Srednicki — Quantum Field Theory · Problem 89.5

Canonical Quantization of Scalar Fields — Mark Srednicki

Physics — QFT · standard-model-quark-sector

Problem

After integrating out the W±W^\pm fields, we get an effective interaction between the hadron and lepton currents that includes Leff=22 ZCC(E‾LγμNeL)(U‾LγμDL) ,(89.35)\mathcal{L}_{\rm eff} = 2\sqrt{2}\,Z_C C(\overline{\mathcal{E}}_{\rm L}\gamma^\mu\mathcal{N}_{e\rm L})(\overline{\mathcal{U}}_{\rm L}\gamma_\mu\mathcal{D}_{\rm L}) \,, \tag{89.35} where we have defined C≡c1GFC \equiv c_1 G_{\rm F} at a renormalization scale μ=MW\mu = M_{\rm W}, and ZCZ_C is a renormalizing factor. This interaction contributes to neutron decay; see section 90. In this problem, following the analysis of problem 88.7, we will compute the anomalous dimension γC\gamma_C of CC due to one-loop photon and gluon exchange. a) Use Fierz identities to show that eq.\,(89.35) can be rewritten as Leff=22 ZCC(E‾LγμDL)(U‾LγμNeL) ,(89.36)\mathcal{L}_{\rm eff} = 2\sqrt{2}\,Z_C C(\overline{\mathcal{E}}_{\rm L}\gamma^\mu\mathcal{D}_{\rm L})(\overline{\mathcal{U}}_{\rm L}\gamma_\mu\mathcal{N}_{e\rm L}) \,, \tag{89.36} and also as Leff=−42 ZCC(D‾cPLNe)(E‾PRUc) .(89.37)\mathcal{L}_{\rm eff} = -4\sqrt{2}\,Z_C C(\overline{\mathcal{D}}^c P_{\rm L}\mathcal{N}_e)(\overline{\mathcal{E}}P_{\rm R}\mathcal{U}^c) \,. \tag{89.37} b) Working in Lorenz gauge and using the results of problem 88.7, show that gluon exchange does not make a one-loop contribution to ZCZ_C. c) Show that only a photon connecting the ee and uu lines makes a one-loop contribution to ZCZ_C. d) Note that E‾PRUc=e†u†\overline{\mathcal{E}}P_{\rm R}\mathcal{U}^c = e^\dagger u^\dagger, and compare this with E‾E=e†eˉ†+h.c.\overline{\mathcal{E}}\mathcal{E} = e^\dagger\bar{e}^\dagger + \text{h.c.}. Argue that the photon-exchange contribution to ZCZ_C is given by the one-loop contribution to ZmZ_m in spinor electrodynamics in Lorenz gauge, with the replacement (−1)(+1)e2→(−1)(+23)e2(-1)(+1)e^2 \to (-1)(+\frac{2}{3})e^2. e) Let γC(α)=c1α+…\gamma_C(\alpha) = c_1\alpha + \ldots, where α=e2/4π\alpha = e^2/4\pi, and find c1c_1. (This c1c_1 should not be confused with the cosine of the Cabibbo angle.)

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