2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Your third link (to "Tong's Lecture") is 22 pages long. Can you be more specific which part of it (specific equations or sections) you're referring to? $\endgroup$
    – The Photon
    Jul 20, 2020 at 16:53
  • $\begingroup$ I suspect there is a confusion between conductivity and susceptibility ($\chi_e$) which is also a response function (if I understand what that means) and does have the properties you mention: the real part indicates a lossless response and the imaginary part indicates a dissipative response. $\endgroup$
    – The Photon
    Jul 20, 2020 at 16:57
  • $\begingroup$ page 82,sorry for the late reply $\endgroup$
    – thone
    Jul 21, 2020 at 9:43

2 Answers 2

2
$\begingroup$

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 $\chi$).

$\endgroup$
2
  • $\begingroup$ nice work! Thank you! $\endgroup$
    – thone
    Jul 23, 2020 at 0:16
  • $\begingroup$ @thone You’re very welcome! May I recommend changing the title of your question to be a little more descriptive? This might allow it to be more useful to others. Perhaps, “Why is the real part of optical conductivity dissipative although the real part of susceptibility is dispersive?” $\endgroup$
    – Gilbert
    Jul 23, 2020 at 2:32
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.