In the theory of angular momentum, we wish to study the projective representations of the rotation group $\text{SO}(3)$, for which we turn to the representation theory of the double cover $\text{SU}(2)$. I understand the finite dimensional representation theory of the Lie algebra $\mathfrak{su}(2)$ where we find either integer or half integer weights depending on the dimension of the representation. However I have not been able to find a satisfying treatment of the infinite dimensional case. Let $\mathcal{H} = L^2(\mathbb{R^3})$. It is well known that the eigenvalues of the angular momentum operators on this Hilbert space will be integer multiples of $\hbar$, not half integers. How can we see this using representation theory?

Edit: I found an answer with the help of the commenters (thank you!). $L^2(\mathbb{R}^3)$ decomposes as an orthogonal direct sum of vector spaces $V_l$, each of which is invariant under the action of the rotation group and thus irreducible under this action. Furthermore, one can show that each of these vector spaces $V_l$ has dimension $2l+1$, where $l$ is an integer. Thus, each of these spaces is odd. Hence, the projective representation of $\text{SO}(3)$ on each $V_l$ will have integer eigenvalues. See Hall - Quantum Theory for Mathematicians for the proof.

  • $\begingroup$ $SO(3)$ is compact so all unitary representations are equivalent to finite dimensional ones... or are you thinking of some limit of large representation or non-unitary irreps? $\endgroup$ Apr 11, 2017 at 19:44
  • $\begingroup$ @ZeroTheHero What do you mean all unitary representations are equivalent to finite dimensional ones? If the Hilbert space I want to represent $\text{SU}(2)$ on is infinite-dimensional, how can I realize this as a finite dimensional rep.? $\endgroup$ Apr 11, 2017 at 19:49
  • $\begingroup$ Decompose the Hilbert space in finite dimensional irreps. See en.wikipedia.org/wiki/… $\endgroup$ Apr 11, 2017 at 19:56
  • $\begingroup$ @ZeroTheHero this doesn't explain why we only see integer eigenvalues. It appears that in the infinite dimensional case we are only looking at ordinary representations of $\text{SO}(3)$ rather than projective ones, but I do not understand why. $\endgroup$ Apr 11, 2017 at 20:17
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    $\begingroup$ No. The Hilbert space is highly reducible, and decomposes into a direct sum of finite-dimensional (in some cases very large) representations. @ValterMoretti just answered as I was writing my comment. $\endgroup$ Apr 11, 2017 at 20:27

1 Answer 1


Since $SO(3)$ is compact, in view of Peter-Weyl's theorem, every unitary strongly continuous representation of $SO(3)$ in a Hilbert space is a direct sum (not a direct integral) of finite-dimensional irreducible representations which, in turn, are finite dimensional representations of $SU(2)$. So, once you know all finite dimensional of $SU(2)$ representations you know everything.

  • $\begingroup$ You just beat me to it.. $\endgroup$ Apr 11, 2017 at 20:26
  • $\begingroup$ Sorry...It happens :) $\endgroup$ Apr 11, 2017 at 20:27
  • $\begingroup$ Not a big deal. Stay well! $\endgroup$ Apr 11, 2017 at 20:27
  • $\begingroup$ @ValterMoretti I understand this but it still doesn't explain why we only see integer values eigenvalues when representing $\text{SU}(2)$ on $L^2(\mathbb{R}^3)$. However, I have just found an explanation as to why this is that I will post as an update. $\endgroup$ Apr 11, 2017 at 20:29
  • $\begingroup$ @JacksonBurzynski please do so! $\endgroup$ Apr 11, 2017 at 20:30

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