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Unformatted text preview: Solid State Theory Solution Sheet 10 FS 11 Prof. M. Sigrist Exercise 10.1 Uniaxial Compressibility We first write the deformation of the Fermi surface in terms of spherical harmonics: k F ( φ,θ ) = k F + γk F [3 cos 2 θ 1] = k F + 4 r π 5 γk F Y 20 ( φ,θ ) . (1) a) The change in volume enclosed by the Fermi surface is thus V = Z dθdφ Z k F ( φ,θ ) k 2 dk = 1 3 Z dθdφ k F + 4 r π 5 γk F Y 20 ( φ,θ ) 3 = 4 π 3 ( k F ) 3 + O ( γ 2 ) . (2) where the vanishing of the first order term follows directly from the orthogonality of the spherical harmonics. b) To calculate the uniaxial compressibility, we first need to calculate the difference of the energy functional accounting for the deformed Fermi surface up to second order. For this purpose, we use the result of ex. 10.3 (Optional) , since we are here just dealing with a special case. We first write the deformation of the Fermi surface as k F ( φ,θ ) = k F + u s 20 Y 20 ( φ,θ ) (3) with u 20 = 4 p π/ 5 γk F , according to the next exercise. With this, we can immedi ately use eq. (38) to find E E = V (2 π ) 3 ( k F ) 3 m * ~ 2 ( u s 20 ) 2 (1 + F s 2 5 ) = 16 π 5 V (2 π ) 3 ( k F ) 5 m * ~ 2 γ 2 (1 + F s 2 5 ) (4) It is now straight forward to calculate the uniaxial compressibility: κ u = 1 V ∂ 2 E ∂P 2 z = 1 V 1 P 2 ∂ 2 E ∂γ 2 = 8 5 π 2 E F P 2 ( k F ) 3 (1 + F s 2 5 ) = 24 5 n E F P 2 (1 + F s 2 5 ) . (5) Here we have used n = ( k F ) 3 3 π 2 . Exercise 10.2 Polarization of a neutral Fermi liquid We introduce the notation σ = 1 , σ 1 = σ x , σ 2 = σ y and σ 3 = σ z ....
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This note was uploaded on 10/04/2011 for the course PHYS fs11 taught by Professor Sigrist during the Spring '11 term at Swiss Federal Institute of Technology Zurich.
 Spring '11
 Sigrist

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