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# definition - fermi energy

definition of Wikipedia

# Fermi energy

The Fermi energy is a concept in quantum mechanics usually referring to the energy of the highest occupied quantum state in a system of fermions at absolute zero temperature.

Confusingly, the term "Fermi energy" is often used to describe a different but closely related concept, the Fermi level (also called chemical potential).[1] The Fermi energy and Fermi level are the same at absolute zero, but differ at other temperatures, as described below.

## Introduction

### Context

In quantum mechanics, a group of particles known as fermions (for example, electrons, protons and neutrons) obey the Pauli exclusion principle. This states that two fermions can not occupy the same (one-particle) quantum state. The states are labeled by a set of quantum numbers. In a system containing many fermions (like electrons in a metal), each fermion will have a different set of quantum numbers. To determine the lowest energy a system of fermions can have, we first group the states into sets with equal energy, and order these sets by increasing energy. Starting with an empty system, we then add particles one at a time, consecutively filling up the unoccupied quantum states with the lowest energy. When all the particles have been put in, the Fermi energy is the energy of the highest occupied state. What this means is that even if we have extracted all possible energy from a metal by cooling it to near absolute zero temperature (0 kelvin), the electrons in the metal are still moving around. The fastest ones are moving at a velocity corresponding to a kinetic energy equal to the Fermi energy. This is the Fermi velocity. The Fermi energy is one of the important concepts of condensed matter physics. It is used, for example, to describe metals, insulators, and semiconductors. It is a very important quantity in the physics of superconductors, in the physics of quantum liquids like low temperature helium (both normal and superfluid 3He), and it is quite important to nuclear physics and to understand the stability of white dwarf stars against gravitational collapse.

The Fermi energy (EF) of a system of non-interacting fermions is the increase in the ground state energy when exactly one particle is added to the system. It can also be interpreted as the maximum energy of an individual fermion in this ground state. The chemical potential at zero temperature is equal to the Fermi energy.

## Illustration of the concept for a one dimensional square well

The one dimensional infinite square well of length L is a model for a one dimensional box. It is a standard model-system in quantum mechanics for which the solution for a single particle is well known. The levels are labeled by a single quantum number n and the energies are given by

$E_n = \frac{\hbar^2 \pi^2}{2 m L^2} n^2. \,$

Suppose now that instead of one particle in this box we have N particles in the box and that these particles are fermions with spin 1/2. Then only two particles can have the same energy, i.e., two particles can have the energy of $E_1=\frac{\hbar^2 \pi^2}{2 m L^2}$, or two particles can have energy $E_2=4 E_1$ and so forth. The reason that two particles can have the same energy is that a particle can have a spin of 1/2 (spin up) or a spin of -1/2 (spin down), leading to two states for each energy level. In the configuration for which the total energy is lowest (the ground state), all the energy levels up to n=N/2 are occupied and all the higher levels are empty. The Fermi energy is therefore

$E_F=E_{N/2}=\frac{\hbar^2 \pi^2}{2 m L^2} (N/2)^2. \,$

### Alternative method for calculating Fermi-energy

Calculate the density of states using the ground state energy eigenstate for a free particle $\epsilon_0=\frac{\hbar^2 k^2}{2m}$

$g(\epsilon)=\sum_{states} \delta(\epsilon-\epsilon_0)$ where $\delta$ is the dirac delta function

$g(\epsilon)=\sum_{spin states}\sum_{k} \delta \left( \epsilon - \frac{\hbar^2 k^2}{2m} \right) {\Delta k}$ $=\frac{2L}{2\pi}\int \delta \left( \epsilon - \frac{\hbar^2 k^2}{2m} \right) d k$ $=\frac{L}{\pi}\int_{-\infty}^{\infty} \delta \left( \epsilon - \frac{\hbar^2 k^2}{2m} \right) dk$

$=\frac{L}{\pi}\int_{-\infty}^{\infty} \frac{\delta(k-k_0)+\delta(k+k_0)}{|f'(k_0)|} dk$

Note that we need two delta functions since there are two roots to $f(k_0)=0$

$g(\epsilon)=\frac{L}{\pi}\frac{2}{|f'(k_{0})|}$

where $f(k)=\epsilon - \frac{\hbar^2 k^2}{2m} \rightarrow f'(k)=-\frac{\hbar^2 k}{m}$ and $f(k_0)=0 \rightarrow k_0=\pm\sqrt\frac{2m\epsilon}{\hbar^2}$.

$g(\epsilon)=\frac{L \sqrt{2m}\epsilon^{-1/2}}{\hbar \pi}$

The integral of the density of states up to the Fermi-energy yields the number of particles

$N=\int_{0}^{\epsilon_{F}} g(\epsilon) d\epsilon$ $=\frac{2L\sqrt{2m\epsilon_{F}}}{\hbar \pi}$

$\epsilon_{F}=\frac{\hbar^2}{2m}\left(\frac{N\pi}{2L}\right)^2$

## Three-dimensional case

The three-dimensional isotropic case is known as the Fermi sphere.

Let us now consider a three-dimensional cubical box that has a side length L (see infinite square well). This turns out to be a very good approximation for describing electrons in a metal. The states are now labeled by three quantum numbers nx, ny, and nz. The single particle energies are

$E_{n_x,n_y,n_z} = \frac{\hbar^2 \pi^2}{2m L^2} \left( n_x^2 + n_y^2 + n_z^2\right) \,$
nx, ny, nz are positive integers.

There are multiple states with the same energy, for example $E_{211}=E_{121}=E_{112}$. Now let's put N non-interacting fermions of spin 1/2 into this box. To calculate the Fermi energy, we look at the case where N is large.

If we introduce a vector $\vec{n}=\{n_x,n_y,n_z\}$ then each quantum state corresponds to a point in 'n-space' with energy

$E_{\vec{n}} = \frac{\hbar^2 \pi^2}{2m L^2} |\vec{n}|^2 \,$

The number of states with energy less than Ef is equal to the number of states that lie within a sphere of radius $|\vec{n}_F|$ in the region of n-space where nx, ny, nz are positive. In the ground state this number equals the number of fermions in the system.

$N =2\times\frac{1}{8}\times\frac{4}{3} \pi n_F^3 \,$
The free fermions that occupy the lowest energy states form a sphere in momentum space. The surface of this sphere is the Fermi surface.

the factor of two is once again because there are two spin states, the factor of 1/8 is because only 1/8 of the sphere lies in the region where all n are positive. We find

$n_F=\left(\frac{3 N}{\pi}\right)^{1/3}$

so the Fermi energy is given by

$E_F =\frac{\hbar^2 \pi^2}{2m L^2} n_F^2 = \frac{\hbar^2 \pi^2}{2m L^2} \left( \frac{3 N}{\pi} \right)^{2/3}$

Which results in a relationship between the Fermi energy and the number of particles per volume (when we replace L2 with V2/3):

 $E_F = \frac{\hbar^2}{2m} \left( \frac{3 \pi^2 N}{V} \right)^{2/3} \,$

The total energy of a Fermi sphere of $N$ fermions is given by

$E_t = {\int_0}^{N} E_F dN^\prime = {3\over 5} N E_F$

Therefore, the average energy of an electron is given by:

$E_{av} = \frac{3}{5} E_F$

#### Alternative method for calculating the Fermi-energy

Calculate the density of states using the ground state energy eigenstate for a free particle $\epsilon_0=\frac{\hbar^2 k^2}{2m}$

$g(\epsilon)=\sum_{\text{states}} \delta(\epsilon-\epsilon_0)$ where $\delta$ is the dirac delta function

$g(\epsilon)=\sum_{\text{spin states}}\sum_{k} \delta \left( \epsilon - \frac{\hbar^2 k^2}{2m} \right)\frac{\Delta k^3}{\Delta k^3}$ $=\frac{2V}{(2\pi)^3}\int \delta \left( \epsilon - \frac{\hbar^2 k^2}{2m} \right) d^{3} k$ $=\frac{V}{4\pi^3}\int_{0}^{\infty} k^2 \delta \left( \epsilon - \frac{\hbar^2 k^2}{2m} \right) dk \int_{0}^{2\pi} d\phi \int_{-1}^{1} d(\cos(\theta))$

$=\frac{4\pi V}{4\pi^3}\int_{0}^{\infty} k^2 \frac{\delta(k-k_0)}{|f'(k_0)|} dk$ $=\frac{V}{\pi^2}\frac{k_{0}^{2}}{|f'(k_{0})|}$

where $f(k)=\epsilon - \frac{\hbar^2 k^2}{2m} \rightarrow f'(k)=-\frac{\hbar^2 k}{m}$ and $f(k_0)=0 \rightarrow k_0=\sqrt\frac{2m\epsilon}{\hbar^2}$.

$g(\epsilon)=\frac{V(2m)^{3/2} \sqrt{\epsilon}}{2\pi^2 \hbar^3}$

The integral of the density of states up to the Fermi-energy yields the number of particles

$N=\int_{0}^{\epsilon_{F}} g(\epsilon) d\epsilon$ $=\frac{2}{3}\frac{V(2m)^{3/2}\epsilon_{F}^{3/2}}{2\pi^2\hbar^3}$

$\epsilon_{F}=\frac{\hbar^2}{2m}\left(\frac{3\pi^2 N}{V}\right)^{2/3}$

## Related quantities

A related quantity is Fermi temperature $T_F$, defined as $E_F/k_B$, where $k_B$ is the Boltzmann constant and $E_F$ the Fermi energy. Other quantities defined in this context are Fermi momentum, $p_F$, and Fermi velocity, $v_F$, the momentum and velocity, respectively, of a fermion at the Fermi surface. (These quantities are not well-defined in cases where the Fermi surface is non-spherical). In the case of the Fermi sphere, they are given by:[2]

$p_F = \sqrt{2 m_e E_F}$
$v_F = \frac{p_F}{m_e}$

where $m_e$ is the mass of the electron.

The Fermi momentum can also be described as $p_F = \hbar k_F$, where $k_F$ is the radius of the Fermi sphere and is called the Fermi wave vector. [3]

## Typical Fermi energies

### Nucleus

Another typical example is that of the particles in a nucleus of an atom. The radius of the nucleus is roughly:

$R = \left(1.25 \times 10^{-15} \mathrm{m} \right) \times A^{1/3}$
where A is the number of nucleons.

The number density of nucleons in a nucleus is therefore:

$n = \frac{A}{\begin{matrix} \frac{4}{3} \end{matrix} \pi R^3 } \approx 1.2 \times 10^{44} \ \mathrm{m}^{-3}$

Now since the Fermi energy only applies to fermions of the same type, one must divide this density in two. This is because the presence of neutrons does not affect the Fermi energy of the protons in the nucleus, and vice versa.

So the Fermi energy of a nucleus is about:

$E_F = \frac{\hbar^2}{2m_p} \left( \frac{3 \pi^2 (6 \times 10^{43})}{1 \ \mathrm{m}^3} \right)^{2/3} \approx 3 \times 10^7 \ \mathrm{eV} = 30 \ \mathrm{MeV}$

The radius of the nucleus admits deviations around the value mentioned above, so a typical value for the Fermi energy usually given is 38 MeV.

## References

1. ^ The use of the term "Fermi energy" as synonymous with Fermi level (a.k.a. chemical potential) is widespread in semiconductor physics. For example: Electronics (fundamentals And Applications) by D. Chattopadhyay, Semiconductor Physics and Applications by Balkanski and Wallis.
2. ^
3. ^ Ashcroft, Neil W.; Mermin, N. David (1976). Solid State Physics. Holt, Rinehart and Winston. ISBN 0-03-083993-9.

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