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Neglecting spin effects, the energy levels of multi-electron atoms are characterized by states of definite total orbital ($L^2$) and spin angular momentum ($S^2$).

From this it seems that the symmetry group of an atomic system is:

$SO(3) \times SU(2)$

I have found nowhere a definite confirmation of this but it is consistent with the organization of atomic levels by term symbols ${}^{2S+1}L$.

Naively I would think the symmetry group would be larger since the Hamiltonian doesn't care about the spin at all (we are ignoring spin-orbit coupling). If the electrons were distinguishable, a spatial rotation could be joined by a set of unitary transformations, one for each spin, to produce a symmetry operation. i.e. for an $N$-electron system a symmetry element would be defined by a rotation $R$ and $N$ unitary operators $u_n \in U(2), n=1,\ldots,N$:

$|x_1\rangle|s_1\rangle\ldots|x_i\rangle|s_i\rangle\ldots|x_N\rangle|s_N\rangle\to R|x_1\rangle u_1|s_1\rangle\ldots R|x_i\rangle u_i|s_i\rangle \ldots R|x_N\rangle u_N|s_N\rangle$

In this case, the symmetry group would be:

$SO(3) \times SU(2)^N$

It appears then that the condition of state-vector antisymmetry restricts our symmetry operations to spatial rotations joined by a simultaneous transformation of each spin state by some unitary transformation, i.e.:

$|x_1\rangle|s_1\rangle\ldots|x_i\rangle|s_i\rangle\ldots|x_N\rangle|s_N\rangle\to R|x_1\rangle u|s_1\rangle\ldots R|x_i\rangle u|s_i\rangle \ldots R|x_N\rangle u|s_N\rangle$

Certainly such symmetry operations preserve the antisymmetry of the state-vector.

my questions are the following:

  • Is the (largest) symmetry group of a general multi-electron atom (without spin-orbit coupling) indeed $SO(3) \times SU(2)$?
  • Do the operations $(R,u)$ described above exhaust the possible symmetry elements for the atomic hamiltonian?
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