Elementary question about charge of doublet

This is probably extremely elementary, but I am not sure how to look for an answer online. I'm reading a physics text that uses the following notation for a doublet of particles:

$$q_L \rightarrow (3,2)_{1/6}$$

What does this notation mean? I have been able to deduce that the subindex $$1/6$$ is the weak hypercharge $$Y$$. But what about the numbers between the parentheses? The text then proceeds to calculate the charge of the doublet as:

$$Q=T^3+Y$$

but as far as I know, $$T^3$$ should be $$T_3$$, the third component of the isospin, right? How should this be computed based on that notation?

• I'm reading a physics text Which one? Commented Nov 27, 2022 at 18:41
• @Ghoster the class notes of a physics professor at my university. They are not available online so I can't link them here, and the professor has since retired, so I don't feel comfortable emailing him to ask about this. I'm getting into particle physics coming from a pure mathematics background, so I'm using all kinds of materials I can find to try and learn quantum field theory by myself before starting my PhD. This is why I'm a bit lost when it comes to some elementary stuff, sadly. Commented Nov 27, 2022 at 19:00

The notation $$(3,2)_{1/6}$$ denotes the dimensionalities of the representations of your particle under SU(3)×SU(2), respectively, in the parenthesis, and the eigenvalue of the hypercharge U(1) as a subscript. So, the 2 here means doublet: weak isospin 1/2, acted upon by 2$$\times$$2 matrices.
Indeed, in your normalization for the weak hypercharge (it is not unique!), $$Q=T_3+Y$$, so that, for the left-chiral quarks of the SM, the weak doublet is $$u \mapsto \begin{pmatrix} 1\\ 0\end{pmatrix} , \qquad d \mapsto \begin{pmatrix} 0\\ 1\end{pmatrix},$$ so that, for $$T_3=\operatorname{diag}(1/2,-1/2)$$ , $$Qu=(1/2 + 1/6)u= 2/3 ~ u, \qquad Qd= (-1/2+1/6)d= -1/3~ d.$$
For the right-chiral quarks, the reps are $$(3,1)_{2/3}$$ and $$(3,1)_{-1/3}$$ respectively. The Higgs isodoublet presents as $$(1,2)_{1/2}$$.
• So, as far as I understand, the value of the charge $Q$ depends only on $Y$, since $T_3$ is always equal to the same matrix, $T_3=diag(1/2,-1/2)$. Right? I'm a bit confused, since I thought $T_3$ was the third component of the isospin, but here it appears to be one of the generators of the $SU(3)$ group. Commented Nov 28, 2022 at 11:56
• Correct, for left-chiral fermions (doublets), but not for right-chiral singlets, where $T_3=0$. I am baffled by your unsound "appears". It has nothing to do with the SU(3) group, which is a red herring here: color commutes with the SU(2), the U(1), and therefore charge. It is there for completeness of description, a completely inert bystander in your context. Commented Nov 28, 2022 at 12:08
• Could you please explain how do I know the value of $T_3$ from this type of notation? I think it's the only thing I'm missing here. How to know if I'm dealing with a doublet or a singlet? Sorry if it's a very basic question. Commented Nov 28, 2022 at 12:19
• The 2 in the second spot in the parenthesis says SU(2) doublet. The 1 says singlet. So in the first case $T_3$ is a diagonal matrix, and in the second trivial. Commented Nov 28, 2022 at 12:45