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 $SU(3)\times SU(2) \times U(1)$ symmetry. Strictly speaking it is a $U(1)$ subgroup of $SU(2)\times U(1)$, 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 $T_3$ and $T_0$. These generators correspond to the diagonal generators of $SU(2)$ and $U(1)$ subgroups of $SU(5)$. We can thus write
$$
Q = T_3+ c T_0
$$
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 $T_a=\lambda_a/2$ with $\mathrm{tr}{\lambda_a \lambda_b} = 2 \delta_{ab}$. It then follows that $T_0$ and $T_3$ are given by
\begin{align}
T_0 = \frac{1}{2\sqrt{15}} \begin{pmatrix} 2 &0& 0 &0& 0 \\ 0 &2& 0 &0& 0 \\ 0 &0& 2 &0& 0 \\ 0 &0& 0 &-3& 0 \\ 0 &0& 0 &0& -3 \end{pmatrix}
\quad \mathrm{and} \quad T_3 =\frac{1}{2} \begin{pmatrix} 0 &0& 0 &0& 0 \\ 0 &0& 0 &0& 0 \\ 0 &0& 0 &0& 0 \\ 0 &0& 0 &-1& 0 \\ 0 &0& 0 &0& +1 \end{pmatrix} \label{eq:smt0su5}
\end{align}
To find $c$ we e.g consider a left-handed d quark with electric charge $Q= -1/3$.

Acting on the $\bf{\bar 5}$ representation $\psi^i = \begin{pmatrix} d^\mathrm{r} & d^\mathrm{g} & d ^\mathrm{y} & \bar{u} & e^+ \end{pmatrix}^T$ we have
$T_0 d = 1/\sqrt{15}d^\alpha$ and $T_3 d^\alpha =0$. Therefore
$$
-\frac{1}{3} d^\alpha = 0+ \frac{1}{\sqrt{15}} c d^\alpha \quad \Rightarrow c=- \sqrt{\frac{5}{3}}
$$
This is also valid for the other particles. E.g. for the antineutrino we have
$$
Q \bar{\nu} = \Big( T_3-\sqrt{\frac{5}{3}} T_0 \Big) \bar{\nu} = \Big[-\frac{1}{2}-\sqrt{\frac{5}{3}} \Big( - \frac{3}{2\sqrt{15} }\Big)\Big] \bar{\nu} = 0
$$
corresponding to the electrically neutral antineutrino. Likewise, we recover that the positron has charge $+1$:
$$
Q {e^+} = \Big( T_3-\sqrt{\frac{5}{3}} T_0 \Big) e^+ = \Big[+\frac{1}{2}-\sqrt{\frac{5}{3}} \Big( - \frac{3}{2\sqrt{15} }\Big)\Big] e^+ = e^+
$$
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