Chemistry Reference
In-Depth Information
This expression is reasonable: if we associate
γ
with the wavenumber
k
and 1
a
0
with the spatial extent of the trial function then the first term
on the right-hand side can be interpreted as the estimated kinetic energy,
and the second as the estimated potential energy:
/γ
=
2
k
2
2
m
−
e
2
=
E
(A.8)
4
πε
0
a
0
where
a
0
is referred to as the Bohr radius. To find the minimum esti-
mated ground state energy, we calculate that d
E
/
d
γ
=
0 when
e
2
m
2
γ
=
/(
4
πε
)
, with the Bohr radius
a
0
=
1
/γ
=
0.53Å. Substitut-
0
ing the calculated
value in eq. (A.7) gives the estimated electron ground
state energy in the hydrogen atom as
γ
me
4
8
E
0
=−
=−
13.6 eV
(A.9)
2
0
h
2
ε
In this instance, the calculated minimum variational energy is equal to the
ground state energy calculated by solving Schrödinger's equation exactly.
This example demonstrates that the variational method can work very
effectively given a suitable choice of starting function, particularly if a free
parameter is included in the function. We also observe, as derived in
Chapter 1, that the calculated variational energy,
≥
E
E
0
, where
E
0
is the
true ground state energy, and that in fact
E
=
E
0
onlywhen the variational
trial function,
f
(γ
,
r
)
equals the true ground state wavefunction,
ψ
(
r
)
.
0
