and it is because frequency and this mass term have the same unity matrix structure. Note that both corrections naturally appear in higher-loop contributions.

On the other hand, non-trivial renormalizations of isotropic mass ($m$) and anisotropic mass ($M_c$) appear even at one-loop level. For example, the renormalization of isotropic mass term can be read off by calculating the third component ($\Gamma^3$) of the self energy, $$\begin{aligned}
&&\gamma^3(k) \equiv {\rm Tr}\left(\Gamma^3 \frac{\partial^2}{\partial k_x \partial k_y} \Sigma_{\psi}(k, 0) \right)  / {\rm Tr} (\Gamma^3 \Gamma^3) \nonumber \\
&=& -\frac{g^2}{c_0}\frac{\partial}{\partial k_x \partial k_y} \int_q \frac{\epsilon_3(k+q)}{ \sqrt{\left(\frac{(k+q)^2}{2m}\right)^2+ \frac{m+2 M_c}{4 m M_c^2} p_c(k+q) }} \frac{1}{q^2}. \nonumber \end{aligned}$$ It can be understood as the self energy corrrection to the original Hamiltonian, $$\begin{aligned}
\epsilon_3(k) &\rightarrow& \epsilon_3(k)- \gamma^3(0) k_x k_y 
% {\rm Tr}\left(\Gamma^3 \frac{\partial^2}{\partial k_x \partial k_y} \Sigma_{\psi}(k, 0) \right)  / {\rm Tr} (\Gamma^3 \Gamma^3) 
\nonumber \\
&=&\epsilon_3(k)\left(1+ u f_1(\frac{m}{M_c}) \log(\frac{\Lambda}{\mu}) \right).\end{aligned}$$ In the last line, the momentum integration is done for $\mu < q < \Lambda$. The correction to the isotropic mass term, $\delta (1/m) = \frac{u}{m}  f_1(\frac{m}{M_c}) \log(\frac{\Lambda}{\mu})$ can be regularized in the standard way. Along the same line, one can determine the renormalization of the anisotropic mass term by calculating the fourth component ($\Gamma^4$) of the 