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MIT 8 512 - Lecture 3: Properties of the Response Function

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Lecture 3: Properties of the Response FunctionKramers-KronigMIT OpenCourseWarehttp://ocw.mit.edu 8.512 Theory of Solids II Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 3: Properties of the Response Function In this lecture we will discuss some general properties of the response functions X, and some uselul relations that they satisfy. 3.1 General Properties of ~(q',w) Recall that wit11 n(F, i) E IR and U(F, i) E IR.. Under Fourier transform, this implies n , w) = n*(q',w) U(-f, -w) = US(gl,ww) As a result, where X" is the imaginary part of the response function X(gl,ww). Consider the extension of w to the complex plane. We can then rewrite the expression for x(q'74 as Without the +iq term in the energy denominator, there would be siryularities (poles) on the real axis whenever w is equa.1 to the spacing between the ground state and some excited state. The presence of +iq pushes these poles just into the lower l/Bplane, emuring that x(q7w) is analytic in the entire upper-l/2 w-plane including the real axis. Analyticity of x($ w) in the upper-1/2 plane is needed to build causality into the theory. Consider the response function in time, ~(t): To evaluate this integral, we perform a contour integral in the complex w plane. For i < 0, closing the contour in the upper 1/2 plane ensures that leJiwtl +0 on the curved portion of the� � � � �� � � Kramers-Kronig 2 contour. Since we have ensured that χ(ω) is analytic in the upper half plane, Cauchy’s residue theorem guarantees that the integral over the entire contour is 0. As a result, the piece we need, i.e. the integral from −∞ to ∞ along the real axis, must also be 0. Thus χ(t) = 0 for t < 0, which means that the system cannot respond to a perturbation until after the perturbation has occurred. What about the t > 0 case? In this case, the contour must be closed in the lower 1/2 plane to prevent the exponential from blowing up. However, the iη in the energy denominator has pushed the singularities into this region of the complex ω plane. Thus the value of the contour integral will be nonzero, and the system will respond to perturbations for t > 0. 3.2 Kramers-Kronig Consider the integral χ(�q, ω�)dω� = 0 (3.8)ω − ω� As we have just shown in the previous section, our definition of χ(� q, ω)q, ω) ensures that χ(� is analytic in the upper-1/2 complex ω plane. Although the integrand here has a pole on the real axis due to the ω − ω� in the denominator, by making a hump over this pole we can ensure that the value of the contour integral itself vanishes by Cauchy’s residue theorem. Assuming that χ(�q, ω) → 0 as |ω| → ∞, ∞ 10 = dω�χ(q, ω� �) Pr + iπχ(�q, ω) (3.9) −∞ ω − ω� where the additional term iπχ(� = ω that q, ω) is one half of the contribution from the pole at ω� we picked up by making a hump over the pole. Thus for fixed �q, 1 q, ω)q, ω) = − ∞ dω� χ��(� χ�(� Pr (3.10)π ��−∞ ω − ω� � q, ω)q, ω) = − π 1 Pr ∞ dω� χω �(− �ω� (3.11)χ��(� −∞ or equivalently ∞ dω� χ(q, ω � χ(q, ω� ) = η lim 0+ − π ω − ω� + iη (3.12) → −∞ The important message from all of this is that the entire response function χ(�q, ω) can be reconstructed from its imaginary (or real) part


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