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I was reading Introduction to Solid State Physics (8th Edition) by Charles Kittel and the static dielectric constant is defined as $$\varepsilon(0) = \varepsilon(\omega=0) $$ and the optical dielectric constant is defined as $$\varepsilon(\infty) = \varepsilon(\omega=\infty).$$ Then the book shows the values for silicon (a semiconductor) where $\varepsilon(0)=11$ and $\varepsilon(\infty)=11$.

My question is what does this mean? How can the dielectric constant of a semiconductor like silicon be the same at such different frequencies?

I would intuitively think that the dielectric constant at large frequencies is smaller because the semiconductor has no way to respond to the external fields. Which seems to be the case for many semiconductors.

Another question. If I look online for figures of Silicon dielectric function, it seems that $\varepsilon(\omega=0)$ is indeed close to $11$ but it seems that $\varepsilon(\infty)$ is converging to 0. I assume this is because these plots don't cover frequencies that are large enough to go over 0?

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    $\begingroup$ Using the symbol $\infty$ for optical frequencies is a bit odd. Are you sure this definition was meant to be universal and not just a convenient notation for one particular problem or proof? (For example if you were showing how dielectric constant varies, assuming a system with a single resonance, it might make sense) $\endgroup$
    – The Photon
    May 8, 2022 at 16:50
  • $\begingroup$ @ThePhoton In the text it is defined as the dielectric constant at high frequencies ∞. Then in the table it is defined as the dielectric as the optical dielectric constant (with the ∞). I assume that they mean the same thing because they mention the Lyddane–Sachs–Teller relation. But are you saying that they might be using a value for \epsilon_∞ that doesn't correspond to high frequencies? $\endgroup$
    – AA10
    May 8, 2022 at 19:35
  • $\begingroup$ Here "$\infty$" mean "really large compared to phonon frequencies." The dielectric function is definitely not constant when you get up near the plasma frequency (its imaginary part has a huge peak near there). Once you get up to x-ray frequencies you see all sorts of crazy behavior (x-ray edges, etc). $\endgroup$
    – hft
    Sep 3, 2022 at 2:06
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    $\begingroup$ It sure would be handy if there was a material that kept its static dielectric constant at x-ray frequencies... $\endgroup$
    – John Doty
    May 21, 2023 at 1:34
  • $\begingroup$ Silicon is not even transparent at optical frequencies, i.e. its dielectric constant is not even close to a real number. Does Kittel really contain such nonsense? $\endgroup$ May 21, 2023 at 2:13

2 Answers 2

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The optical complex refractive index, n-ik, of Si, and many other materials can be found here. Optical RI of Si At optical frequencies n-ik is rather constant. At (deep) UV frequencies n-ik varies quite a bit because that is where discrete electronic excitation énergies are found. You can pick an optical frequency where $$\epsilon = (n-ik)^2 = n^2-k^2=11\,.$$ At very high frequencies n approaches 1 and k approaches 0.

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  • $\begingroup$ But textbooks say that at high frequencies for Silicon \epsilon = 11. Which means that epsilon(0) = epsilon(infinity). This is the case for some semiconductors like Ge. $\endgroup$
    – AA10
    May 15, 2022 at 22:00
  • $\begingroup$ If you have better data than the sources on the his website, please share, but I believe this to be rather accurate. $\endgroup$
    – my2cts
    May 15, 2022 at 23:17
  • $\begingroup$ You can find the data for Ge on the same site. $\endgroup$
    – my2cts
    May 15, 2022 at 23:26
  • $\begingroup$ @AA10 Textbooks are evidently rather sloppy on this issue. $\endgroup$
    – my2cts
    Jan 20 at 8:23
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I think it is correct that $\varepsilon(\infty) \sim \varepsilon(0)$.

The dielectric constant of an insulator will not vary that much as a function of frequency because the electrons are tightly bound. This is completely different from the case of a metal where there are free charges that can screen the electric field. See this article for theoretical calculation of epsilon for silicon and comparison to experiment.

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  • $\begingroup$ For Ge also happens that epsilon(infinity) ~ epsilon(0). But in most semiconductors epsilon(0)>epsilon(infinity) which makes sense because at high frequencies there is less time to respond to external fields. $\endgroup$
    – AA10
    May 15, 2022 at 22:06
  • $\begingroup$ It may help to check what you think against the measured values on refractiveindex.info. $\endgroup$
    – my2cts
    Jan 20 at 8:15

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