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# hw1_sol - Physics 240B Spring 2008 Homework 1 Solutions...

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Physics 240B, Spring 2008 Homework 1 Solutions Jesse Noffsinger February 14, 2008 Solutions are in general not the original work of the author. Problem 1) Chapter 6 of Kittel is very helpful for this problem. Portions not worked out explicitly here appear in the recommended texts. a) The total number of electrons up to wavevector k is: N e = 2 × πk 3 (2 π/L ) 3 = V 3 π 2 k 3 (1) The density of states, D ( k ) is: D ( k ) = dN dk = V π 2 k 2 (2) while: ǫ = ¯ h 2 k 2 2 m (3) Yielding D ( ǫ ) = dN = V 2 π 2 parenleftbigg 2 m ¯ h 2 parenrightbigg 3 / 2 ǫ 1 / 2 (4) b) The total number of electrons is: N = V 3 π 2 parenleftbigg 2 F ¯ h 2 parenrightbigg 3 / 2 ǫ 1 / 2 F = 3 N ǫ F bracketleftBigg V π 2 parenleftbigg 2 m ¯ h 2 parenrightbigg 3 / 2 bracketrightBigg 1 (5) Entering this expression for ǫ 1 / 2 F into Eqn (4) gives: D ( ǫ F ) = 3 N 2 ǫ F (6) 1

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c) C v = ∂U ∂T (7) The total energy of the sytem being U = integraldisplay ǫf ( ǫ,T ) D ( ǫ ) (8) which depends on temperature only through the Fermi-Dirac distribution. The Sommerfield expansion [Eqn 2.70 in Ashcroft and Mermin] can be applied, or a direct following of Kittel’s derivation gives: C v = π 2 3 k 2 b TD ( ǫ F ) (9) d) If we associate an energy ± μH with spin up/down electrons, we can find the paramagnetic susceptibility of the free electron gas, defined as χ = ∂M ∂H (10) The total magnetic moment per volume is M = μ ( n + n ), arising from the difference in the number of spin up and spin down electrons. M = 1 2 μ integraldisplay [ f ( ǫ μH ) f ( ǫ + μH )] D ( ǫ ) (11) The 1/2 accounts for the spin in the standard density of states formula. Taking μH to be small with respect to the electronic energies [Ziman pg.332], and the energy derivative of the Fermi-Dirac distribution to be a delta function peaked at ǫ F : M = μ 2 HD ( ǫ F ) (12) or χ = μ 2 D ( ǫ F ) (13) e) The thermal conductivity is defined as the negative ratio between the temperatue gradient and the total flux of thermal energy.
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