we have to take a different approach resides in the fact that on the lattice measurements are made for various individual components of the free indices. Therefore, we have to take a more general approach and decompose our Green’s functions into a basis of Lorentz tensors which have the [*same*]{} symmetry structure as the operator which is inserted into the Green’s function. For the tensor current this means that the basis has to be antisymmetric in the two free indices since $\sigma^{\mu\nu}$ $=$ ${\mbox{\small{$\frac{1}{2}$}}}[\gamma^\mu ,\gamma^\nu]$. In the case of the two Wilson operators each operator in the respective levels are totally symmetric in the indices and are traceless in $d$-dimensions, [@1]. To be specific we have $$\begin{aligned}
{\cal S} {\cal O}^{W_2}_{\mu\nu} &=& {\cal O}^{W_2}_{\mu\nu} ~+~
{\cal O}^{W_2}_{\nu\mu} ~-~ \frac{2}{d} \eta_{\mu\nu}
{\cal O}^{W_2\,\sigma}_{\sigma} ~, \nonumber \\
{\cal S} {\cal O}^{W_3}_{\mu\nu\sigma} &=&
{\cal O}^{W_3}_{S~\mu\nu\sigma} ~-~ \frac{1}{(d+2)} \left[
\eta_{\mu\nu} {\cal O}^{W_3~~\rho}_{S~\sigma\rho} ~+~
\eta_{\nu\sigma} {\cal O}^{W_3~~\rho}_{S~\mu\rho} ~+~
\eta_{\sigma\mu} {\cal O}^{W_3~~\rho}_{S~\nu\rho} \right] \end{aligned}$$ with $$\begin{aligned}
{\cal O}^{W_3}_{S~\mu\nu\sigma} &=&
\frac{1}{6} \left[ {\cal O}^{W_3}_{\mu\nu\sigma} ~+~
{\cal O}^{W_3}_{\nu\sigma\mu} ~+~
{\cal O}^{W_3}_{\sigma\mu\nu} ~+~
{\cal O}^{W_3}_{\mu\sigma\nu} ~+~
{\cal O}^{W_3}_{\sigma\nu\mu} ~+~
{\cal O}^{W_3}_{\nu\mu\sigma} \right] \end{aligned}$$ where the basic operators are $$\begin{aligned}
{\cal O}^{W_2}_{\mu\nu} &=& \bar{\psi} \gamma_\mu D_\nu \psi ~, \nonumber \\
