Is there a way to derive the representations of $SO(3)$ without the usual method with the ladder operators which also gives the ones of $SU(2)$?

The usual way to do these calculations is to start from the commutation relations of the Lie algebra associated with $SU(2)$ (or that of $SO(3)$, which is the same given that $\mathfrak{so}(3) \approx \mathfrak{su}(2)$ as far as I understand) and from there to go throught the ladder-operators-thing to obtain all of the representations of $SU(2)$. Is there another way to derive the representations of $SO(3)$ which is specific of $SO(3)$ and not also applicable to $SU(2)$?

  • $\begingroup$ It is inefficient to derive the reps of $\mathrm{SO}(3)$ alone. It is a general principle that, given a Lie group, all its linear reps are induced by linear reps of its universal cover. Therefore, it always suffices to look at the universal cover, which in the case of $\mathrm{SO}(3)$ is $\mathrm{SU}(2)$. Why would you want to make your life harder by not looking at that? $\endgroup$ – ACuriousMind Nov 9 '14 at 22:39
  • $\begingroup$ @ACuriousMind because I wanted to understand what properties of $SO(3)$ restrict its representations to be integer-spin ones $\endgroup$ – glS Nov 9 '14 at 22:57
  • $\begingroup$ The relevant property is that $\mathrm{SU}(2)$ double covers $\mathrm{SO}(3)$, and if you look which reps project down nicely along the covering map, you find they are precisely the integer spin ones. $\endgroup$ – ACuriousMind Nov 9 '14 at 23:01
  • $\begingroup$ @ACuriousMind well I'm sure that that is extremely insightful... for who understands what it means :) $\endgroup$ – glS Nov 10 '14 at 21:27

Be careful. It may be the case that $\mathfrak{su}(2)=\mathfrak{so}(3)$, but it is not the case that $SU(2)=SO(3)$. $SU(2)=\mathrm{Spin}(3)$ and $\rho :SU(2)\rightarrow SO(3)$ is the two-sheeted universal cover of $SO(3)$. It thus turns out that only the integer spin representations of $SU(2)$ factor through $\rho$ to give well-defined representations of $SO(3)$.

Concretely, the spherical harmonics $Y_{\ell m}(\theta ,\phi )$ span an irreducible representation of $SO(3)$ in $L^2(S^2)$. This representation has spin $\ell$ (dimension $2\ell +1$), and this yields all the finite-dimensional irreducible complex representations of $SO(3)$. Note that $\ell$ must be an integer in this context (and so it does not give you all the irreducible representations of $SU(2)$).

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  • $\begingroup$ Ok, but that is the "usual" way isn't it? Being that the representations of $SU(2)$ are those of $SO(3)$ plus more, the method with ladder operators $J_\pm$ starting from the commutation relations of the Lie algebra gives all the (finite-dimensional, irreducible) representations of $SU(2)$ and hence between those the ones of $SO(3)$. What I am asking is if there is some method to specifically derive the irreps of $SO(3)$, without also getting those of $SU(2)$. Which means probably without using the associated Lie algebra, but I am not sure about this $\endgroup$ – glS Nov 9 '14 at 21:43
  • $\begingroup$ The spherical harmonics give you all the irreducible representations of $SO(3)$, but it does not give you all those of $SU(2)$. You'll note this does not make any direct reference to the Lie algebra of $SO(3)$ $\endgroup$ – Jonathan Gleason Nov 9 '14 at 21:45
  • $\begingroup$ You're absolutely right, I realized that just after I added the last comment. This makes me wonder: what, from the group theory point of view, leads to the restriction of the values of the orbital angular momentum? I guess it comes from the fact that $L_i = \varepsilon_{ijk} x_j p_k$, but how is this expressed in the language of group theory? $\endgroup$ – glS Nov 9 '14 at 21:57
  • $\begingroup$ Suppose that the representation has spin $n/2$ with $n\in \mathbb{Z}^+$. Then, in particular, this representation will contain an element $v$ such that $S_zv=\frac{n}{2}v$. This is at the level of the Lie algebra. At the level of the Lie group, $R_z(\theta )=\exp (-i\theta S_z)$ is a rotation about the $z$-axis by an angle $\theta$. Hence, $R_z(\theta )v=\exp (-i\theta \frac{n}{2})$. For $\theta =2\pi$, we had better have that $R_z(2\pi )=R_z(0)$; however, using the formula above, $R_z(2\pi )=\exp (-i\pi)=-1$. Thus, this does not give us a representation of $SO(3)$ for half-integer spin. $\endgroup$ – Jonathan Gleason Nov 9 '14 at 22:38
  • $\begingroup$ So what you are suggesting is to exclude half-integer angular momentum representations because they require 720° of rotation to get the state back to itself? We accept this property for spin angular momentum though, why should we a priori consider it unacceptable for the orbital one? However I was thinking of a reason coming from the mathematical properties of the definition of the orbital angular momentum that further restrict its representations. For example, could this be related to $L_i$ be an antisymmetric tensor? $\endgroup$ – glS Nov 9 '14 at 22:52

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