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Hot answers tagged standard-model

7

Color charge in the sense of "being blue, red, green" is not a quantum mechanical observable because the $\mathrm{SU}(3)$ gauge transformations mix the colors. This means it is meaningless to say "We have a blue particle", because we can perform a gauge transformation and then we "have a red particle". Since physical descriptions related by gauge ...

4

Is there anyone knowledgable enough in this area who would be able to comment on some of the possible theoretical/ hypothetical implications of the existence of spin 3 particles? Is there any thought that their existence could imply additional fundamental forces? If you look at the presentation linked in the link you gave , in page five, you will see ...

3

Is there something wrong with this process? (it will admittedly be suppressed by $|V_{su}|^2\approx \frac{1}{20}$, i.e. "doubly cabbibo suppressed") or maybe even replace the Z boson with gluons.

2

Yes, wikipedia has a table which lists the 19 free parameters that need to be tuned by experiments. These include, as you already said, the masses of the elementary particles including the Higgs Boson, and some other notable ones are: CKM Mixing angles and CP-violation phase. Gauge coupling of he three symmetries (U(1), SU(2), SU(3)). Higgs VEV

1

Baez actually has another paper (with Huerta) that goes into more detail about this. In particular, Sec. 3.1 is where it's explained, along with some nice examples. The upshot is that the hypercharges of known particles work out just right so that the action of that generator is trivial. Specifically, we have Left-handed quark Y = even integer + 1/3 ...

1

How can we see that the group $N$ generated by $$g = (e^{2\pi i/3} I, -I, e^{i\pi /3}) \in SU(3)\times SU(2)\times U(1)$$ acts trivially on all fields in the Standard Model? First of all, note that $g$ is in the center of $SU(3)\times SU(2)\times U(1)$. Therefore its representative in the adjoint representation is the identity. Since gauge bosons ...

1

In addition to TwoBs' comments, in the 90s, there had been many collisions at SLAC (USA) with polarized electron/positron beams of the SLC collider running at the $Z$ pole. Therefore, the right handed component of the electroweak interaction has been extensively tested. See the wikipage: http://en.wikipedia.org/wiki/SLAC_National_Accelerator_Laboratory

1

The VEV is quantum mechanical, it can not be read off from the Lagrangian. To find the VEV from the potential requires one to quantise the theory, then calculate the effective action at strong coupling. What should happen is that at strong coupling the quarks form hadrons; which are quark condensates. To perform the calculations is very tough and not yet ...

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