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This Wikipedia article: http://en.wikipedia.org/wiki/Left%E2%80%93right_symmetry states that the weak part of the SM gauge group is not $SU(2)_L \times U(1)_Y$ but $ \frac{ SU(2)_L \times U(1)_Y}{\bf{Z}_2}$. If this is correct, what is the significance of the $\bf{Z}_2$ ?

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While the question is based on physics, the answer is almost entirely pure math. If you've had a course on group theory (and I highly recommend taking one in the math department in addition to whatever has been mentioned in physics classes), you'll recognize this as taking the quotient of the original group by $\mathbb{Z}_2$. What that means is we started with our original symmetry group $SU(2)_\mathrm{L} \times U(1)_\mathrm{Y}$, which I'll call $G$, and found a normal subgroup within it isomorphic to $\mathbb{Z}_2$.

At this point we say that we want to identify one of the nontrivial symmetries with the identity as a single, cohesive unit. What does this do to the rest of the group? Well, we want to preserve as much of the group structure as possible, and so we pair off all the other elements in $G$ in the same way. Each of these pairs is a coset. The set of cosets inherits its group structure from $G$, and we call this new group the quotient group $G/\mathbb{Z}_2$. In the case of finite groups where cardinality has meaning, we note that $$\lvert G/\mathbb{Z}_2 \rvert = \frac{\lvert G \rvert}{\lvert \mathbb{Z}_2 \rvert} = \frac{\lvert G \rvert}{2}.$$

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The weak part of the standard model gauge group is SU(2)xU(1), but you need to check that every one of these gauge transformations acts on the fields nontrivially. If one of these transformations does nothing to the standard model fields, then you can consider this gauge tranformation to be identified with the identity.

The only elements of SU(2) you can identify with the identity is the one corresponding to a rotation by 360 degrees, since only this one commutes with everything else. In SU(2), the 360 degree rotation is the 2 by 2 matrix -1, and since there are weak doublets, these are like spin-fermions, they get negated by this transformation. But you also have a U(1), so you need to undo the negation by an appropriate U(1) transformation in order to keep the fields the same. If you can do this for all the fields in the standard model, then you can say that the true gauge group is SU(2)xU(1)/Z_2.

The Higgs and lepton doublets are all SU(2) doublet with U(1) charge 1/2, so you need to do a 360 degree SU(2) rotation, and a U(1) rotation by angle $2\pi n$ where n is an odd integer, so that 1/2 times $2\pi n$ is $\pi n$ and gives you a -1 phase.

The right handed lepton has charge 1, so it is invariant under the $2\pi n$. This means the pure lepton sector is invariant.

The left handed quark doublet is has charge 1/6, so in order to be invariant, you need n to be a multiple of 3, which means n=3, and the U(1) rotation is by $6\pi$.

The right handed quarks have U(1) charge 2/3 and -1/3, so they get a phase of $4\pi$ and $2\pi$ from the rotation, or nothing at all.

So the $Z_2$ Wikipedia is talking about does a 360 degree rotation on the doublet fields (it changes their sign) and then does a half-turn in the U(1), so that the 1/6 and 1/2 fields change sign, but the charge 1,1/3,2/3 fields do not. This transformation keeps all the fields the same, so it is consistent to say that the standard model gauge group is reduced by this symmetry.

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Ron, are there repercussions about being sloppy about this factor of $\bf{Z}_2$? That is, if I plod along ignorantly calculating SM cross sections and decay rates what is the result of ignoring this? –  DJBunk Mar 15 '13 at 21:18
    
Not for SM perturbative cross sections--- these are unaffected by the global gauge group--- they don't see the large field configurations at all, the SU(2) fields are near the identity all the time. You would only change the allowed representations, the monopole spectrum and the instanton configurations. There might be a way to discriminate between different gauge groups in a confining theory, where the long distance gauge configurations would be reduced in the quotient groups, I don't know, I'll have to think about it. –  Ron Maimon Mar 16 '13 at 1:09
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