95%(22)21 out of 22 people found this document helpful
This preview shows page 4 - 7 out of 13 pages.
simplify the propagators. Note that by momentum conservation,‘2-2p1·‘=r2-2p2·randk2-2p4·k=q2-2p3·q.Now, we could go ahead and write down all the diagrams, expand everything out,and try to simplify.That’s actually pretty hard (congrats to those who man-aged it!). So let’s write things more symbolically at first and do some judiciousrearranging. LetΠμν(‘) =-igμν+ (1-ξ)‘μ‘ν‘2‘2be the photon propagator. We always have two photon momenta in our Feynmangraphs, which we’ll take to be‘andr, both going from left to right.There’sonly one loop momentum, so‘andrare related by momentum conservation:‘+r=p1+p2=p3+p4However, to keep expressions simpler, it’s useful not to solve forrin terms of‘(or vice versa), but instead write the loop integral as a double integral, with adelta function to enforce momentum conservation:Zd4‘ d4r(2π)4δ(p1+p2-‘-r)For convenience, we’ll write this as simplyR. Note that‘andrappear symmet-rically in the integral, so for any functionf(‘, r), we haveRf(‘, r) =Rf(r, ‘).
Phys 253a6our sum of diagrams should beX?=12ZVμνL(‘, r)Πμρ(‘)Πνσ(r)Vρ,σR(‘, r)=12Z(UμνL(‘, r) +UνμL(r, ‘) + 2ie2gμν)Πμρ(‘)Πνσ(r)×(UρσR(‘, r) +UσρR(r, ‘) + 2ie2gρσ)=12ZUμνL(‘, r)Πμρ(‘)Πνσ(r)UρσR(‘, r) +UνμL(r, ‘)Πμρ(‘)Πνσ(r)UσρR(r, ‘)+12ZUμνL(‘, r)Πμρ(‘)Πνσ(r)UσρR(r, ‘) +UνμL(r, ‘)Πμρ(‘)Πνσ(r)UρσR(‘, r)+12ZUμνL(‘, r) +UνμL(r, ‘) Πμρ(‘)Πνσ(r)(2ie2gρσ)+12Z(2ie2gμν)Πμρ(‘)Πνσ(r)UρσR(‘, r) +UσρR(r, ‘)+12Z(2ie2gμν)Πμρ(‘)Πνσ(r)(2ie2gρσ)=12[I + I] +12[II + II] +12[III + III] +12[IV + IV] + V=I + II + III + IV + Vwhere in getting to the third equality, we expanded the second line and groupedterms in square brackets that are the same by index redefinitions and switchingr↔‘under the integral sign. This is exactly the sum of diagrams we wanted,but we’ve managed to write it in a form that will make gauge invariance obvious.Indeed, note thatVμνL(‘, r)=-ie2(2p1-‘)μ(r-2p2)ν‘2-2p1·‘+(2p1-r)ν(‘-2p2)μr2-2p1·r-2gμνrνVμν(l, r)=-ie2((2p1-‘)μ-(‘-2p2)μ-2rμ)=-2ie2(p1+p2-‘-r)μ=0since‘+r=p1+p2by conservation of momentum. Similarly we find that