# 9.7: Helium


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.

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