Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2009
Laura Gagliardi
Lecture 18, October 23, 2009
Solved Homework
If the approximate energy is given by
<
H
> = 3
a
4
– 4
a
3
–36
a
2
+ 10
where
a
is a variational parameter, local minima and maxima of the energy will
correspond to stationary points of <
H
> with respect to
a
, i.e., points where
d H
da
=
0
Setting the derivative equal to zero gives
0 = 12
a
3
– 12
a
2
–72
a
which may be simplified as
0 =
a
(
a
2
–
a
– 6) =
a
(
a
–3)(
a
+ 2)
The three solutions to this cubic equation are
a
= –2, 0, and 3. It is a trivial matter to plug
these values back into the expression for the energy and find that the corresponding
values of <
H
> =
E
are –54, 10, and –219. Analysis of the sign of the second derivative of
<
H
> with respect to
a
indicates these values to be a local minimum, a local maximum,
and a local minimum, respectively (although their characters are obvious in this case).
The variational principle states that our lowest energy (–219) will be an upper
bound to the true lowest energy (which would be the groundstate energy), so the highest
energy the system can possibly have in its ground state is –219 (in whatever units
a
is
expressed).
Some Simple Variational Calculations
Last lecture, we considered the variational principle in the context of basis
functions and their uses for molecular calculations. The variational approach provides a
prescription for computing molecular orbitals as linear combinations of atomic orbitals,
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182
where the coefficients of the linear combinations are the variational parameters subject to
optimization. However, to better appreciate some of the concepts associated with the
variational principle, let’s now take a step back and focus on a vastly simpler system with
only a single variational parameter.
Consider the ground state for a particle having mass 1 a.u. in a box of length
L
= 1
a.u. We know the eigenfunction and eigenvalue (in a.u.) for this particleinabox system
exactly, namely
!
1
x
( )
=
2
sin
"
x
(
)
0
#
x
#
1
E
1
=
"
2
2
(181)
Now, let us imagine that we had no idea how to derive the correct eigenfunction.
But, we certainly imagine that the function should have no nodes (it’s the ground state,
after all), and we also know from the boundary conditions it must be zero at both ends of
the box. One very simple choice for a trial function might be
!
x
( )
=
x
1
"
x
(
)
(182)
This function has the correct behavior at
x
= 0 and
x
= 1 and it is everywhere positive in
between. Of course, there’s no variational parameter in
ξ
, so there is nothing there to
optimize. A simple way to introduce such a parameter is to choose instead
!
x
;
a
(
)
=
x
a
1
"
x
(
)
(183)
where
a
is the variational parameter to be optimized. The optimization condition
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 Spring '08
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 Energy, Atomic orbital, wave function, Laura Gagliardi Lecture

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