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# 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 non-interacting) 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 k-vector in k-space.
This k-space 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 k-space (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
side-surface), 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:
$$E_{F}=\frac{\hbar^{2}}{2m}|k_{F}|^{2}=\frac{\hbar^{2}}{2m}(3\rho \pi)^{2/3}$$
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