A. Zee, Quantum Field Theory in a Nutshell · Problem IV.7.4

Field Theory Redux

Physics — QFT

Problem

Take the Pauli-Villars regulated Λλμν(k1,k2)\Lambda^{\lambda\mu\nu}(k_1, k_2) and contract it with qλq_\lambda. The analog of the trick in chapter II.7 is to write qγ5q\gamma^5 in the second term as [2M+(p̸−M)−(p̸−q̸+M)]γ5[2M + (\not{p} - M) - (\not{p} - \not{q} + M)]\gamma^5. Now you can freely shift integration variables. Show that qλΛλμν(k1,k2)=−2MΔμν(k1,k2)q_\lambda\Lambda^{\lambda\mu\nu}(k_1, k_2) = -2M\Delta^{\mu\nu}(k_1, k_2) where Δμν(k1,k2)≡(−1)i3∫d4p(2π)4tr γ51p̸−q̸−Mγν1p̸−̸k1−Mγμ1p̸−M+{μ,k1↔ν,k2}\Delta^{\mu\nu}(k_1, k_2) \equiv (-1)i^3 \int \frac{d^4p}{(2\pi)^4} \text{tr } \gamma^5 \frac{1}{\not{p} - \not{q} - M} \gamma^\nu \frac{1}{\not{p} - \not{k}_1 - M} \gamma^\mu \frac{1}{\not{p} - M} + \{\mu, k_1 \leftrightarrow \nu, k_2\} Evaluate Δμν\Delta^{\mu\nu} and show that Δμν\Delta^{\mu\nu} goes as 1/M1/M in the limit M→∞M \to \infty and so the right hand side goes to a finite limit. The anomaly is what the regulator leaves behind as it disappears from the high energy spectrum: It is like the smile of the Cheshire cat. [We can actually argue that Δμν\Delta^{\mu\nu} goes as 1/M1/M without doing a detailed calculation. By Lorentz invariance and because of the presence of γ5\gamma^5, Δμν\Delta^{\mu\nu} must be proportional to ϵμνλρk1λk2ρ\epsilon^{\mu\nu\lambda\rho}k_{1\lambda}k_{2\rho}, but by dimensional analysis, Δμν\Delta^{\mu\nu} must be some constant times ϵμνλρk1λk2ρ/M\epsilon^{\mu\nu\lambda\rho}k_{1\lambda}k_{2\rho}/M. You might ask why we can't use something like 1/(k12)1/21/(k_1^2)^{1/2} instead of 1/M1/M to make the dimension come out right. The answer is that from your experience in evaluating Feynman diagrams in (3+1)(3 + 1)-dimensional spacetime you can never get a factor like 1/(k12)1/21/(k_1^2)^{1/2}.]

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