Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The argument, which I reproduce here from Ramond's `Journies BSM', is originally by Witten and Vafa in ($\it{Phys}$. $\it{Rev}$. $\it{Lett}$. 53, 535(1984)). The argument is that for $\theta = 0 $ (mod $2\pi$) the QCD vacuum energy $E(\theta)$ is minimized. Starting from the Euclidean path integral for QCD ( with just massive fermions charged under QCD, nothing else) in a volume V

$ e^{- V E (\theta)} = \int \mathcal{D}A \mathcal{D} q \mathcal{D} \bar{q} exp \left( - \int d^4 x \mathcal{L}\right) $


$\mathcal{L} = - \frac{1}{4 g^2} Tr (G_{\mu \nu}G_{\mu \nu}) + \bar{q}_i (\gamma^\mu D_\mu+ m_i) q_i + \frac{i \theta}{32 \pi^2} Tr(G_{\mu \nu} \tilde{G}_{\mu \nu} ). $

Integrating out the quarks we obtain:

$ e^{- V E (\theta)} = \int \mathcal{D}A \det{(\gamma^\mu D_\mu+ m_i)} \mathcal{D} q \mathcal{D} \bar{q} exp \int d^4 x\left( \frac{1}{4 g^2} Tr (G_{\mu \nu}G_{\mu \nu}) - \frac{i \theta}{32 \pi^2} Tr(G_{\mu \nu} \tilde{G}_{\mu \nu} )\right). $

In pure QCD the quarks have vector-like couplings and so $\det{ (\gamma^\mu \mathcal{D}_\mu+ m_i) }$ is positive and real. For each eigenvalue $\lambda$ of $\gamma^\mu \mathcal{D}_\mu$ there is another of opposite sign. Thus

$\det{ (\gamma^\mu D_\mu+ m_i) } = \prod_\lambda (i \lambda +M) = \prod_{\lambda>0} (i \lambda +M)(-i \lambda +M) = \prod_{\lambda>0} ( \lambda^2 +M^2 )^2 >0 $

Thus if $\theta$ were zero, the integrand would be made up of purely real and positive quantities. Now, the inclusion of the $\theta$ term with its i can only reduce the value of the path integral, which is the same as increasing the value of $E(\theta)$. It follows that $E(\theta)$ is minimized at $\theta = 0$. This is the motivation for axions where $\theta$ is promoted to a dynamical field which then relaxes to a vev of 0.

Ramond adds there is `the slight caveat' that with Yukawa couplings, the fermion determinant may not longer be positive nor real. Witten and Vafa don't seem to make any similar caveat. Ramond's note about yukawas seems to invalidate the whole argument. Where does this leave axions as a candidate to solve the Strong CP problem?

share|cite|improve this question

I know this is a year old question, but I am going to attempt an answer. As far as I can tell, this is not really a caveat. The reason for this is that I can always set the overall phase of the quark mass determinant to be zero with a chiral U(1) transformation. For a discussion of this see for example the chapter on theta vacua in Weinberg's QFT book. The main point is that:

  1. If you do a chiral U(1) transformation in the path integral, you will get a non trivial transformation of the integral measure because of the anomaly, but the anomaly function appearing is exactly the same form as the theta term. The result is that the chiral U(1) transformation $\Psi_f \to e^{i \alpha_f \gamma_5}\Psi_f$ changes $\theta$ to $\theta - 2 \sum_f \alpha_f$
  2. On the other hand, this transformation also changes the phases of the quark mass matrix as $m_f \to m_f e^{i 2 \alpha_f}$
  3. But if you treat this transformation as just a change of variables in the path integral, than it shouldn't change the physics. This means that the phase of the quark mass matrix and the $\theta$ parameter are not really independent, in the sense that physics can only depend on the combination: $e^ {-2 i\theta} \prod_f m_f$, since this is the combination left invariant under the change of variables.

In particular this means you can rotate the quark mass matrix phase to be zero, and in this way shift the phase to the $\theta$ term. Therefore I don't believe this is a caveat, it is just a step in the argument left out at most.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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