Consider a free electron gas (Section 5.3.1) with unequal numbers of spin-up and spin-down particles (N₊ and N_- and respectively). Such a gas would have a net magnetization (magnetic dipole moment per unit volume)

where μ_B = eћ/2m_e is the Bohr magneton. (The minus sign is there, of course, because the charge of the electron is negative.)
(a) Assuming that the electrons occupy the lowest energy levels consistent with the number of particles in each spin orientation, find E_tot. Check that your answer reduces to Equation 5.56 when N₊ = N_-.
(b) Show that for M/μ_B + + N_-)/V (which is to say, |N₊ - N_-| ,+ + N_-)), the energy density is

The energy is a minimum for M = 0, so the ground state will have zero magnetization. However, if the gas is placed in a magnetic field (or in the presence of interactions between the particles) it may be energetically favorable for the gas to magnetize. This is explored in Problems 5.33 and 5.34.

Equation 5.56

M = (N-N_) V -MBK = Mk. (5.74)