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9 Angular Momentum I
So far we haven’t examined QM’s biggest success–atomic structure and
the explanation of atomic spectra–in detail. To do this need better under-
standing of angular momentum. In brief: we’ll find that eigenfunctions of
atomic problem have structure
ψ(r) = u n (r)Y lm (θ, φ) (1)
where u n
(r) is soln. to radial S.-eqn. we examined in Week 4, and
Y
lm
(θ, φ) is eigenfunction of angular momentum operator L. Degeneracy
of atomic states within a given shell corresponding to principal quantum
number n explained by angular momentum quantum numbers l, m.
9.1 Orbital Angular Momentum
Classical analogy, take
L = r × pˆ (2)
In other words
L
x = y pˆ z − z pˆ y
L
y = z pˆ x − xpˆ z
L
z = xpˆ y − y pˆ x
Note that since [y, pˆ x
] = 0, etc., antisymmetry property r × pˆ = −pˆ × r
is ok, meaning there is no ambiguity in taking the classical expression and
replacing r and p by their operator forms everywhere.
Before solving H-atom, we need to derive formal properties of L. First
summarize these, then prove a few of the interesting ones. Need definition
of antisymmetric “permutation symbol”
ijk
1 if ijk are in “cyclic” order, e.g. 123, 312,etc.
−1 if ijk are in “anticyclic” order, e.g. 321, 132, etc.
0 if any two indices are equal
So the components of L in this notation are
L
i
ijk x j pˆ k
with implied summation convention (if I write a two repeated indices
below, it means sum over them, even if I leave out the Σ).
Summary of useful relations involving L
L
†
i
L
i
All components of L Hermitian.
2. [
L
i
L
j
] = i¯h≤ ijk
L
k
e.g. [
L
x
L
y
] = i¯h
L
z
3. [
L
i
, x j
] = ih≤¯ ijk
x k
e.g. [
L
y
, x] = −ihz¯
4. [
L
i
, pˆ j
] = i¯h≤ ijk
pˆ k
e.g. [
L
y
, pˆ x
] = −ih¯pˆ z
5. [
L
i
, pˆ
2 ] = 0
∑
j
(
[
L
i
, pˆ j
]ˆp j
− pˆ j
[
L
i
, pˆ j
]
)
= −ih≤¯ ijk pˆ k pˆ j
Pf. of [
L
2 ,
L
i ] = 0. Use (9):
[
L
2
,
L
i
] = −[
L
i
L
2
] = −[
L
i
L
2
x
] − [
L
i
L
2
y
] − [
L
i
L
2
z
]
= −([
L
i
L
x
]
L
x
L
x
[
L
i
L
x
]) + (x → y) + (x → z)
= −ih≤¯ i 1 k
L
k
L
x
L
x
L
k
since ≤ ijk
is antisymmetric.
9.2 Eigenfunctions of
L
z
For a central force problem, V (r) = V (r),
L
2
,
L
z , and Π all commute
with H, so we can find a complete set of eigenfctns. of all 4 ops. First
construct eigenfctns of
L
z in polar coordinates,
x = r sin θ cos φ
y = r sin θ sin φ
z = cos θ.
Notice that
∂ψ
∂φ
∣ ∣ ∣ ∣ ∣ ∣
r,θ
∂ψ
∂x
∂x
∂φ
∂ψ
∂y
∂y
∂φ
∂ψ
∂z
∂z
∂φ
∂ψ
∂x
(−)r sin θ sin φ +
∂ψ
∂y
r sin θ sin φ
= x
∂ψ
∂y
− y
∂ψ
∂x
Mutiply by ih¯ to find −ih¯
∂ψ
∂φ
= (xpˆ y − y pˆ x )ψ, or
L
z
= −ih¯
∂
∂φ
Eigenfctns of
L
z , i.e. soln. to
L
z ψ = αψ is
ψ ∝ e
iαφ/h¯
(14)
Require ψ be single-valued fctn. of position, i.e. when φ → φ+2π, better
get same value back again. Thus we find new quantum number α = mh¯,
or
ψ ∝ e
imφ
, m = 0, ± 1 , ± 2... (15)
Eigenvalues of
L
z therefore mh¯.
9.3 L as generator of rotations
Q: How does wave fctn. ψ(r) change when we rotate coordinate system
to new coordinates r
′ ? Define rotation to be around axis nˆ, through angle
θ.
A:
ψ
′
= e
−iθnˆ·L/¯h
ψ (16)
where U = e
−iθˆn·L/¯h is an operator to be understood in terms of its Taylor
expansion, U = 1 − iθnˆ · L/¯h +
1
2
(iθnˆ · L/h¯)
2
Hermitian operator, so U is unitary, U
† U = 1. The operator L is referred
to as the generator of intfinesimal rotations, see below.
Check in special case: rotate around ˆz, 1st by infinitesimal angle δφ:
ψ
′
(r, θ, φ) = ψ(r, θ, φ − δφ)
' ψ(r, θ, φ) − δφ
∂φ
ψ(r, θ, φ) (17)
Now use representation of
L
z we just worked out:
So we see a ≥ b
2 .
Ladder operators for angular momentum
Define
L
L
x
L
y
L
−
L
x
− i
L
y
Note
L
†
L
− , etc. Since [
L
2
,
L
i ] = 0 and [
L
i
L
j ] = ih≤¯ ijk
L
k , find
(check!)
[L
2
, L ±
] = 0 (30)
[
L
z
L
± ] = ±¯h
L
±
Now proceed `a la harmonic oscillator case—apply
L
to Eq.(23):
L
L
2
)ψ = a
L
ψ =
L
2
(
L
ψ) (32)
so
L
ψ is an eigenfctn. of
L
2 with eigenvalue a. Now apply to Eq.(24):
L
L
z ψ = b(
L
ψ) (33)
[
L
L
z
]ψ +
L
z
L
ψ) = −h¯
L
ψ +
L
z
L
ψ) (35)
Rearrange to get
L
z
L
ψ) = (b + ¯h)(
L
ψ), (36)
which =⇒
L
ψ is eigenfctn of
L
z
with eigenvalue b + ¯h. Label simult.
e’fctns of
L
2 and
L
z
by ψ ab
, then
L
ψ ab = ψ a,b+¯h
L
−
ψ ab
= ψ a,b−¯h
in general.
? So we begin to get the picture—
L
±
move us up and down the ladder of
L
z
quantum numbers. Recalling the SHO, need to find out where the top
and bottom of the ladder are!
Know b
2 ≤ a from above, so sequence b, b±h, b¯ ±2¯h · · · must terminate
above and below, i.e. there exist b max and b min for each choice of a:
L
ψ a,b max
L
z
ψ a,b max
= b max
ψ a,b max
L
−
ψ a,b min
L
z
ψ a,b min
= b min
ψ a,b min
Now apply
L
2 to ψ a,b max
to find a. Convenient to have form
L
2
=
L
2
x
L
2
y
L
2
z
L
−
L
L
y
L
x
] + L
2
z
L
2
=
L
−
L
L
2
z
L
z
and similarly
L
2
=
L
L
−
L
2
z
− h¯
L
z
Therefore using (42)
L
2
ψ a,b max
= aψ a,b max
= (b
2
max
)ψ a,b max
and same argument applied to ψ a,b min
using (43) gives
L
2
ψ a,b min
= aψ a,b min
= (b
2
min
− ¯hb min
)ψ a,b min
Summarize:
a = b
2
max
a = b
2
min
− hb¯ min
Difference of (46) and (47) is
0 = (b max
− b min
)(b max
) + ¯h(b max
which has soln. only when b max
= −b min
Now recall b is eigenvalue of
L
z
, showed b = m¯h, m = 0, ± 1 , ± 2 · · · so
must have
b max − b min = nh¯ (49)
