9-2
If we consider the vibrational motion of the molecule, we expect there to be
solutions to the time-independent Schrödinger equation that will correspond to stationary
vibrational states of the system behaving as a quantum mechanical system. That is, there
will be a wave function from which the probability of finding any particular range of
bond lengths can be computed. To find such stationary states, we need to solve
!
h
2
2
μ
d
2
dr
2
+
V r
( )
"
#
$
$
%
&
’
’
(
r
( ) =
E
(
r
( )
(9-1)
where
μ
is some sort of effective mass for the vibration (called the “reduced mass”; it’s
not simply the mass of atom A + atom B:
that mass would be appropriate if we were
thinking of the molecule AB as a free particle translating through space, but vibration is a
different kind of motion, and we’ll get back to
μ
later), the one dimension we care about
is the bond length
r
(we could as easily use
x
in our notation, but
r
is more conventional),
and the potential energy
V
is just the curve in the above figure.
Later in the class, we will get to how one might use quantum mechanics to
predict
the bonding potential, but for now let’s just accept that it is a nice physical observable
that can be measured. However, to plug it into our eq. 9-1, we need some convenient
mathematical representation of that curve. We could somehow develop a spline fit or
other elegant form, but that might make the solution of the differential equation quite
difficult. For the moment, let’s keep things really simple. One way to represent
any
function is from knowledge of that function’s value, and the value of its derivatives, at a
single point. We may then use a so-called Taylor expansion to compute the function’s
value at any other point. If we choose as our function our bond potential V, our
coordinate as
r
, and our known point as the equilibrium bond distance, we have
V r
( ) =
V r
eq
( )
+
dV
dr
r
=
r
eq
r
!
r
( )
+
1
2!
d
2
V
dr
2
r
=
r
eq
r
!
r
( )
2
+
1
3!
d
3
V
dr
3
r
=
r
eq
r
!
r
( )
3
+
L
(9-2)
For the sake of simplicity, we now do several things. First, define the zero of energy to be
V
(
r
eq
). That makes term 1 on the r.h.s. equal to zero. Next, notice that
r
eq
defines the
local minimum on the bond potential curve. Since it is a critical point on the curve, we
know that the first derivative of
V
must be zero at that point. So, now the second term on
the r.h.s. is zero. The next term, the quadratic term, is not zero. So, let’s throw away all