This preview shows pages 1–3. Sign up to view the full content.
Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2009
Laura Gagliardi
Answers to Homework Set 3
From lecture 9: (a) Rewrite eq. 96 in the standard form of an eigenvalue equation. (b)
Show that by proper choice of
a
, the function
e
"
ar
2
is an eigenfunction of the operator
d
2
dr
2
"
qr
2
#
$
%
&
’
(
where
q
is a constant. (c)
Show how you can use the results from parts (a) and (b) to
determine the energy of the ground state of the QMHO. (d)
How do you know that the
eigenfunction corresponds to the ground state?
(a)
Eq. 96 can be rewritten as
d
2
dr
2
"
#
2
r
2
$
%
&
’
(
)
*
r
( ) =
"
+
*
r
( )
which is in standard form where the eigenvalue is negative
α
.
(b) To show that
e
"
ar
2
is an eigenfunction with proper choice of
a
we require
d
2
dr
2
"
qr
2
#
$
%
&
’
(
e
"
ar
2
=
ze
"
ar
2
Evaluating the l.h.s. we have
d
2
dr
2
"
qr
2
#
$
%
&
’
(
e
"
ar
2
=
"
2
ae
"
ar
2
+
4
a
2
r
2
e
"
ar
2
"
qr
2
e
"
ar
2
=
"
2
a
+
q
"
4
a
2
( )
r
2
[ ]
e
"
ar
2
and for the prefactor on the r.h.s. to be a constant (so as to satisfy the eigenvalue
condition) it must be true that
a
is
/2
in which case the eigenvalue will be –2
a
, which
is simply
"
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentHW32
(c)
Given that eq. 96 was rewritten as above, it is clear that
q
corresponds to
β
2
and the
eigenvalue
a
will then simply be –
β
. From eq. 95 we know
β
and thus we have
a
=
"
μ
k
h
but
a
is also –
α
based on our rewriting of eq. 96 above. So, noting from eq. 95 what
α
is defined to be, we now have
"
2
E
h
2
=
"
k
h
if we now solve for
E
we have
E
=
h
2
k
which by comparison to eq. 922 is indeed the correct result for
n
= 0 (remember that
h
is
h
/2
π
).
(d)
The trial function we have been examining is a gaussian function without nodes (i.e.,
it is nonnegative everywhere) so it must be the ground state wave function. Note that this
process above could be repeated to get energies for the various excited states by including
powers of
r
(that would introduce nodes) and enforcing orthogonality, but it would get to
be rather painful rather quickly…
From lecture 10: HBr and DBr (H =
1
H, D =
2
H, and Br =
79
Br) are observed to absorb
infrared radiation at 2439 and 1750 cm
–1
, respectively. Answer the following questions:
(a)
In the ground vibrational state, which has more average kinetic energy, HBr or DBr?
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 Staff
 Physical chemistry, pH

Click to edit the document details