Physics 413: Statistical Mechanics - Homework Solutions 8 Problem 1: Quantum corrections to classical ideal gas The classical (Boltzmann) limit corresponds to ⟨n⟩ = 1 e β(ϵ-μ) + δ ≪ 1 Expanding the denominator gives ⟨n⟩ = e -β(ϵ-μ) (1 - δe -β(ϵ-μ) )+ O[(e -β(ϵ-μ) ) 3 ] Average particle number and internal energy read N = ∫ dϵ a(ϵ) e -β(ϵ-μ) (1 - δe -β(ϵ-μ) ) U = ∫ dϵ ϵ a(ϵ) e -β(ϵ-μ) (1 - δe -β(ϵ-μ) ) with a(ϵ)= V 2π 2 ( m ¯ h 2 ) 3/2 √ 2ϵ The energy per particle is U N = ∫ dϵ ϵ 3/2 e -βϵ (1 - δe -β(ϵ-μ) ) ∫ dϵ ϵ 1/2 e -βϵ (1 - δe -β(ϵ-μ) ) = k B T (3/4) √ π (1 - δe βμ /2 5/2 ) (1/2) √ π (1 - δe βμ /2 3/2 ) = 3 2 k B T (1 + δe βμ √ 2/8) The factor e βμ in the correction term can be calculated in lowest order, i.e. for the classical ideal gas. e βμ = N V ( 2π¯ h 2 mk B T ) 3/2 = N V λ 3 where λ is the thermal wave length. Therefore U N = 3 2 k B T (1 + δNλ 3 √ 2/(8V )) The corrections to the pressure most easily follow from the general relation p = (2U )/(3V ) which can be proven as follows: p = - ( ∂ Ω ∂V ) T,μ = - Ω V = k B T V ln Z μ = k B T V 1 δ ∫ dϵ a(ϵ) ln[1 + δe -β(ϵ-μ) ] partial integration = 2 3 k B T V 1 δ ∫ dϵ a(ϵ)ϵ βδe -β(ϵ-μ) 1+ δe -β(ϵ-μ) = 2 3 U V