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In figure 3 of this document, there is data relating $\Re(\sigma(\omega))$ to the Fermi energy. It is claimed that $\Re(\sigma(\omega))$ is determined via reflectivity measurements. How is this done? What is the formula relating the two?

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Without knowing anything about this system, the terms scream out "optical theorem" to me, noting that the Fermi energy limits low energy scatterings because there is no where for gently scattered electrons to go. –  dmckee Jan 23 '13 at 0:27
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The optical conductivity $\sigma$ is basically equivalent to the dielectric function $\epsilon$:

$$ \sigma(\omega) = i\omega\epsilon_0 (1-\epsilon(\omega)) $$

So the real part of the conductivity contains the same information as the imaginary part of the dielectric function:

$$ \sigma'(\omega) = \epsilon_0 \epsilon''(\omega) \omega $$

You can determine the dielectric function from the reflectance; it is the square root of the complex index of refraction, which you can determine from angle-dependent reflectance measurements for s and p polarization (basically, ellipsometry.)

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