Renormalization in HQET

Last updated Aug 5, 2023

• Rework

To renormalize a composite operator of the form $$O = \bar{\psi},\Gamma,\psi$$ we renormalize the field $$\psi=Z^{1/2}\psi \psi_R$$ with $\psi_R$ the renomalized field and $Z\psi$ the renormalization constant. For the renormalized operator $O_R$ we have $$O_R = \frac{1}{Z_O}O = \bar{\psi}R\Gamma\psi_R +
​​ \underbrace{\left(\frac{Z\psi}{Z_O} - 1\right)\bar{\psi}R\Gamma\psi_R}\text{counterterm}$$

Example: heavy-light current in HQET

$$(\bar{q}\Gamma h_v)_R = \bar{q}_R\Gamma {h_v}R + \left(\frac{\sqrt{Z_q}\sqrt{Z_h}}{Z\Gamma} - 1\right)\bar{q}_R\Gamma {h_v}_R$$

Plan for renormalizing the heavy-light current $\bar{q};\Gamma ;h_v$

1. Find light quark field renormalization constant $Z_q$
2. Find heavy (static) quark field renormalization constant $Z_h$
3. Find the counterterm for the heavy-light current
4. Extract heavy-light current renormalization constant $Z_\Gamma$

The Lagrangian in the light quark sector is $$\mathcal{L}\big|_\text{light} = \bar{q}i\cancel{\partial}q,$$ and with $q = Z^{1/2}q q_R$ we get $$\mathcal{L}\big|\text{light} = \bar{q}_Ri\cancel{\partial}q_R + \underbrace{(Z_q - 1)\bar{q}Ri\cancel{\partial}q_R}\text{counterterm},$$ Remember, the light d.o.f. are massless!

To calculate the renormalization at one loop, we consider the one-loop self energy of the quark. The renormalization condition requires the counterterm to regulate the infinities from the self energy. Namely

The explicit calculation leads to $$Z_q = 1 - C_F \frac{\mu^{-2\epsilon}\alpha_s}{4\pi\epsilon}$$

Derivation of $Z_q$

First we calculate the amplitude for the one loop self energy $\mathcal{M}1$, where we just insert the massless propagators for the quark and gluon. We use Feynman gauge, in which the gluon propagator is $$\frac{-i\delta{ab}g_{\mu\nu}}{k^2+i\epsilon},$$ and using the QCD Feynman rules the amplitude for the loop part (dropping external fields) after integrating over the internal momentum is $$\mathcal{M}1 = T^aT^a (ig)^2 \int \frac{d^dq}{(2\pi)^d}\gamma\mu i \frac{\cancel{p}-\cancel{q}}{(p-q)^2 + i\epsilon}\gamma^\mu\frac{-i}{q^2+i\epsilon}.$$ We then use the Feynman parametrization for the integral $$\mathcal{M}1 = -1_c C_F g^2 \int\frac{d^dq}{(2\pi)^d}\gamma\mu(\cancel{p}-\cancel{q})\gamma^\mu \int^1_0 dx \frac{1}{[x(p-q)^2+(1-x)q^2]^2}$$