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What is the connection between complete set of commuting observables and generators of the Lie group?

I have a Hamiltonian written down in second quantized formalism and I also checked that it commutes with the generators of the Lie group - $SU(3)$. How would I construct algebraically eigenstates of this Hamiltonian? How many quantum numbers do I need?

Is there any procedure to determine full symmetry of the Hamiltonian? Like in hydrogen atom, it is not $SO(3)$ but $SO(4)$.

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  • $\begingroup$ Which Hamiltonian? $\endgroup$ – Qmechanic Nov 16 '14 at 17:25
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If the Hamiltonian has $SU(2)$ symmetry then the Hilbert space can be spanned by eigenstates of Cartan subalgebra (one dimensional) operator and Casimir operator. As we know from elementary quantum mechanics

$$\begin{align} \hat{L}^{2}\left|l,m \right\rangle &=l(l+1)\left|l,m \right\rangle,\\[3mm] \hat{L}_{z}\left|l,m \right\rangle &= m\left|l,m \right\rangle,\\[3mm] \left[\hat{L}^{2}, \hat{L}_{z}\right] &= 0. \end{align}$$
So, we need two quantum numbers $l,m$.

$SU(3)$ group have two-dimensional Cartan subalgebra and two Casimir operators. Hilbert space can be spanned by eigenstates of Cartan subalgebra - two operators and one Casimir operator. In sum, $3$ quantum numbers are necessary.

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