Z Now consider integrating this function around the contour C 4 i 1 C i defined

# Z now consider integrating this function around the

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Z . Now consider integrating this function around the contour C = 4 i =1 C i , defined in Fig. 2. As the contour C is traversed both θ and φ decrease as the contour is traversed in a CCW fashion, and the phase of f varies Mathematical Methods of Physics v3.1 14 November 2015, 12:51 Noon Autumn 2015 The University of Chicago, Department of Physics Page 8 of 9 accordingly; we give some sample phases at A-F in Table I. From the foregoing considerations, in the limit where 2 , 4 0 , we can see that on the contour segment C 1 , we have arg [ f ( z )] = 0 , so Z C 1 f ( z ) dz = Z C 1 | z | - 2 / 3 | z - 1 | - 1 / 3 dz Z 1 0 dx ( x 2 - x 3 ) 1 / 3 = I, (53) where I used that z = x here and | z - 1 | = (1 - x ) on this piece of contour. We chose the branch so that the overall argument here is zero. Meanwhile, on the contour segment C 3 , we have arg [ f ( z )] = 2 π/ 3 , so Z C 3 f ( z ) dz = Z C 3 | z | - 2 / 3 | z - 1 | - 1 / 3 e 2 πi/ 3 dz e 2 πi/ 3 Z 0 1 dx ( x 2 - x 3 ) 1 / 3 = - e 2 πi/ 3 Z 1 0 dx ( x 2 - x 3 ) 1 / 3 = - e 2 πi/ 3 I. (54) Meanwhile, on the circular contour segment C 2 , we have r = 1+ O ( 2 ) and ρ = 2 , with θ = 0 and φ decreasing from π to - π , so the integral is (dropping sub-dominant powers of 2 ) Z C 2 f ( z ) dz Z - π π ( i 2 e ) e - iφ/ 3+ iπ/ 3 - 1 / 3 2 2 / 3 2 0 , (55) the latter in the limit where 2 0 . Similarly, on the circular contour segment C 4 , we have r = 4 and ρ = 1+ O ( 4 ) , with φ = - π and θ decreasing from 0 to - 2 π , so the integral is (dropping sub-dominant powers of 4 ) Z C 4 f ( z ) dz Z - 2 π 0 ( i 4 e ) - 2 / 3 4 e - 2 iθ/ 3+2 πi/ 3 1 / 3 4 0 , (56) the latter in the limit where 4 0 . So in the limit 2 , 4 0 , we have I C f ( z ) dz = (1 - e 2 πi/ 3 ) I. (57) Now suppose I construct a single contour ˜ C = C P C 0 ( - P ) where I start some point on C , move all the way around C and back to just before this point, then pass to C 0 by some path P (which does not cross or run near the branch cut), then traverse all of C 0 and finally then return to C in the opposite direction along a path which I will call - P which is infinitesimally displaced from P . Then Cauchy’s Theorem together with the observation that f ( z ) is analytic in the interior of ˜ C tells us that H ˜ C = 0 = R C + R P + R C 0 + R - P . On P and - P , the argument of f must differ by the amount of phase f ( z ) picks up on C or C 0 , but since P and - P are not running near branch cuts, the phase shift must be a multiple of 2 π (this is almost by definition and construction: if it did not, there TABLE I Phase of the integrand at selected points A-F as defined in Fig. 2. Note that a full traversal of the contour (change in Δ θ = Δ φ = - 2 π ) yields Δ arg f [ z ] = +2 π as well, which is important: f is single-valued. Point θ φ arg[ f ( z )] A 0 π 0 B 0 π 0 C 0 0 π/ 3 D 0 - π 2 π/ 3 E 0 - π 2 π/ 3 F - π - π 4 π/ 3 A (after traversing C once) - 2 π - π 2 π Mathematical Methods of Physics v3.1 14 November 2015, 12:51 Noon Autumn 2015 The University of Chicago, Department of Physics Page 9 of 9 would be a branch cut between P and - P ); this is confirmed by Table I. Therefore, the integrals over P and - P will cancel out in the limit where the infinitesimal displacement between them vanishes, and we have I C f ( z ) dz = - I C 0 f ( z ) dz I C 0 f ( z ) dz = ( e 2 πi/ 3 - 1) I = 2 ie iπ/ 3 sin( π/ 3) I = i 3 e iπ/ 3 I. (58) Now consider that in the limit where the radius R of the contour C 0 #### You've reached the end of your free preview.

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