Consider the screened Coulomb interaction in electron liquid, which in the random phase approximation (RPA) takes the form $$ V(q,\omega)=\frac{v(q)}{1-v(q)\Pi(q,\omega)}, $$ where $v(q)$ is the unscreened Coulomb interaction, $\Pi(q,\omega)$ is the electron gas polarizability.

It is known the exact screened interaction $V_\mathrm{exact}(q,\omega)$ obeys the Kramers-Kronig relations $$ \mathrm{Re}\,V_\mathrm{exact}(q,\omega)=v(q)-\frac1\pi\mathcal{P}\int d\omega'\frac{\mathrm{Im}\,V_\mathrm{exact}(q,\omega')}{\omega-\omega'}, $$ $$ \mathrm{Im}\,V_\mathrm{exact}(q,\omega)=\frac1\pi\mathcal{P}\int d\omega'\frac{\mathrm{Re}\,V_\mathrm{exact}(q,\omega')-v(q)}{\omega-\omega'}. $$ The polarizability $\Pi(q,\omega)$, being a retarded response function, also obeys similar relations (although without $v(q)$ in the right hand sides).

Does the RPA interaction $V(q,\omega)$ obey the Kramers-Kronig relations? If yes, how it can be proved?

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    $\begingroup$ I am not certain, but I think the answer is yes. The RPA form is basically a subset of the Feynman diagrams in the exact case. Each Feynman diagram should satisfy the analyticity required by Kramers-Kronig relations. $\endgroup$
    – leongz
    May 22, 2017 at 19:07
  • $\begingroup$ @leongz We can, for example, truncate the RPA series up to the third order $V=v+v\Pi v+v\Pi v\Pi v$ (taking, in fact, a subset of Feynman diagrams) and then we get the second order poles of $V$ where $\Pi$ has the first order ones. The presence of second order poles will probably ruin the required analytical properties, isn't it? $\endgroup$ May 23, 2017 at 14:27
  • $\begingroup$ I don't think the order of the poles matters in Kramers-Kronig relations. Please see derivation on wiki. $\endgroup$
    – leongz
    May 24, 2017 at 9:08

1 Answer 1


Yes. The RPA response function $V(\omega)$ still obeys the Kramers-Kronig (KK) relation, as long as the polarization function $\Pi(\omega)$ obeys the KK relation. The key point is to show that all the higher-order poles that appear in the RPA expansion can be reduced to first-order using the KK relation of $\Pi(\omega)$, such that they do not cause a problem.

The KK relation of the polarization function implies that we can express $\Pi(\omega)$ as $$\Pi(\omega)=\int\frac{\mathrm{d}\omega'}{2\pi}\frac{A_\Pi(\omega')}{\omega-\omega'+\mathrm{i}0_+},$$ where $A_\Pi(\omega)\equiv -2\Im\Pi(\omega)$ is the spectral function of the polarization. Then let us focus on the term $v\Pi(\omega)v\Pi(\omega)v$ in the RPA expansion. We want to show that its sencond-order poles actually can be resolved by first-order poles and hence not problematic. To see this, we start from $$\begin{split}\Pi(\omega)^2&=\int\frac{\mathrm{d}\omega_1}{2\pi}\frac{\mathrm{d}\omega_2}{2\pi}\frac{A_\Pi(\omega_1)}{\omega-\omega_1+\mathrm{i}0_+}\frac{A_\Pi(\omega_2)}{\omega-\omega_2+\mathrm{i}0_+}\\ &=\int\frac{\mathrm{d}\omega_1}{2\pi}\frac{\mathrm{d}\omega_2}{2\pi}\Big(\frac{1}{\omega-\omega_1+\mathrm{i}0_+}-\frac{1}{\omega-\omega_2+\mathrm{i}0_+}\Big)\frac{A_\Pi(\omega_1)A_\Pi(\omega_2)}{\omega_1-\omega_2}\end{split}.$$ Then we can define a new spectral function $$A_\Pi^{(2)}(\omega)=2\int\frac{\mathrm{d}\omega'}{2\pi}\frac{A_\Pi(\omega)A_\Pi(\omega')}{\omega-\omega'}=2A_\Pi(\omega)\Re\Pi(\omega)=-4\Im\Pi(\omega)\Re\Pi(\omega),$$ which resolves the pole $(\omega-\omega')^{-1}$ in the integrant by the KK relation of $\Pi(\omega)$, hence reducing the total order of the poles by one. The spectral function $A_\Pi^{(2)}$ will be as analytic as $\Pi(\omega)$, which is exactly the spectral function of $\Pi(\omega)^2$ with out sencond-order poles: $$\Pi(\omega)^2=\int\frac{\mathrm{d}\omega'}{2\pi}\frac{A^{(2)}_\Pi(\omega')}{\omega-\omega'+\mathrm{i}0_+}.$$ Following the above approach, it is straight forward to show that all the higher-order terms in the RPA expansion has the spectral resolution of the same form in terms of the first-order poles only. $$\Pi(\omega)^n=\int\frac{\mathrm{d}\omega'}{2\pi}\frac{A^{(n)}_\Pi(\omega')}{\omega-\omega'+\mathrm{i}0_+}.$$ All the higher-order spectral functions $A_\Pi^{(n)}(\omega)$ can be express as a polynomial of $\Re\Pi(\omega)$ and $\Im\Pi(\omega)$. So as long as $\Pi(\omega)$ obeys the KK relation, all terms in the RPA expansion will also obeys the KK relation, as well as the RPA response function $V(\omega)$.

  • $\begingroup$ Thank you! But I don't understand what do you mean by "resolving second-order pole by first-order poles". For example, if $A_\Pi(\omega)=(a/\pi)/[(\omega-\omega_0)^2+a^2]$ (Lorentzian spectral function), then $\Pi(\omega)=(1/2\pi)(\omega-\omega_0+ia)^{-1}$, i.e. $\Pi(\omega)$ has the first-order pole at $\omega=\omega_0-ia$, which means existence of a damped excitation. At the same time, $\Pi(\omega)=(1/2\pi)^2(\omega-\omega_0+ia)^{-2}$ still has the second-order pole at $\omega=\omega_0-ia$, as can be obtained either by direct squaring of $\Pi(\omega)$ or by using the function $A_\Pi^{(2)}$. $\endgroup$ May 28, 2017 at 12:06

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