sol06 - Solid State Theory Solution Sheet 6 FS 2011 Prof....

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Unformatted text preview: Solid State Theory Solution Sheet 6 FS 2011 Prof. M. Sigrist Exercise 6.1 Lindhard function At T = 0 the Fermi-Dirac distribution function n F ( k ) reduces to θ ( F- k ). As usual, we go from the discrete summation to a d-dimensional integral. Then, the static Lindhard function is given by χ ( q ) ≡ χ ( q ,ω = 0) = 1 Ω X k n F ( k + q )- n F ( k ) k + q- k- i ~ η = 1 (2 π ) d Z d d k θ ( F- k + q )- θ ( F- q ) k + q- k- i ~ η (1) with the Fermi energy F and the Heaviside step function θ ( x ) = ( 1 ,x ≥ ,x < . (2) Next we split the integral and perform a change of variables in the second integral ( k → k- q ) such that χ ( q ) =- 1 (2 π ) d Z d d k θ ( F- k ) 1 k + q- k- i ~ η- 1 k- k- q- i ~ η . (3) The dispersion relation for free electrons is given by k = ~ k 2 / 2 m . We can therefore define the Fermi wave vector k F = √ 2 m F / ~ and the integration can be simplified further to χ ( q ) =- 1 (2 π ) d 2 m ~ 2 Z | k | <k F d d k 1 q ( q + 2 k )- i ~ η + 1 q ( q- 2 k ) + i ~ η . (4) where we introduced the abbreviation ~ = 2 m/ ~ . i) In the 1 dimensional case the integral is then simply χ ( q ) =- 1 2 π 2 m ~ 2 k F Z- k F d k 1 q ( q + 2 k )- i ~ η + 1 q ( q- 2 k ) + i ~ η . (5) We remark that for | q | < 2 k F there is a singular point in the integral which is ’cured’ by η meaning that in the limit η → 0 we have to calculate the principle value since lim η → ( z- iη )- 1...
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sol06 - Solid State Theory Solution Sheet 6 FS 2011 Prof....

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