3$, while the interaction contributes $-Un^2/4$. With number densities, $U$ has units of energy times volume and the displayed interaction is an energy density; multiplying by the sample volume does not change the threshold.
Back to article page
