4
$\begingroup$

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...

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – JamalS
    Dec 27, 2014 at 19:44

1 Answer 1

1
$\begingroup$

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.

The coefficients in this linear combination vary with normalizations and signs chosen in the Lagrangian.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Bulkilol
    Dec 28, 2014 at 22:28
  • $\begingroup$ 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. $\endgroup$ Dec 28, 2014 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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