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?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.