# How to find the matrix generators of higher dimensional irreducible representations of $\operatorname{SU}(2)$?

Using the most general form of a $$2\times 2$$ unitary matrix $$U$$ of determinant $$+1$$ and using the formula $$T^a=-i\frac{\partial U}{\partial\theta^a}|_{\{\theta_a\}=0} ~{\rm with}~ a=1,2,3.$$ In this way, I can find out the representations of three generators of $$\operatorname{SU}(2)$$. It gives three Pauli matrices $$\sigma^{1},\sigma^{2}$$ and $$\sigma^{3}$$. But how do I find the generators of $$\operatorname{SU}(2)$$ for the three or higher dimensional irreducible representations of $$\operatorname{SU}(2)$$?

• Do you know about raising and lowering operators, also called "ladder" operators? – paul garrett Mar 12 at 15:40
• @paulgarrett Yes. I know – mithusengupta123 Mar 12 at 16:29
• In that case, WP provides the complete answer. The exponentials (generic group elements) are not that bad either. – Cosmas Zachos Apr 6 at 14:08

The complexified Lie algebra is $$\mathfrak{sl}(2)$$, which is a little easier to talk about. Let $$h=\pmatrix{1 & 0 \\0 & -1}$$, $$R=\pmatrix{0 & 1 \\ 0 & 0}$$ and $$L=\pmatrix{0 & 0 \\ 1 & 0}$$ be the usual triple. Granting that we know that the weights/eigenvalues of $$h$$ on the unique irreducible of a given dimension are $$n, n-2, n-4, \ldots, 4-n, 2-n, -n$$, we can map $$h$$ to the diagonal matrix with those entries. The "raising" operator $$R$$ can be mapped to the matrix with 1's at the $$(i,i+1)$$ entries, and zeros elsewhere. The "lowering" operator $$L$$ can be mapped to a matrix with non-zero entries just at the $$(i,i-1)$$ entries, and zeros elsewhere. The non-zero entries of the image of $$L$$ are completely determined by the Lie bracket requirement that $$[\rho(R),\rho(L)]=\rho(h)$$, where $$\rho$$ is the representation.