By Forshaw J.

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Thus, for qed, if the original Lagrangian is (ignoring the gaugefixing term), 1 L = − Fµν F µν + iψ ∂/ψ − eψ Aψ / − mψψ, 4 then redefine everything by: 1/2 ψ = Z2 ψR , Z1 eˆ, e = Ze eˆ = 1/2 Z2 Z3 1/2 Aµ = Z3 AµR , m = Zm m, ˆ where the subscript R stands for “renormalised”. In terms of the renormalised fields 1 L = − Z3 FR µν FRµν + iZ2 ψ R ∂/ψR − Z1 eˆψ R A / R ψR − Zm Z2 mψ ˆ R ψR . 4 Writing each Z as Z = 1 + δZ, re-express the Lagrangian one more time as 1 L = − FR µν FRµν + iψ R ∂/ψR − eˆψ R A / R ψR − mψ ˆ R ψR + (δZ terms).

We can’t trust our calculations if α s (µ) > 1. In practice, you can perhaps use scales for µ down to about 1 GeV, but not much lower, and 2 GeV is probably safer. 8) to two loops gives gˆ(µ) = g + g 3 a1 ln 2 M2 M2 2 M 5 + b + g a ln + b ln + c2 , 1 2 2 µ2 µ2 µ2 with a similar equation for gˆ(µ 0 ) in terms of g. Renormalisability implies that gˆ(µ) can be expanded in terms of gˆ(µ0 ), ∞ gˆ2n+1 (µ0 )Xn , gˆ(µ) = n=0 where the Xn are finite coefficients. Show that this implies that a 2 is determined once the one loop coefficient a1 is known.

However, we have seen that both SU(2) and the rotation group SO(3) have the same, angular momentum, algebra. What is going on? It must be that SO(3) is not simply connected. In fact, there is a mapping, called a covering, from SU(2) to SO(3) which preserves the group property: that is if U ∈ SU(2) is mapped to f (U) ∈ SO(3), then f (UV ) = f (U)f (V ). In the SU(2) → SO(3) case, two elements of SU(2) are mapped on to every element of SO(3). Whenever a group G has the same Lie algebra as a simply connected group S there must be such a covering S → G.

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