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.

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    $\begingroup$ Actually the solution to this is via the ladder operators $\hat L_\pm$. Acting on the top state $\vert j,j\rangle$, laddering down will generate a finite dimensional space that is by definition irreducible. $\endgroup$ Jan 9, 2020 at 21:03
  • $\begingroup$ To vulgarize the above comment of @ZeroTheHero even more, for your j=1 triplet posited, if it were reducible, it would contain a singlet, so an eigenvector of all $J_z,J_+,J_-$, which you can see is not achievable. $\endgroup$ Jan 9, 2020 at 21:08
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    $\begingroup$ Neither of the comments are fully rigorous. A finite-dimensional representation is not necessarily irreducible, nor is any representation not containing a singlet necessarily irreducible. An example contradicting both claims is the representation formed from the direct product of a $j=1/2$ rep with a $j = 1$ rep. $\endgroup$ Jan 9, 2020 at 21:24
  • $\begingroup$ I don't understand why the fact that ladder operators generate a finite dimensional space implies that the representation is irreducible. Shouldn't I prove that there aren't any non-trivial subrepresentations? Intuitively, I see that if I only took two eigenstates (for example with $m=1$ and $m=0$), I could obtain, through a certain rotation, a vector that has a $m=-1$ component and that therefore doesn't belong to the space generated by the other two. Is this what you meant or is it something even more trivial? $\endgroup$
    – Gnegne
    Jan 9, 2020 at 21:52
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    $\begingroup$ I just realized ZeroTheHero's comment was probably not claiming that finite-dimensional representations are irreducible, but that a representation generated by acting with all the ladder operators starting from a single state is irreducible. That is pretty much how the proof goes. $\endgroup$ Jan 10, 2020 at 3:24

1 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.


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