A5 - the ground state of helium

the unperturbed operator:

{\hat{H}}_{0} = - \frac{ℏ}{2 m} (\nabla_{1}^{2} + \nabla_{2}^{2}) - \frac{Z e^{2}}{4 π ε_{0}} (\frac{1}{r_{1}} + \frac{1}{r_{2}})

this assumes an infinitely heavy nucleus, considering electronic motion only
the ground state wave function and energy of each electron are:

\begin{matrix} Ψ_{100}^{0} = {(\frac{Z^{3}}{π a_{0}^{3}})}^{\frac{1}{2}} \exp - (\frac{Z r}{a_{0}}) \\ E_{100}^{0} = Z^{2} Rydbergs \end{matrix}

the $1 s$ state has $n = 1, l = 0, m_{l} = 0, m_{s} = \pm 1 / 2$ or $α, β$
the perturbation is:

λ \hat{V} = \frac{r^{2}}{4 π ε_{0} | r_{1} - r_{2} |} = \frac{r^{2}}{4 π ε_{0} r_{12}}

the ground state of the two electron system is the product of the two ground state wave functions for the individual electrons: $Φ_{100}^{0} = Ψ_{100}^{0} (r_{1}) Ψ_{100}^{0} (r_{2})$ , and energy $2 E_{100}^{0} = - 2 Z^{2}$ Ry
the complete antisymmetric ground state wave function is $^{1} S_{2}$ ( $1 s^{2}$ ) state, given by:

Φ_{100}^{0} (12) = {[{(\frac{Z^{3}}{π a_{0}^{3}})}^{\frac{1}{2}}]}^{2} e^{- {\frac{Z r_{1}}{a}}_{0}} e^{- {\frac{Z r_{2}}{a}}_{0}} \times \frac{1}{\sqrt{2}} (| α_{1} ⟩ | β_{2} ⟩ - | β_{1} ⟩ | α_{2} ⟩) = \frac{Z^{3}}{π a_{0}^{3}} e^{- Z (r_{1} + r_{2}) / a_{0}} \times \frac{1}{\sqrt{2}} (| α_{1} ⟩ | β_{2} ⟩ - | β_{1} ⟩ | α_{2} ⟩)

the first-order correction to the ground energy:

E_{100}^{1} = [\frac{1}{\sqrt{2}} (⟨ α_{1} | ⟨ β_{2} | - ⟨ β_{1} | ⟨ α_{2} |) (\frac{e^{2}}{4 π ϵ_{0}}) {(\frac{Z^{3}}{π a_{0}^{3}})}^{2}] \times [\iint d^{3} r_{1} d^{3} r_{2} \frac{1}{r_{12}} e^{- 2 Z (r_{1} + r_{2}) / a_{0}} \frac{1}{\sqrt{2}} (| α_{1} ⟩ | β_{2} ⟩ - | β_{1} ⟩ | α_{2} ⟩)]

collecting the spin and position terms:

E_{100}^{1} = \frac{1}{2} (⟨ α_{1} | ⟨ β_{2} | α_{1} ⟩ | β_{2} ⟩ + ⟨ β_{1} | ⟨ α_{2} | β_{1} ⟩ | α_{2} ⟩) \times (\frac{e^{2}}{4 π ϵ_{0}}) {(\frac{Z^{3}}{π a_{0}^{3}})}^{2} \iint \frac{d^{3} r_{1} d^{3} r_{2}}{r_{12}} e^{- 2 Z (r_{1} + r_{2}) / a_{0}}

the second term can be analytically evaluated, giving: $E_{100}^{1} = \frac{5}{4} Z$ Ry
so, the ground state energy $= 2 E_{100}^{0} + E_{100}^{1} = (- 2 Z^{2} + \frac{5}{4} Z)$ Ry
for Helium $(Z = 2) : - 5.5$ Ry
cf. $- 8$ Ry using just $H_{0}$ , and experimentally $- 5.8$ Ry
second-order PT will lower this estimate