# 2D Fermi Gas

by cepheid
Tags: fermi
 Emeritus Sci Advisor PF Gold P: 5,196 Although I have some major conceptual problems with the Fermi gas as treated in my solid state physics notes (see this thread: http://www.physicsforums.com/showthread.php?t=161222, I have attempted to solve this homework problem in an analogous manner to the solution for the 3D Fermi gas given in the notes, ignoring these conceptual hangups. 1. The problem statement, all variables and given/known data Find the density of states for a 2D electron gas. 2. Relevant equations See attempted solution below 3. The attempt at a solution Assume that in real space the gas is confined to an area $$A = l_xl_y$$ Write the components of the electron wavevector in terms of the respective principle quantum numbers: $$k_x = \frac{2\pi}{l_x}n_x \ \ \ \ k_y = \frac{2\pi}{l_y}n_y$$ Therefore the number of states associated with an element $d^2\mathbf{k}$ is (supposedly) $$2dn_xdn_y = \frac{A}{4\pi^2}2dk_xdk_y$$ Again, in an analogous way to what was done in the notes in 3D, I switch to polar coordinates, so that I can get the density of states as a function of $k = |\mathbf{k}|$. # of states between k and k + dk $$Z(k)dk = \frac{A}{4\pi^2}2(2 \pi k dk)$$ Change variables to convert Z(k) to D(E), the density of states as a function of energy. $$E = \frac{\hbar^2k^2}{2m} \Rightarrow dE = \frac{\hbar^2}{2m}2kdk \Rightarrow dk = \frac{m}{\hbar^2 k}dE$$ $$D(E)dE = \frac{A}{4\pi^2} 2(2 \pi k) \frac{m}{\hbar^2 k}dE = \frac{Am}{\pi \hbar^2} dE$$ $$\Rightarrow D(E) = \textrm{const.} = \frac{Am}{\pi \hbar^2}$$ This result is drastically different from the 3D case, and later on in the same problem, it leads me to the conclusion that the Fermi energy is independent of temperature for the 2D gas. Is this solution correct, or have I made some egregious error somewhere?
 Emeritus Sci Advisor PF Gold P: 11,155 Yes, your derivation and final result are correct.

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