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If I know the values for the complex dielectric ($\epsilon$$_1$ + $i$$\epsilon$$_2$) as a function of photon energy, how can I calculate the optical conductivity from this data? I'm mainly focused on the real part of conductivity.

From what I've read, the simplest equation to use would be: $$\sigma(\omega)=\frac{\omega\epsilon_2}{4\pi}$$ When using this equation, I've found mostly satisfactory results and agreement with experiment, however, there does seem to be some problems with this equation.

First off, since the conductivity results from direct multiplication of angular frequency ($\omega$), the conductivity will always approach zero as the photon energy approaches zero, which should not be the case for metallic materials.

Secondly, I have seen many very different equations for the optical conductivity which depend on things other than the dielectric values (and are also much more complex). Although, there doesn't seem to be one equation which papers consistently use, which is why I'm rather confused on which method is the proper way of calculating optical conductivity.

If anyone has any experience or advice with this, please let me know. Any help would be appreciated.

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    $\begingroup$ As omega ->0, epsilon -> infinity. So the product need not vanish. Try plugging in the Drude model for permittivity for example. $\endgroup$ – Raziman T V Jan 22 '17 at 18:20
  • $\begingroup$ Thanks, this is a good point. Although in the material I'm working with, the dielectric doesn't go to infinity, but this is probably due to an effect of the computational method in use. Although, I'm still confused as to why different authors use such different equations depending on the paper. $\endgroup$ – Nathan Jan 22 '17 at 18:44
  • $\begingroup$ Can you give some examples of different formulas? Perhaps they all reduce to the same thing. $\endgroup$ – Raziman T V Jan 22 '17 at 19:15
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    $\begingroup$ Eq. 4 of the second paper is exactly what you get if you multiply the Drude epsilon by omega and take the limit as omega tends to zero. The permittivity of materials itself is a function of frequency and depends on various parameters. All those complicated expressions will appear in the conductivity formulae as well. $\endgroup$ – Raziman T V Jan 22 '17 at 20:35
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    $\begingroup$ Yes. Just that some would consider the conductivity to include both real and imaginary parts of permittivity instead of just the imaginary part. $\endgroup$ – Raziman T V Jan 22 '17 at 21:33
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I will use the SI system here. The right hand side of Maxwell's equation $$ curl \vec H=\vec j_{tot}=\sigma \vec E+ \epsilon \frac {∂ \vec E}{∂t} $$ is the sum of the convection current $\sigma \vec E$ and the displacement current $\epsilon \frac {∂ \vec E}{∂t}=i\omega \epsilon \vec E$ where $\sigma$ is the conductivity, $\epsilon$ is the absolute permittivity, and $\omega$ is the angular frequency. Thus a complex permittivity $\epsilon =\epsilon_1+i\epsilon_2$ means that the negative imaginary part $\epsilon_2$ corresponds to a positive conductivity $$ \sigma=-\omega \epsilon_2$$ Conversely, if you express a finite (metal) conductivity $\sigma$ as the negative imaginary part $\epsilon_2$ of the permittivity, you have $\epsilon_2$ going to $-∞$ for $\omega$ going to zero so that $-\omega \epsilon_2$ goes to $\sigma$.

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All the optical constants are complex in the frequency domain. Table 2.1 in Dressel and Gruner (Electrodynamics of Solids) gives a nice conversion for the real and imaginary components of the equivalent permittivity, conductivity and refractive index. They use CGS, where the real part of $\sigma$ is $\omega/4\pi$ times the imaginary part of $\epsilon$. The second part of your question, basically how you compute the conductivity or permittivity, depends on the microscopic model you use. Alex Yuffa and I go through a number of these calculations in both time (where everything is real) and frequency domain in: Linear response laws and causality in electrodynamics AJ Yuffa, JA Scales European Journal of Physics 33 (6), 1635.

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  • $\begingroup$ I tried to answer both parts of the OP without making any assumptions as to what was "really" meant. I'd love to hear why people downvoted this. I'm sure I can learn something. Are there mistakes? $\endgroup$ – JohnS Apr 14 '18 at 18:15

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