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4

I addressed this a little at your other question but this one is more like physics. Yes, they're decay product lists. Beware that if all the modes are only "seen" or "not seen," you are sort of looking at the hairy edge of what's experimentally accessible. The indented lines are subtypes of the same reaction. Using your example, a neutral particle can ...

2

Quark oscillations don't exist by the definition of a quark. A quark is defined as a mass eigenstate and thus it is not going to oscillate with time (energy eigenstates don't charge with time of course!). To see how this works consider the relevant quark interaction terms without any choice of basis, - m _d \bar{d} d - m _u \bar{u} u - i ...

2

The important point is the fact that such a mass term breaks the gauge symmetry (Edit: I am assming that you want to build the Majorana mass term using SM available fields -- no extension considered -- of which there is only $\nu_L$). Namely, the desired term is (one generation suffices): $$\frac{1}{2}\,M\, \nu_L^T \,\mathcal{C}^\dagger\,\nu_L\, +\, ... 4 The Standard Model Yukawa interactions must be SU(3)\times SU(2) \times U(1)_Y gauge invariant. The down-type Yukawa interaction is$$ \mathcal{L} \supset -y_d \bar Q \phi d_R + \text{h.c.}. $$This is indeed gauge invariant. The \bar Q d_R form a colour singlet (3^* \times 3), the \bar Q \phi form an SU(2) singlet (2^*\times2), and the whole ... 5 The reason is that the SU(2) invariant in \mathbf{2}\otimes\mathbf{2} (or in their complex conjugate \mathbf{2}^*\otimes \mathbf{2}^*) is given by contracting the two \mathbf{2} with the anti-symmetric 2\times 2 matrix \epsilon_{ab}, as i\tau_2 is. In the case at hand the two \mathbf{2}^* are \bar{Q} and the \Phi^*. You could form another ... 0 I suggest you to read http://arxiv.org/pdf/1408.0287.pdf for a nice discussion on the RG dependence of the effective potentials and the RG invariant statements that one can make. About your point, I think you are simply misunderstanding what people mean: they are calculating at what point in \Phi, not in \mu, the potential cross zero. 3 Lets analyse the Majorana condition and the Majorana mass term. A massive Majorana neutrino \chi_j (a Majorana spin 1/2 fermion) having mass m_j>0 can be described in a local quantum field theory (eg. the standard model) by a four component spin 1/2 field \chi_j(x) which satisfies the Dirac equation and the Majorana condition which reads:$$ ...

4

The difference between the Higgs boson and the bosons of the three/four fundamental (depending whether you include gravity as a quantized theory or not) actions is that the latter are associated with gauge symmetries, while the Higgs plays a role in spontaneous symmetry breaking. Photons, W- and Z-bosons, gluons and gravitons arise from the requirement that ...

0

The Standard Model Lagrangian before and after spontaneous symmetry breaking (SSB) is renormalizable. To see that recall that the rule is (though it may not be immediately obvious as to why this rule holds) that a theory is renormalizable if all the terms in the Lagrangian are of dimension 4 or less. This is true by design for the Standard Model in which all ...

1

The easiest way to form $SU(2)$ singlets in the most general way is to use the techniques of Young Tableau. The method is discussed from a physicists perspective in many lecture notes online. One such example is given here. Using such method its easy to show that 2 lepton doublets make a singlet and a triplet under $SU(2)$, 2 \otimes 2 = 3 ...

5

Yes, massive particles such as W-bosons, Z-bosons, quarks, and leptons couple to the Higgs field via the cubic (Yukawa) interaction, so they may also exchange the virtual Higgs. Yes, because the virtual particle is massive, one gets the Yukawa potential that includes the exponential dumping with distance. This "Higgs force" is much less fundamental and ...

3

Indeed, the Standard Model is consistent in perturbative expansions, which means that we do not know if the Standard Model is consistent or not. So it is possible that the original Standard Model with 15 Weyl fermions per family is not consistent. In other words, there may not exist any well defined quantum model, whose low energy effective theory reproduce ...

15

The Standard Model is consistent in perturbative expansions. It is inconsistent non-perturbatively but all these inconsistencies only show up "qualitatively" at energies well above the Planck energy – where we know the non-gravitational Standard Model to be inapplicable, anyway. The inconsistencies of the Standard Model involve the Landau poles – the ...

2

You should take it completely literally. (Quibbles about the Higgs field vs the Higgs boson are misguided. Particles don't acquire masses until the point at which the Higgs boson appears, so attributing the particle masses to the Higgs boson is just as correct.) However, there is a simple way to picture this. The concept of a Higgs boson is completely ...

10

"Binding a massless particle into a small space" is a good phrase for a popular discussion, but it is not the only way to picture the Higgs mechanism. Another perspective comes from the fact that every particle inside some interaction field behaves exactly like its energy or momentum has changed. This concept is called canonical momentum, in contrast to the ...

4

A standard simple answer (for the standard Higgs boson field) is that a particle acquires mass by passing through this field, which changes the particle's inertia (thus appearing as acquiring mass which is a measure of inertia among others) Of course the standard Higgs boson is still investigated (if it is the standard one and not some variation of other ...

16

Short answer: do not take it literally, without further context. In order to understand the Higgs boson's role in the Standard model, it is necessary to take a closer look at the framework in which we describe elementary particles: quantum field theory. In this approach, particles are described as excitations of fields that spans all spacetime. The ground ...

98

The Higgs field (note it is the field that is important here, not the Higgs boson itself, which is just a ripple in the Higgs field) gives particles mass in the same sense that the strong force gives the proton mass (context: $99\%$ of the mass of the proton comes not from the mass of its constituent quarks, but from the fact that roughly speaking the quarks ...

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