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I am very confused that some atoms called high spin or magnetic atoms have spin level more than $\frac{1}{2}$ but are still said to have $SU(2)$ symmetry.

Why not $SU(N)$?

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You can apply a rotation to many objects, see…. If I understand your confusion correctly, this thread might clarify the situation, if you take into account that you have countable infinite representations of the rotation group. – NikolajK Nov 17 '12 at 16:31
This is because the group $SU(2)$ is essentially the same as the rotation group $SO(3)$ (more specifically, $SU(2)$ is a double cover of $SO(3)$, with two $SU(2)$ matrices, differing by a sign, corresponding to each rotation). The assertion then says that high-spin atoms have rotational symmetry, with the proper treatment of signs (i.e. possibly accumulating a global $-1$ after a $2\pi$ rotation). – Emilio Pisanty Nov 18 '12 at 13:59
up vote 2 down vote accepted

$SU(2)$ has irreducible unitary representation of every spin $0,1/2,1,3/2,\dots$. Indeed, the spin $j$ is just the historical way of recording the dimension $1+2j$ of the representation space of an irreducible unitary representation.

On the other hand, the quantum numbers of $SU(N)$ (characterizing its irreducible unitary representations) are significantly more complicated than a single spin. For example, $SU(3)$ is physcally associated with flavor or color, not with spin.

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Yes, SU(2) can represent all spins or angular momentum. SU(2) symmetry is God given as long as it is real spin. But I want to point out that there could be SU(N) states, say, scattering of alkali earth atoms.

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