44
$\begingroup$

Why is there no analog for $\Theta_\text{QCD}$ for the weak interaction? Is this topological term generated? If not, why not? Is this related to the fact that $SU(2)_L$ is broken?

$\endgroup$
4
  • 3
    $\begingroup$ Good question, and looking forward to the answers if any. ;-) $\endgroup$ Jan 26, 2012 at 6:57
  • 1
    $\begingroup$ There is an electroweak vacuum angle $\theta_{EW}$, but as mentioned below it can be rotated away in the standard model thanks to the chiral symmetry in the electroweak sector. For a nice recent discussion see e.g. arxiv.org/abs/1402.6340. However, when we go beyond the SM this is not necessarily the case. Shifman and Vainshtein recently showed that, for example, in GUTs $\theta_{EW}$ is physical and has the same value as $\theta_{QCD}$: arxiv.org/abs/1701.00467 $\endgroup$
    – jak
    Apr 12, 2018 at 13:01
  • 1
    $\begingroup$ As a final remark, the standard reference regarding this question is Can the electroweak θ-term be observable? by A. A. Anselm and A. A. Johansen sciencedirect.com/science/article/pii/0550321394903921 $\endgroup$
    – jak
    Apr 12, 2018 at 13:28
  • $\begingroup$ related physics.stackexchange.com/q/91535/12813 $\endgroup$
    – wonderich
    May 25, 2018 at 16:29

1 Answer 1

32
$\begingroup$

In the presence of massless chiral fermions, a $\theta$ term in can be rotated away by an appropriate chiral transformation of the fermion fields, because due to the chiral anomaly, this transformation induces a contribution to the fermion path integral measure proportional to the $\theta$ term Lagrangian.

$$\psi_L \rightarrow e^{i\alpha }\psi_L$$

$${\mathcal D}\psi_L {\mathcal D}\overline{\psi_L}\rightarrow {\mathcal D} \psi_L {\mathcal D}\overline{\psi_L} \exp\left(\frac{i\alpha g N_f}{64 \pi^2}\int F \wedge F\right)$$

So the transformation changes $\theta$ by $C \alpha g N_f $ ($g$ is the coupling constant, $N_f$ the number of flavors).

The gluons have the same coupling to the right and left handed quarks, and a chiral rotation does not leave the mass matrix invariant. Thus the QCD $\theta$ term cannot be rotated away.

The $SU(2)_L$ fields however, are coupled to the left handed components of the fermions only, thus both the left and right handed components can be rotated with the same angle, rotating away the $\theta$ term without altering the mass matrix.

$\endgroup$
13
  • 1
    $\begingroup$ Nice, would you add one or two formulae? What is the parameter of the transformation (and which one) to remove the $\theta\cdot F\wedge F$ term? And a related question: is there some simple way to add some chiral couplings of new fermions to $SU(3)_{color}$ to solve the strong CP-problem? $\endgroup$ Jan 26, 2012 at 13:56
  • 3
    $\begingroup$ @Luboš I am not an expert, from reading only, I think that your suggestion is quite close to one solution to the strong CP problem assuming the mass of the u-quark is exactly zero,though not widely accepted. $\endgroup$ Jan 26, 2012 at 14:55
  • 3
    $\begingroup$ What about the Yukawa couplings? You absorb the phase into the Higgs? Or into right handed fermions? $\endgroup$
    – Thomas
    Jan 26, 2012 at 20:10
  • 1
    $\begingroup$ @karlzr In the weak sector the situation is almost the same. However, the weak bosons do not care about the RH fermions and hence there is no shift of $\theta_{weak}$ when we rotate them. This extra freedom allows us to make the mass terms real (or leave them real) while rotating at the same time $\theta_{weak}$ to zero. $\endgroup$
    – jak
    May 31, 2017 at 9:08
  • 1
    $\begingroup$ @MichaelAngelo To be specific, I'll consider the SU(5) Georgi-Glashow model. Here the fermions belong to the $\bar{5}$ a $10$ representations. The net anomaly of these representations vanishes, thus there is no way to rotate away the whole the theta term. After symmetry breaking the theta term will decompose into theta terms of the $W$-bosons, QCD, heavy gauge bosons and possibly combinations of them. The $W$-bosons theta term can be rotated away as above by a chiral transformation, the QCD theta term will remain as well as the other terms in the decomposition. ... $\endgroup$ Apr 24, 2019 at 8:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy