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.


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 $\psi^i$, $C_{(\bar{\psi}\gamma^5\psi)\,(\bar{\psi}\psi)\,\psi}=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.