Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2010
Laura Gagliardi
Lecture 7, September 22, 2010
Solved Homework
We are given that
A
is a Hermitian operator such that
A
φ
1
=
a
φ
1
,
A
φ
2
=
b
φ
2
,
A
φ
3
=
b
φ
3
, and
A
φ
4
=
c
φ
4
,
a
≠
b
≠
c
, and
Z
is some other operator for which [
A
,
Z
] = 0. We may
then state whether the following integrals are definitely zero, or may be nonzero
(a)
<
φ
1

φ
4
>
Equals zero, because nondegenerate eigenfunctions of a
Hermitian operator are orthogonal.
(b)
<
φ
2

φ
4
>
Equals zero, because nondegenerate eigenfunctions of a
Hermitian operator are orthogonal.
(c)
<
φ
2

φ
3
>
May or may not be zero. Degenerate eigenfunctions are not
necessarily orthogonal, even though orthogonal functions
having identical eigenvalues
can
be constructed from linear
combinations of them.
(d)
<
φ
3

φ
4
>
Equals zero, because nondegenerate eigenfunctions of a
Hermitian operator are orthogonal.
(e)
<
φ
1

A

φ
3
>
Equals zero. <
φ
1

A

φ
3
> = <
φ
1

b
φ
3
> =
b
<
φ
1

φ
3
> and we
know <
φ
1

φ
3
> = 0 because nondegenerate eigenfunctions of
a Hermitian operator are orthogonal.
(f)
<
φ
2

A

φ
3
>
May or may not be zero. <
φ
2

A

φ
3
> = <
φ
2

b
φ
3
> =
b
<
φ
2

φ
3
> and we know from (c) above that this overlap
integral need not be zero.
(g)
<
φ
1

Z

φ
4
>
Equals zero. If the commutator of two operators is zero, then
they share common eigenfunctions and we were told that the
functions
φ
belong to this set. Thus, <
φ
1

Z

φ
4
> =
<
φ
1

d
φ
4
> =
d
<
φ
1

φ
4
> where
d
is an eigenvalue of
Z
and
we know from (a) above that this overlap integral is zero.
(h)
<
φ
2

Z

φ
3
>
May or may not be zero. Continuing from our above logic,
<
φ
2

Z

φ
3
> = <
φ
2

e
φ
3
> =
e
<
φ
2

φ
3
> and we know from
(c) above that this overlap integral need not be zero.
(i)
<
φ
2
+
φ
3

Z

φ
2
–
φ
3
>
May or may not be zero. This one is a bit tricky. If
we expand the integral, we get <
φ
2

Z

φ
2
> – <
φ
2

Z

φ
3
> +
<
φ
3

Z

φ
2
> – <
φ
3

Z

φ
3
>. The first question is, just
because
φ
2
and
φ
3
are degenerate for the Hamiltonian, are they
necessarily degenerate for the operator
Z
? The answer is, yes.
We can prove this by noting that, by the turnover rule,
<
φ
2

Z

φ
3
> = <
Z
φ
2

φ
3
> =
f
<
φ
2

φ
3
> where
f
is the
appropriate eigenvalue for
Z
. But, we already showed in (h)
above that <
φ
2

Z

φ
3
> =
e
<
φ
2

φ
3
>, so
e
must be equal to
f
.
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72
So, if we pull all of the eigenvalues out from our sum of 4
integrals we have
e
<
φ
2

φ
2
> –
e
<
φ
2

φ
3
> +
e
<
φ
3

φ
2
> –
e
<
φ
3

φ
3
>. Assuming that
φ
2
and
φ
3
are normalized, the first
and last terms cancel as
e
–
e
. However, since
φ
2
and
φ
3
are
degenerate, they are not necessarily orthogonal. Let us say that
the overlap integral <
φ
2

φ
3
> =
c
. Then <
φ
3

φ
2
> =
<
φ
2

φ
3
>* =
c
*. For a complex number,
c
≠
c
* unless
c
is
simply a real number. So, the second and third terms of the
sum do
not
necessarily cancel and we cannot say that the
integral is necessarily zero.
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