Why is the QCD vacuum important in non-abelian gauge theories?

[QuantumCosmo]

Ah yeah I think I got it :)

But how does this B+L U(1) symmetry look?
I think U(1)_B is the vector symmetry of QCD:
$$\psi_L \rightarrow e^{i\alpha}\psi_L$$
and
$$\psi_R \rightarrow e^{i\alpha}\psi_R$$

and B is the conserved charge of that symmetry.

But I do not know the symmetries corresponding to L, B+L or B-L...

 

[blechman]

The Standard Model has quarks and leptons. The quarks have baryon number +1/3, and the leptons have lepton number +1, with their antiparticles the negative. This is a U(1) symmetry of the SM Lagrangian.

HOWEVER: both of these symmetries are anomalous: if you compute the triangle diagrams where you insert a B current on one vertex, and W bosons on the other two vertices, that triangle diagram is nonzero. Same with inserting an L current instead of a B current.

But it is also true that these anomalies (the B and the L) are equal. So that if you take the DIFFERENCE of the two, they cancel and that current is not anomalous. That is the B-L current. By a similar analysis, the sum of the two currents is 2x the anomaly of either one of them.

Notice however, that there is NO QCD anomaly: if you replace the W bosons with gluons, the triangle with the B current vanishes on its own. The one for L vanishes trivially, since leptons have no color charge.

So we have the equations:

$$\partial_\mu J_{B+L}^\mu = \frac{Ag^2}{16\pi^2}W\tilde{W}$$
$$\partial_\mu J_{B-L}^\mu =0$$

Notice that THERE ARE NO GLUONS IN THIS EQUATION! Just W bosons.

So by the arguments given above: performing a phase redefinition on all the fermions (quarks AND leptons) in accord with the B+L charge (using Weyl notation where the R-fermions are ANTIfermions; you can translate to Dirac easily by inserting $\gamma^5$s in the appropriate places):

$$q_L\rightarrow e^{i\alpha/3}q_L$$
$$q_R\rightarrow e^{-i\alpha/3}q_R$$
$$l_L\rightarrow e^{i\alpha}l_L$$
$$l_R\rightarrow e^{-i\alpha}l_R$$

will not change the lagrangian sans the theta terms (since it's a symmetry), and it won't change the QCD theta term, since there is no (B+L)-G-G anomaly. But it WILL shift the W-theta term by an amount proportional to $\alpha$. So if you choose $\alpha$ just right, you can cancel the SU(2) theta angle without doing any more damage.

If you try to shift the QCD angle, you cannot use this phase redefinition I wrote down. You have to use another one (such as rotating both qL and qR by the SAME angle rather than opposite angle). Such a redefinition will reintroduce phases into the masses of the quarks, and so THAT is why $\theta_{QCD}$ is physical. There will always be a phase there somewhere.

UNLESS one of the quark masses vanish. In that case, we're golden. Some people believe the up quark might have vanishing mass. It would mean that $\theta_{QCD}$ is unphysical, and therefore solve the strong CP problem; but it also would go against various other results such as lattice calculations. Personally I am inclined to believe the up mass is nonzero, but plenty of perfectly respectable physicists disagree with me.

Anyway, I hope that helps. Does that make sense?

 

[QuantumCosmo]

Anyway, I hope that helps. Does that make sense?

Hm, yeah, I think I might have understood it :)

My whole problem is that my QFT basics aren't that good... and that I often have problems with (I guess) very easy parts, for example those U(1) transformations.

I now suspect (after what you just wrote) that
$$q_L\rightarrow e^{i\alpha/3}q_L$$
$$q_R\rightarrow e^{-i\alpha/3}q_R$$
is a (global) symmetry of the Lagrangian and it's corresponding charge is B.

And
$$l_L\rightarrow e^{i\alpha}l_L$$
$$l_R\rightarrow e^{-i\alpha}l_R$$
is another symmetry of the Lagrangian with charge L.

And if I want the symmetry with charge B+L, I simply apply both of them. Is that correct?

As for the results of applying that symmetry (the change in the weak $$\theta$$ term), I think I have understood it :)

Thanks a million for that!

 

[QuantumCosmo]

Hi, I have found an article that explains why the $$\theta_{weak}$$ term can be eliminated.
It seems, though, this is different from what we discussed here (although I don't really understood what they were doing...)
A. A. Anselm and A. A. Johansen, Nucl. Phys. B407(1993) 652

 

[ansgar]

see my presentation about this topic:

http://www.isv.uu.se/~wouda/axion-beamer-GW.pdf

 

[blechman]

And if I want the symmetry with charge B+L, I simply apply both of them. Is that correct?

yup!

As for the results of applying that symmetry (the change in the weak $$\theta$$ term), I think I have understood it :)

Thanks a million for that!

As to that reference: that's just filling in all the details by doing the instanton calculation carefully, finding the relevant "zero modes" of the fermions, etc. I skipped all those details for the sake of sanity!

As to ansgar's presentation: thanks for sharing. It looks like a very interesting talk.