[next] [prev] [up] [top]

2.3 Variational methods: the helium atom

Summary: The most famous example of the variational principle is the ground state of the two-electron helium atom.

If we could switch off the interactions between the electrons, we would know what the ground state of the helium atom would be: Ψ(r1,r2) = ϕ100Z=2(r 1)ϕ100Z=2(r 2), where ϕnlm Z is a single-particle wave function of the hydrogenic atom with nuclear charge Z. For the ground state n = 1 and l = m = 0 (spherical symmetry). The energy of the two electrons would be 2Z2 ERy = 108.8 eV. But the experimental energy is only 78.6 eV (ie it takes 78.6 eV to fully ionise neutral helium). The difference is obviously due to the fact that the electrons repel one another.

The full Hamiltonian (ignoring the motion of the proton - a good approximation for the accuracy to which we will be working) is

2 2m(12 + 22) 2cα 1 r1 + 1 r2 + cα 1 r1 r2

where 1 2 involves differentiation with respect to the components of r1, and α = e2 (4πε 0c) 1137. (See here for a note on units in EM.)

A really simple guess at a trial wave function for this problem would just be Ψ(r 1 , r 2 ) as written above. The expectation value of the repulsive interaction term is (5Z4)ERy giving a total energy of 74.8 eV. (Gasiorowicz demonstrates the integral, as do Fitzpatrick and Branson.)

It turns out we can do even better if we use the atomic number Z in the wave function Ψ as a variationalparameter (that in the Hamiltonian, of course, must be left at 2). The best value turns out to be Z = 2716 and that gives a better upper bound of 77.5 eV – just slightly higher than the experimental value. (Watch the sign – we get an lower bound for the ionization energy.) This effective nuclear charge of less than 2 presumably reflects the fact that to some extent each electron screens the nuclear charge from the other.

[next] [prev] [up] [top]