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I`m studying solid state physics and I have a question about this. I know that the Fermi energy is the energy which can be defined at $0K$.

In $k$ space electrons occupy each k state from zero at 0K so, if we draw picture in $k$ space it should be left side of attached picture.

In this, Fermi level is located at $K_F$ and Fermi energy is summation of all of $k$ state.

When we apply electric field, electron can get additional energy $-eV$, so sphere is larger than before and has a hole which radius is $-eV.$ In this, right side of the attached picture, Fermi energy and Fermi energy are shifted. Is it right? enter image description here

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Short Answer

This is wrong. The whole Fermi surface shifts to the right (assuming this is opposite the direction of the $E$-field).

Long (mathematical) Answer

Imagine we have a free electron gas of $N$ electrons in an electric field $ \newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\veb}[1]{\mathbf{#1}} \veb{E}=-\e\hat{\veb{e}}_x$. In the case where $\e=0$ we have an occupation number: $$n(\veb{k})=\f{1}{\exp\l\beta\l \hbar\veb{k}^2/2m-\mu\r\r+1} $$ The question we need to ask ourselfs is how does this change when $\e\ne 0$. It turns out that we get: $$n(\veb{k})=\f{1}{\exp\l\beta\l \hbar(k_x-v_dm/\hbar)^2/2m+\hbar k_y^2/2m-\mu\r\r+1} $$ i.e. $k_x \rightarrow k_x-v_dm/\hbar$ where $v_d$ is the drift velocity. I am going to refrain from an explicit proof of this since a decent one can be found here. In the limit of $T=0$ we would usually say that $n(\veb{k})=1$ for $\l \hbar\veb{k}^2/2m-\mu\r \lt 0$ and $n(\veb{k})=0$ otherwise. In our case this condition is (trivially) changed to: $$\l \hbar(k_x-v_dm/\hbar)^2/2m+\hbar k_y^2/2m-\mu\r\lt 0$$ $$(k_x-v_dm/\hbar)^2+k_y^2\lt \f{2\mu m}{\hbar}$$ which describes a circle in $k$-space shifted by an amount $v_dm/\hbar$.

Other Resources

A physical explanation should be in most introductory solid state physics books e.g. The Oxford Solid State Basics by S.H.Simon pg 35.

Key words to search for: Drude Theory, Sommerfeld theory, conductivity.

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  • $\begingroup$ The proof you link to seems to have been removed (or is otherwise inaccessibility). I think it is also worth mentioning that this assumes the existence of some sort of extra scattering process that creates drag on the electrons. $\endgroup$ – By Symmetry Oct 24 '18 at 9:34
  • $\begingroup$ @BySymmetry Thanks, I have replaced it with the Wayback version. $\endgroup$ – Quantum spaghettification Oct 24 '18 at 12:45

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