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| -rw-r--r-- | physics/quantum/fermigas.page | 39 |
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diff --git a/physics/quantum/fermigas.page b/physics/quantum/fermigas.page index 0114b43..de66ee1 100644 --- a/physics/quantum/fermigas.page +++ b/physics/quantum/fermigas.page @@ -1,35 +1,34 @@ -=============== -Fermi Gas -=============== +# 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 +square well with dimensions $l_{x}, l_{y}, l_z$. 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 +as $\psi(x,y)=\psi_{x}(x)\psi_{y}(y)\psi_{z}(z)$. The solutions will be the usual ones to the Schrodinger equation: -:m:`$$\frac{-\hbar^2}{2m}\frac{d^2 \psi_x}{dx}=E_x \psi_x$$` +$$\frac{-\hbar^2}{2m}\frac{d^2 \psi_x}{dx}=E_x \psi_x$$ -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 +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: -:m:`$$\psi_{n_{x}n_{y}n_{z}}(x,y,z)=\sqrt{\frac{4}{l_{x}l_{y}}}\sin\left(\frac{n_{x}\pi}{l_{x}}x\right)\sin\left(\frac{n_{y}\pi}{l_{y}}y\right)\sin\left(\frac{n_{z}\pi}{l_{z}}z\right)$$` +$$\psi_{n_{x}n_{y}n_{z}}(x,y,z)=\sqrt{\frac{4}{l_{x}l_{y}}}\sin\left(\frac{n_{x}\pi}{l_{x}}x\right)\sin\left(\frac{n_{y}\pi}{l_{y}}y\right)\sin\left(\frac{n_{z}\pi}{l_{z}}z\right)$$ -and the associated energies (with :m:`$E = E_x + E_y + E_z$`): +and the associated energies (with $E = E_x + E_y + E_z$): -:m:`$$E_{n_{x}n_{y}n_z}=\frac{\hbar^{2}\pi^{2}}{2m}\left(\frac{n_{x}^{2}}{l_{x}^{2}}+\frac{n_{y}^{2}}{l_{y}^{2}}+\frac{n_{z}^{2}}{l_{z}^{2}}\right)=\frac{\hbar^2|\vec{k}|^2}{2m}$$` +$$E_{n_{x}n_{y}n_z}=\frac{\hbar^{2}\pi^{2}}{2m}\left(\frac{n_{x}^{2}}{l_{x}^{2}}+\frac{n_{y}^{2}}{l_{y}^{2}}+\frac{n_{z}^{2}}{l_{z}^{2}}\right)=\frac{\hbar^2|\vec{k}|^2}{2m}$$ -where :m:`$|\vec{k}|^2$` is the magnitude of the particle's k-vector in k-space. +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 :m:`$(k_{x},k_{y},k_z)$` corresponding to the positive integer +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 :m:`$|\vec{k_F}|^2$`. +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 @@ -41,11 +40,11 @@ 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: +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: -:m:`$$\frac{1}{8}(\frac{4\pi}{3} |k_{F}|^{3})&=&\frac{Nq}{2}(\frac{\pi^{3}}{V})\\|k_{F}|&=&\sqrt{\frac{3Nq\pi^2}{V}}^3=\sqrt{3\pi^2\rho}^3$$` +$$\frac{1}{8}(\frac{4\pi}{3} |k_{F}|^{3})&=&\frac{Nq}{2}(\frac{\pi^{3}}{V})\\|k_{F}|&=&\sqrt{\frac{3Nq\pi^2}{V}}^3=\sqrt{3\pi^2\rho}^3$$ -:m:`$\rho$` is the "free fermion density". The corresponding energy is: +$\rho$ is the "free fermion density". The corresponding energy is: -:m:`$$E_{F}=\frac{\hbar^{2}}{2m}|k_{F}|^{2}=\frac{\hbar^{2}}{2m}(3\rho \pi)^{2/3}$$` +$$E_{F}=\frac{\hbar^{2}}{2m}|k_{F}|^{2}=\frac{\hbar^{2}}{2m}(3\rho \pi)^{2/3}$$ |