Why is the real part of optical conductivity dissipative although the real part of susceptibility is dispersive? 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?
 A: I understand your confusion between the apparent opposite roles of the optical conductivity, $\sigma$, and the electric susceptibility, $\chi$. Both are response functions:
$$ P(t)=\epsilon_0\int_{-\infty}^t \chi(t-t’)E(t’)dt’$$
$$ J(t)=\int_{-\infty}^t \sigma(t-t’)E(t’)dt’$$
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:
$$\nabla\times H=J+\frac{\partial D}{\partial t}.$$
Let’s replace the source terms with the Fourier transforms of the response functions (and using $D=\epsilon_0E+P=\epsilon E$):
$$\nabla\times H=\sigma E+\frac{\partial \epsilon E}{\partial t}.$$
For a monochromatic wave, a time derivative amounts to multiplying by $i\omega$. So we have
$$\nabla\times H=(\sigma+i\omega\epsilon) E.$$
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 $\chi$).
A: The optical conductivity and susceptibility both measure the response of the system to the electric field, but they are measuring different responses. The current density is related to the velocity of the carriers, and the polarization is related to the positions of the carriers:
$\vec{J}=\rho(\vec{x})\dot{\vec{x}}=\sigma\vec{E}$
$\vec{P}=\rho(\vec{x})\vec{x}=\epsilon_0\chi\vec{E}$
The Fourier components of the velocity and position are related by
$\dot{\vec{x}}=-i\omega \vec{x}$,
so there is a $\pi/2$ phase shift between the quantities, which exchanges the information content of the real and imaginary parts of the two response functions.
