I am reading Section $3$ of this review titled The Strong CP Problem and Axions by R. D. Peccei and also the post here.

A famous solution of the Strong CP problem in QCD is offered by the proposal of the Peccei-Quinn symmetry and its spontaneous breakdown. In a nutshell, the idea is that a global anomalous chiral ${\rm U(1)}$ symmetry, called the Peccei-Quinn symmetry, exists at very high energies. It is also broken at very high energies to produce a low-energy effective Lagrangian $$\mathscr{L}_{\rm eff}=\mathscr{L}_{{\rm SM}}+\Big(\bar{\theta}+\xi\frac{a}{f_a}\Big)\frac{g^2}{32\pi^2}G^a_{\mu\nu}\tilde{G}^{\mu\nu}_a-\frac{1}{2}\partial_\mu a\partial^\mu a+\mathscr{L}_{\rm int}(\partial_\mu a,\psi)\tag{1}$$ where $\bar{\theta}$ is called the effective theta parameter of QCD, $G^a_{\mu\nu}$ are the gluon field strength tensor, $a$ is a Goldstone boson field (called axion), $f_a$ is the scale at which the symmetry is broken and $\xi$ is a dimensionless coupling.

I wonder how the Lagrangian of Eq.$(1)$ can be derived. I tried to sketch a possible way to motivate a derivation but it remains incomplete. Any suggestions about how to proceed further?

My attempt We start by defining the ${\rm U(1)}_{\rm PQ}$ transformation on the Standard Model (SM) fields and also extend the SM with an additional complex scalar field $\phi$ (with a potential $V(\phi)$ and some interaction terms with the SM fields).

Assuming $V(\phi)$ is minimized at $|\phi|=f_a\big/\sqrt{2}$, we can write $\phi$ as $$\phi=\frac{1}{\sqrt{2}}(f_a+\rho)\exp\Big[i\frac{a}{f_a}\Big].\tag{2}$$ When expanded, the gradient term for $\phi$ gives rise to $$(\partial_\mu\phi^*)(\partial^\mu\phi)=\frac{1}{2}(\partial_\mu\rho)(\partial^\mu\rho)+\frac{1}{2}(\partial_\mu a)(\partial^\mu a)+\frac{1}{2f_a^2}(\rho^2+2f_a\rho)(\partial_\mu a)(\partial^\mu a).\tag{3}$$ A typical potential becomes $$V(\phi)=\frac{\lambda}{4}\Big(|\phi|^2-\frac{f_a^2}{2}\Big)^2=\frac{\lambda}{16}\Big(\rho^2+2f_a\rho\Big)^2\tag{4}$$

Question Since $a$ be the axion field, we get the desired kinetic term $-\frac{1}{2}\partial_\mu a\partial^\mu a$ (apart from a sign). But we should also obtain another physical field $\rho$. Why is the field $\rho$ absent in the Lagrangian $(1)$?

Note I do not know of any reference that starts with the high energy Lagrangian and from that systematically derives the effective Lagrangian $(1)$. If there is any reference containing a detailed derivation of $(1)$, please suggest.

  • $\begingroup$ About the last question: the mass of the radial component is of the order of the symmetry breaking scale. The effective theory at lower energies does not contain it. $\endgroup$
    – coconut
    Dec 28, 2019 at 21:10
  • $\begingroup$ This is not very clear to me. Could you please supply some steps? @coconut $\endgroup$
    – SRS
    Dec 30, 2019 at 10:38
  • $\begingroup$ why must it be a complex field, and not just the axion? And where do you want to derive it from, strings, GUT, supersymmetry? $\endgroup$
    – Kosm
    Aug 7, 2020 at 7:19
  • $\begingroup$ Peccei-Quinn symmetry is a U(1) symmetry. Without a complex scalar field, how will you have a U(1) symmetry? en.wikipedia.org/wiki/Peccei%E2%80%93Quinn_theory . I have also given a relevant reference in the question. Not SUSY but from PQ symmetry. Axions arise after the spontaneous breakdown of PQ symmetry. I have changed the misleading title. @Kosm $\endgroup$
    – SRS
    Aug 7, 2020 at 7:33
  • $\begingroup$ @SRS it's a global symmetry, not gauged. Axion is shifted by a constant, so its kinetic term is invariant. $\endgroup$
    – Kosm
    Aug 7, 2020 at 7:38

1 Answer 1


There are many ways to generate the topological term $$(\bar\theta + \theta_a)G\tilde G\,.$$ Let's consider the simplest example, known as the KSVZ mechanism. Let's introduce a new vector-like quark $q$ to the standard model Lagrangian. "Vector-like" means that the quark does not have chiral symmetry, and so it is an electroweak singlet. It is a dirac fermion, so it may have a bare mass, allowing us to evade collider constraints on new strongly coupled particles.

Let $\Phi$ be the axion parent field, so it will obey a (global) $U(1)$ axial symmetry, known as Peccei Quinn symmetry. This new quark will similarly be charged under the global $U(1)$, leading to the following Lagrangian for $\Phi$ and $q$ $$\mathscr{L} = \partial_\mu\Phi\partial^\mu\Phi + \lambda\left(|\Phi|^2 - \frac12f_a\right)^2 + {\rm i}\bar q\gamma^\mu\partial_\mu q - m \bar q q + (y\Phi \bar q q + \text{h.c.})\,.$$ The scale $f_a$ must be extremely large to avoid constraints, so $\Phi$ may safely be assumed to have acquired its VEV $$\Phi\to \frac{1}{\sqrt{2}}f_a e^{{\rm i}\theta}\,.$$ In this expression, I've neglected the fact that the radial mode still is a degree of freedom. This choice amounts to saying its excitations will be suppressed in our effective theory of the axion, which is accurate as long as $f_a$ is much bigger than the energy scales we wish to deal with.

Thus, the angular degree of freedom couples to the quark with the effective Lagrangian $$\mathscr{L} = \frac{1}{2}f_a^2\partial_\mu\theta\partial^\mu\theta + {\rm i}\bar q\gamma^\mu\partial_\mu q - m \bar q q + \frac{f_a}{\sqrt{2}}(y e^{{\rm i} \theta}\bar q q + \text{h.c.})\,.$$

Now, we recall the following non-trivial fact about QCD. A chiral rotation of the quarks $q\to e^{{\rm i}\gamma^5\alpha}q$ introduces the following CP violating term into the QCD Lagrangian $$-\alpha \frac{g_s^2}{32\pi^2}G\tilde G\,.$$ This originally caused a lot of confusion because this term is a total derivative, and therefore it cannot be derived at any order in perturbation theory. However, in the path integral formulation, it is not hard to check that the chiral rotation of quarks introduces this term through the path integral measure (see this reference).

Thus, if we want the quark mass matrix to be real, and thus conserve CP, we must rotate away the imaginary part of the quark Yukawas. The term $\bar\theta$ is the CP violating phase of the Standard Model, and it can be written $$\bar\theta = \arg\det M$$ where $M$ is the quark mass matrix.

Earlier, however, we introduced a new source of CP violation: the axion parent field phase $\theta,$ coupled to QCD through a new vector-like quark. Introducing a diagonal pure-imaginary entry to the quark mass matrix leads to $$\arg\det M\to \bar\theta + \theta\,,$$ and thus we arrive at the final axion Lagrangian $$\mathscr{L} = \frac{1}{2}f_a^2\partial_\mu\theta\partial^\mu\theta +(\bar\theta + \theta)G\tilde G + ...\,.$$

  • $\begingroup$ Thanks for a detailed answer! I have several questions though. If the scale $f_a$ is large, the radial mode in the expansion of the kinetic term can perhaps be neglected (see Eq. (3)) but can it also be neglected in the potential term (see Eq. 4)? Also your starting Lagragian should start with the topological term $\theta G\tilde{G}$. Right? @David $\endgroup$
    – SRS
    Aug 9, 2020 at 6:15
  • 1
    $\begingroup$ The radial modes are extremely heavy (their mass is order $f_a$) so any effect they have at low energy will be suppressed by the ratio $E/f_a$, which is safely negligible except possibly in the early universe, were PQ symmetry may be restored. Regarding your second point, my starting Lagrangian had not yet rotated away the Yukawa phase $\arg\det M + \theta$, so the starting Lagrangian had this term hidden in the path integral measure. $\endgroup$
    – Guy
    Aug 9, 2020 at 14:45
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    $\begingroup$ It's still not clear to me how the $\xi\frac{a}{f_a}G\tilde{G}$ term arises. Could you please elaborate the last paragraph to make this explicit? (Note to self: $θ = a/f_a$). $\endgroup$
    – Jollywatt
    Aug 25, 2020 at 10:41
  • $\begingroup$ @Jollywatt It will be a bit more than a paragraph, but thanks for reminding me: I'll look for that source I mentioned now. $\endgroup$
    – Guy
    Aug 26, 2020 at 15:04
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    $\begingroup$ @Jollywatt I've located the reference I promised. I think it does an excellent job explaining how to get the anomaly from the path integral, but if anything remains unclear, let me know! $\endgroup$
    – Guy
    Aug 26, 2020 at 15:13

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