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The masses of W-bosons and Z-bosons are known – and the mass of W-bosons multiplied by some coupling constants was known indirectly through the force they mediate, as the Fermi's constant. But the mass of e.g. the W-boson comes from the gauge-invariant kinetic term of the Higgs boson, $$ \frac 12 D_\mu H \cdot D^\mu H $$ The covariant derivative includes ...


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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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@SAS answered most of the questions, however I believe there's a crucial point which still needs to be addressed: the chirality. Indeed, it is not obvious a priori why $$\Psi^T C \Psi\,\Phi\,,$$ (where $\Phi$ is some Higgs representation) leads to a Dirac-type masses instead of Majorana masses. Why not the common $\bar\Psi \Psi$? It turns out to be the ...


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The general problem that the Higgs mechanism solves is giving mass to spin-one particles. It turns out that finding relativistic, unitary theories of spin-one massive particles is non-trivial. There are a few known ways of doing it (this paper has a pretty good list of sources), but the oldest and easiest is probably the Higgs mechanism. In contrast, there ...


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Of course the SM Higgs gives mass to both fermions as well as the gauge bosons. However, the latter is much more fundamental and predictive than the former. Point is, it is enough for a scalar to transform non-trivially under a gauge symmetry to contribute to its associated gauge bosons masses (after taking vev). This contribution is constrained by the ...


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I think the question was already answered here, look at either my answer of the one by Adam. The punch line is that that the expectation value of a field (such as the field $\phi$ at the bottom of the mexican hat potential) is fixed by the way the source $j$ that couples to $\phi$ in the path-integral for the functional generator is sent to zero. As there ...


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Many answers discuss the transition from the unstable state to the stable one. Let me thus discuss the issue of choosing the ground state itself. I will suppose a two dimensional Mexican hat potential. As Numrock realised, it has degeneracy. There is nothing which lift this degeneracy in principle. Then you can change the ground state without energy. This ...


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In relativistic QFT, this cannot be a process in time. The unstable initial state does not exist at all: An unstable ground state is impossible in relativistic QFT at temperature T=0 (i.e., the textbook theory in which scattering calculations are done) since it would be a tachyonic state with imaginary mass, while the Kallen-Lehmann formulas require $m^2\ge ...


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The fermion masses result from Yukawa interactions after EWSB: $$ m_f = \frac1{\sqrt 2} y_f v $$ Thus the Yukawa couplings govern lepton and quark masses. Of course the masses should be diagonalized. For the leptons, as there no neutrino masses in the SM, the lepton interactions and masses can be simultaneously diagonlized, whereas differences in up- and ...


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In fact in some sense the full state of the quantum system shouldn't break any symmetry. It's just that the overlap of the different vacuum states (in field theory) vanishes. So in other words the full quantum state is a superposition of all the different vacua, but when we make observations we "collapse the wave function" (feel free to insert your ...


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Physically, you look for the ground state of your theory in order to make a correct predictive calculus around the ground state. The "random" choice of the ground state depends on the dimension of the theory, in fact, you can obtain a "hat" shape or a simple 2-dimensional shape with just 2 possibilities for the ground state (look the scalar quantum ...


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I'm not 100% sure of your level so just as a heads up, I put some comments in parentheses that are meant to give technical caveats. If they don't make sense just ignore everything in parentheses, the zeroth order answer You can look at it that way, but actually it turns out to be much more complicated to understand what's going on in detail. (Basically you ...



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