0198506961.pdf

54 Helium

The spherical symmetry of s-electrons means that the charge in the

regionr 1 <r 2 acts like a point charge at the origin, so that

V 12 (r 2 )=

Q(r 2 )

4 π 0 r 2

,

whereQ(r 2 ) is the charge within a radius ofr 2 from the origin, which

(^18) HereQ(∞)=−e. is given by 18
Q(r 2 )=
∫r 2
0
ρ(r 1 )4πr^21 dr 1. (3.22)
The electrostatic energy that arises from the repulsion equals

E 12 =

∫∞

0

V 12 (r 2 )ρ(r 2 )4πr^22 dr 2. (3.23)

For the 1s^2 configuration there is an exactly equal contribution to the

energy fromV 21 (r 1 ), the (partial) potential atr 1 produced by electron 2.

Thus the total energy of the repulsion between the electrons is twice that

(^19) As is usual in calculations of the in- in eqn 3.23. (^19) Using the radial wavefunction for a 1s-electron, we find
teraction between electric charge dis-
tributions, one must be careful to
avoid double counting. This method
of calculation avoids this pitfall, as
shown by the general argument in Ap-
pendix B. An alternative method is
used in Woodgate (1980), Problem 5.5.
J1s 2 =2×
e^2
4 π 0

∫∞

0

{∫r 2

0

1

r 1

4 Z^3 e−(Z/a^0 )2r^1 r^21 dr 1

}

4 Z^3 e−(Z/a^0 )2r^2 r^22 dr 2

=

e^2 / 4 π 0

2 a 0

5

4

Z=(13.6eV)×

5

4

Z. (3.24)

For helium this givesJ1sZ=2 2 =34eV.

3.3.2 Excited states: the direct integral

A1snlconfiguration of helium has an energy close to that of annl-

electron in hydrogen, e.g. in the 1s2p configuration the 2p-electron has

a similar binding energy to then= 2 shell of hydrogen. The obvious

explanation, in Bohr’s model, is that the 2p-electron lies outside the 1s-

orbit so that the inner electron screens the outer one from the full nuclear

charge. Applying an analogous argument to the quantum treatment of

(^20) The effect of the repulsion propor- helium leads to the HamiltonianH=H 0 a+Ha′,where 20
tional to 1/r 12 canbeconsideredin
terms of potentials like that in eqn 3.21
(and Appendix B). The potential at
the position of the outer electronr 2
arising from the charge distribution of
electron 1 accounts for a large por-
tion of the total repulsion:V 12 (r 2 )
e^2 / 4 π 0 r 2 in the region whereρnl(r 2 )
has an appreciable value. Hence it
makes sense to includee^2 / 4 π 0 r 2 in
the zeroth-order HamiltonianH 0 aand
treat the (small) part left over as a per-
turbationH′a.
H 0 a=−

^2

2 m

(

∇^21 +∇^22

)

−

e^2

4 π 0

(

2

r 1

+

1

r 2

)

(3.25)

and

H′a=

e^2

4 π 0

(

1

r 12

−

1

r 2

)

. (3.26)

In the expression forH 0 a, electron 2 experiences the Coulomb attraction

of a charge +1e.InHa′ the subtraction ofe^2 / 4 π 0 r 2 from the mutual

repulsion means that the perturbation tends to zero at a large distance

from the nucleus (which is intuitively reasonable). This decomposition

differs from that in Section 3.1. The different treatment of the two

electrons makes the perturbation theory a little tricky, but Heisenberg