By Forshaw J.
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Extra info for An introduction to QED and QCD
30) implies that aλ (k), a†λ′ (k ′ ) = −gλλ′ 2E (2π)3 δ 3 (k − k′ ). e. we interpret a †λ (k) as an operator which creates quanta of the electromagnetic field (photons) with polarisation λ and momentum k. 32) is opposite to that of the other 3 polarisations. This shows up in the fact that these timelike photons make a negative contribution to the energy: H= d3 k 1 E a†1 a1 + a†2 a2 + a†3 a3 − a†0 a0 . 33) Fortunately, although we might not realize it yet, we have already solved the problem! The Lorentz gauge condition implies that, for all physical observables, the contributions from the timelike and longitudinal photons always cancels.
E. without loss of generality we can fix ∂µ Aµ = 0. e. Aµ (x) → Aµ (x) + ∂µ χ(x) where χ(x) must satisfy the wave equation, ∂µ ∂ µ χ = 0. Immediately we try to quantise the electromagnetic field we hit a problem. 28) and from this it follows that Π 0 = 0. This means we have no possibility to impose a non-zero commutation relation between Π 0 and A0 , which we would need if we are to quantise the field. Fortunately, all is not lost. e. write 1 1 L = − F µν Fµν − jµ Aµ − (∂µ Aµ )2 . 29) The new term fixes the gauge and ξ is a dimensionless Lagrange multiplier (as such it can take on any value we choose).
Scattering, or dis, a photon strikes a quark in a proton, say, imparting a large momentum to it. Some strong interaction corrections to this part of the process can be calculated perturbatively. As the quark heads off out of the proton, however, the low energy strong interactions cut in again and “hadronise” the quark into the particles you actually detect. 5. We now try to repeat the procedure we used for renormalising the coupling in qed, but this time in qcd, which is also a renormalisable theory.