Fall Term Exam, N^{o}. 2 (solution by D. Freed)

Problem:

Compute in ϕ^{4} theory in four dimensions the anomalous dimension of the operator ϕ^{2}, in one–loop approximation.

Solution:

Consider the Lagrangian density in Euclidean space

L = \frac{1}{2} |dϕ|^{2} + \frac{m^{2}}{2} ϕ^{2} + \frac{λ}{4!} ϕ^{4} − jϕ^{2}.

The coefficient j is renormalized to 1-loop by the diagram shown,

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which contributes to the correlation function ⟨ϕ^{2}ϕϕ⟩. Evaluating in momentum space, we obtain

\frac{−jλ}{2} ∫ \frac{d^{4}q}{(2π)^{4}} \frac{1}{(q^{2} + m^{2}) ((k − q)^{2} + m^{2})},

where the factor of 2 comes from the symmetry of the diagram. This is logarithmically divergent, and cutting off at Λ we compute the divergence as Λ ⟶ ∞ to be

\frac{−jλ}{2} · \frac{1}{(2π)^{4}} · vol S^{3} · ℓn Λ/μ = \frac{−jλ}{16π^{2}} ℓn Λ/μ,

where μ is a fixed constant.

To cancel this divergence, we renormalize j to

ĵ_{0} = j (1 + \frac{λ}{16π^{2}} ℓn Λ/μ) .

The anomalous dimension of j is computed as γ_{j} = \frac{μ}{j} \frac{∂j}{∂μ} with ĵ_{0} fixed and Λ ⟶ ∞. Thus

0 = γ_{j} (1 + \frac{λ}{16π^{2}} ℓn Λ/μ) − \frac{λ}{16π^{2}} + \frac{β}{16π^{2}} ℓnΛ/μ,

where β = μ \frac{∂λ}{∂μ} is the β-function. Since γ_{j} = O(λ) and β = O(λ^{2}) we find γ_{j} = \frac{λ}{16π^{2}} . Now the anomalous dimension of the Lagrangian vanishes, and so the anomalous dimension γ_{ϕ^{2}} of ϕ^{2} equals −γ_{j}, or

γ_{ϕ^{2}} = −\frac{λ}{16π^{2}} .