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Changing one Landau quasiparticle occupation changes the energies of others. To second order around equilibrium,Here $\mathcal{V}$ is volume, occupations are per momentum and spin state, and $f$ has units of energy times volume.^[ref]

For a toy two-state change at fixed volume, take $\delta n_a=\delta n_b=\eta$, $f_{aa}=f_{bb}=0$, $f_{ab}=f_{ba}=g\mathcal{V}$, and unperturbed energies ${\xi }_a={\xi }_b=e$. The expression gives $2e\eta +g{\eta }^2$. The factor $1/2$ counts the interaction of the same pair once; it does not halve each excitation's energy. This finite-state illustration only explains the bookkeeping of the Landau quasiparticle energy functional.

The quasiparticle residue is the coherent pole weight of the single-particle Green's function. Its coherent spectral term $Z\delta (E-{\xi }_p)$ contributes $Z\Theta (-{\xi }_p)$ to the zero-temperature momentum occupation. A smooth remainder leaves a jump $Z$ at the Fermi surface; Landau quasiparticle occupations have a unit step.^[ref]

For the local illustrative distribution in the figure, put $x=(p-p_F)/p_F$ and$\Theta$ is the Heaviside step function. Its one-sided limits are $0.825$ and $0.175$, giving $Z=0.65$. The smooth term is stipulated; this local curve does not specify a normalized distribution over all momenta. In particular, $Z$ is neither the fraction of atoms above the Fermi surface nor $m/m^{\ast }$.