Not sure how helpful it will be, but this comes from my notes on GUT charge quantisation:
In the Standard Model, the electric charge is the quantum number related to the U(1) of SU(3)×SU(2)×U(1) symmetry. Strictly speaking it is a U(1) subgroup of SU(2)×U(1), but that doesn't make a difference in this case. The Eigenvalues of U(1) are continuous - it is just a phase - and so there is no reason why we should expect the electric charge to be quantised. However, if the overall symmetry is SU(5) then we know that all the Eigenvalues of its generators are discrete, as this is the case for any simple non-Abelian group. The electric charge is then the Eigenvalue of one of the generators of SU(5) and can hence only take on discrete values.
The electric charge is additive, i.e. the charge of a two-particle state is the sum of the charges of the two one-particle states. This is only possible if Q is diagonal when acting on the one-particle states (otherwise they get mixed up). SU(5) has four independent diagonal generators and we know that the electric charge generator commutes with the color generators, of which there are two, i.e. the two generators corresponding to the two independent Eigenvalues of SU(3), i.e. the rank of SU(3). There are thus two generators left in SU(5) two construct the electric charge generator from and we arbitrarily chose to call these T3 and T0. These generators correspond to the diagonal generators of SU(2) and U(1) subgroups of SU(5). We can thus write
Q=T3+cT0
This shows that in the SU(5) GUT the formula for the electric charge is not enlarged to include any other terms than from the GWS theory.
We can estimate the value of c. First, we normalise the generators of SU(5) in such a way that Ta=λa/2 with trλaλb=2δab. It then follows that T0 and T3 are given by
T0=12√15(200000200000200000−300000−3)andT3=12(000000000000000000−100000+1)
To find c we e.g consider a left-handed d quark with electric charge Q=−1/3.
Acting on the ˉ5 representation ψi=(drdgdyˉue+)T we have
T0d=1/√15dα and T3dα=0. Therefore
−13dα=0+1√15cdα⇒c=−√53
This is also valid for the other particles. E.g. for the antineutrino we have
Qˉν=(T3−√53T0)ˉν=[−12−√53(−32√15)]ˉν=0
corresponding to the electrically neutral antineutrino. Likewise, we recover that the positron has charge +1:
Qe+=(T3−√53T0)e+=[+12−√53(−32√15)]e+=e+
We have derived a relation between the electric charges of the quarks and the leptons. Indeed, the structure of SU(5) automatically results in the fact that the charge of the d quark is −1/3 times the electric charge of the positron.
This post imported from StackExchange Physics at 2020-11-30 18:54 (UTC), posted by SE-user Oбжорoв