# Highest occupied orbitals or bands

I am not a student of physics, but a student of nano-technology, hence I might sound extremely stupid, but I just want to clarify my doubt even if sounds very trivial.

The no. of free charge carriers, $$n$$, around an energy $$E_{0}$$ could be given as

$$n=\int_{E_{0}-\Delta/2}^{E_{0}+\Delta/2}\rho\left(E\right)f(E)\,\mathrm{d}E$$ where $$\rho(E)$$ is the density of states and $$f(E)$$ is the Fermi distribution. Now, because of the shape of the Fermi distribution, I would assume that using the integral below, $$n$$ would be zero only at $$E_0=\infty$$, and hence at finite temperatures, no state $$E_0$$ would be completely filled or completely empty and have a non-zero probability distribution value $$f(E)\rho(E)$$ of being occupied. But at $$T=0K$$, the story is different, since the fermi distribution becomes a kind of step. Does that mean that at finite temperatures, we cannot logically assign a highest filled orbital and that the concept of Highest or lowest occupied states exist only in context of the zero kelvin story??

• yes, but at higher temperatures, does the conductivity, of the insulator or semiconductor, increase because the fermi distribution evolves to push finite carrier probabilities for $E \geq E_c$, or do you suggest that the fermi level $E_F$ ALSO shifts in accordance to the sommerfeld expansion? – ubuntu_noob Dec 28 '18 at 10:05