BE Problem
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I am currently working on modelling the density of states and optical conductivity of graphene utilizing the GW algorithm. In calculating the exchange self energy of the system, the formula I am currently using is
$$f(w,T) = \frac{1}{e^{\frac{\omega}{T}}-1}$$
where the planck constant and the boltzmann constant $k$ is set to 1. To conservative bosonic particles, such as the core of helium-4, then it is believed to form a Bose-Einstein Condensate. I am dealing with the non-conservative bosonic particles, such as photons and phonons.
What would happen if I set the temperature $T$ equals to zero for the non-conservative bosons? My advisor believes that there would be no Bose-Einstein Condensate because the boson can pop in and out of the system. Is this true? If it is, what happens to the bosonic particles at the--or at least near--zero temperature?
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Update and Edit
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I've consulted my adviser and my colleagues, and this is the result. In no way that this is the solution, but it is one step further nonetheless. I've made a model of the BE distribution and it follows similarly to the image in the middle:
[![MB, BE, and FD Distributions][1]][1]
<sub>(source: [universe-review.ca](https://universe-review.ca/I13-23-dist.jpg))</sub>
What I did is vary T = 1 K, 0.1 K, and 0.01 K in Scidavis (a numerical software on linux, if you are wondering). As the T decreases, the graph gets steeper and steeper; analytically, inserting T = 0 in the formula would equal to infinity. This of course, is the puzzling question.
Since I am required to put this in my calculation, my adviser suggested that at T = 0 the distribution equals to 0, where we assumed that the photons disappear after being absorbed by the electrons.
Why does this matter so much? Because I am now calculating the self energy of the system and the final formula requires a Hilbert transform integration from -inf to +inf. If there are anyone working on this problem or something similar, this would really help.
[1]: https://i.sstatic.net/zBre2.jpg