Chemistry Reference
In-Depth Information
Here,
A
and
are constants that give us some flexibility in defining the function
y
.This
property gives us a promising lead, because the second derivative of
y
is just a number
times
y
. However, such a simple exponential of
x
for
β
in Equation (A6.15) will not work,
because
x
can take on both positive and negative values (extension and compression of
the bond) and the boundary conditions require
ψ
to tend to zero at either extreme. An
exponential function does tend to zero at large negative
x
, but it increases indefinitely for
positive
x
.
An exponential-type function that does have the correct behaviour for the boundary
conditions is the Gaussian:
ψ
N
exp
−
β
x
2
2
ψ
tr
=
(A6.17)
The subscript 'tr' is added here to indicate that this is a 'trial' wavefunction, i.e. it has a
promising functional form, but there is some way to go yet. The constants
N
and
control
the height and width respectively of the Gaussian, and we need to see if there are values
of these constants that allow
β
ψ
tr
to satisfy Equation (A6.15).
ψ
tr
in Equation (A6.15), its second derivative is needed. Here, the rules for
differentiating a function of a function can help, so that if we write
To use
=
β
x
2
2
u
(A6.18)
then we can take the derivative of
ψ
tr
as follows:
d
ψ
tr
d
x
=
d
u
d
x
d
d
u
N
exp(
−
u
)
=−
N
β
x
exp(
−
u
)
=−
x
βψ
tr
(A6.19)
The second derivative can then be obtained:
d
2
ψ
tr
d
x
2
d
d
x
(
d
ψ
tr
d
x
=
=
−
x
βψ
tr
)
=−
βψ
tr
−
x
β
(
β
2
x
2
−
β
)
ψ
tr
(A6.20)
The left-hand side of Equation (A6.15) now becomes
d
2
ψ
tr
d
x
2
2
x
2
2
x
2
2
x
2
−
α
ψ
tr
=
(
β
−
β
)
ψ
tr
−
α
ψ
tr
(A6.21)
The right-hand side of Equation (A6.15) contains only numbers multiplying
ψ
tr
, and so for
this solution to work we must not have any
x
2
terms remaining. The only way to achieve
this is to set
β
=
α
(A6.22)
With this value for
, our trial wavefunction has satisfied the equation because the oper-
ators on the left-hand side have produced the trial wavefunction multiplied by a constant,
and Equation (A6.15) becomes
β
2
μ
m
E
−
αψ
tr
=−
ψ
tr
(A6.23)
2
