which is the claimed result. You can similarly check that the same substitution is legalin the diagrampqpq//∼e2[¯v(vectorq)γμu(vectorp)]Dμν(k)[¯u(vectorp′)γνv(vectorq′)](6.90)In fact, although we won’t show it here, it’s a general fact that in every Feynman dia-gram we may use the very nice, Lorentz invariant propagatorDμν=−iημν/p2.squareNote:This is the propagator we found when quantizing in Lorentz gauge (using theFeynman gauge parameter). In general, quantizing the Lagrangian (6.37) in Lorentzgauge, we have the propagatorDμν=−ip2parenleftbiggημν+ (α−1)pμpνp2parenrightbigg(6.91)Using similar arguments to those given above, you can show that thepμpν/p2termcancels in all diagrams. For example, in the following diagrams thepμpνpiece of thepropagator contributes as∼¯u(p′)γμu(p)kμ= ¯u(p′)( /p−/p′)u(p) = 0∼¯v(p)γμu(q)kμ= ¯u(p)( /p+ /q′)u(q) = 0(6.92)6.5 Feynman RulesFinally, we have the Feynman rules for QED. For vertices and internal lines, we write•Vertex:−ieγμ•Photon Propagator:−iημνp2+iǫ•Fermion Propagator:i( /p+m)p2−m2+iǫFor external lines in the diagram, we attach•Photons: We add a polarization vectorǫμin/ǫμoutfor incoming/outgoing photons.In Coulomb gauge,ǫ0= 0 andvectorǫ·vectorp= 0.•Fermions: We add a spinorur(vectorp)/¯ur(vectorp) for incoming/outgoing fermions. We adda spinor ¯vr(vectorp)/vr(vectorp) for incoming/outgoing anti-fermions.– 143 –
6.5.1 Charged Scalars“Pauli asked me to calculate the cross section for pair creation of scalarparticles by photons.It was only a short time after Bethe and Heitlerhad solved the same problem for electrons and positrons.I met Bethein Copenhagen at a conference and asked him to tell me how he did thecalculations. I also inquired how long it would take to perform this task;he answered, “It would take me three days, but you will need about threeweeks.” He was right, as usual; furthermore, the published cross sectionswere wrong by a factor of four.”Viki WeisskopfThe interaction terms in the Lagrangian for charged scalars come from the covariantderivative terms,L=Dμψ†Dμψ=∂μψ†∂μψ−ieAμ(ψ†∂μψ−ψ∂μψ†) +e2AμAμψ†ψ(6.93)This gives rise to two interaction vertices. But the cubic vertex is something we haven’tseen before: it contains kinetic terms.How do these appear in the Feynman rules?After a Fourier transform, the derivative term means that the interaction is strongerfor fermions with higher momentum, so we include a momentum factor in the Feynmanrule. There is also a second, “seagull” graph. The two Feynman rules arepq−ie(p+q)μand+ 2ie2ημνThe factor of two in the seagull diagram arises because of the two identical particlesappearing in the vertex.(It’s the same reason that the 1/4! didn’t appear in theFeynman rules forφ4theory).