Then the above expression reduces to R xx \u03c9 1 Z X \u03bc\u03bd h \u03bc j x P \u03bd i 2 f \u03b5 \u03bd f \u03b5

# Then the above expression reduces to r xx ω 1 z x

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Then, the above expression reduces to, R xx ( ω ) = - 1 Z 0 X μ,ν |h μ | j x P | ν i| 2 f ( ε ν ) - f ( ε μ ) ~ ω + ε μ - ε ν + i 0 + , (32) where the expression inside the [ ... ] brackets is known as the Lindhard-function and f ( ... ) is the Fermi-function. (The factor f ( ε ν ) - f ( ε μ ) arises from f ( ε ν )[1 - f ( ε μ )] - f ( ε μ )[1 - f ( ε ν )].)
5 Problem 3: We already saw in the previous problem that the conductivity is given by, σ xx ( ω ) = 1 1 Ω X μ,ν |h μ | p x | ν i 2 ε μ - ε ν + ω + i 0 + [ f ( ε ν ) - f ( ε μ )] - 1 Nq 2 mc 2 . (33) Let us now look at the real part of σ xx ( ω ), Re σ xx ( ω ) = π ω X μ,ν |h μ | p x | ν i| 2 δ ( ε μ - ε ν + ω )[ f ( ε ν ) - f ( ε μ )] . (34) We now use a trick to express the matrix elements of p x in terms of matrix elements of x as follows: h μ | ~ p | ν i = m i ~ h μ | [ ~ r, H 0 ] | ν i = m i ~ ( ε ν - ε μ ) h μ | ~ r | ν i . (35) Therefore, the expression for Re σ xx , in the limit of small ω (so that we can taylor expand f ( ε ν ) about ε μ ), simplifies to, Re σ xx ( ω ) = πm 2 ~ 2 ω 2 X μ f 0 ( ε μ ) h μ | x 2 | μ i - h μ | x | μ i 2 . (36) However, we have been told in the problem that the last quantity in the brackets above remains finite in the limit of L → ∞ . Therefore, in the limit ω 0, Re σ xx ( ω ) 0. Now note that Re σ xx ( - ω ) =Re σ xx ( ω ). If we compute the imaginary part, Im σ xx ( ω ), via a Kramers-Kronig transformation, we obtain, Im σ xx ( ω ) = 1 π Z -∞ P Re σ xx ( ω 0 ) ω 0 - ω 0 = 1 π 2 ω Z 0 P Re σ xx ( ω 0 ) ω 0 2 - ω 2 0 . (37) Therefore, Im σ xx ( ω 0) = 0. Problem 4: We would like to compute the scattering amplitude, which is going to be a matrix in spin-space, using first-order Born approximation for the following Hamiltonian: H = ~ p 2 2 m + U ( ~ r ) + λ U ( ~ r ) · ( ~ p × ~ S ) . (38) The scattered wave solution at large r is given by, ψ ˆ n,τ ( ~ r, σ ) e e ik ˆ n.~ r δ στ + e ikr r f στ r, ˆ n ) , (39) where ˆ n and τ = ± 1 denote the incident direction and spin orientation of the incoming particle and f is a 2 × 2 matrix which depends on ˆ r , the direction in which the amplitude is measured. To lowest non-vanishing order, f is given by (see Sakurai, chap.7), f στ r, ˆ n ) = - 1 4 π 2 m ~ 2 Z d 3 r 0 e - ik ˆ r.~ r 0 [ U ( ~ r 0 ) δ στ + λ r 0 U ( ~ r 0 ) · ( ~ p × ~ S ) στ ] e ik ˆ n.~ r 0 . (40)
6 If we now integrate the second term by parts and use the fact that it is a short-ranged potential (vanishes fast enough as ~ r 0 → ∞ ), then f στ r, ˆ n ) = - 1 4 π 2 m ~ 2 Z d 3 r 0 e ik n - ˆ r ) .~ r 0 U ( ~ r 0 )[ δ στ - iλk 2 n - ˆ r ) · n × ~ S ) στ ] . (41) It is now useful to write the above amplitude as f = f 0 u , where, f 0 = - 1 4 π 2 m ~ 2 Z d 3 r 0 e ik n - ˆ r ) .~ r 0 U ( ~ r 0 ) , (42) u = e iλk 2 r × ˆ n ) . ~ S . (43) We have exponentiated the matrix in the [ ... ] brackets for small λ . The matrix u is clearly an SU(2) matrix since ~ S represent the spin-1/2 operators.

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• Physics, Trigraph, Jx, matrix elements, py − Ay