Timeline for Chemical potential for ideal fermi gas with single particle energies $\epsilon_k^{\pm}=\pm \hbar c |\mathbf k|$
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| Dec 14, 2017 at 14:42 | comment | added | Kite.Y | @Lenol Yup, eranreches's already answers your question. Or, as I mentioned above, if you really just wanted to stare at the question itself, and deem it completely as a "single electron energy level filling" problem without considering holes in solids or energy constant shifting, then the problem, strictly speaking, is not perfectly defined --- there should be always a lower bound of energy for any system. So to solve it, you must explain it properly first. | |
| Dec 14, 2017 at 11:23 | comment | added | eranreches | @Lenol By the same reasoning you can construct a general expression for the energy of the system $$U=\int_{-\infty}^{0}(-\epsilon)f_{\rm h}(\epsilon)D(\epsilon){\rm d}\epsilon+\int_{0}^{\infty}\epsilon f_{\rm e}(\epsilon)D(\epsilon){\rm d}\epsilon$$ This expression converges because $$f_{\rm h}(\epsilon\rightarrow-\infty)\rightarrow 0$$ and $$f_{\rm e}(\epsilon\rightarrow\infty)\rightarrow 0$$ exponentially fast. | |
| Dec 14, 2017 at 11:22 | comment | added | eranreches | @Lenol You just don't do that this way. Instead, lets fix $U(T=0)=0$. We can do this because energy is defined up to a constant. Now imagine you excite an electron from level $-\epsilon$ into level $\epsilon$. It means that you create a hole at $-\epsilon$ with energy $-\left(-\epsilon\right)=\epsilon$ and an electron at $\epsilon$ with energy $\epsilon$. The total energy of the system has increased by $2\epsilon$. | |
| Dec 14, 2017 at 4:01 | comment | added | Lenol | If I to calculate the internal energy of the system, $U=\int_{-\infty}^{\infty}\epsilon f(\epsilon,\mu=0) D(\epsilon) d\epsilon $, the integral would be divergent. How can I make sense of this? | |
| Dec 14, 2017 at 1:38 | history | answered | Kite.Y | CC BY-SA 3.0 |