Polarization, the Lindhard Function, and Particle–Hole Continua

The Lindhard function is the exact density response of a noninteracting Fermi gas. Its real part describes reactive screening, while its imaginary part measures the particle–hole states available at a specified momentum and energy. It is therefore both a solvable response function and the kinematic input to screening, collective-mode, and weak-coupling calculations; the original dielectric calculation is Lindhard 1954, pp. 16–25, PDF.

Required background. Finite-Density Diagrammatics and Medium Insertions fixes the propagator and occupation conventions.

Helpful background. Sources, Linear Response, and Kubo Formulae fixes the retarded-response convention, and The Fermi Gas and Fermi-Surface Kinematics supplies the zero-temperature phase space.

The retarded polarization bubble

For particles with dispersion $\xi_{\mathbf p}=\varepsilon_{\mathbf p}-\mu$ , degeneracy $g$ , and density vertex one, define the retarded density response by

\Pi_0^R(\mathbf q,\omega) =g\int\!\frac{\mathrm d^d p}{(2\pi)^d} \frac{f(\xi_{\mathbf p})-f(\xi_{\mathbf p+\mathbf q})} {\omega+\xi_{\mathbf p}-\xi_{\mathbf p+\mathbf q}+i0^+}.

This sign convention gives $\Pi_0^R(\mathbf 0,0)=-\partial n/\partial\mu$ . It follows either by doing the fermionic Matsubara sum and continuing $i\Omega_m\to\omega+i0^+$ , or directly from the Lehmann representation of the density operator. The imaginary part is

\operatorname{Im}\Pi_0^R(\mathbf q,\omega) =-\pi g\int\!\frac{\mathrm d^d p}{(2\pi)^d} \bigl[f(\xi_{\mathbf p})-f(\xi_{\mathbf p+\mathbf q})\bigr] \delta\!\left(\omega+\xi_{\mathbf p}-\xi_{\mathbf p+\mathbf q}\right).

For $\omega>0$ it is nonpositive: a passive system has the positive dynamic structure factor $S=-2\operatorname{Im}\Pi^R$ at zero temperature in this normalization.

Particle–hole continuum

Take $T=0$ and $\varepsilon_p=p^2/(2m)$ . A positive-frequency excitation starts inside the Fermi sphere and ends outside it. Since

\varepsilon_{\mathbf p+\mathbf q}-\varepsilon_{\mathbf p} =\frac{q^2}{2m}+\frac{pq}{m}\cos\theta,

the allowed energies in three dimensions lie between

\omega_-(q)=\max\!\left(0,\frac{q^2}{2m}-v_Fq\right), \qquad \omega_+(q)=\frac{q^2}{2m}+v_Fq.

The lower boundary is zero for $q<2k_F$ because the two Fermi spheres intersect. Outside this region, $\operatorname{Im}\Pi_0^R=0$ at $T=0$ . Finite temperature rounds the boundary, and collisions broaden it; neither effect changes the underlying energy–momentum constraint.

Static response in three dimensions

Let $x=q/(2k_F)$ and let $\nu_F=gmk_F/(2\pi^2)$ be the total density of states at the Fermi energy. Direct angular and radial integration gives

\Pi_0^R(q,0) =-\nu_F\left[ \frac12+\frac{1-x^2}{4x} \ln\left\lvert\frac{1+x}{1-x}\right\rvert \right].

The expression is continuous at $q=2k_F$ but its derivative is nonanalytic. That Kohn anomaly is the momentum-space origin of long-distance Friedel oscillations. The checks

\Pi_0^R(0,0)=-\nu_F, \qquad \Pi_0^R(q,0)\sim-\frac{4\nu_Fk_F^2}{3q^2} \quad(q\gg k_F)

verify the compressibility sign and the ultraviolet decay.

Dynamic and static limits do not commute

At fixed nonzero $\omega$ and $q\to0$ , number conservation forces the density response to vanish. Expanding above the continuum, $|\omega|\gg v_Fq$ , gives

\Pi_0^R(q,\omega) =\frac{nq^2}{m(\omega+i0^+)^2}+O(q^4/\omega^4).

In contrast, taking $\omega\to0$ first gives $-\nu_F$ . Thus

\lim_{\omega\to0}\lim_{q\to0}\Pi_0^R(q,\omega)=0, \qquad \lim_{q\to0}\lim_{\omega\to0}\Pi_0^R(q,\omega)=-\nu_F.

The first limit probes a spatially uniform, time-dependent scalar potential, which is removable by a gauge transformation in a number-conserving system. The second probes static compressibility.

Checks and limits

The exact bubble obeys the free-gas $f$ -sum rule and the Kramers–Kronig relation. Its continuum is kinematic, not a quasiparticle lifetime: an interacting self-energy changes the single-particle lines, while a compatible vertex is also required for a conserving density response. Band structure changes the continuum boundaries, nesting singularities, and static nonanalyticities; the spherical formulas above should not be used unchanged on a lattice.

Common pitfalls

Dropping the degeneracy convention. The density of states and density must use the same spin or flavor factor $g$ .

Reversing the response sign. With $\delta n=\Pi^R\delta U$ , a positive external potential lowers the equilibrium density, so $\Pi^R(0,0)<0$ .

Treating a continuum edge as a collective pole. The branch cut represents independent particle–hole excitations. A collective mode requires a zero of an interacting inverse response on the appropriate analytic sheet.

Exercises

Recover the static compressibility

Show directly that $\Pi_0^R(0,0)=-\partial n/\partial\mu$ .

Solution

Set $\mathbf q$ small before taking the ratio. Since $f(\xi_{\mathbf p})-f(\xi_{\mathbf p+\mathbf q})\simeq-f'(\xi_{\mathbf p})(\xi_{\mathbf p+\mathbf q}-\xi_{\mathbf p})$ , the ratio tends to $f'(\xi_{\mathbf p})$ . Hence $\Pi_0=g\int \mathrm d^dp/(2\pi)^d\,f'(\xi_p)=-\partial n/\partial\mu$ . At $T=0$ this is $-\nu_F$ .

Locate the upper continuum edge

For $q<2k_F$ , maximize the excitation energy subject to $p\le k_F$ and $|\mathbf p+\mathbf q|\ge k_F$ .