# anomalous chiral symmetry and the $\bar\theta$ parameter

I am studying anomalous $U(1)$'s, related to the strong CP problem, and I have some trouble with the origin of the parameter $\bar{\theta}$.

We start with the QCD Lagrangian with the topological term: $$\mathcal{L}_{QCD}= -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\theta \epsilon^{\mu\nu\rho\lambda}F_{\mu\nu}F_{\rho\lambda}+\big(i\bar{\chi}_L^f\bar{\sigma}^\mu D_\mu\chi_L^f-\bar{\chi}_L^f M_{ff'}\chi_R^{f'}+\mathrm{h.c.}\big)$$ where $\chi_{L,R}^f$ are chiral Weyl fermions (a sum over the flavours $f$ is understood). At this point, we have a classical $U(1)_{chiral}$ symmetry: $$\chi_L^f\to e^{i\alpha}\chi_L^f\qquad\chi_R^f\to e^{-i\alpha}\chi_R^f\qquad \forall f \text{ and } \alpha\in\mathbb{R}$$ At the quantum level, it is anomalous, and we have $$\partial_\mu J^\mu=\mathcal{A}=C\epsilon^{\mu\nu\rho\lambda}F_{\mu\nu}F_{\rho\lambda}$$ In the path integral formalism, it translates to a non-invariant measure: $$\prod_f\mathcal{D}\chi_L^f\mathcal{D}\chi_R^f\to \prod_f\mathcal{D}\chi_L^f\mathcal{D}\chi_R^f ~ \exp \left( i\int \! d^4x \, \, \alpha \, C\epsilon^{\mu\nu\rho\lambda}F_{\mu\nu}F_{\rho\lambda} \right)$$ This means that effectively, a $U(1)$ transformation shifts the theta parameter: $$\theta \to \theta+\alpha$$

To get the physical $\bar{\theta}=\theta-\arg(\det M)$ parameter, I then guess that I'm supposed to do a transformation with parameter $\alpha=-\arg(\det M)$, however, I don't see how that gives me a diagonalised real mass matrix . I think I'm missing something rather simple, but I fail to see it...

• LaTeX tip: when you have complicated or large expressions in exponentials, it's best to place them inside \left( and \right) brackets preceded by \exp. – JamalS Dec 27 '14 at 19:44

When you apply the chiral $U(1)$, you also need to transform the matrix $M_{ff'}$ to preserve the Lagrangian. Furthermore, each flavor may transform with a different phase factor. The matrix $M_{ff'}$ will transform to $M' = UMU$ where $U = \operatorname{diag}(e^{i\alpha_1}, \ldots, e^{i\alpha_n})$. The parameter $\alpha$ shifts by something proportional to $\sum_i \alpha_i$. Note that $\det M' = e^{2i\sum_i \alpha_i} \det M$, so that some linear combination of $\arg\det M$ and $\alpha$ is invariant.
• I think I misunderstood the point of $\bar\theta$, but I get it now. You have a $U(1)$ symmetry only in the limit of vanishing mass. Here you do a field redefinition to diagonalise the matrix. But since you don't have invariance anymore, you must assign a transformation to the mass matrix (à la spurion). Since you don't have invariance, shouldn't the anomaly also be modified? – Bulkilol Dec 28 '14 at 22:28
• The Lagrangian is still invariant. It's just that with the transformed fields, the components of the mass matrix are different. Since the symmetry is anomalous, the parameter $\theta$ isn't physical, but $\overline\theta$ is invariant, and so can be physical. – Robin Ekman Dec 28 '14 at 22:47