c15 JWBS043-Rogers September 13, 2010 11:28 Printer Name: Yet to Come
240 EARLY QUANTUM THEORY: A SUMMARYThe antisymmetric wave function is the same as an expanded 2×2 determinant:
∣
∣
∣
∣φα(1) φβ(1)
φα(2) φβ(2)∣
∣
∣
∣=φα(1)φβ(2)−φα(2)φβ(1)To impose the condition of antisymmetry on two otherwise identical orbitals, the
simple linear combinations of one-electron wave functions must be replaced by a
single-determinantwave function of the form for the two-electron case (e.g., the
helium atom):ψ(1,2)=∣
∣
∣
∣
φα(1) φβ(1)
φα(2) φβ(2)∣
∣
∣
∣
Generalizing to several electrons led to more complicateddeterminantalwave
functions. The use ofantisymmetrized orbitals, as they were called, leads to the
Hartree–Fock computational method. This method was applied to atoms and was
tentatively applied to molecules larger than H 2.15.4 THE HARTREE INDEPENDENT ELECTRON METHOD
One can treat the electrons of a many-electron atom as though they were entirely
independent of one another moving under the influence of the nuclear chargeZ,
whereZis the atomic number, an integer greater than 1. This amounts to assuming
higher-levelhydrogen-like orbitalswith principal quantum numbersn=Z/2 because
each filled orbital contains 2 independent electrons. The resulting atomic energy will
be wrong because electrons are, in fact, not independent.
This crude model can be substantially improved by taking aneffective nuclear
chargesmaller thanZ, motivated by our intuitive understanding that, no matter what
the overall electronic probability distribution may be, there is some probability that
some electrons will carry a negative charge between the nucleus and the other elec-
trons, thereby reducing the influence of the nuclear charge on them. Any designated
electron will beshieldedfrom the nuclear charge by the other electrons. Simply
plugging in arbitrary numbers for the effective nuclear chargeZeff<Zgives an im-
provement in agreement of the calculated energy of the system with the experimental
value. It turns out thatZeff= 1 .6 is a pretty good guess in the example of helium, a
two-electron problem with a nuclear charge of 2.0. Unfortunately, substitution of ar-
bitrary numbers is theoretically meaningless. One would prefer to be able to calculate
a radial electron probability distribution resulting in a theoretically derived energy
andZefffor any single electron from a calculated energy distribution of all the others.
In treating helium, we shall begin with the reasonable first guess that each of its
two independent one-electron wave functions resembles that of the hydrogen atom:∼=ψ 1 ψ 2