Take a trial function for the helium atom as \(\psi=\) \(\psi(1) \psi(2),\) with \(\psi(1)=\left(\zeta^{3} / \pi\right)^{1 / 2} \mathrm{e}^{-\zeta r_{1}}\) and \(\psi(2)=\left(\zeta^{3} / \pi\right)^{1 / 2} \mathrm{e}^{-\zeta_{2}}, \zeta\) being a parameter, and find the best ground-state energy for a function of this form, and the corresponding value of \(\zeta\). Calculate the first and second ionization energies. Hint. Use the variation theorem. All the integrals are standard; the electron repulsion term is calculated in Example 7.2 Interpret \(Z\) in terms of a shielding constant. The experimental ionization energies are \(24.58 \mathrm{eV}\) and \(54.40 \mathrm{eV}\).

The ground-state energy of a helium atom is given by the expectation value of the Hamiltonian. Note that \(\zeta\) is a parameter representing the shielding constant. This is given by \(\langle H \rangle = \langle \psi | H | \psi \rangle. \) The Hamiltonian of helium atom in atomic units \(H = -\frac{1}{2}\Delta _1 -\frac{1}{2}\Delta _2 - \frac{2}{|\vec{r_1}|} - \frac{2}{|\vec{r_2}|} + \frac{1}{|\vec{r_2}-\vec{r_1}|}\) where the Laplacians \(\Delta _1, \Delta _2\) represent the kinetic energy of electrons 1 and 2, the terms \(-\frac{2}{|\vec{r_1}|}, -\frac{2}{|\vec{r_2}|}\) represent the potential energy of the interaction of electrons with the nucleus, and \(\frac{1}{|\vec{r_2}-\vec{r_1}|}\) represents electron-electron repulsion. Substitute the trial function \(\psi(1) \psi(2)\) into the equation, and calculate the integral to obtain the ground-state energy.

Apply the variation theorem, which states that for a normalized wave function \(\psi\), the expectation value for the energy is always greater than the true ground state energy. Differentiate the expression of \(\langle H \rangle\) with respect to \(\zeta\) and set it as 0 to find the minimum value of \(\langle H \rangle\), which results in the best approximation to the ground state energy. This gives the optimal value of \(\zeta\) and the best estimation of the ground-state energy.

The first ionization energy is the energy difference between the ground state of He atom and that of He+ ion, \(\Delta E = E_{He+} - E_{He}\). The second ionization energy is the difference between the energy of the He+ ion and that of He++ (which is a bare nucleus), \(\Delta E = E_{He++} - E_{He+}\). Calculate these energy differences.

In terms of atomic structure, \(\zeta\) can be interpreted as the effective nuclear charge experienced by an electron; it's less than the actual nuclear charge because of the shielding or screening effect by other electrons. The greater the value of \(\zeta\) (up to the number of protons in the nucleus), the stronger the attraction between the nucleus and the electron, leading to a lower ground-state energy.