Helium Atom

Do a variational calculation to estimate the ground state energy of the electrons in the helium atom. The Hamiltonian for two electrons, assuming the nucleus is fixed, is

(i)

Assume a wave function of the form

(ii)

where a₀ is the Bohr radius, α is the variational parameter, and χ_s is the spin state of the two electrons.

SOLUTION

In the ground state of the two-electron system, both orbitals are in 1s states. So the spin state must be a singlet χ_s with S = 0. The spin plays no role in the minimization procedure, except for causing the orbital state to have even parity under the interchange of spatial coordinates. The two-electron wave function can be written as the product of the two orbital parts times the spin part:

(1)

(2)

where a₀ is the Bohr radius and α is the variational parameter. The orbitals ѱ(r) are normalized to unity. Each electron has kinetic (K) and potential (U) energy terms which can be evaluated:

(3)

(4)

where E_R = 13.6 eV is the Rydberg energy. The difficult integral is that due to the electron–electron interaction, which we call V:

(5)

First we must do the angular integral over the denominator. If r_m is the larger of r₁ and r₂ then the integral over a 4π solid angle gives

(6)

In the second integral we have set x = 2αr₁/a₀ and y = 2αr₂/a₀, which makes the integrals dimensionless. Then we have split the y-integral into two parts, depending on whether y is smaller or greater than x. The first has a factor 1/x from the angular integrals, and the second has a factor 1/y. One can exchange the order of integration in one of the integrals and demonstrate that it is identical to the other. We evaluate only one and multiply the result by 2:

(7)

(8)

(9)

This completes the integrals. The total ground state energy E(α) in Rydbergs is

(10)

We find the minimum energy by varying α. Denote by α₀ the value of α at which E(α) is a minimum. Setting to zero the derivative of E(α) with respect to α yields the result α₀= 27/16. The ground state energy is

(11)