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If I consider an OPE of some operators, which belong to a theory where parity is not violated, will I have a constraint on the kind of operators appearing in the right hand side ?

For example, I have a Gross-Neveu-Yukawa model which is known to be parity even, if I consider an OPE of $(\overline{\psi}.\psi)^n \times (\overline{\psi}.\psi)^n \psi^i$ operators of this theory will there be any constraints from parity ? Here $i$ is O(N) index. I am defining my theory in some general $d$ dimensions.

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The OPE coefficients respect the symmetries. For example, consider the OPE $$ \mathcal{O}^i(x) \times \mathcal{O}^j(0)=\sum_{k} C^{ij}_{k}\left[|x|^{\Delta_k-\Delta_i-\Delta_j}\mathcal{O}^k(0)+\mathrm{descendants}\right] $$ where for the time being I am suppressing the spin index. Then the $C^{ij}_{k}$ transform as $$ C^{ij}_{k}\rightarrow U^{i}_a U^{i}_b U^{-1\, c}_{k}C^{ab}_{c} $$ where the operators are transforming (for simplicity all in the same irreducible representation) as $\mathcal{O}^k\rightarrow U^{k}_{a} \mathcal{O}^a$ under some internal symmetry group transforrmation.

In the case at hand, parity is both simpler and harder than the case above because it has very simple representations, although it is a spacetime symmetry that reflects the point $$x^\mu \rightarrow \mathcal{P}^{\mu}_{\nu}x^\nu$$ where $\mathcal{P}^{\nu_1}_{\mu_1}=\mathrm{diag}(1,-1,\ldots,-1)$. The operators are either even or odd, that is $$P\mathcal{O}^{\pm}(0)P^{-1}=\pm \mathcal{}\mathcal{O}^{\pm}(0)$$ for scalars and $$ P\mathcal{O}^{\pm}_{\mu_1\ldots\mu_\ell}(0)P^{-1}=\pm \mathcal{P}^{\nu_1}_{\mu_1}\cdots \mathcal{P}^{\nu_\ell}_{\mu_\ell}\mathcal{O}^{\pm}_{\nu_1\ldots\nu_\ell}(0) $$ for integer spin-$\ell$ fields. Analogous formulas hold for half-integer spins. Now, let's consider for simplicity the OPE of two scalars with parity $\eta_{1}$ and $\eta_{2}$ respectively. $$ \mathcal{O}^{(\eta_1)}_{1}(x) \times \mathcal{O}^{(\eta_2)}_2(0)=\sum_{\mathcal{O}} C_{12\mathcal{O}}\left[|x|^{\Delta_k-\Delta_1-\Delta_2-\ell}x^{\mu_1}\ldots x^{\mu_\ell}\mathcal{O}_{\mu_1\ldots\mu_\ell}(0)+\mathrm{descendants}\right]\,. $$ Applying with the parity transformation on both side of the OPE we see a simple selection rule, namely that $C_{12\mathcal{O}}$ vanishes unless the the parity $\eta_\mathcal{O}=\pm$ of $\mathcal{O}$ is such that $$\eta_1\cdot\eta_2\cdot\eta_\mathcal{O}=1\,.$$

Let's consider now your example. If I understand correctly your notation you want to know the selection rule for the OPE coefficient among $\bar{\psi}\psi$, itself, and $\psi^i$, that is the property under parity of $C_{(\bar{\psi}\psi)\,(\bar{\psi}\psi)\, \psi^i}$. Since $(\bar{\psi}\psi)$ has parity $+$, the OPE coefficient is non-vanishing only if $\psi^i$ has also parity $+$ (which is the case). Therefore, there is no constraint coming from parity on $C_{(\bar{\psi}\psi)\,(\bar{\psi}\psi)\,\psi}$. If you had instead considered the OPE among operators with opposite parity (say $(\bar{\psi}\gamma^5\psi)$ and $(\bar{\psi}\psi)$) it would have forbidden the appearance of $\psi^i$, $C_{(\bar{\psi}\gamma^5\psi)\,(\bar{\psi}\psi)\,\psi}=0$.

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