PHYS427fa08_HW09Solution

# PHYS427fa08_HW09Solution - UIUC Physics 427 Homework 9...

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UIUC Physics 427 Homework 9 Solutions Philip Powell October 27, 2008 1. Problem 1 (K&K 7.3) (a) Consider a Fermi electron gas in its ground state. The energy of the gas is given by K&K U 0 = Z F 0 D ( ) d (1) where the density of states is given by D ( ) = V 2 π 2 2 m ~ 2 3 / 2 (2) As noted on page 217 of K&K we obtain the result U 0 = 3 5 N F F = ~ 2 2 m 3 π 2 N V 2 / 3 (3) or equivalently U 0 = 3 ~ 2 10 m 3 π 2 N V 2 / 3 (4) The pressure is given, in general, by p = - ∂F ∂V τ,V (5) Since F = U - τσ , at τ = 0 we have F = U 0 so that p = - ∂U 0 ∂V τ,V (6) Substituting (4) we obtain p = (3 π 2 ) 2 / 3 5 ~ 2 m N V 5 / 3 (7) (b) Next, we seek to find the entropy of a Fermi electron gas for τ F . Recall that the definition of heat capacity is C = dU (8) The thermodynamic identity is dU = τdσ - pdV (9) 1

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Combining these two expressions and requiring that dV = 0 gives C = τ ∂σ ∂τ V (10) From K&K page 218 we have C = π 2 2 N τ F + · · · (11) Combining these two expressions gives ∂σ ∂τ V = π 2 2 N 1 F + · · · (12) Integrating, we obtain σ = π 2 2 N τ F + · · · (13) where we have noted that the ground state entropy (i.e. σ (0)) is zero, since there is a unique ground state, so the multiplicity is 1. 2. Problem 2 (K&K 7.4) Consider the chemical potential μ ( T ) in d = 1 and d = 3 dimensions. We wish to determine the general behavior of μ ( T ) for small non-zero temperatures. We start by noting that the total number of particles in a system may be expressed in the form N = Z 0 D ( ) f ( , μ ) d (14) Recalling that D 1 - 1 / 2 and D 3 1 / 2 we can write N Z 0 1 1 e ( - μ 1 ) + 1 d N Z
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