Biomedical Engineering Reference
In-Depth Information
infinity as the two electrons approach each other). It is shown in the classic texts such as
Eyring
et al.
(1944) that
5
Z
8
e
2
4πε
0
a
0
ε
(1)
=
ε
(1)
hc
0
=
5
Z
8
hc
0
e
2
4πε
0
a
0
=
274 343 cm
−
1
and my revised estimate of the first ionization energy is now 164 606 cm
−
1
, in better
agreement with experiment.
14.5 Variation Method
Another, completely different method of finding approximate solutions to the Schrödinger
equation is based on the following theorem.
Theorem
If the lowest eigenvalue of a Hamiltonian
H
is ε and Φ is a function with the correct
boundary conditions, then
Φ
∗
H
Φ dτ
Φ
∗
Φ dτ
≥
ε
Once again, the proof is given in all the classic texts such as Eyring
et al.
(1944). I can
illustrate the use of this technique by reference to the helium atom. The spatial part of the
zero-order wavefunction is
exp
Z
3
π
a
0
Z
(
r
1
+
r
2
)
Ψ (
r
1
,
r
2
)
=
−
a
0
where
Z
is the atomic number (2 in this case). We now allow for the possibility that one
electron partially screens the nucleus and so the second electron sees a reduced nuclear
charge
Z
. This suggests that we use a trial wavefunction that comprises a product of
modified helium 1s orbitals with effective nuclear charge
Z
e
exp
(
Z
)
3
π
a
0
Z
(
r
1
+
r
2
)
Φ
(
r
1
,
r
2
)
=
−
a
0
We would expect
Z
to lie between 1 and 2 and we look for the value of
Z
that makes
the
variational integral
Φ (
r
1
,
r
2
)
g
(
r
1
,
r
2
)
Φ (
r
1
,
r
2
) dτ
1
dτ
2
Φ
2
(
r
1
,
r
2
) dτ
1
dτ
2
h
(1)
(
r
1
)
+
h
(1)
(
r
2
)
+ˆ
ε
=
(14.16)
a minimum. I have dropped the complex conjugate signs * because of the real nature of the
wavefunction. The denominator is
Φ
2
(
r
1
,
r
2
) dτ
1
dτ
2
=
1
s
2
(
r
1
)dτ
1
1
s
2
(
r
2
) dτ
2
=
1
