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This question already has an answer here:

When studying the irreducible representations of SO(3) one usually looks at the irreps of the infinitesimal rotations instead, i.e. the ones of so(3), the Lie Algebra of SO(3). The Irreps of so(3) can be parametrized by a single number $j \in {0, 1/2, 1, 3/2, ...}$ These irreps of so(3) can be raised to irreps of SO(3), via the map that maps elements of so(3) to SO(3).

My question now is: How does SU(2) come into play? IIRC the irreps of SO(3) only correspond to the full-integer irreps of so(3). How is this possible when every so(3) irrep can get raised to one of SO(3) as described above? Do several irreps of so(3) get raised to a single one of SO(3)?

If what I said so far is correct, then the discovery of the Zeeman effect called for an enhancement of this theory, as some spectral lines showed a degeneracy of $2j + 1 \in N$, meaning j had to be a half-integer: only the full integer irreps of SO(3) aren't enough!

SO(3) and SU(2) are isomorphic to each other, up to the Rotations $\pm id$. How does this solve the issue of SO(3) not accounting for half-integer representations?

Cheers and thanks in advance!

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marked as duplicate by ACuriousMind, Brandon Enright, Ali, Qmechanic Aug 3 '14 at 23:20

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The Lie algebras and $\mathfrak{so}(3)$ $\mathrm{su}(2)$ are isomorphic, but the Lie groups $\mathrm{SO}(3)$ and $\mathrm{SU}(2)$ are not. In fact $\mathrm{SU}(2)$ is the double cover of $\mathrm{SO}(3)$; there is a 2-1 homomorphism from the former to the latter.

How is this possible when every so(3) irrep can get raised to one of SO(3) as described above?

The half-integer irreps of $\mathfrak{so}(3)$ do not have corresponding $\mathrm{SO}(3)$ irreps, but they do have corresponding $\mathrm{SU}(2)$ irreps. When you try to exponentiate the half-integer irreps, you don't get a representation of $\mathrm{SO}(3)$.

An explanation of why we miss physically relevant stuff when we consider only $\mathrm{SO}(3)$ and not its double cover is given here:

Idea of Covering Group

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