Irreducible representation of SU(2)

[davidge]

I'm reading a paper on physics where it's said it can be shown that every irreducible representation of ##SU(2)## is equivalent to the one which uses the Ladder Operators. I am a noob when it comes to this subject, but I'd like to know whether or not the proof is easy to carry out.

 

[fresh_42]

It takes a few steps but is not really difficult. Usually the irreducible representations of ##\mathfrak{sl}_\mathbb{R}(2,\mathbb{C})## are classified and ##\mathfrak{su}_\mathbb{R}(2,\mathbb{C}) \cong \mathfrak{sl}_\mathbb{R}(2,\mathbb{C})##. I'm not sure how they are translated to the group representations, but I assume you meant those of the tangent spaces, normally referred to as generators.

Let ##v_m## be a maximal vector of the ##(m+1)## dimensional representation space. Then for ##k=0, \ldots , m## we define
$$
v_{m-2k-2} := \frac{1}{(k+1)!}\;Y^{k+1}.v_m\; \text{ and } \;v_{-m-2}=v_{m+2}=0
$$
and get the following operation rules
$$
\begin{array}{ccc}
H.v_{m-2k}&=&(m-2k)\;v_{m-2k}\\
X.v_{m-2k}&=&(m-k+1)\;v_{m-2k+2}\\
Y.v_{m-2k}&=&(k+1)\;v_{m-2k-2}
\end{array}
$$
##H## is the semisimple part which defines the eigen values, ##X## is ascending, and ##Y## descending.

 

[davidge]

Thanks.

Maybe an other, more easy way, to show that a representation is irreducible, would be to show that it can't be rewritten using similarity transformation in block diagonal form.

What are your thoughts?

 

[fresh_42]

In a single case a brute force calculation is probably doable for a given representation, especially for such small groups. But in general, one wants to know the rest of the series, here ##A_n## or even more generally all semisimple cases as well plus the exact action, i.e. the eigenvalues. The advantage of the theorem above is, that we know that all (finite dimensional) irreducible representations look like this.