0198506961.pdf

3.3 Evaluation of the integrals in helium 55

did the calculation as described in Bethe and Salpeter (1957) or Bethe
and Salpeter (1977); he found the direct integral

J1snl=

e^2

4 π 0

∫∫(

1

r 12

−

1

r 2

)

|u1s(1)|^2 |unlm(2)|^2 d^3 r 1 d^3 r 2. (3.27)

This must be evaluated with the appropriate wavefunctions, i.e.uZnlm=1
rather thanuZnlm=2,anduZ1s=2as before.^21 For the excited electronunlm=^21 We have not derived this integral rig-
orously but it has an intuitively reason-
able form.

Rnl(r)Ylm(θ, φ), whereRnl(r) is the radial function forZ=1. Wewrite
the direct integral as

J1snl=

e^2

4 π 0

∫∞

0

∫∞

0

J(r 1 ,r 2 )R^210 (r 1 )R^2 nl(r 2 )r^21 dr 1 r^22 dr 2 , (3.28)

where the angular parts are contained in the function^2222 Y 00 (θ 1 ,φ 1 )=1/

√

4 π.

J(r 1 ,r 2 )=

∫ 2 π

0

∫π

0

∫ 2 π

0

∫π

0

(

1

r 12

−

1

r 2

)

1

4 π

|Ylm(θ 2 ,φ 2 )|^2

×sinθ 1 dθ 1 dφ 1 sinθ 2 dθ 2 dφ 2.

(3.29)

The calculation of this integral requires the expansion of 1/r 12 in terms
of spherical harmonics:^2323 Yk,q∗ (θ 1 ,φ 1 )=(−1)qYk,−q(θ 1 ,φ 1 ).

1

r 12

=

1

r 2

∑∞

k=0

(

r 1

r 2

)k

4 π

2 k+1

∑k

q=−k

Yk,q∗(θ 1 ,φ 1 )Yk,q(θ 2 ,φ 2 ) (3.30)

forr 2 >r 1 (andr 1 ↔r 2 whenr 1 >r 2 ). Only the term fork=0
survives in the integration over angles in eqn 3.29 to give^2424 Whenk = 0 the integral of the func-
tionYk,q∗(θ 1 ,φ 1 )overθ 1 andφ 1 equals
zero.
J(r 1 ,r 2 )=

{

0forr 1 <r 2 ,

1 /r 1 − 1 /r 2 forr 1 >r 2.

Whenr 1 <r 2 the original screening argument applies and eqn 3.25
gives a good description. Whenr 1 >r 2 the appropriate potential is
proportional to− 2 /r 2 − 1 /r 1 andJ(r 1 ,r 2 ) accounts for the difference
between this and− 2 /r 1 − 1 /r 2 used inH 0 a. Thus we find

J1snl=

e^2

4 π 0

∫∞

0

{∫∞

r 2

(

1

r 1

−

1

r 2

)

R^210 (r 1 )r^21 dr 1

}

R^2 nl(r 2 )r 22 dr 2.
(3.31)
Evaluation of this integral for the 1s2p configuration (in Exercise 3.6)
givesJ1s2p=− 2. 8 × 10 −^2 eV—three orders of magnitude smaller than
J1sZ=2 2 in eqn 3.7 (evaluated from eqn 3.24). The unperturbed wavefunc-
tion forZ= 1 has energy equal to that of the corresponding level in
hydrogen and the small negative direct integral accounts for the incom-
pleteness of the screening of thenl-electron by the inner electron.

3.3.3 Excited states: the exchange integral

The exchange integral has the same form as eqn 3.16 but withuZnlm=1
rather thanuZnlm=2(anduZ1s=2as before). Within the spatial wavefunction