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integral equations in quantum mechanics 54920.4.2 Numerical Principal ValuesA numerical evaluation of the principal value limit (20.24) is awkward becausecomputers have limited precision. A better algorithm follows from the theoremP∫ +∞−∞dkk − k 0=0. (20.26)This equation says that the curve of 1/(k − k 0 ) as a function of k has equal andopposite areas on both sides of the singular point k 0 . If we break the integral upinto one over positive k and one over negative k, a change of variable k →−kpermits us to rewrite (20.26) asP∫ +∞0dkk 2 − k02 =0. (20.27)We observe that the principal-value exclusion of the singular point’s contributionto the integral is equivalent to a simple subtraction of the zero integral (20.27):P∫ +∞0f(k) dkk 2 − k02 =∫ +∞0[f(k) − f(k 0 )] dkk 2 − k02 . (20.28)Notice that there is no P on the RHS of (20.28) because the integrand is no longersingular at k = k 0 (it is proportional to the df/dk) and can therefore be evaluatednumerically using the usual rules. The integral (20.28) is called the Hilbert transformof f and also arises in inverse problems.20.4.3 Reducing Integral Equationsto Matrix Equations (Algorithm)Now that we can handle singular integrals, we can go about reducing the integralequation (20.20) to a set of linear equations that can be solved with matrix methods.We start by rewriting the principal-value prescription as a definite integral[H&T 70]:R(k ′ ,k)=V (k ′ ,k)+ 2 π∫ ∞0dp p2 V (k ′ ,p)R(p, k) − k 2 0V (k ′ ,k 0 )R(k 0 ,k)(k 2 0 − p2 )/2µ. (20.29)We convert this integral equation to linear equations by approximating the integralas a sum over N integration points (usually Gaussian) k j with weights w j :R(k, k 0 ) ≃ V (k, k 0 )+ 2 πN∑j=1− 2 π k2 0V (k, k 0 )R(k 0 ,k 0 )k 2 j V (k, k j)R(k j ,k 0 )w j(k 2 0 − k2 j )/2µN∑m=1w m(k 2 0 − k2 m)/2µ . (20.30)−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 549