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4

You could ignore over-all factors, but definitely not couplings. The (not quite purely) theoretical mass term in the generic Weinberg-Salam model is, instead, proportional to $$ (W_\mu^1)^2+ (W_\mu^2)^2+\left (W_\mu^3-\frac{g'}{g} B_\mu\right ) ^2, $$ (not quite purely, as the form was all but suggested by a skein of experimental facts--a hugely long story ...


3

Using concrete realization of this generators, you can easily check that $T_1, T_2, T_3$ are exactly $SU(2)$ and $T_8$ commute with them and so gives you $U(1)$.


3

Yes and yes to your first two questions. Fermions are massless at sufficiently high temperature when the VEV is zero and the symmetry unbroken. They acquire mass when the universe expands and cools, the VEV becomes nonzero, and the symmetry breaks.


2

The natural density of Higgs bosons is zero. This is why we had to use an enormous particle accelerator to briefly create some. They quickly decayed as soon as we made them, because they are very unstable. In the same way that you can have a static electric or magnetic field without having any photons, you can have a static Higgs field without any Higgs ...


1

A quantum Higgs field is everywhere, filling up the vacuum, and normally there are no real Higgs particles to be seen--just as the quantum field of the electron is all over the place, but, in the ground state, the vacuum, there are no real electrons to be seen. If you had energy available, 1MeV, you could "pluck" that electron field, and create an electron-...


1

You have the usual misunderstanding of confusing the Higgs field with the Higgs boson. In quantum field theory every elementary particle in the standard model is a field covering all space and time: an electron field, an electron neutrino field etc. The Higgs boson is one of the elementary particles. The fields are like a coordinate system or a Lorenz ...


1

When we talk about a false vacuum decay we mean there is some field that is stuck in a metastable state and then decays to its ground state i.e. the true vacuum. The energy of the field in the metastable state cannot just disappea,r so when the field escapes from the metastable state it will initially have the same energy density. We expect the field to ...


1

The Higgs potential receives quantum corrections from the Yukawa couplings with fermion doublets. The magnitude of the correction is determined by the Yukawa coupling constant, which is itself proportional to the fermion mass. As a result, the heaviest fermions make the greatest contribution to the effective action. Because the top quark is by far the ...


1

You can use a program I wrote, Package-X for Mathematica, which allows you to directly calculate diagrams like this. By the way, I don't think you wrote down your amplitude correctly. Use LoopIntegrate in conjunction with FermionLine to initiate the computation of the integral Then replace the dot products with the relevant on shell conditions (p.p->mμ^...


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