blob: 9cdf2c60241572a89444d9a9cb32295220503676 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55


format: rst
toc: no
title: Fermi Gas
...
Derivation of the Fermi Energy

Consider a crystal lattice with an electron gas as a 3 dimensional infinite
square well with dimensions $l_{x}, l_{y}, l_z$. The wavefunctions of
individual fermions (pretending they are noninteracting) can be seperated
as $\psi(x,y)=\psi_{x}(x)\psi_{y}(y)\psi_{z}(z)$. The solutions will be
the usual ones to the Schrodinger equation:
$$\frac{\hbar^2}{2m}\frac{d^2 \psi_x}{dx}=E_x \psi_x$$
with the usual wave numbers $k_x=\frac{\sqrt{2mE_x}}{\hbar}$, and quantum
numbers satisfying the boundry conditions $k_x l_x = n_x \pi$. The full
wavefunction for each particle will be:
$$\psi_{n_{x}n_{y}n_{z}}(x,y,z)=\sqrt{\frac{4}{l_{x}l_{y}}}\sin\left(\frac{n_{x}\pi}{l_{x}}x\right)\sin\left(\frac{n_{y}\pi}{l_{y}}y\right)\sin\left(\frac{n_{z}\pi}{l_{z}}z\right)$$
and the associated energies (with $E = E_x + E_y + E_z$):
$$E_{n_{x}n_{y}n_z}=\frac{\hbar^{2}\pi^{2}}{2m}\left(\frac{n_{x}^{2}}{l_{x}^{2}}+\frac{n_{y}^{2}}{l_{y}^{2}}+\frac{n_{z}^{2}}{l_{z}^{2}}\right)=\frac{\hbar^2\vec{k}^2}{2m}$$
where $\vec{k}^2$ is the magnitude of the particle's kvector in kspace.
This kspace can be imagined as a grid of blocks, each representing a possible
particle state (with a double degeneracy for spin). Positions on this grid have
coordinates $(k_{x},k_{y},k_z)$ corresponding to the positive integer
quantum numbers. These blocks will be filled
from the lowest energy upwards: for large numbers of occupying particles,
the filling pattern can be approximated as an expanding spherical shell with
radius $\vec{k_F}^2$.
Note that we're "over counting" the number of occupied states because the
"sides" of the quarter sphere in kspace (where one of the associated quantum
numbers is zero) do not represent valid states. These surfaces can be ignored
for very large N because the surface area to volume ratio is so low, but the
correction can be important. There will then be a second correction due to
removing the states along the individual axes twice (once for each
sidesurface), u.s.w.
The surface of this shell is called the Fermi surface and represents the most
excited states in the gas. The radius can be derived by calculating the total
volume enclosed: each block has volume $\frac{\pi^3}{l_x l_y
l_z}=\frac{\pi^3}{V}$ and there are N/2 blocks occupied by N fermions, so:
$$\frac{1}{8}(\frac{4\pi}{3} k_{F}^{3}) = \frac{Nq}{2}(\frac{\pi^{3}}{V}) $$
$$k_{F} = \sqrt{\frac{3Nq\pi^2}{V}}^3=\sqrt{3\pi^2\rho}^3$$
$\rho$ is the "free fermion density". The corresponding energy is:
$$E_{F}=\frac{\hbar^{2}}{2m}k_{F}^{2}=\frac{\hbar^{2}}{2m}(3\rho \pi)^{2/3}$$
