9.7: Helium
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Helium is the simplest system for which we are unable to accurately calculate the energy.
For a single electron moving in the field of a helium nucleus, the spatial wavefunctions are similar to those of hydrogen |u_{nlm} \rangle.
When a second electron is added, a reasonable basis set is exchange-symmetrised wavefunctions consisting of spin states multiplying hydrogenic spatial parts:
(u_{nlm}({\bf r_1})u_{n' l' m'}({\bf r_2}) \pm u_{n' l' m'} ({\bf r_1})u_{nlm}({\bf r_2})) \nonumber
Since the overall wavefunction must be antisymmetric, the singlet (exchange-antisymmetric) spin states must combine with symmetric spatial states, and the triplet (exchange-symmetric) spin states must combine with antisymmetric spatial states.
If both electrons were in the same spatial state, the antisymmetric spatial wavefunction would be:
|(a({\bf r_1})a({\bf r_2}) − a({\bf r_2})a({\bf r_1})) \rangle = 0 \nonumber
Hence there is no triplet for the ground state.