Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2010
Laura Gagliardi
Lecture 10, October 1, 2010
Solved Homework
We are asked to find <
x
> and <
x
2
> for the first two harmonic oscillator wave
functions, which are
Ψ
0
x
( ) =
μ
π
1/ 4
e
−
μ
x
2
/ 2
and
Ψ
1
x
( )
=
k
μ
4
π
1/4
2
k
μ
1/2
x
e
−
k
μ
x
2
/2
So, for <
x
> we have
x
n
=
0
=
Ψ
0
x
( )
x
Ψ
0
x
( )
x
n
=
1
=
Ψ
1
x
( )
x
Ψ
1
x
( )
Before we do any heavy lifting with the integrals, note that
Ψ
0
is an even function,
Ψ
1
is
an odd function, and
x
is an odd function. So, we have integrals of either even
x
odd
x
even or odd
x
odd
x
odd. Either case results in an odd function, so the expectation value
in both cases is zero. Remember that in our derivation
x
(or
r
) represents the
displacement
from the equilibrium bond length between the two balls. Thus, a value of zero implies the
equilibrium distance (i.e., the bottom of the potential energy well). Since
all
harmonic
oscillator wave functions are either even or odd,
all
of the wave functions have <
x
> = 0.
That is, they are all symmetric about the equilibrium length, and the average of all of our
experiments to measure bond length will be the equilibrium bond length
irrespective of
what vibrational state we measure
.
As for <
x
2
>, now life is not so easy, since parity says the integrals need not be
zero. Let’s start with
Ψ
0
.

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x
2
n
=
0
=
μ
π
1/ 4
e
−
μ
x
2
/2
x
2
−∞
∞
∫
μ
π
e
−
μ
x
2
/ 2
dx
=
μ
π
1/ 2
x
2
e
−
μ
x
2
/
dx
−∞
∞
∫
An integral table provides
x
2
e
−
ax
2
dx
−∞
∞
∫
=
1
2
a
π
a
1/2
If we plug in the appropriate value of
a
we have
x
2
n
=
0
=
μ
π
2
μ
π
μ
=
2
μ
Let’s do a quick sanity check on this answer. It says that as the force constant
k
gets
larger, the average of the squared displacement from the equilibrium bond length will get
smaller. That makes sense:
stiffer spring, less displacement as it vibrates. It also says that
as the reduced mass goes up, the average displacement will be smaller. That too makes
intuitive sense. Kinetic energy goes up with mass, so in the same vibrational period we
will have to slow down (and thus go less far) to maintain the same energy if the mass
increases.
Now, what about the case for
Ψ
1
? There we have
x
2
n
=
1
=
μ
4
π
2
μ
x
e
−
μ
x
2
/ 2
x
2
−∞
∞
∫
μ
4
π
×
2
μ
x
e
−
μ
x
2
/ 2
dx
=
μ
4
π
4
μ
x
4
e
−
μ
x
2
/
dx

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