Fermi energy

The Fermi energy is a concept in quantum mechanics referring to the energy of the highest occupied quantum state in a system of temperature. This article requires a basic knowledge of quantum mechanics.

Introduction

Context

In quantum mechanics, a group of particles known as neutrons are fermions) obey the Pauli exclusion principle. This principle states that two identical fermions can not be in the same 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 in 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 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 helium (both normal ³He and superfluid ⁴He), and it is quite important to nuclear physics and to understand the stability of white dwarf stars against gravitational collapse.

Advanced context

The Fermi energy (E_F) of a system of non-interacting 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 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 spin-1/2 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. When we look at the total energy of this system, the configuration for which the total energy is lowest (the ground state), is the configuration where all the energy levels up to n=N/2 are occupied and all the higher levels are empty. The Fermi energy is therefore

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

The three-dimensional case

The three-dimensional isotropic case is known as the fermi ball.

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 n_x, n_y, and n_z. The single particle energies are

$E_{n_x,n_y,n_z} = \frac{\hbar^2 \pi^2}{2m L^2} \left( n_x^2 + n_y^2 + n_z^2\right) \,$

n_x, n_y, n_z are positive integers.

There are multiple states with the same energy, for example $E 100 = E 010 = E 001$ . 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 for 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 E_f is equal to the number of states that lie within a sphere of radius $|\vec{n}_f|$ in the region of n-space where n_x, n_y, n_z are positive. In the ground state this number equals the number of fermions in the system.

$N =2 \frac{1}{8} \frac{4}{3} \pi n_f^3 \,$

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 V^2/3):

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

The total energy of a fermi ball of $N 0$ fermions is given by

$E = {\int_0}^{N_0} E_f(N) dN = {3\over 5} N_0 E_f$

Typical fermi energies

White dwarfs

Stars known as White dwarfs have mass comparable to our electron gas. The number density of electrons in a White dwarf are on the order of 10³⁶ electrons/m³. This means their fermi energy is:

$E_f = \frac{\hbar^2}{2m_e} \left( \frac{3 \pi^2 (10^{36})}{1 \ \mathrm{m}^3} \right)^{2/3} \approx 3 \times 10^5 \ \mathrm{eV} \,$

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 protons in the nucleus, and vice versa.

So the fermi energy of a nucleus is about:

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

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.

Fermi level

The Fermi level is the highest occupied energy level at absolute zero, that is, all energy levels up to the Fermi level are occupied by electrons. Since fermions cannot exist in identical energy states (see the exclusion principle), at absolute zero, electrons pack into the lowest available energy states and build up a "Fermi sea" of electron energy states. [1] In this state (at 0 K), the average energy of an electron is given by:

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

where $E f$ is the Fermi energy.

The Fermi momentum is the Fermi surface. The Fermi momentum is given by:

$p_F = \sqrt{2 m_e E_f}$

where $m e$ is the mass of the electron.

This concept is usually applied in the case of dispersion relations between the energy and momentum that do not depend on the direction. In more general cases, one must consider the Fermi energy.

The Fermi velocity is the average velocity of an electron in an atom at absolute zero. This average velocity corresponds to the average energy given above. The Fermi velocity is defined by:

$V_f = \sqrt{\frac{2 E_f}{m_e}}$

where $m e$ is the mass of the electron.

Below the Fermi temperature, a substance gradually expresses more and more quantum effects of cooling. The Fermi temperature is defined by:

$T_f = \frac{E_f}{k}$

where k is the Boltzmann constant.

Quantum mechanics

According to quantum mechanics, fermions -- particles with a half-integer spin, usually 1/2, such as Fermi-Dirac statistics. The ground state of a non-interacting fermion system is constructed by starting with an empty system and adding particles one at a time, consecutively filling up the lowest-energy unoccupied quantum states. When the desired number of particles has been reached, the Fermi energy is the energy of the highest occupied molecular orbital (HOMO). Within conductive materials, this is equivalent to the lowest unoccupied molecular orbital (LUMO); however, within other materials there will be a significant gap between the HOMO and LUMO on the order of 2-3 eV.

Free electron gas

In the electrical conductivity. The study of the Fermi surface is sometimes called Fermiology. The Fermi surfaces of most metals are well studied both theoretically and experimentally.

The Fermi energy of the free electron gas is related to the chemical potential by the equation

$\mu = E_F \left[ 1- \frac{\pi ^2}{12} \left(\frac{kT}{E_F}\right) ^2 - \frac{\pi^4}{80} \left(\frac{kT}{E_F}\right)^4 + \cdots \right]$

where E_F is the Fermi energy, k is the Fermi-Dirac statistics.

References

Kroemer, Herbert; Kittel, Charles (1980). Thermal Physics (2nd ed.). W. H. Freeman Company. ISBN 0-7167-1088-9.
Table of fermi energies, velocities, and temperatures for various elements.
a discussion of fermi gases and fermi temperatures.

Categories: Statistical mechanics

This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Fermi_energy". A list of authors is available in Wikipedia.