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sol11 - Solid State Theory Solution Sheet 11 SS 11 Prof M...

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Solid State Theory Solution Sheet 11 SS 11 Prof. M. Sigrist Exercise 11.1 Relaxation time approximation a) The transition rates are computed by use of Fermi’s golden rule, relating the proba- bility for a process to occur to the quantum mechanical matrix elements of scattering off a single impurity a weighting with the impurity density (cf. section 6.3 of the lecture notes) W ( k , k 0 ) = 2 π n imp ~ h k 0 | V imp | k i 2 δ ( ε k - ε k 0 ) (1) The matrix elements 1 h k 0 | V | k i = V ( k - k 0 ) are given by the Fourier transform of the potential V ( r ) δ ( r ) V ( k - k 0 ) = 1 2 π Z d 2 r e i ( k - k 0 ) · r V 0 δ ( r ) = 1 2 π V 0 . (2) Thus, the transition rates are given as W ( k , k 0 ) = n imp 2 π ~ V 2 0 | {z } W 0 δ ( ε k - ε k 0 ) , (3) and given the isotropy of the energy ε k k 2 they do not depend on k or k 0 as long as the energy is conserved. b) In the static limit, the left hand side of the Boltzmann equation ∂f ( r , k , t ) ∂t + r f ( r , k , t ) · d r d t + k f ( r , k , t ) · ( e E ) (4) reduces with f ( r , k , t ) = f ( k ) to the third term proportional to the driving external field. We restrict our considerations to linear order in E since we are interested in linear response to the driving force. In consequence, we must replace k f with k f 0 because δf is already of order O ( E ). This term represents the so-called drift term. The right hand side of the Boltzmann equation is the collision integral, which for impurity scattering acquires the form (cf. Eq. (6.23) in the lecture notes) ∂f ( k ) ∂t coll = - Z d 2 k (2 π ) 2 W ( k , k 0 )( f ( k ) - f ( k 0 )) = - Z d 2 k (2 π ) 2 W 0 δ ( ε k - ε k 0 ) ( f 0 ( k ) + δf ( k ) - f 0 ( k 0 ) - δf ( k 0 )) = - Z d θ d k k (2 π ) 2 W 0 m ~ 2 k δ ( k - k 0 ) ( f 0 ( k ) + δf ( k ) - f 0 ( k 0 ) - δf ( k 0 )) = - m ~ 2 Z d θ (2 π ) 2 W 0 [ δf ( θ ) - δf ( θ 0 )] . (5) 1 Note that the potential only depends on the difference between k and k 0 due to the homogeneity of the system. 1
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Taking (4) and (5) together we arrive at k f 0 · ( e E ) = - Z d θ 0 ˜ W 0 [ δf ( θ ) - δf ( θ 0 )] , (6) where ˜ W 0 = mW 0 / (2 π ~ ) 2 . c) We start with the Boltzmann equation, X l d l e ilθ = - Z d θ 0 ˜ W 0 X l f l e ilθ - X l f l e ilθ 0 ! , (7) which we project onto the m -th Fourier mode with multiplication of both sides with exp( - imθ ) and integration over θ Z d θe - imθ X l d l e ilθ = - Z d θe - imθ Z d θ 0 ˜ W 0 X l f l e ilθ - X l f l e ilθ 0 !
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