Chapter 4-3: Angular Momentum
I.
ANGULAR MOMENTUM
In the classical theory of central forces, both energy and angular momentum L = r × p are conserved. Correspond∂
. Namely,
ingly, operators can be defined in quantum mechanics L̂ = r̂ × p̂, where p̂ = ~i ∂r
h ∂
h ∂
−z
i ∂z
i ∂y
h ∂
h ∂
L̂y = ẑ p̂x − x̂p̂z = z
−x
i ∂x
i ∂z
h ∂
h ∂
L̂z = x̂p̂y − ŷ p̂x = x
−y
i ∂y
i ∂x
(1)
[L̂i , L̂j ] = ijk i~L̂k
(2)
L̂x = ŷ p̂z − ẑ p̂y = y
The commutation relations are
where ijk is the Levi-Civita symbol.
This implies that any two angular momentum components can not be measured simultaneously according to the
def
uncertainty principle, e.g., ∆L̂x ∆L̂y ≥ ~2 | hL̂z i |. Nevertheless, L̂2 = L̂2x + L̂2y + L̂2z is commute with L̂i (i = x, y, z),
i.e., [L̂2 , L̂i ] = 0.
def
Moreover, L̂± = L̂x ± iL̂y , then [L̂2 , L̂± ] = 0, but [L̂± , L̂i ] 6= 0
Based on such commutation relations, simultaneous eigenstates Y can be defined. Namely,
• Because [L̂2 , L̂i ] = 0, a common complete set {Y} can be obtained, where L̂2 Y = λY and L̂z Y = µY with λ
and µ real (why?) numbers.
• Because [L̂2 , L̂± ] = 0,
L̂2 (L̂± Y) = L̂± (L̂2 Y) = L̂± (λY) = λ(L̂± Y)
.
• Because [L̂± , L̂i ] 6= 0,
L̂z (L̂± Y) = [L̂z , L̂± ]Y + L̂± L̂z Y = ±~L̂± Y + L̂± (µY) = (µ ± ~)(L̂± Y).
| {z }
cc. above
In this respect, L̂± are ladder operators as we defined for the harmonic oscillators.
Unlike the ladder operators for harmonic oscillators, where only the ground state can be obtained by â− ψ0 = 0, for
ladder operators L̂± defined for angular moment, both upper (“top”) and lower (“bottom”) limit can be obtained.
• The “top” state is defined by L̂+ Ytop = 0. Suppose L̂z Ytop = µYtop = ~lYtop , then
L̂2 Ytop = (L̂− L̂+ + L̂2z + ~L̂z )Ytop = (0 + ~2 l2 + ~2 l)Ytop = ~2 l(l + 1)Ytop = λYtop
⇒ λ = ~2 l(l + 1)
• The “bottom” state is defined by L̂− Ybottom = 0. In the same way (L̂z Ybottom = ~l0 Ybottom ), it can be shown
that
L̂2 Ybottom = ~2 l0 (l0 − 1)
.
2
• Because L̂2 (L̂± Y) = λ(L̂± Y), ~2 l(l + 1) = ~2 l0 (l0 − 1). Since there is no special reason for the bottom level to be
of higher expectation value for L̂z than the top level, l0 = −l. This sets the lower (−l~) and upper (l~) bound
for hL̂z i.
• If L̂− (L̂+ ) is kept being applied to Y“top” (Y“bottom” ), by certain integer N number of steps, Y“bottome” (Y“top” )
will be reached. That is l = −l + N → l = N2 .
• All in all,
L̂2 Ylm = ~2 l(l + 1)Ylm , L̂z Ylm = ~mYlm
where
l = 0, 1/2, 1, 3/2, ..., m = −l, −l + 1, ..., l − 1, l
.
• Discussion
p
l(l + 1) ≥ l.
II.
ORBITAL ANGULAR MOMENTUM
In spherical coordinates, L̂ = r̂ × ~i ∇ = rer ×
∂
∂r er
+
1 ∂
r ∂θ eθ
+
1
∂
r sin θ ∂φ eφ
=
~
i
∂
∂θ eφ
−
1 ∂
sin θ ∂φ eφ
. Remember,
eθ = (cos θ cos φ)ex + (cos θ sin φ)ey − sin θek and eφ = − sin φex + cos θey . Thus,
∂
∂
~
− sin φ
− cos φ cot θ
L̂x =
i
∂θ
∂φ
~
∂
∂
L̂y =
cos φ
− sin φ cot θ
i
∂θ
∂φ
~ ∂
L̂z =
i ∂φ
∂
∂
L̂± = L̂x ± iL̂y = ±~e±iφ
± i cot θ
∂θ
∂φ
1 ∂
∂
1 ∂2
L̂2 = −~2
(sin θ ) +
sin θ ∂θ
∂θ
sin2 θ ∂φ2
Following L̂2 Ylm = ~2 l(l + 1)Ylm and L̂z Ylm = ~mYlm , we get
1 ∂
∂
1 ∂2
2 m
2
L̂ Yl = −~
(sin θ ) +
Ylm = ~2 l(l + 1)Ylm
sin θ ∂θ
∂θ
sin2 θ ∂φ2
~ ∂ m
L̂z Ylm =
Y = ~mYlm
i ∂φ l
These equations are exactly (equivalent to) those eigenvalue problems of L̂2 and L̂z . That is, spherical harmonics
are eigen-wavefunctions of L̂2 and L̂z .
Homework:
1. prove [L̂i , L̂j ] = ijk i~L̂k .
2. prove [L̂2 , L̂i ] = 0, i = (x, y, z).
3. show [L̂2 , L̂± ] = 0, and find out [L̂± , L̂i ].
4. prove L̂2 = L̂± L̂∓ + L̂2z ∓ ~L̂z .