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7

I think that the only honest answer to this "how" question is (for the electromagnetic interaction -- the strong and weak interactions are analogous) with a mathematical formula: $$\mathcal{L}_\text{int} = e\overline{\psi} \gamma^\mu A_\mu \psi.$$ This is the term that we add to our model to describe the interaction of electrons ($\overline\psi,\psi$) with ...


5

Let us start with $U(1)$ electromagnetism and see why it does not have such interactions. The field strength tensor is given by $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$, and the relevant part of the QED Lagrangian is proportional to $F_{\mu\nu}F^{\mu\nu}$. This means that the Lagrangian has only terms that are at most quadratic in the gauge field ...


3

My answer to my own question here may be helpful. No, there is no anti-isospin as there is no anti-spin, because $SU(2)$ has no complex representations. In contrast, $SU(3)$ has complex representations and therefore the conjugate charge is different from normal charge, which means in the case of $SU(3)$ color A complex representation $R$, is a group ...


2

If symmetry conserves some important properties of theory (like unitarity) then its explicit breaking makes theory inconsistent. For example, all gauge symmetries must be unbroken due to unitarity of theory. When they become explicitly broken, unphysical states with indefinite metric in Hilbert space arise, and unitarity is lost. That's why we worry about ...


2

Let $\mathcal{M}$ be our spacetime. Then, a gauge theory is given by a connection form $A$ on a principal bundle over it (that locally projects onto the spacetime in a way compatible with gauge transformations), which is the gauge potential. Maxwell's equations1 (in vacuum) are the equations of motion for the gauge field for the Yang-Mills action coupled to ...


2

General comment to the question (v3): Non-abelian YM [such as, e.g., YM with gauge group $SU(2)$ or $SU(3)$] has besides quartic gauge boson interactions also cubic gauge boson interactions, while abelian YM (aka. QED) has neither. This is because the Feynman-rules for the cubic (quartic) gauge boson vertices are linear (quadratic) in the Lie algebra ...


2

A bosonic symmetry that acts differently on the different components of a supermultiplet is an R-symmetry. Such a symmetry does not commute with the supercharges. Since the commutator between an R-symmetry and a supercharge gives something Grassmann odd, it has to given another supercharge. Schematically $$ [R,Q] = Q $$ or in terms of variations acting on ...


2

The "first-quantized setup" is the setup of quantum mechanics, where single particles are considered quantum objects, but fields like the electromagnetic field are still treated classically. However, the derivation of the action in the following is wholly classical (and the first-quantized setup arises when considering $x^\mu$ as quantum fields in the sense ...


1

$\pi^0$ is a real field, and uncharged. $\pi^{\pm}$ are both complex fields, and satisfy $\pi^- = (\pi^+)^*$. So we can rewrite the pion kinetic terms (focusing only on the electromagnetic interaction, ignoring completely the weak interactions) as \begin{equation} \frac{1}{2} (D \pi^0)^2 + \frac{1}{2}(D_\mu\pi^-)^\dagger (D^\mu \pi^-) + \frac{1}{2} (D_\mu ...


1

A priori, it is hard to know without having any experience with dimensional reductions. One has to get a feeling for how certain quantities change under the process, e.g. how components of the higher-dimensional gauge fields may turn into adjoint scalars, how spinors behave. An interesting thing to note is that symmetries of the lower-dimensional theory ...



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