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3

It's a convention which of these reps is called ${\bf 10}$ and which is called $\overline{\bf 10}$. The convention that the people choose is arguably the simpler one among the two: ${\bf 10}$ is the antisymmetric product of two ${\bf 5}$, i.e. ${\bf 5}\wedge{\bf 5}$, which are also without bars. With that choice, one can prove that $\overline{\bf 10}$ which ...

2

Whenever you have a symmetry group $G$, it means that for each $g\in G$ there is an operator $U(g)$ (usually unitary) in the system corresponding to the action of $g$. "states behave like irrep of $G$" means that the state space can be organized into subspaces, and in each subspace $U(g)$ form an irrep of G (i.e. the matrix representation of $U(g)$).

8

The relevant Lie group isomorphism reads $$\tag{1a} U(2)~\cong~[U(1)\times SU(2)]/\mathbb{Z}_2.$$ In detail, the Lie group isomorphism (1a) is given by $$U(2)~\ni~ g\quad\mapsto\quad$$ $$\left(\sqrt{\det g}, \frac{g}{\sqrt{\det g}}\right) ~\sim~ \left(-\sqrt{\det g}, -\frac{g}{\sqrt{\det g}}\right)$$ $$\tag{1b}~\in ~[U(1)\times SU(2)]/\mathbb{Z}_2.$$ ...

1

I'll use a notation that probably you know. I'll denote group representations by Dynkin labels. Consider the following: \begin{equation*} [1,0]_3 = [1]_2 q^1 + [0]_2q^0 \ , \end{equation*} and \begin{equation*} [0,1]_3 = [1]_2 q^{-1} + [0]_2q^0 \ . \end{equation*} Where $q$ is the $U(1)$-charge of the representation. By Dynkin diagrams the $SU(3)$ algebra ...

1

Since this question looks like homework we will be somewhat brief. OP's notes are apparently describing the symmetry of the corresponding Young diagram for each $SU(3)$ irrep. Each box corresponds to an index. Roughly speaking, indices in same row (column) are symmetric (antisymmetric), respectively. Examples: A single box $[~~]$ corresponds to the ...

1

$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no "reason" you can use the ladder operators. Rather, they are the reason angular momentum is quantized in integer steps. You can define them, there's no inconsistency, so you can use them, and using them leads you to conclude that the angular momentum is raised/lowered in integer steps by them, in the way ...

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