We apply the variational theorem, $W=\int\varphi^{\ast}\hat{H}\varphi dq\geq E_{fund.}$, using as proof function: \begin{equation} \varphi=\frac {1}{\sqrt{\pi}}\zeta^{3/2}e^{-\zeta r_1}\frac{1}{\sqrt{\pi}}\zeta^{3/2}e^ {-\zeta r_2}=\phi(1)\phi(2) \end{equation} We have obtained this test function from hydrogenoid 1s orbitals for both electrons, changing the atomic number Z by the variational parameter $\zeta$ (zeta).
The physical interpretation of $\zeta$ is simple, one electron tends to shield the other in front of the nucleus, each electron is subjected to an effective nuclear charge less than the total nuclear charge Z.
The Hamiltonian for the Helium atom, in atomic units, has the form: \begin{equation} \hat{H}=\frac{1}{2}\nabla_1^2-\frac{1}{2}\nabla_2 ^2-\frac{Z}{r_1}-\frac{Z}{r_2}+\frac{1}{r_{12}} \end{equation} Substituting into the variational integral:

\begin{equation} W= \left\langle\varphi\left|\hat{H}\right|\varphi\right\rangle=\left\langle\phi(1)\phi(2)\left|\frac{1}{2}\nabla_1^2-\frac{1}{2}\nabla_2^2-\frac{Z}{r_1}-\frac{Z}{r_2}+\frac{1}{r_{12}}\right|\phi(1)\phi(2)\right\rangle \end{equation}

Applying the linearity of the integral and separating into five integrals:

\begin{eqnarray} W=\underbrace{\left\langle\phi(1)\left|-\frac{1}{2}\nabla_1^2\right|\phi(1)\right\rangle}_{\zeta^2/2} \underbrace{\left\langle \phi(2)| \phi(2)\right\rangle}_{1} + \underbrace{\left\langle\phi(2)\left|-\frac{1}{2}\nabla_2^2\right|\phi(2 )\right\rangle}_{\zeta^2/2} \underbrace{\left\langle \phi(1)|\phi(1)\right\rangle}_{1} + \nonumber\\ \underbrace{ \left\langle\phi(1)\left|-\frac{-Z}{r_1}\right|\phi(1)\right\rangl e}_{-Z\zeta} \underbrace{\left\langle \phi(2)|\phi(2)\right\rangle}_{1} + \underbrace{\left\langle\phi(2)\left|-\frac{-Z}{r_2}\right|\phi(2)\right\rangle}_{-Z\zeta} \underbrace{\left\langle \phi(1)|\phi(1) \right\rangle}_{1}+\nonumber\\ \underbrace{\left\langle\phi(1)\phi(2)\left|-\frac{1}{r_{12}}\right|\phi(1)\phi(2)\right\rangle}_{5\zeta/8} \end{eqnarray}

Adding the different contributions to the variational energy, we get:

\begin{equation} W=\zeta^2- 2Z\zeta+\frac{5\zeta}{8} \end{equation}

The minimum energy, $W{min}$, is obtained by differentiating W with respect to the variational parameter and setting it equal to zero. In this way we obtain the optimal parameter, ($\zeta_{opt}$), which substituted in W gives $W_{min}$.

\begin{equation} \frac{dW}{d\zeta}=0=2\zeta-2Z+5/8 \end{equation} Solving for the optimal parameter,

\begin{equation} \zeta_{op}=Z- 5/16 \end{equation}

Taking the optimal parameter to the variational energy, we get: \begin{equation} W_{min}=\left(Z-5/16\right)^2-2Z\left(Z- 5/16\right)+5/8\left(Z-5/16\right) \end{equation}

Substituting Z=2 into the energy, we get a value (in electron volts) of, $W_{min }=-77\;eV$, which compared to the experimental value of 79 eV does not yield an error of 1.9$\%$.

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