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1

The coupling for $f_L \to f_R$ is exactly the same as the one for $f_R \to f_L$ (the Yukawa coupling source of the fermion mass term). So yes, on average a fermion is "as left-handed as right-handed" when propagating freely.


2

We don't need "spontaneous symmetry breaking in Lagrangian formalism" in general. We can use spontaneous symmetry breaking to generate masses for some fields. It so turns out that spontaneous symmetry breaking introduces those masses without breaking some other nice features of the model at hand: mostly renormalizability and that symmetry that we are ...


0

I might be totally wrong, but my understanding is that the condensate by definition cannot be written as a state created by a creation operator. Afaik, the condensate isn't even a state at all but an expectation value of the field operator in the ground/vacuum state (vacuum expectation value or "vev"). Given a field operator $\Phi$ you get: ...


4

Yes, your expectations seem reasonable when thinking of the Higgs mechanism induced mass, as explained by @nmoy. However, note that one needs to be careful in defining what one means by "mass" at high temperatures. A theory at finite (non-zero) temperature breaks Lorentz invariance. There are multiple ways to think of this: There is a preferred frame, ...


4

In the Standard Model, fermions are given their mass through yukawa terms, described by the Lagrangian: $$\mathcal{L}_{\mathrm{F}} = \overline{\psi} \gamma^{\mu} D_{\mu} \psi + y_{\psi} \overline{\psi} \phi \psi$$ Where $y_\psi$ is the yukawa coupling and $\phi$ is the Higgs field. At this stage, much like the gauge bosons, the fermions do not yet have ...


3

The Higgs mass does not stem from eating Goldstone bosons, since the Higgs is not a gauge field. Since we are breaking an $\mathrm{SU}(2) \subset \mathrm{SU}(2)_L \times \mathrm{U}(1)_Y$ completely, we have three Goldstone bosons, which are eaten by three of the four electroweak gauge bosons to form the massive $W^\pm,Z$ with the photon remaining massless. ...


6

Yes, it is correct to say that the Higgs boson, just like other elementary particles, get its mass from the interaction with the Higgs boson – which means "with itself" in the case of this particle. More concretely, the mass may be derived from the Higgs potential (energy density) $$ V(h) = \frac a4 h^4 - \frac b2 h^2 + c$$ where the additive shift $c$ ...


1

elementary particles do not have any particular speed. It could be could down untill almost zero motion, or it can be speeed up until becoming a black hole, who knows.


1

It is mainly measurement and detector errors that make up the width in the plots you show. The Monte Carlo simulates the detector resolution and folds in the theoretical values when it says that the width agrees. The real width is expected to be much smaller. In this we see that the real width is only given as a bound by the experiments the CMS ...


0

The decay width of a particle is antiproportional to its lifetime. Looking at the partial width of the $H \rightarrow \mu \mu$ decay, one could expect that the lifetime of the Higgs is large. This would be correct, if the Higgs could only decay to muons. In other words: The Higgs decaying to muons has a low probability (a low branching ratio). This comes ...



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