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authorUser <>2009-10-13 02:52:09 +0000
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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:
+and the associated energies (with :m:`$E = E_x + E_y + E_z$`):
+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:`$$E_{F}=\frac{\hbar^{2}}{2m}|k_{F}|^{2}=\frac{\hbar^{2}}{2m}(3\rho \pi)^{2/3}$$`