Angular momentum and rotation group representations

Question

In Sakurai's book it's written that the operator $D_{m',m}^{(j)}=\left\langle{j,m'}\Big|\exp{\frac{-i \mathbf{ J\cdot \hat{n} } \phi}{\hbar}}\Big|{j,m}\right\rangle$ is the "$2j+1$-dimensional irreducible representation of the rotation operator". I was wondering whether there is a simple way to prove that it is irreducible (if it's not obvious for some reasons I'm missing).

For example, considering $j=1$ and the vector space generated by the eigenkets of $J^2$ and $J_z$, how can I show that the components of the angular momentum generate an irreducible representation?

My idea would be to use the fact that any other operator that commutes with $J_x$, $J_y$ and $J_z$ is a multiple of the identity (which is proved easily) and then use some sort of converse of Schur's lemma, if it even exists.

I'm not even sure this is the best way to tackle the problem, as I only know the most basic things about group theory.

Answer 1

Here’s a sketch of proof.

Any transformation can be factored in a sequence of upper triangular matrix (UT), a diagonal matrix (D) and a lower triangular (LT). Thus an arbitrary element $R\in SU(2)$ will have the form \begin{align} R(\boldsymbol{\phi})e^{-i \vec \phi\cdot \vec L}= UT(\phi)\cdot D(\phi)\cdot LT(\phi)\, . \end{align} This decomposition is basically the same as writing the transformation as \begin{align} R(\boldsymbol{\phi})=e^{-i\vec\phi\cdot\vec L}=e^{\zeta(\phi)L_-} \cdot e^{-i\beta(\phi) L_z}\cdot e^{\xi(\phi)L_+} \tag{1} \end{align} i.e. in antinormal-ordered form, where $\zeta,\xi $ and $\beta$ are functions of the original parameters $\boldsymbol{\phi}$ of the transformation. If one uses the Euler parametrization of elements in $SU(2)$ then you can find these $\zeta$ etc in multiple sources (v.g. Perelomov’s book on coherent states)

It is then clear that, acting on the highest weight states, (1) amounts to repeatedly acting on this state by the lowering operator $L_-$. By construction, if $\boldsymbol{\phi}$ is a generic element then there cannot be any invariant subspace.

Note that this relies on the action of (1) on the unique highest weight state in your space. For instance, if you consider the Hilbert space for two spin-1/2 systems, then each of the irreducible subpieces $S=0$ and $S=1$ has a highest weight so acting on the $\vert{+}\rangle\vert{+}\rangle$ (the highest weight of the $S=1$ subspace, generates a $3$-dimensional irreducible subspaces which does not “access” the $S=0$ subspace, and a rotation of the uncoupled $\vert{+}\rangle \vert -\rangle$ does not generate an irreducible subspace as this state is not a highest weight.