Helium has two electrons and a nucleus with a +2e charge. We place the origin of coordinates in the nucleus, giving coordinates ($x_1,y_1,z_1$), ($x_2,y_2,z_2$) to electrons 1 and 2. From nuclear charge we take +Ze to include helioid ions, $H ^-$, $Li+$, $Be^{2+}$...
We write the Hamiltonian of Helium.
\begin{equation} \hat{H}=-\frac{\hbar^2}{2m}\nabla_1^2-\frac{\hbar^2}{2m}\nabla_2^2-\frac{Ze^2 }{r_1}-\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \end{equation}
Due to the term $\frac{e^2}{r_{12} }$ the Schrödinger equation is not separable in any coordinate system and we must use approximate methods to solve it.
We apply the perturbation theory, choosing as the unperturbed system the atom with two electrons without interelectronic interaction, whose Hamiltonian is given by:
\begin{equation} \hat{H}^{0}=-\frac{\hbar^2} {2m}\nabla_1^2-\frac{\hbar^2}{2m}\nabla_2^2-\frac{Ze^2}{r_1}-\frac{Ze^2}{r_2} \end{equation}
Since there is no interaction between the electrons, they behave like hydrogenoids. The wave function resulting from solving the equation, $\hat{H}^0\psi^{(0)}=E^{(0)}\psi^{0}$, is the result of multiplying the 1s functions of the hydrogen atom for each electron.
\begin{equation} \psi^{(0)}=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}e^{ -Zr_1/a_0}\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}e^{-Zr_2/a_0} \end{equation}
This function can also be represented by: $\psi=1s(1)1s(2)$, where $1s(1)=\frac{1}{\sqrt{\pi}}\left(\frac{Z} {a_0}\right)^{3/2}e^{-Zr_1/a_0}$ y $1s(2)=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{ a_0}\right)^{3/2}e^{-Zr_2/a_0}$.
On the other hand, the energy of the unperturbed system is obtained as the sum of the energies of both hydrogenoid electrons.
\begin{equation} E^{(0)}=\frac{-Z^2e^2}{2a_0}+\frac{-Z^2e^2}{2a_0}=\frac{-Z^2e^2 }{a_0}=-108.8\;eV \end{equation}
If we take into account that the experimental value of the Helium energy is -79 eV, with this first approximation we have an error of 38$\%$, as a consequence of Neglect the term $\frac{e^2}{r_{12}}$.
We now proceed to calculate the disturbance, $\hat{H'}=\hat{H}-\hat{H}^{0}=\frac{e^2}{r_{12}}$, which we will use to obtain the first order correction in energy.
\begin{equation} E^{(1)}=\left\langle \psi^{\ast (0)}\left|\hat{H'}\right|\psi^{(0)}\right\rangle =\frac{5Ze^2}{8a_0}=34\;eV \end{equation}
This first order correction gives us an energy for the ground state of the Helium atom of: \begin{equation} E=E^ {(0)}+E^{(1)}=-108.8+34=-74\;eV \end{equation} The error committed drops to 5.3$\%$.
Applying second and third order corrections, an energy very close to the experimental one is obtained. \begin{equation} E=E^{(0)}+E^{(1)}+E^{(2)}+E^{(3)}=108.8+34-4.3+0.1=79\; eV \end{equation}