The chiral $U(1)$ anomaly in QCD can be calculated exactly by one-loop Feynman diagrams, for example by the famous triangle diagram. I am currently performing the computation to get a better understanding of the QCD $\theta$-term, $\mathcal{L}\propto\theta G\tilde{G}$, where $G$ is the gluon field strength tensor and $\tilde{G}_{\mu\nu}=\frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}G^{\rho\sigma}$ is its Hodge dual.

However, I am stuck in the last step of the computation. After evaluating the triangle diagram with the usual approach of Feynman parametrization, shift of the momentum integral, exploiting symmetry properties of the momentum integral, ..., I finally obtain the expression

$$g^2 Tr(T^aT^b)\varepsilon_{\mu\nu\rho\sigma}q_1^\mu q_2^\nu \varepsilon_1^\rho \varepsilon_2^\sigma,$$

where $g$ is the QCD coupling strength, $T^a$ are the generators of the Lie group, $\varepsilon_{\mu\nu\rho\sigma}$ is the epsilon tensor, $q_1$ and $q_2$ are the incoming gluon momenta, and $\varepsilon_1$ and $\varepsilon_2$ are the gluon helicity states.

The trace $Tr(T^aT^b)$ is simply $\delta^{ab}$, but I don't know how to handle the momenta and the helicity states. How do I rewrite this expression into the final result,


to obtain the above-mentioned $\theta$-term?


1 Answer 1


You now want to go back to position space. I'll do that here very schematically which gives the answer without keeping track of the overall factor.

Essentially, under the Fourier transform, $q_{i\mu} \to \partial_\mu$ and $\epsilon_{i\mu} \to A_\mu$. Then

$$ g^2 \text{Tr}(T^aT^b) \varepsilon_{\mu\nu\rho\sigma} q_1^\mu q_2^\nu \epsilon_1^\rho \epsilon_2^\sigma \to g^2 \text{Tr}(T^aT^b) \varepsilon_{\mu\nu\rho\sigma} \partial^\mu A^\nu \partial^\rho A^\sigma \sim g^2 \text{Tr}(T^aT^b) \varepsilon_{\mu\nu\rho\sigma} F^{\mu\nu} F^{\rho\sigma} \sim g^2 \text{Tr}(T^aT^b) \ast ( F \wedge \ast F ) \sim g^2 \ast \text{Tr} (F \wedge \ast F ) $$

  • 2
    $\begingroup$ Could you please explain to me why your proportionalities still hold true for non-Abelian gauge fields? I see that $\partial^\mu A^\nu$ is proportional to $F^{\mu\nu}$ if $F$ is the Abelian photon field strength. But a non-Abelian gluon field strength $G$ also contains the additional term $f^{abc}A^b A^c$, which is bilinear in the gluon fields $A$. As far as I see, these additional terms somehow have to vanish -- are they canceling out somewhere in the computation? $\endgroup$
    – Thomas
    Oct 6, 2016 at 10:10

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.