You can similarly check that the same substitution is legal in the diagram p q

You can similarly check that the same substitution is

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which is the claimed result. You can similarly check that the same substitution is legal in the diagram p q p q / / e 2 v ( vector q ) γ μ u ( vector p )] D μν ( k )[¯ u ( vector p ) γ ν v ( vector q )] (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 propagator D μν = μν /p 2 . square Note: This is the propagator we found when quantizing in Lorentz gauge (using the Feynman gauge parameter). In general, quantizing the Lagrangian (6.37) in Lorentz gauge, we have the propagator D μν = i p 2 parenleftbigg η μν + ( α 1) p μ p ν p 2 parenrightbigg (6.91) Using similar arguments to those given above, you can show that the p μ p ν /p 2 term cancels in all diagrams. For example, in the following diagrams the p μ p ν piece of the propagator contributes as ¯ u ( p ) γ μ u ( p ) k μ = ¯ u ( p )( / p / p ) u ( p ) = 0 ¯ v ( p ) γ μ u ( q ) k μ = ¯ u ( p )( / p + / q ) u ( q ) = 0 (6.92) 6.5 Feynman Rules Finally, we have the Feynman rules for QED. For vertices and internal lines, we write Vertex: ieγ μ Photon Propagator: μν p 2 + Fermion Propagator: i ( / p + m ) p 2 m 2 + For external lines in the diagram, we attach Photons: We add a polarization vector ǫ μ in / ǫ μ out for incoming/outgoing photons. In Coulomb gauge, ǫ 0 = 0 and vector ǫ · vector p = 0. Fermions: We add a spinor u r ( vector p )/¯ u r ( vector p ) for incoming/outgoing fermions. We add a spinor ¯ v r ( vector p )/ v r ( vector p ) for incoming/outgoing anti-fermions. – 143 –
6.5.1 Charged Scalars “Pauli asked me to calculate the cross section for pair creation of scalar particles by photons. It was only a short time after Bethe and Heitler had solved the same problem for electrons and positrons. I met Bethe in Copenhagen at a conference and asked him to tell me how he did the calculations. 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 three weeks.” He was right, as usual; furthermore, the published cross sections were wrong by a factor of four.” Viki Weisskopf The interaction terms in the Lagrangian for charged scalars come from the covariant derivative terms, L = D μ ψ D μ ψ = μ ψ μ ψ ieA μ ( ψ μ ψ ψ∂ μ ψ ) + e 2 A μ A μ ψ ψ (6.93) This gives rise to two interaction vertices. But the cubic vertex is something we haven’t seen 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 stronger for fermions with higher momentum, so we include a momentum factor in the Feynman rule. There is also a second, “seagull” graph. The two Feynman rules are p q ie ( p + q ) μ and + 2 ie 2 η μν The factor of two in the seagull diagram arises because of the two identical particles appearing in the vertex. (It’s the same reason that the 1 / 4! didn’t appear in the Feynman rules for φ 4 theory).

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