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To preface, I have little background in representation and Lie theory.

My understanding is as such: Given any finite dimensional Hilbert space $\mathcal{H}$ and representation $\rho: SU(2) \rightarrow GL(\mathcal{H})$, due to the compactness of $SU(2)$ the representation $\rho$ is completely reducible. That is $\rho \cong \bigoplus_i \rho_i$ where each $\rho_i$ is an irreducible representation of $SU(2)$ over some invariant subspace $\mathcal{H}_i$. These irreducible representations are characterized by the dimension of their (invariant) representation space and so are unique up to isomorphism. We identify each irreducible representation with a spin value.

My question is: what is the physical evidence that tells us $SU(2)$ is the right group to use to model non-relativistic spin?

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    $\begingroup$ Well, $SU(2)$ is isomorphic to $Spin(3)$. $\endgroup$
    – Qmechanic ♦
    Commented May 27, 2023 at 7:52
  • $\begingroup$ I do not understand the physical meaning of this statement. Wikipedia says a similar thing; SU(2) is (isomorphic to) the double cover of SO(3). However, I am missing the information which connects this statement with SU(2) being a proper description of non-relativistic spin. @Qmechanic $\endgroup$
    – Silly Goose
    Commented May 27, 2023 at 8:11
  • $\begingroup$ Okay I'm looking at your answer from a prior post. Reworded, is the logic: Define angular momentum to be the generator of rotations. We care about angular momentum. We would use SO(3), but Quantum mechanics allows for projective representations, so we use the double cover of SO(3), Spin(3). What's easier is using SU(2)$ \cong$ Spin(3). Hence, SU(2) is the "correct" group to model angular momentum in general, not just spin. It falls out of a special case that spin exists. @Qmechanic $\endgroup$
    – Silly Goose
    Commented May 27, 2023 at 9:00
  • $\begingroup$ Yes, that's essentially the logic. $\endgroup$
    – Qmechanic ♦
    Commented May 27, 2023 at 9:32
  • $\begingroup$ Hm, upon further thinking this doesn't explain spin I don't think. It seems like one splits a Hilbert space into $\mathcal{H}_s \otimes \mathcal{L}^2$. The projective representation of $SO(3)$ on finite dimensional $\mathcal{H}_s$ gives us spin. The projective representation of $SO(3)$ on the infinite dimensional $\mathcal{L}^2$ gives us orbital angular momentum. But this doesn't explain the partition of the Hilbert space into finite and infinite degrees of freedom. @Qmechanic $\endgroup$
    – Silly Goose
    Commented May 27, 2023 at 13:25

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