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Why is the real part of optical conductivity dissipative although the real part of susceptibility is dispersive?

Physics Asked on August 23, 2021

I have a puzzle about the optical conductivity. In Drude model, it is often said that the real part of tells us about the dissipation of energy in the system; the imaginary part of the conductivity tells us about the response of the system. for example in Tong’s EM lecture or Tsymbal’s lecture. On the other hand, the optical conductivity is reponse function. From the linear reponse theory, for example Tong’s lecture, we know that the imaginary part arises due to dissipative processes. There seems a contradiction here. How to understand it?

One Answer

I understand your confusion between the apparent opposite roles of the optical conductivity, $sigma$, and the electric susceptibility, $chi$. Both are response functions: $$ P(t)=epsilon_0int_{-infty}^t chi(t-t’)E(t’)dt’$$ $$ J(t)=int_{-infty}^t sigma(t-t’)E(t’)dt’$$ where $P$ is the polarization density, and $J$ is the current density.

Their difference comes from how $P$ and $J$ interact with an electromagnetic wave. Check out Ampere’s Law: $$nablatimes H=J+frac{partial D}{partial t}.$$

Let’s replace the source terms with the Fourier transforms of the response functions (and using $D=epsilon_0E+P=epsilon E$): $$nablatimes H=sigma E+frac{partial epsilon E}{partial t}.$$

For a monochromatic wave, a time derivative amounts to multiplying by $iomega$. So we have $$nablatimes H=(sigma+iomegaepsilon) E.$$

So there you have it! $E$ generates $H$ in an electromagnetic wave through $sigma$ and $epsilon$. They play exactly the same role in Ampere’s Law, except due to the time derivative in Maxwell’s addition, $epsilon$ has an extra factor of $i$. Voila! The imaginary part of $sigma$ acts like the real part of $epsilon$ (or $chi$).

Correct answer by Gilbert on August 23, 2021

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