Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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