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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 ...


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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 $SU(2)$, \begin{equation} 2 \otimes 2 = 3 ...


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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 ...


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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 ...


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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 ...


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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 ...


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"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 ...


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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 ...


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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 ...


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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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Notation $W^{-}, W^{+}$ may confuse in a sense that it may seem that here are two different particles which aren't connected by charge conjugation. But of course, $W^{+}$ is only $(W^{-})^{\dagger}$, so it is an antiparticle to $W^{-}$. So term $( W^{-} \cdot W^{+} )$ is simple $|W|^{2}$ (which is standard for the mass-term), and, of course, both of particle ...


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I) An axial (vector) symmetry transformation acts opposite (the same) under the left-handed and right-handed parts of a Dirac spinor, cf. chiral symmetry. II) The full symmetry group is the product group $G=SU(3)_F\times SO(3,1)$. The quark $q$ transforms in the representation $\underline{3} \otimes \underline{4}$, i.e. under the fundamental ...


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A Dirac spinor, as your $q$ is, has four components, corresponding to one left-handed and one right-handed Weyl (two-component) spinor, $$q = q_L + q_R.$$ $\gamma_5$ is the $4\times4$ matrix that is $1$ on the right-handed part and $-1$ on the left-handed part. The expression $$q\mapsto q' = \exp(i\Phi_a \lambda^a /2 \gamma_5)q$$ means $$q_{R(L)} \mapsto ...


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The Particle Data Book, published by the Particle Data Group is probably the nearest to what you're asking for. The actual book is a massive tome costing a fortune. I'm not sure how much of and in what form it's online, but some intrepid Googling should find you most of what you want. As fqq points out the book is available from Phys. Rev. D vol. 86 Issue 1 ...



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