Biomedical Engineering Reference
In-Depth Information
In order that Equation (14.11) may be true for all values of λ, the coefficients of λ on
either side of the equation must be equal. Equating and rearranging we find
H
(0)
Ψ
(0)
i
ε
(0
i
Ψ
(0)
=
i
Ψ
(1)
i
ε
(1)
i
−
H
(1)
Ψ
(0)
i
H
(0)
ε
(0)
i
−
=
(14.12)
Ψ
(2)
i
H
(0)
−
H
(1)
Ψ
(1)
i
ε
(0)
i
ε
(2
i
Ψ
(0)
ε
(1
i
Ψ
(1)
−
=
+
i
i
Solution of these equations gives
λ
2
j
H
(1)
ij
H
(1)
ji
ε
(0)
i
λ
H
(1)
ii
ε
i
=
+
+
+···
ε
(0)
i
ε
(0)
j
−
=
i
(14.13)
λ
j
H
(
1
)
ji
Ψ
(0)
i
Ψ
(0)
j
Ψ
i
=
+
+···
ε
(0)
i
ε
(0)
j
−
=
i
I have used the following shorthand for the integrals:
H
(1)
Ψ
(0
j
dτ
H
(1)
ij
Ψ
(0)
i
=
(14.14)
The first-order correction to the energy can therefore be calculated from the unperturbed
wavefunction and the perturbing Hamiltonian:
H
(1)
ii
ε
(1)
i
=
Ψ
(
0
)
i
H
(1)
Ψ
(
0
i
dτ
=
On the other hand, the second-order correction to the energy requires knowledge of the
remaining states
Ψ
(0)
i
H
(1)
Ψ
(0
j
dτ
2
ε
(0)
i
ε
(2)
i
=
(14.15)
ε
(0)
j
−
j
=
i
If we return to the helium atom where the zero-order problem is two noninteracting
atomic electrons, the zero-order wavefunctions and energies are shown in Table 14.1. The
perturbation is the Coulomb repulsion between the two electrons
1
r
12
and the first-order correction to the ground-state energy is
e
2
4πε
0
H
(1)
=
Ψ
1
e
2
4πε
0
Ψ
1
dτ
1
r
12
ε
(1)
=
where Ψ
1
is given in Table 14.1.
Evaluation of the integral is far from easy, since it involves the coordinates of both
electrons (it is therefore a six-dimensional integral), and it has a singularity (it tends to
