Speaker
Description
The one-loop matching between the overlinee coupling and the MS coupling is done in two steps. First, we
calculate the relation of the overlinee coupling to a renormalised lattice coupling. Combination with
the known one-loop relation between the lattice and the MS couplings then yields the desired result.
\subsection{The basic calculation}
Expand $\overline{g}(L)$ the running coupling in terms of $g_0$ the overlinee coupling:
\begin{equation}
\overline{g}^2 = g_0^2 + p_1 g_0^4 + O(g_0^6),
\end{equation}
where the coefficient $p_1$ depends on the number of flavors $n_f$ the overlinee quark mass in lattice units $m_0$ and the lattice size $$l\equiv \frac{L}{a} .$$
For later convenience this dependence is split into the pure gauge part and the quark contribution:
\begin{equation}
p_1 = p_1(n_f,z,l)= p_{1,0}(l) + n_f p_{1,1}(z,l),
\end{equation}
with:
\begin{equation}
z = \overline{m} L , \qquad \overline{m} = \frac{1}{a}\ln(1+m_0a).
\end{equation}
Note that expansion of $\overline{m}$ w.r.t. small lattice spacing $a$ gives the first order result $\overline{m}=m_0$.
\\\\
Now expand $g_0$ in terms of renormalised coupling $g_{lat}(\mu)$ at the $\mu$ scale gives:
\begin{equation}
g_0^2 = g_{lat}^2 + z_1 g_{lat}^4 + O(g_{lat}^6),
\end{equation}
\begin{equation}
z_1\equiv z_{1,0}(a\mu) + n_f z_{1,1}(a\mu) = 2b_0(n_f,0)\ln(a\mu),
\end{equation}
where $b_0(n_f,0)$ is known:
\begin{equation}
b_0(n_f,0) = \frac{1}{4\pi^2}(11 - \frac{2}{3} n_f)
\end{equation}
Then we have the relation between renormalised coupling constants $\overline{g}$ and $g_{lat}$:
\begin{equation}
\overline{g}^2 = g_{lat}^2 + (p_1 + z_1) g_{lat}^4 + O(g_{lat}^6).
\end{equation}
The goal is to calculate the quark field contribution $p_{1,1}$. Using the standard Wilson plaquette action, we have:
\begin{equation}
p_{1,1}(z,l)= (k n_f)^{-1} \frac{\partial}{\partial{\eta}} \ln \det(D_5)|{\eta=\nu=0},
\end{equation}
where:
\begin{align}
k =& 12 l^2 (\sin{\frac{\pi}{3l^2}}+\sin{\frac{2\pi}{3l^2}} ) = 12 (L/a)^2 \left[\sin(\frac{1}{3}\pi(a/L)^2) + \sin(\frac{2}{3}\pi(a/L)^2) \right]
\
D_5 =& \gamma_5 (D + m_0)
\
D =& \sum{\mu}(\frac{\gamma_\mu}{2}(\nabla_\mu^ + \nabla_\mu ) - \frac{a}{2}\nabla_\mu^\nabla_\mu) + \sum_{\mu\nu} c_{sw} \frac{ia}{4}\sigma_{\mu\nu}P_{\mu\nu}
\
\sigma_{\mu\nu} =& \frac{i}{2} [\gamma_\mu,\gamma_\nu]
\end{align}
