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The $\Delta$ is a quartet of particles with isospin 3/2: $$ \Delta^-, \Delta^0, \Delta^+, \Delta^{++} $$ I would expect the anti-$\Delta$ to be written $\bar\Delta$, with the four isospin projections $$ \bar\Delta^{--}, \bar\Delta^-, \bar\Delta^0, \bar\Delta^+ $$ In this case the antiparticle of the $\Delta^+$ would be the $\bar\Delta^-$. If you'd like a ...


9

Gluons and photons are similar in the sense that they are both massless gauge bosons. They do, however, correspond to different gauge symmetries: photons arise due to $\mathrm{U}(1)$ symmetry, while gluons follow from $\mathrm{SU}(3)$. This leads to a different number of particles: there is only one photon, while there are eight different gluons, ...


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From the perspective of fundamental quantum field theory, gluons and photons are quite similar. Both of them are gauge bosons, meaning that their existence is required by a mathematical mechanism called local gauge invariance. However, as particles, there isn't any particular connection between them. For instance, there's no reason they both have to be ...


1

There are two things that define a particle physics model (at low energies). The first one is the gauge group G we want the model to be symmetric under. For the Standard Model (SM) we set this to $G=SU(3)\times SU(2)\times U(1)$ (for good experimental reasons!). This will uniquely determine the number of gauge bosons needed to make the model consistent. The ...


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The four quantum field theories (QCD, QED, QFD, and EWT) unite quantum mechanics and special relativity. They are all fully understood, complete, and proven. In your quote for the standard model, there are not four distinct field theories, the electroweak has united the electromagnetic and the weak in one field theory, the electroweak theory. The ...


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1mol of protons is $6\times10^{23}$ particles. If proton decay has a half life of $6\times10^{33}$ years, then there should be one proton decaying per year per $10^{10}$mol. Since one mol of protons weighs one gram, that's a mere $10^7$kg of protons. Even at the density of liquid hydrogen (70kg/m$^3$), that's only $1.4\times10^6$m$^5$ of liquid volume. Take ...


2

The number of gauge bosons is restricted by symmetry: a given theory with a certain gauge invariance admits as many gauge bosons as there are generators of the corresponding gauge group. For example, there is one generator for $\mathrm{U}(1)$, resulting in the existence of a photon. $\mathrm{SU}(3)$ admits eight generators, which yield eight gluons. This is ...


1

After reading through the corresponding chapters in several books I think I'm now able to give a "semi-satisfactory" answer to my own question (and to understand Lubos first comment ;) ). I write semi-satisfactory, because I hope someone with a deeper understanding of these topics will give a better answer. My explanation is still a little bit heuristic, ...


0

For massless gauge bosons the only meaningful way to transform under parity is like polar vectors since they have to transform as $i\partial_\mu$ in the covariant derivative. For massive gauge boson, one in principle doesn't have this constraint, although one could assign a parity as for massless vectors assuming the mass comes from spontaneous symmetry ...


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Strictly speaking the prefix semi means half, but it's often used in the sense of partial. A good example of this would be semiconductor. So semileptonic just means partially leptonic.


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I'd like to give a different point of view than the one provided in Qmechanic's answer. The reason is not because of gauge invariance. Indeed, gauge invariance is just a statement of redundancy and it can't possibly have any physical consequences. My answer is instead the following: the photon is massless because it has just 2 degrees of freedom while ...


8

I) At the perturbative/diagrammatic level of photon self-energy/vacuum-polarization $\Pi^{\mu\nu}$ , the photon masslessness is protected by the Ward identity, which in turn is a consequence of - you guessed it - gauge invariance. For the explanation in the setting of QED, see e.g. Ref. 1. Fig. 1: A one-loop contribution to the photon ...


1

Your last equation does not make sense. On the left had side you have a scalar 1, while on the right hand side there is supposed to be the unit matrix. However, it is $$ 1=\begin{pmatrix}d' & s' & b'\end{pmatrix}\begin{pmatrix}d' \\ s' \\ b'\end{pmatrix} = \begin{pmatrix}d & s & b\end{pmatrix}V_{CKM}^\dagger V_{CKM}\begin{pmatrix}d \\ s \\ ...


2

It does not mediate flavor changing because the couplings in the gauge basis are of the form $ c_L Z_\mu \bar{\psi}^i_L \gamma^\mu \psi^i_L$ and $c_R Z_\mu \bar{\psi}^i_R \gamma^\mu \psi^i_R$ with $i$ flavor index and $c_{L,R}$ flavor independent by gauge invariance (that is, $c_L$ and $c_R$ are proportional to the identity in flavor space). In this basis ...


2

In general, the covariant derivative acting upon a field in a representation $\rho : G \to V_\rho$, is given by $$ D = \mathrm{d} - \mathrm{i} g \rho(A) \wedge $$ or, in coordinates/generators, indeed $$ D_\mu = \partial_\mu - \mathrm{i} g A_\mu^a \rho(T^a)$$ where $\rho(T^a)$ are the generators of $G$ in the chosen representation. Now, for the triplet ...



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