Fermi energy definition

Question

Ok, so I'm having a hard time understanding the definition of Fermi Energy. Several sites basically repeat each other, saying that it is the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero 1, and others say that it is the highest energy that the electrons assumes at 0K 2. Are these 2 concepts the same, and I'm just not getting it? Is the top level of an electron at 0K equal to the difference in energy between highest and lowest occupied states? Any clarification would be greatly appreciated.

Answer 1

One can derive the fermi-energy using quntumstatistics and is a more complex derivation. But in condensed matter you find a more vivid explanation about the meaning.

Let us assume a free fermi gas or electron gas. We make the following assumptions

1. $N\gg 1$ conducting electrons move on a homogeneous charge background
2. No interaction between particles
3. Pauli-Principle, meaning $2$ electrons per energy state
4. Describing crystal as cube with edge length $L$
5. Choose periodic boundary conditions

We now start with the one particle Schroedinger equation

$$-\dfrac{\hbar^2}{2m_e}\Delta \psi_{\vec{k}} = \epsilon_{\vec{k}} \psi_{\vec{k}}.$$

We find plane waves as solutions

$$\psi_{\vec{k}} = \dfrac{1}{\sqrt{L^3}} \text{exp}(i{\vec{k}}\cdot {\vec{r}}).$$

Where $\epsilon_{\vec{k}} = \frac{\hbar^2k^2}{2m}$. The periodic boundary conditions are

$$\psi(x,y,z) = \psi(x+L,y+L,z+L)$$

and it must follow that for $a \in \{x,y,z\}$

$$k_a \cdot (a+L) = k_aa+2\pi n_a \Leftrightarrow k_a = \dfrac{2\pi}{L}$$

with $n_a \in \mathbb{N} $. As we can see the values of the wavevector are discrete, which is due to the finite volume. The k-space is also made up of discrete points where we have one point per volume $(2\pi/L)^3$. We now want to fill all these electrons in these states. Each state can contain two electrons due to the Pauli-Principle (one with spin-up, one with spin-down).

We start by fillig the states with the lowest energy at ${\vec{k}=0}$. For many particles the electrons will fill a sphere in the k-space, the so called Fermi-Sphere. The radius of this sphere is called the Fermi-Wavevector. One can calculate the radius as follows

1. We have one state per volume $V_1 = (2\pi/L)^3$, which is filled by $2$ electrons
2. The sphere of volume $V_2 = \dfrac{4}{3}\pi k^3_F$ contains all $N$ electrons
3. The number $N$ of electrons then is:$$N = 2\cdot \dfrac{\dfrac{4}{3}\pi k^3_F}{\bigg(\dfrac{2\pi}{L}\bigg)^3} \Leftrightarrow k_F = \bigg(\dfrac{3\pi^2N}{V}\bigg)^{1/3} = (3\pi^2n)^{1/3}$$ with $V = L^3$ the volume of the cube in which the electrons are located and $n = N/V$ the charge density.

Now to the interesting part. In case of $T = 0\,$K the maximum occupied energy is given as the Fermi-Energy

$$\epsilon_F = \dfrac{\hbar^2k^2_F}{2m}.$$

As you can see the surface of the Fermi-Sphere is a surface of constant energy.

Answer 2

From the Schrödinger equation, energy for a bound electron is quantized, such that only certain energy levels are allowed. Because electrons are fermions, they obey the Pauli exclusion principle, which states that no two electrons can have all their quantum numbers (such as energy level, orbital, spin) equal.

This means that on each energy level, there are only a certain amount of electrons that can occupy that energy level (depending on how many orbitals that level has). If another electron would be added, that electron would have to occupy a different energy state (usually a higher one). The Fermi energy is then the energy of the highest occupied state, when the system is in the ground state.

When the system is in the ground state implies that all levels under the highest occupied state are also occupied.