# What is the importance of Fermi sphere?

I am very confused with the meaning of Fermi sphere. I understand that it is exactly the same as the energy levels and Fermi energy in the real space and Fermi sphere is in K space but I don't know why is this important. What I have understood is the relation of Fermi sphere and real energy level is similar to that of crystal lattice in real space and reciprocal lattice but still anyone who could explain the significance of Fermi sphere in a very clear way.

Consider a system of electrons confined in a cube of side length $L=V^{1/3}$. Let's assume that this "electron gas" is diluted enough that we can neglect the electron-electron interactions. This means that we have to solve the Schroedinger's equation for a free particle:

$$-\frac{\hbar^2}{2m} \nabla^2 \psi(\mathbf r) = E\ \psi(\mathbf r)\tag{1}\label{1}$$

Moreover, since we are interested in the bulk properties of the material, we will assume periodic boundary conditions (PBC):

$$\psi(x+L,y,z) = \psi(x,y,z)\\ \psi(x,y+L,z) = \psi(x,y,z)\\ \psi(x,y,z+L) = \psi(x,y,z) \tag{2}\label{2}$$

It is well known that the solution of Eq. \ref{1} is a plane wave:

$$\psi_{\mathbf k}(\mathbf r) = \frac 1 {\sqrt{V}} e^{i \mathbf k \cdot \mathbf r} \tag{3}\label{3}$$

with energy

$$E(\mathbf k) = \frac{\hbar^2k^2}{2m} \tag{4}\label{4}$$

If you apply the conditions \ref{2} to the solution \ref{3}, you will get

$$e^{i k_x} L=e^{i k_y} L=e^{i k_z} L=1 \tag{5}\label{5}$$

and therefore

$$k_\alpha = \frac{2 \pi n_\alpha} L \ \ \ (\alpha =x,y,z) \tag{6}\label{6}$$

where $n_\alpha$ are integers. Therefore, the allowed wavevectors form a discrete "grid" in reciprocal space (figure below - from D.J. Griffiths, Introduction to Quantum Mechanics). Notice how reciprocal space comes out naturally because of the relation \ref{4} between the energy of an electron and its wavevector.

At $T=0$, the electrons will occupy the lowest available energy levels, starting with with $\mathbf k = \mathbf 0$. Since electrons are fermions with spin $1/2$, to satisfy Pauli's exclusion principle we can accommodate only two of them in every energy level. Therefore, starting from $\mathbf k=\mathbf 0$, we can imagine to place 2 electrons in every point in reciprocal space allowed from eq. \ref{6}. If the number of electrons is very large, it is easy to see that the result of this filling looks very much like a sphere: this is what we call the Fermi sphere.

The radius of this sphere, $k_F$, is related to energy by equation \ref{4}. The energy of the electrons on the surface of the Fermi sphere is the Fermi energy:

$$E_F= \frac{\hbar^2k_F^2}{2m} \tag{7}\label{7}$$

References

For a detailed discussion of Fermi surfaces in general, see for example Ashcroft-Mermin, Solid State Physics.

• I am beginner to solid state physics and for now I think i'd like to understand for free electrons. Before going for computing the Fermi surfaces,I'd like to know about it first. What i don't understand is how is it different from real space energy levels. – Rima Jan 27 '18 at 12:14
• @Rima What do you mean exactly by "real space" energy levels? – valerio Jan 27 '18 at 12:27
• @Rima One can connect to electron velocities. Kinetic energy is $mv^2/2 = p^2/(2m)$. But these electrons must have de-Broglie wavelengths $\lambda = h/p$ that fit into a three-dimensional box of metal in real space. This makes that only discrete values of the wavevector $k = 2\pi/\lambda = \hbar p$ are allowed. – Pieter Jan 27 '18 at 12:50
• fermi surface is in reciprocal space,right? like the crystal lattice has real space or bravais lattice ,is there no any such space for fermi energy? I don't know how do i explain it. – Rima Jan 27 '18 at 16:05
• Funny thing is this was the derivation that i read today for my test and I had not understood it in detail. But now I think i understand it. – Rima Jan 29 '18 at 15:36