0
$\begingroup$

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 $SU(2)$?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $SU(2)$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.