 Solution

Periodic boundary conditions give wavevectors $(2\pi /L)(n_x,n_y)$. In the continuum, large-box limit each state occupies area $(2\pi /L{)}^2$ in wavevector space. A circular annulus therefore contains $[L^2/(2\pi {)}^2]2\pi kdk$ spinless states. Since $ϵ={\hslash }^2{k}^2/(2m)$, the density of states isAt finite $L$ the exact spectrum is discrete; this formula is its continuum density.

With electron spin the two Zeeman branches are ${ϵ}_{\mathrm{kin}}\mp \mu H$. The requested calculation includes this spin coupling and neglects orbital coupling to the field; for a perpendicular field, Landau levels would change the result. It is appropriate for a field parallel to the ideal two-dimensional plane, or for the stated spin-only model. Taking $\mu H\ge 0$, the lower branch starts at $-\mu H$, and the upper one at $\mu H$, soConstruct the ground state by filling the lowest one-electron levels subject to the Pauli exclusion principle. If the Fermi energy ${ϵ}_F$ exceeds $\mu H$, the two populations are $N_{+}=g_0({ϵ}_F+\mu H)$ and $N_{-}=g_0({ϵ}_F-\mu H)$. ConsequentlyThe ground-state energy, including the Zeeman term, follows by integrating the energy over each occupied branch: