792 19 The Electronic States of Atoms. III. Higher-Order Approximations
conjugates because our function is real. We assume that the nucleus is stationary, so
the correct Hamiltonian operator is:Ĥ−h ̄2
2 m(∇^21 +∇ 22 )+
1
4 πε 0(
−
Ze^2
r 2−
Ze^2
r 1−
e^2
r 12)
(19.1-4)
where the number of protons in the nucleus is denoted byZ. We will setZequal to 2
at the end of the calculation.
Since the helium-atom Hamiltonian operator is independent of the spin coordinates,
the integral over the spin coordinates can be factored from the space coordinate integra-
tion. Because of the normalization and orthogonality of the spin functions, integration
over the spin coordinates gives a factor of 2, which cancels the normalizing constant
1/2. We could have omitted the spin factor and the spin integration from the beginning.
We now haveW
∫
ψ 100 (1)ψ 100 (2)(
ĤHL(1)+ĤHL(2)+ e2
4 πε 0 r 12)
ψ 100 (1)ψ 100 (2)d^3 r 1 d^3 r 2 (19.1-5)Exercise 19.1
Show that the spin integration in Eq. (19.1-3) leads to a factor equal to 2.
TheĤHL(1) andĤHL(2) terms in the Hamiltonian operator give ground-state energy eigen-
values for a hydrogen-like atom after operating and integrating.Exercise 19.2
Show that thêHHL(1) term in Eq. (19.1-5) yields a contribution toWequal toE 1 (HL) and that
thêHHL(2) term yields an equal contribution.We now haveW 2 E 1 (HL)+
∫
ψ 100 (1)∗ψ 100 (2)∗(
e^2
4 πε 0 r 12)
ψ 100 (1)ψ 100 (2)dq (19.1-6)Evaluation of the integral in this equation is tedious and we give only the result:^2W 2 E 1 (HL)+
5 Ze^2
8(4πε 0 a)2 E 1 (HL)−
5 Z
8
〈V〉H(1s) (19.1-7)where〈V〉H(1s)is the expectation value of the potential energy for the hydrogen atom
(not hydrogen-like) in its ground state. From Eq. (17.3-19)W− 2
Z^2 e^2
2(4πε 0 a)+
5 Ze^2
8(4πε 0 a)− 108 .8eV+ 34 .0eV− 74 .8 eV (19.1-8)The numerical values in Eq. (19.1-8) are forZ2. The hydrogen-like orbital
energies are proportional toZ^2 while the term due to electron–electron repulsion is(^2) I. N. Levine,Quantum Chemistry, 5th ed., Prentice-Hall, Englewood Cliffs, NJ, 2000, p. 254ff.