Chem 120A
Angular Momentum, continued
03/15/06
Spring 2006
Lecture 23
READING:
Engel (CS): Chapters 7.47.5 and 9; Appendices A.4.2 and A.6
Additional optional reading: McQuarrie and Simon, Physical Chemistry
, pg. 191206
Chapter 3 of Townsend is also on the course website.
Notes regarding the spherical harmonics,
Y
lm
In the last lecture, we introduced the spherical harmonic functions
Y
lm
(
θ
,
φ
)
as the eigenfunctions of
ˆ
L
2
and
ˆ
L
z
in the position (physical space) representation. Here we will further examine their physical significance.
The
Y
lm
(
θ
,
φ
)
are probability amplitudes of finding a particle at a particular value of
θ
and
φ
. We mentioned
in the last lecture that the state
Y
00
is the spherical harmonic that corresponds to the
s
orbitals.
Y
00
is equal
to
1
4
π
, which implies that the probabilities of finding a particle at any point
(
θ
,
φ
)
are equal (spherically
symmetric). For the states
Y
1
,
±
1
, we had complex amplitudes:
Y
1
,
1
=
3
8
π
sin
θ
e
i
φ
Y
1
,
−
1
=
3
8
π
sin
θ
e
−
i
φ
,
for which the corresponding probabilities are
P
1
,
±
1
(
θ
,
φ
) =
3
8
π
sin
2
θ
.
Here, the probability is 0 at
θ
=
0
,
π
(along the
z
axis) and increases until
θ
=
π
2
, i.e. in the
xy
plane. The
probability for
Y
1
,
0
is
P
1
,
0
=
3
4
π
cos
2
θ
,
which is maximal along the
z
axis. When the functions
Y
1
x
,
Y
1
y
, and
Y
1
z
are each combined with the solution
to the radial part of the Schr¨odinger equation for hydrogen,
R
n
,
1
(
r
)
, then these comprise the
np
orbitals of
hydrogen.
Back to the hydrogen atom
For any single particle in 3D space, the kinetic energy depends on the Laplacian,
∇
2
. In spherical polar
coordinates, the Laplacian is
∇
2
=
∂
2
∂
r
2
+
2
r
∂
∂
r
−
1
r
2
¯
h
2
ˆ
L
2
.
(1)
Chem 120A, Spring 2006, Lecture 23
1
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Because
r
,
L
=
0, we can solve the angular part of the Schr¨odinger equation immediately, irrespective of
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 Spring '07
 Whaley
 Physical chemistry, Atom, pH, CHEM 120A, Schr¨ dinger equation

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