Abstract
The magnetic field variation of the exchange energy of a free-electron gas is calculated by deriving the form of the "Fermi hole" around each electron. The wave functions which are used exactly diagonalize the kinetic energy. The resulting charge distribution is integrated to find the exchange energy. The exchange energy has the same periodicity in the reciprocal magnetic field which is displayed by the kinetic energy. Hence, it is concluded that the de Haas-van Alphen effect is unchanged by the field variation of the exchange energy except for a possible shift in the phase of the oscillations.