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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 :m:`$l_{x}, l_{y}, l_z$`. The wavefunctions of 
-individual fermions (pretending they are non-interacting) can be seperated
-as :m:`$\psi(x,y)=\psi_{x}(x)\psi_{y}(y)\psi_{z}(z)$`. The solutions will be
-the usual ones to the Schrodinger equation:
-
-:m:`$$\frac{-\hbar^2}{2m}\frac{d^2 \psi_x}{dx}=E_x \psi_x$$`
-
-with the usual wave numbers :m:`$k_x=\frac{\sqrt{2mE_x}}{\hbar}$`, and quantum
-numbers satisfying the boundry conditions :m:`$k_x l_x = n_x \pi$`. The full
-wavefunction for each particle will be:
-
-:m:`$$\psi_{n_{x}n_{y}n_{z}}(x,y,z)=\sqrt{\frac{4}{l_{x}l_{y}}}\sin\left(\frac{n_{x}\pi}{l_{x}}x\right)\sin\left(\frac{n_{y}\pi}{l_{y}}y\right)\sin\left(\frac{n_{z}\pi}{l_{z}}z\right)$$`
-
-and the associated energies (with :m:`$E = E_x + E_y + E_z$`):
-
-:m:`$$E_{n_{x}n_{y}n_z}=\frac{\hbar^{2}\pi^{2}}{2m}\left(\frac{n_{x}^{2}}{l_{x}^{2}}+\frac{n_{y}^{2}}{l_{y}^{2}}+\frac{n_{z}^{2}}{l_{z}^{2}}\right)=\frac{\hbar^2|\vec{k}|^2}{2m}$$`
-
-where :m:`$|\vec{k}|^2$` 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 :m:`$(k_{x},k_{y},k_z)$` 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 :m:`$|\vec{k_F}|^2$`. 
-
-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 :m:`$\frac{\pi^3}{l_x l_y
-l_z}=\frac{\pi^3}{V}$` and there are N/2 blocks occupied by N fermions, so:
-
-:m:`$$\frac{1}{8}(\frac{4\pi}{3} |k_{F}|^{3})&=&\frac{Nq}{2}(\frac{\pi^{3}}{V})\\|k_{F}|&=&\sqrt{\frac{3Nq\pi^2}{V}}^3=\sqrt{3\pi^2\rho}^3$$`
-
-:m:`$\rho$` is the "free fermion density". The corresponding energy is:
-
-:m:`$$E_{F}=\frac{\hbar^{2}}{2m}|k_{F}|^{2}=\frac{\hbar^{2}}{2m}(3\rho \pi)^{2/3}$$`