# Electric field dependence of Fermi level, Fermi energy

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?

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

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 $$v_dm/\hbar$$.