Chemistry Reference
In-Depth Information
Following the motion of the oscillating bond in a stationary state, as we did for the
classical model, no longer makes any sense in a wavefunction picture. All we can look
for is the probability that an observation of the bond will see a particular extension. The
probability of finding any particular value of
x
should be expected to depend on the value
sought. For example, extensions of the bond which give a potential energy greater than the
total should be very unlikely. The wavefunction
gives the probability that the bond is
between
x
and
x
+d
x
as the square of the wavefunction magnitude,
ψ
|
ψ
|
(
x
)
2
d
x
, and so we
ψ
must expect
to be a function of
x
that approaches zero at large
x
.
To obtain
ψ
we solve the time-independent Schrödinger equation for the system:
2
kx
2
2
d
2
d
x
2
+
1
−
ψ
=
E
ψ
(A6.13)
2
μ
m
This equation is another statement that the total energy
E
is the sum of the kinetic energy
and potential energy, but in this case there is no time dependence. These two contri-
butions must be drawn from the wavefunction, and Schrödinger demonstrated that the
kinetic energy should be a function of its second derivative with respect to position, the
first term on the left of Equation (A6.13). A mathematical tool for extracting informa-
tion from the wavefunction is called an operator, so this first term is the kinetic energy
operator.
The potential energy depends on the particular system. For the harmonic oscillator we
can use the same form as the classical potential, Equation (A6.1), in Equation (A6.13).
This is now the potential energy operator, since it is 'operating' on the wavefunction. The
inclusion of the wavefunction will allow us to average the potential energy across all bond
extension values weighted according to their probability.
In Equation (A6.13),
is related to the Planck constant via
h
2
=
(A6.14)
π
It is useful to rearrange Equation (A6.13) in such a way that the second differential in
x
has no multiplying coefficient and to group the constants on the left-hand side into a single
symbol:
d
2
ψ
d
x
2
2
μ
m
E
=
μ
m
k
−
α
2
x
2
ψ
=−
ψ
where
α
2
(A6.15)
2
2
Solutions of Equation (A6.15) will require us to 'try out' or trial functional forms for
ψ
which have a hope of balancing the left and right sides; on taking the second derivative and
simplifying the left-hand side we must end up with just a number multiplying
. Our trial
function must also conform to any boundary conditions of the problem
.
Here, we know
that at large positive or negative
x
the wavefunction must tend to zero, and so this is the
required boundary condition.
We can make a start using a basic property of the exponential function; the derivative is
the same as the function itself:
ψ
d
y
d
x
=
d
2
y
d
x
2
=
β
β
β
=
β
=
β
2
y
if
y
A
exp(
x
)
then
A
exp(
x
)
y
and
(A6.16)
