Chemistry Reference
In-Depth Information
This very powerful idea is valid for both classical and quantum systems. In cases described
by quantum mechanics it only works when we have correctly identified a stationary state
of the system, and so can be a useful test of results.
Equation (A10.9) shows that
1
2
T
=−
U
, which is the correct result for the Coulomb
power law (
n
=−
1). That the virial theorem is obeyed is also confirmation that the expo-
nent in the
|
s
function (Equation (A10.3)) has the optimum form: we are correct to use
−
−
ζ
ζ
=
exp(
1, as will be
checked in Problem A10.1. The electron is distributed as a function of
r
, so the decay
constant affects the averaging process and so is important in calculating the expectation
values of the energies.
The virial theorem links the form of the potential the electron experiences to the balance
between average kinetic and potential energies in the ground state. Figure A10.1 shows that
the electron is more or less confined to a region 2 bohr from the centre of the atom, where
its kinetic energy is positive. If the potential field for the electron were to change (e.g. by
an increase of the nuclear charge), then both
r
) rather than some other radial decay, such as exp(
r
) with
T
and
U
would alter:
U
would become
more negative due to the increased Coulomb interaction and
would become more
positive because the wavefunction would be more strongly confined by the new potential.
However, the virial theorem says that the balance would still give
T
in the new
ground state and the shape of the wavefunction would have to alter to ensure this.
The sum of
T
=−
1
2
U
0.5 Ha, so the system is more stable
with the electron in the 1s orbital than with the electron and proton separated. Figure A10.1
indicates that at the Bohr radius the total energy has its minimum value, but the total
required for inclusion in the bond formation energy has to take into account the integral of
the energy over the entire wavefunction, as indicated by the use of expectation values in
Equation (A10.8).
T
+
U
from Equation (A10.9) is
−
Problem A10.1:
We can use the virial theorem to obtain the exponential decay constant
in the 1s orbital function. To do this, consider the general form of a trial
s
-function
with a decay constant
1 gives a wavefunc-
tion that satisfies the virial theorem. The Greek letter 'zeta' is widely used to describe
the decay constants for basis sets in computational chemistry, and so we have adopted
that convention here. The normalized trial function
ζ
and then we try to prove that only
ζ
=
|
s
tr
will be
s
tr
=
ζ
2
√
π
3
/
|
exp(
−
ζ
r
)
(A10.11)
1. Confirm that
.
2. Using the approach laid out in Section A9.8, show that the expectation value for the
potential energy is
|
s
tr
is normalized for any choice of
ζ
1
r
U
=−
s
tr
|
|
s
tr
=−
ζ
(A10.12)
You will need to make use of the standard integrals from Table A9.4 for the radial
part of Equation (A10.12), and remember that the integration over
θ
and
φ
will give
a factor of 4
.
π
