Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 2 / c

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- Solution c

Solution

The bare coupling does not depend on the renormalization scale. Expressing it through the renormalized field and coupling gives, to the stated order,

0 = μ \frac{d}{d μ} [μ^{ϵ} (g + δ g) - 2 μ^{ϵ} g δ Z] .

At one loop

δ Z = 0

. Put

a = 3/ (16 π^{2})

and

δ g = a g^{2} / ϵ

. If

β_{g} = μ d g / d μ

, differentiation and expansion through

g^{2}

yield

β_{g} = - ϵ g + \frac{3 g ^{2}}{16 π ^{2}} + O (g^{3}) .

Since

λ = μ^{ϵ} g

,

β_{λ} = μ \frac{d λ}{d μ} = ϵ λ + μ^{ϵ} β_{g} = \frac{3 μ ^{- ϵ} λ ^{2}}{16 π ^{2}} + O (λ^{3}),

and in four dimensions

β_{λ} = \frac{3 λ ^{2}}{16 π ^{2}} + O (λ^{3}) .

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