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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 $\operatorname{SU}(2)$?

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

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