How to use Young diagrams to express the product of an $SU(2)$ doublet and an $SU(2) $singlet?

Question

How to use Young diagrams to express the product of an SU(2) doublet and an SU(2) singlet, i.e. $2 \otimes 1 = 2$? I can only think of it as follows:

\begin{equation} \Box \otimes \Box = \Box\Box \oplus \begin{array}{|r|r|} \hline \\ \hline \\ \hline \end{array} \end{equation} But this seems not right, for these are the diagrams for the product of two SU(2) doublets, i.e. $2 \otimes 2 = 1 \oplus 3$. What should be the correct Young diagrams for $2 \otimes 1 = 2$?

Moreover, what are the Young diagrams for the product of two singlets?

Answer 1

As you should know, the Young tableau for the $\bf{2}$ of SU(2) is of course the following $$\bf{2} =\begin{array}{|r|} \hline \\\hline \end{array}.$$ As Qmechanic had said, the singlet is simply the Young tableau with no boxes. It is then trivial that $\bf{2}\otimes \bf{1} = 2$.

If you were looking to practice performing tensor products of representations using Young tableaux, you could also represent the singlet in this representation as $$\bf{1} =\begin{array}{|r|r|} \hline x\\ \hline y\\ \hline \end{array}.$$ So that $$\bf{2}\otimes\bf{1} =\bf{2} =\begin{array}{|r|} \hline \\\hline \end{array} \otimes \begin{array}{|r|r|} \hline x\\ \hline y\\ \hline \end{array}.$$ The only allowed diagram given by the rules discussed in chapter 12 of Georgi, for reference, is $$\bf{2}\otimes\bf{1} =\begin{array}{|r| r} \hline & x|\\ \hline \underline{y} \end{array}.$$ Here the $x$ and $y$ are boxes but the blank space at the bottom right is not. Using the rules discussed in Georgi, the row on the left vanishes, it is absorbed into a Levi-Civita. The remaining Young tableau is then $$\bf{2}\otimes\bf{1} =\begin{array}{|r|}\hline x \\\hline \end{array} = \bf{2}.$$

Answer 2

An $SU(2)$ singlet ${\bf 1}$ is the trivial representation, which corresponds to the Young tableau of no boxes, and which is the unit element wrt. tensor product, i.e. $V\otimes {\bf 1} = V$, so OP's relation ${\bf 2}\otimes {\bf 1} = {\bf 2}$ is a triviality.