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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?

Dilaton
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    Good question, and looking forward to the answers if any. ;-) – Luboš Motl Jan 26 '12 at 06:57
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    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. https://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 $\theta_{QCD}$: https://arxiv.org/abs/1701.00467 – jak Apr 12 '18 at 13:01
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    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 https://www.sciencedirect.com/science/article/pii/0550321394903921 – jak Apr 12 '18 at 13:28
  • related https://physics.stackexchange.com/q/91535/12813 – wonderich May 25 '18 at 16:29

1 Answers1

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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.

David Bar Moshe
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    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? – Luboš Motl Jan 26 '12 at 13:56
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    @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. – David Bar Moshe Jan 26 '12 at 14:55
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    What about the Yukawa couplings? You absorb the phase into the Higgs? Or into right handed fermions? –  Jan 26 '12 at 20:10
  • @Thomas: To the right handed fermions. They are not coupled to the gauge fields so their transformation does not change the path integral measure – David Bar Moshe Jan 27 '12 at 03:59
  • There are some basic things that I don't get. Why couldn't the LH and RH fermions rotate in the same way in QCD? And why do all flavors have to rotate at the same time? – larueroad Apr 23 '17 at 04:27
  • @karlzr Weak interactions only couple to LH fermions and thus a transformation of the RH has no effect on the measure, i.e. does not shift $\theta_{weak}$. For QCD both, the trafo of the RH and of the trafo of LH lead to separate shifts of $\theta_{QCD}$, because gluons couple to RH and LH quarks. The crucial thing is that we have an extra freedom in the weak sector to rotate the RH fermions, because the weak bosons do not care about them, and this allows us to set $\theta_{weak} \to 0$. In QCD you do not have this freedom. – jak May 31 '17 at 09:03
  • @karlzr To summarize: We have two places where the rotations make a difference: for the mass terms and for the $\theta$ term. In the absence of massless fermions, we can't perform arbitrary chiral rotations, because the mass terms must be real. This means immediately that we can't rotate $\theta_{QCD}$ to zero. (Of course, it could miraculously be that the rotation that makes the mass terms real, cancels at the same time $\theta_{QCD}$. However there is no reason why this should be the case and this is the strong CP puzzle). – jak May 31 '17 at 09:07
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    @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. – jak May 31 '17 at 09:08
  • Would you then not get back a $\theta$ angle in GUT theories? – Michael Angelo Apr 23 '19 at 17:20
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    @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. ... – David Bar Moshe Apr 24 '19 at 08:57
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    @MichaelAngelo I don't know of a detailed work analyzing the implications of these terms, but they can possibly affect the the monopole and dyon spectra of the theory through the Witten effect. Here are two articles analyzing these effects: https://arxiv.org/abs/1711.05721 http://cds.cern.ch/record/455906/files/0008322.pdf – David Bar Moshe Apr 24 '19 at 08:57
  • @JakobH You may only rotate the left and right-handed sector by the same angle (which is necessary to make the phase disappear in the mass term) if you have a $U(1)$ symmetry. The $U(1)_Y$ does this for you, but in QCD you don't have a $U(1)$ symmetry so you always have to rotate the L and R sector with opposite angle. This is essential right? Is this correct? – Michael Angelo Apr 24 '19 at 16:45
  • For the electroweak part, in the unbroken theory, I would think the Higgs field does not couple to the right-handed neutrino(if it exists), and couples to the left-handed neutrino in a $SU(2)_L$ invariant way, and therefore it is possible to do a chiral rotation on them to cancel the $\theta$ term; in the broken theory, the chiral rotation on the massless neutrino does the job. While in the QCD, the Higgs field is a singlet under $SU(3)$, which makes its coupling to the quarks not chirally invariant. – WunderNatur Nov 15 '19 at 22:48