PS6-solutions

PS6-solutions - Phys 253a 1 Problem Set 6 solutions...

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Unformatted text preview: Phys 253a 1 Problem Set 6 solutions November 7, 2010 1. (a) There are many diagrams one can draw (I count 23 at order e 4 ). Many of these are corrections to the two and three point functions for the scalars or the photon and will cancel in small groups. We will begin with some of these. For example, consider the following diagram: p 1 p 2 p 4 p 3 To simplify subscripts we will refer to the initial particles as e s and the final particles as s. Lets call the right moving momenta in the top and bottom photon propagators in the loop l and r respectively. Then the numerator in the expression for this diagram is given by N =(- ie ) 4 ( p 1- p 2 ) (- i ) g - (1- ) ( p 1 + p 2 ) ( p 1 + p 2 ) ( p 1 + p 2 ) 2 ( l- r ) ( l- r ) (- i ) g - (1- ) ( p 3 + p 4 ) ( p 3 + p 4 ) ( p 3 + p 4 ) 2 ( p 3- p 4 ) (2 ) 4 4 ( p 1 + p 2- l- r ) . Since we have ( p 1- p 2 ) ( p 1 + p 2 ) = m 2 e- m 2 e = 0 and ( p 3 + p 4 ) ( p 3- p 4 ) = m 2 - m 2 = 0 on shell, both (1- ) pieces vanish and this diagram is by itself gauge invariant. Now lets get our hands a little more dirty. Consider the following pair of dia- grams: Phys 253a 2 p 1 p 2 p 4 p 3 p 1 p 2 p 4 p 3 The right hand side of both of these diagrams are the same. Lets write an expression for the them, including the middle propagator: R =- i ( p 3 + p 4 ) 2 + i g - (1- ) ( p 3 + p 4 ) ( p 3 + p 4 ) ( p 3 + p 4 ) 2 (- ie )( p 3- p 4 ) = e ( p 4- p 3 ) ( p 3 + p 4 ) 2 + i As before, we have used the fact that p 3 and p 4 are on the mass shell. Now lets write the expressions for the left hand sides. Lets call the right moving photon momentum in the loop l in each case. Then the left part of the first diagram is given by L 1 = Z d 4 l (2 ) 4 i (- ie ) ( p 1- l ) 2- m 2 + i (2 p 1- l ) - i l 2 + i g - (1- ) l l l 2 2 ie 2 g . and of the second by L 2 = Z d 4 l (2 ) 4 i (- ie ) ( p 2- l ) 2- m 2 + i (- 2 p 2 + l ) - i l 2 + i g - (1- ) l l l 2 2 ie 2 g , Lets consider the sum ( L 1 + L 2) = Z d 4 l (2 ) 4- ie l 2 + i g - (1- ) l l l 2 2 ie 2 g 1 ( p 2- l ) 2- m 2 + i (- 2 p 2 + l ) + 1 ( p 1- l ) 2- m 2 + i (2 p 1- l ) Were interested in the (1- ) term. Suppressing the integral and- ie l 2 + i 2 ie 2 g Phys 253a 3 term, this is given by- (1- ) l l 2 l (- 2 p 2 + l ) ( p 2- l ) 2- m 2 + i + l (2 p 1- l ) ( p 1- l ) 2- m 2 + i =- (1- ) l l 2 (- 2 l p 2 + l 2 )(( p 1- l ) 2- m 2 ) + (2 l p 1- l 2 )(( p 2- l ) 2- m 2 ) (( p 2- l ) 2- m 2 + i )(( p 1- l ) 2- m 2 + i ) =- (1- ) l l 2 (- 2 l p 2 + l 2 )(- 2 l p 1 + l 2 ) + (2 l p 1- l 2 )( l 2- 2 l p 2 ) (( p 2- l ) 2- m 2 + i )(( p 1- l ) 2- m 2 + i ) We note that the numerator in the last line vanishes identically, so the (1- ) term in L 1 + L 2 drops out. We saw above that the (1- ) term in R also dropped out, so that the sum of the entire diagrams, (...
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This note was uploaded on 02/04/2012 for the course PHYS 253A at Harvard.

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PS6-solutions - Phys 253a 1 Problem Set 6 solutions...

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