Concise Physical Chemistry

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

242 EARLY QUANTUM THEORY: A SUMMARY

What can be said of the interaction between electron 1 with electron 2 can be said

of electron 2 interacting with electron 1:

V 2 =e^2

∫

|ψ 1 (r 1 )|^2

r 21

dτ 1

The operatorhˆi(i= 1 ,2) is the kinetic energy operator for electron 1 or 2:

−
^2

2 me

∇i^2 =

−
^2

2 me

(

∂^2

∂x^2

+

∂^2

∂y^2

+

∂^2

∂z^2

)

plus the (negative) potential energy of attraction between the nucleus and the electron,

−Ze^2 /ri^2 :

hˆi=−

2

2 me

∇ 22 −

Ze^2

ri^2

With these potential and kinetic energies, we can write Schrodinger equations for ̈

electrons 1 and 2:

[

Tˆ 1 +Vˆ 1 (r 2 )

]

ψ 1 (1)=hˆ 1 ψ 1 (1)=ε 1 ψ 1 (1)

and

[

Tˆ 2 +Vˆ 2 (r 1 )

]

ψ 2 (2)=hˆ 2 ψ 2 (2)=ε 2 ψ 2 (2)

With these approximate Hamiltonians, we have two Schrodinger equations that con- ̈

tain the energy of the two electrons in the helium atomεi(i= 1 ,2):

hˆψi=εijψj (i= 1 ,2)

The two equations are integrodifferential equations because they have an integral

as part ofV(r) and second differentials in∇^2. As written, they arecoupledbecause

one equation depends on the solution of the other through the potential energies. In

the first equation,V 1 (r 2 ) contains the electron density of electron 2; in the second

equation,V 2 (r 1 ) contains the electron density of electron 1. When we assume a wave

function (any wave function) weuncouplethese two equations. Though uncoupled

and therefore solvable, the equations will no doubt lead to the wrong answer for

the energiesεibecause the assumedψ(r) is only a guess. Hartree’s guess was

the most reasonable one he could make; he tookψ 1 andψ 2 to be hydrogen-like

orbitals.

AllowingZto approach an effective nuclear chargeZeff,wehavetheHartree

equations. Uncoupling these equations by assuming functionsψ 1 andψ 2 was a