Dear Brik
What you found is usual and is simply the signature of a high absorption in 
this high energy region. 
Indeed, a simple way to understand the negative value of eps1 is to look at the 
following relation: 

eps1 = n^2 - k^2

with n the real part and k the imaginary part of the complex refractive index. 
k is the extinction coefficient and is related to absorption while n is related 
to diffusion. In a metal the refractive index AT LOW energy (in the IR region) 
is negative due to a very high absorption in this region. As a consequence k is 
large and thus eps1 is negative. 

For a semiconductor, you have no absorption in the IR and part of the visible 
range (depending on the gap value) but you can have strong absorption at higher 
energies leading to a similar feature. 
To conclude the fact that eps1 is negative is not a proof of a metallic 
character by itself. It will depend where you find this negative eps1 value, it 
should be at very low energy. 

Here is a reference in which I discussed this point based on chemist arguments. 
At this time I was working on the concept of optical channels. You will see on 
figure 1 the impact of changing the compound density on eps2 and eps1. Look to 
Figs. 1a and 1c. The absorption is exactly the same because the chromophore are 
the same. But this material is lamellar (2D) and I just increase the interlayer 
distance leading to have more vaccum in the system, thus lower density. The 
consequence is that for low density eps1 is always positive (k is never higher 
than n) while in for higher density k is negative after 5 eV, because k is 
larger than n due to a larger absorption (twice more in this case).

Best Regards

Brik Hamida <> a écrit :

> Hi
> Im working on semiconductor material. I have calculated  the real part of
> dielectric function (epsilon1) as function of energy. For high energy , I
> found epsilon1 negative !
> Please can someone explain me what means this negative values ? I read that
> any material havig an negative epsilon1 , it becomes a metal , it is right
> ??
> Thanks in advance
> Brik
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