In Sakurai's Quantum mechanics book, he says the hydrogen atom has $SO(4)$ symmetry by explicitly exhibiting operators $I_i,K_i$ that satisfy the commutation relation of the Lie algebra $so(4)$. Namely, if we restrict our attention to some eigenspace of the Hamiltonian $H$, then it is possible to find operators $I_i,K_i,i=1,2,3$ such that $$[I_i,I_j]=i\hbar\epsilon_{ijk}I_k,$$$$[K_i,K_j]=i\hbar\epsilon_{ijk}K_k,$$$$[I_i,K_j]=0.$$ Then he claims $SO(4)$ symmetry follows because these form the Lie algebra $so(4)=su(2)\oplus su(2)$. My question is there are many groups with the same Lie algebra as $SO(4)$, its universal cover $SU(2)\times SU(2)$ for instance. So what tells us the symmetry group is $SO(4)$ and not those other groups?

  • $\begingroup$ You really, really, really don't wish to care about topology. All the arguments involved, if you allow for his (and anybody reasonable's) exposition, handle Casimir invariants and representations comporting with the respective Lie algebras, in this case the direct sum of su(2) s including their half-integral spin reps! You would only be interested in SO(4) if you were studying the topology of recondite maps for recondite constructions, a fraught issue for QM. Strong avuncular advice: stick to Lie algebras. You might ask your question in the MSE, but you should not care about the answer! $\endgroup$ Oct 13, 2021 at 14:45
  • $\begingroup$ Related: How can one see that the Hydrogen atom has $SO(4)$ symmetry? $\endgroup$
    – Qmechanic
    Feb 2 at 7:43


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