legs. Thus there is a momentum of $(p$ $+$ $q)$ flowing out through the operator insertion whose location is indicated by the circle containing a cross. In order to determine the renormalization constants for the basic operators in the ${\overline{\mbox{MS}}}$ and RI${}^\prime$ schemes one chooses $q$ $=$ $-$ $p$. However, for the RI${}^\prime$/SMOM scheme the two momenta are left unconstrained. Instead to define the symmetric point of subtraction for the renormalization the square of the momenta satisfy, [@33; @34; @35], $$p^2 ~=~ q^2 ~=~ ( p + q )^2 ~=~ -~ \mu^2$$ which imply $$pq ~=~ \frac{1}{2} \mu^2$$ where $\mu$ is the renormalization scale introduced to ensure the coupling constant is dimensionless in $d$-dimensions. Given this the Green’s function is decomposed into a basis of independent Lorentz tensors, ${\cal P}^i_{(k) \, 
\mu_1 \ldots \mu_{n_i} }(p,q)$, with associated amplitude, $\Sigma^{{\cal O}^i}_{(k)}(p,q)$, which is the value we will compute at the symmetric subtraction point, $$\left. \left\langle \psi(p) {\cal O}^i_{\mu_1 \ldots \mu_{n_i}}(-p-q) 
\bar{\psi}(q) \right\rangle \right|_{p^2 = q^2 = - \mu^2} ~=~ \sum_{k=1}^{n_i} 
{\cal P}^i_{(k) \, \mu_1 \ldots \mu_{n_i} }(p,q) \, 
\Sigma^{{\cal O}^i}_{(k)}(p,q) ~.$$ Here the bracketed subscript $k$ labels the tensor of the basis and the superscript $i$ is the operator level label, 