Concise Physical Chemistry

c15 JWBS043-Rogers September 13, 2010 11:28 Printer Name: Yet to Come

THE HARTREE INDEPENDENT ELECTRON METHOD 241

and from the Schr ̈odinger equation,

Hˆ=E

we expect to find

hˆψi=εijψj

Mathematically, the operatorhˆis a 2×2 transformation matrix that transforms the

vectorψiintoεijψj. This Hamiltonian operator is written in lowercase to show that it

is approximate and not the exact operatorHˆ. In the case of helium with two electrons,

the state vector|ψ〉can be expressed as a two-element column vector

|ψ〉=

(

ψ 1

ψ 2

)

We can improve uponψandεby successive approximations, using aniterative

procedure. We shall assume that we know where one electron is (we don’t) and

calculate its shielding effect on the other electron. We then calculate the probability

density function for the location of the second electron as influenced by our initial

guess as to its shielding from the first. With this information we can improve upon our

initial guess as to the location of the first electron because we now have a probable

location of the second electron. This iterative process can be continued until further

iteration results in no further lowering of the calculated energy, whereupon we have

the best estimate of the energy and the electron distribution we can get from the

procedure.

The Hamiltonian operatorhˆincludes the kinetic energy operator of each electroni

and the attractive electrostatic potential energy exerted on the electron by the nucleus

hˆi=Tˆi−e

2

ri^2

(i= 1 ,2)

The potential energy of one electron, call it electron 1, is influenced by the other,

call it electron 2, according to their chargeeand the distance between them,r 12.

By the uncertainty principle we do not know where electron 2 is, but we do have a

probability densityfunction|ψ(r 2 )|^2 governing its location in a volume element of

spacedτ 2. Hence we have a most probablecharge distributionfunctione|ψ 2 (r 2 )|^2.

The probable potential energy of electron 1 shielded from the nucleus by electron 2

is, in suitable units, the product of the chargesedivided by the distance between them

V 1 =e

∫

e|ψ 2 (r 2 )|^2

r 12

dτ 2 =e^2

∫

|ψ 2 (r 2 )|^2

r 12

dτ 2

The integration is over the space occupied by electron 2 because we cannot locate it

as a point charge.