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Short answer: to accurately model reality. Long answer: The weak interaction has several peculiar properties: The $W$ bosons are vector bosons (so the weak theory is likely a gauge theory) The $W$ bosons have electric charge The $W$ bosons have mass. (The $Z$ boson hadn't been observed experimentally; it was a prediction of the SM) The $W$ bosons couple ...


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Detailed answer The thing is that you cannot really force in an a priori interpretation of the gauge group with which you extend your existing theory. You can only decide on the symmetries of your theory. So keeping things very general you start by trying to gauge the symmetry group $SU(2)$. This has three generators and so gives rise to three independent ...


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The question that you have asked have some vague arguments as well as some partially true facts regarding Standard Model (SM). First, Yes SM describes physics up to some energy scale which is 14 TeV. On the other hand, if we accept Plank energy ($~10^{18}$GeV) as a fundamental energy scale, then we can possibly expect new beyond the SM energy scale. A ...


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What exactly happens to these massless degrees of freedom in the Higgs potential after symmetry breaking, i.e. after we expand the Higgs fields about the vev? You will make a redefinition of the vector fields $V_{\mu}' = V_{\mu} - \partial_{\mu} \eta $, where $\eta$ was your goldstone, Doing that you can see that a term like $$ \nu^{2} (V_{\mu} + ...


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Conventionally in QFT, particles and antiparticles are defined with positive energy $k^0\geq 0$ only. (Recall that would-be negative energy states are reinterpreted as matter/antimatter of the opposite kind in order to make the vacuum stable.)


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You are correct that Goldstone's theorem does not apply to the 1-D Heisenberg chain, or indeed to any local 1-D system, because there is no continuous symmetry breaking in 1-D for local Hamiltonians. But Goldstone modes are not the only possible kinds of gapless excitations - you don't need continuous SSB for a system to be gapless. There are plenty of ...


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The definition of a spin liquid as a spin system "with no spontaneously broken symmetries" is out of date and no longer used, partially for the reason you describe. If you perturb as spin-liquid Hamiltonian by adding small terms that break all the symmetries, then the ground state will still be a spin liquid even though there are no longer any symmetries ...


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The scale where some symmetry gets broken can be computed using the renormalization group equations for the gauge couplings. It's the other way around. Once you know the scales of the model (masses of the fields) you can compute the RGE. Since what ultimately matters is the mass of the representation, not its vev, if you accept some tuning (and have ...



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