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I'll give a slightly more mathematical answer. First let me expand upon what chirality is all about. Quantum fields transform on specific representations of the Lorentz group. The irreducible representations are known as the $(A,B)$ representations and they are labelled by two integers or half-integers $A$ and $B$. If you have never seem this, please see ...


11

Parity involves a transformation that changes the algebraic sign of the coordinate system. Parity is an important idea in quantum mechanics because the wavefunctions which represent particles can behave in different ways upon transformation of the coordinate system which describes them. Under the parity transformation: The parity transformation changes a ...


3

Since they are uncharged they cannot interact with photons. In the standard model photons can only couple to charged elementary particles or charged gauge bosons, e.g charged leptons such as electrons, quarks such as up, down and charged gauge bosons such as W+. Since they cannot couple to photons, they cannot absorb light. Hope this helps.


3

The W bosons have magnetic moments according to the fact that they have a spin and are charged. Indeed in the standard model there is an interaction term of the form: \begin{equation} \mathcal{L}_{WW\gamma}=-ieF_{\mu \nu} W^{+,\mu} W^{-,\nu} \end{equation} Where $F$ is the Faraday tensor, and $W^+$ and $W^-$ are the fields associated to the bosons. This term ...


2

What distinguishes the chiral projections from other projections is that they are invariant under (commute with) continuous Lorentz transformations. Therefore, it's possible to have a theory in which only one projected "half" of the spinor exists, but which is still Lorentz invariant. That's interesting. And the real world turns out to be that way, ...


2

One reason why neutrinos are so hard to detect is that they have no electromagnetic charge so they don’t feel the electromagnetic force. This means that they don’t absorb light or interact with photons in any way. There are other particles with no charge - the most common is the neutron. But the neutron feels the strong force. If it gets close enough to a ...


2

For a fermion-antifermion pair, you can easily see that $$ C= (-)^{L+S}, $$ so, e.g., $$ C(^1S_0)=+ , ~~~\leadsto ~~~~ \to 2\gamma,\\ C(^3S_1)=- , ~~~\leadsto ~~~~ \to 3\gamma, $$ the photons being odd under C. The first case covers the lowest-lying pseudoscalars, like the pion, and parapositronium; while the second covers the ρ, ψ, ..., and ...


2

In short, no. One of the answers to the other question does say that the strong force gauge group can't have an extra factor of $U(1)$ because "total phase rotations of the quark wave function are already part of the model", referring to the $U(1)$ factor in the Standard Model gauge group. The $U(1)$ factor in that answer isn't electromagnetic $U(1)...


2

Recall that flavor, color, and S are commuting operators. Ultimately, you may be interested in the hadron spectrum (the eigenstates of the strong Hamiltonian, which is symmetric, more or less, under the corresponding transformations generated by them), or couplings among such eigenstates. The idea of the quark model is to build irreducible representations ...


1

If you are asking whether there can be a colorless 4-gluon vertex, group theoretically, the answer is "of course", since $8\otimes 8\otimes 8\otimes 8$ contains a singlet. It is the celebrated term in the QCD lagrangian, $$ g^2 f^{abc}f^{aeh}A^b_{[\mu} A^c_{\nu]}A^{e~~[\mu}A^{h~~\nu]}. $$ Gluons are always virtual, so, indeed, the four-gluon vertex ...


1

It is just a convention that is easy to remember, so it was adopted: The flavor charge has the same sign as the electric charge. With the advent of quarks, one simply writes down the quark content to avoid confusion and dyslexia... For d and u quarks, people heed isospin, which yields more meaningful labels than pure U and D additive numbers; isospin ...


1

If you google "strange matter in cosmological models" , you will see a number of references for papers calculating this with various models. From the latest: In this paper, we have examined the Kantowski–Sachs space-time in the presence of strange quark matter with the appearance and non-appearance of strings in f(R) gravity. Here, R is the Ricci ...


1

Very late to the party, but an explicit calculation of $\mu\rightarrow e\gamma$ can be found in the chapter Flavour Changing processes in my QFT Notes that are available on https://hepnotes.com.


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