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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?

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Good question, and looking forward to the answers if any. ;-) – Luboš Motl Jan 26 '12 at 6:57
up vote 18 down vote accepted

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.

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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
@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
Thanks a lot, David! – Luboš Motl Jan 26 '12 at 16:33
What about the Yukawa couplings? You absorb the phase into the Higgs? Or into right handed fermions? – Thomas 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 3:59

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