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- Thread starter happy42er
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nrqed

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Right. To have a renormalizable theory, one must have a U(1) x SU(2) gauge invariance. After spontaneous symmetry breaking, two of the gauge fields of the SU(2) symmetry as well as one linear combination of the U(1) generator and tau_3 generator acquire a mass..Those are the Z,W+,W-. The other, orthogonal, linear combination of the generators give the massless photon. If one simply throws in a W+ and W-, there is no gauge symmetry and the theory can't be made renormalizable.

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arivero

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nrqed

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very interesting reference. Thanks!

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I thought though that the reason why the Z and W bosons where included was because without them the theory violated unitarity.

Isn't that so?

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I don't think that unitarity enters in any way.

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I don't think that unitarity enters in any way.

When I referred to unitarity I meant that in the Fermi theory there are cross sections that grow with the energy.

For example:

[tex]\sigma(e \nu \rightarrow e \nu) \propto {G_F}^2 s[/tex]

Since cross sections express the likelihood of interaction between particles, what happens is that at sufficient high energies the probability of some process happening is greater than 1. In the Fermi theory this energies are around [tex]\sqrt{s}[/tex]=300 GeV.

That's why I thought that there was a problem with unitarity.

After looking into it now I'd say that the theory had both problems, it wasn't renormalizable and it violated unitarity.

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