If the term V ee = e /(4 πε 0 | r 1 − r 2 |), representing the repulsion between the two electrons, were excluded, the Hamiltonian would become the sum of two hydrogen-like atom Hamiltonians with nuclear charge +2 e. The ground state energy would then be 8 E 1 = −109 eV, where E 1 is the Rydberg constant, and its ground state wavefunction would be the product of two wavefunctions for the ground state of hydrogen-like atoms: ψ ψ ( r 1 , r 2 ) = Z 3 π π a 0 3 e − − Z ( r 1 + r 2 ) / a 0 . where a 0 is the Bohr radius and Z = 2, helium's nuclear charge. The expectation value of the total Hamiltonian H (including the term V ee) in the state described by ψ 0 will be an upper bound for its ground state energy. ⟨ V ee ⟩ is −5 E 1 /2 = 34 eV, so ⟨ H ⟩ is 8 E 1 − 5 E 1 /2 = −75 eV.