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 understanding of angular momentum. In brief: we’ll find that eigenfunctions of
atomic problem have structure
ψ(r) = un(r)Ylm(θ, φ)
(1)
where un(r) is soln. to radial S.-eqn. we examined in Week 4, and
Ylm(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
1
L = r × p̂
(2)
L̂x = y p̂z − z p̂y
L̂y = z p̂x − xp̂z
L̂z = xp̂y − y p̂x
(3)
(4)
In other words
Note that since [y, p̂x] = 0, etc., antisymmetry property r × p̂ = −p̂ × 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”




1 if ijk are in “cyclic” order, e.g. 123, 312,etc.



²ijk = 
(5)
−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
L̂i = ²ijk xj p̂k
(6)
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
1. L̂†i = L̂i
All components of L Hermitian.
2. [L̂i, L̂j ] = ih̄²ijk L̂k
e.g. [L̂x, L̂y ] = ih̄L̂z
3. [L̂i, xj ] = ih̄²ijk xk
e.g. [L̂y , x] = −ih̄z
4. [L̂i, p̂j ] = ih̄²ijk p̂k
e.g. [L̂y , p̂x] = −ih̄p̂z
5. [L̂i, p̂2] = 0
2
6. [L̂i, V (r)] = 0
7. [L̂2, L̂i] = 0 where L̂2 ≡ L̂2x + L̂2y + L̂2z
? Major point will be that due to 5),6),7), the eigenfctns of central force
h̄2
problems can be chosen as simultaneous eigenfunctions of H = − 2m
∇2 +
V (r) and L̂2 and any component of L, e.g. L̂z .
Pf. of L̂†i = L̂i: e.g.,
L̂†x = (y p̂z − z p̂y )†
= p̂z y − p̂y z
= y p̂z − z p̂y = L̂x
(7)
Pf. of [L̂i, L̂j ] = ih̄²ijk L̂k , e.g.
[L̂z , L̂y ] = [(xp̂y − y p̂x), (z p̂x − xp̂z )]
= [xp̂y , z p̂x] − [xp̂y , xp̂z ]
−[y p̂x, z p̂x] + [y p̂x, xp̂z ]
= p̂y z[x, p̂x] − 0 − 0 + y p̂z [p̂x, x]
= ih̄(p̂y z − y p̂z ) = −ih̄L̂x
(8)
Does it work? Check sign ²321 = −1 OK!
Pf. of [L̂i, p̂2] = 0: First need general relation
[A, BC] = ABC − BCA
= ABC − BAC + BAC − BCA
= [A, B]C + B[A, C]
Now evaluate using 4):
[Li, p̂2] =
X
j
[Li, p̂2j ]
3
(9)
=
Xµ
j
[L̂i, p̂j ]p̂j − p̂j [L̂i, p̂j ]
¶
= −ih̄²ijk p̂k p̂j + ih̄²ijk p̂j p̂k = 0
(10)
Pf. of [L̂2, L̂i] = 0. Use (9):
[L̂2, L̂i] = −[L̂i, L̂2] = −[L̂i, L̂2x] − [L̂i, L̂2y ] − [L̂i, L̂2z ]
= −([L̂i, L̂x]L̂x + L̂x[L̂i, L̂x]) + (x → y) + (x → z)
= −ih̄²i1k (L̂k L̂x + L̂xL̂k ) + . . . = 0
(11)
since ²ijk is antisymmetric.
9.2
Eigenfunctions of L̂z
For a central force problem, V (r) = V (r), L̂2, L̂z , and Π all commute
with H, so we can find a complete set of eigenfctns. of all 4 ops. First
construct eigenfctns of L̂z in polar coordinates,
x = r sin θ cos φ
y = r sin θ sin φ
z = cos θ.
Notice that
¯
∂ψ ¯¯¯
∂ψ ∂x ∂ψ ∂y ∂ψ ∂z
=
+
+
¯
∂φ ¯r,θ
∂x ∂φ ∂y ∂φ ∂z ∂φ
∂ψ
∂ψ
=
(−)r sin θ sin φ +
r sin θ sin φ
∂x
∂y
∂ψ
∂ψ
= x
−y
∂y
∂x
Mutiply by ih̄ to find −ih̄ ∂ψ
∂φ = (xp̂y − y p̂x )ψ, or
4
(12)
∂
Lz = −ih̄ ∂φ
(13)
Eigenfctns of L̂z , i.e. soln. to L̂z ψ = αψ is
ψ ∝ eiαφ/h̄
(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 = mh̄,
or
ψ ∝ eimφ,
m = 0, ±1, ±2 . . .
(15)
Eigenvalues of L̂z therefore mh̄.
9.3
L as generator of rotations
Q: How does wave fctn. ψ(r) change when we rotate coordinate system
to new coordinates r0? Define rotation to be around axis n̂, through angle
θ.
A:
ψ 0 = e−iθn̂·L/h̄ψ
(16)
where U = e−iθn̂·L/h̄ is an operator to be understood in terms of its Taylor
expansion, U = 1 − iθn̂ · L/h̄ + 12 (iθn̂ · L/h̄)2 + · · ·. Note θn̂ · L is a
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 ẑ, 1st by infinitesimal angle δφ:
ψ 0(r, θ, φ) = ψ(r, θ, φ − δφ)
' ψ(r, θ, φ) − δφ
∂
ψ(r, θ, φ)
∂φ
Now use representation of L̂z we just worked out:
5
(17)
∂
)ψ
∂φ
= (1 − iδφL̂z /h̄)ψ
ψ 0 = (1 − δφ
(18)
(19)
Now rotate by finite angle φ0: apply above tranformation n À 1 times
for small φ0/n:
ψ 0 = (1 −
iφ0L̂z n
) ψ
nh̄
(20)
Recall
n
x
lim
(1
+
x/n)
=
e
n→∞
(21)
=⇒ ψ 0 = e−iφ0L̂z /h̄
(22)
Generalize to infinitesimal rotation δφ around n̂ to get Eq. (16)
9.4
Eigenvalues of L̂2
Program is to find complete set of eigenfunctions of both L̂z and L̂2.
(Could pick any component of L, L̂z is conventional):
L̂2ψ = aψ
L̂z ψ = bψ.
(23)
(24)
(ψ, L̂2ψ) = (ψ, L̂2xψ) + (ψ, L̂2y ψ) + (ψ, L̂2z ψ)
≥ 0
(25)
(26)
First note
R
since each term (ψ, L̂2i ψ) = (L̂iψ, L̂iψ) = dx|L̂iψ|2. Now can figure out
something about a and b:
(ψ, L2ψ) = (ψ, L2xψ) + (ψ, L2y ψ) + (ψ, L2z ψ)
↓
↓
↓
↓
a(ψ, ψ)
≥0
≥0
b2(ψ, ψ)
6
(27)
So we see a ≥ b2.
Ladder operators for angular momentum
Define
L̂+ = L̂x + iL̂y
L̂− = L̂x − iL̂y
(28)
(29)
Note L̂†+ = L̂−, etc. Since [L̂2, L̂i] = 0 and [L̂i, L̂j ] = ih̄²ijk L̂k , find
(check!)
[L2, L±] = 0
[L̂z , L̂±] = ±h̄L̂± .
(30)
(31)
Now proceed à la harmonic oscillator case—apply L̂+ to Eq.(23):
L̂+(L̂2)ψ = aL̂+ψ = L̂2(L̂+ψ)
(32)
so L̂+ψ is an eigenfctn. of L̂2 with eigenvalue a. Now apply to Eq.(24):
L̂+L̂z ψ = b(L̂+ψ)
↓
[L̂+, L̂z ]ψ + L̂z (L̂+ψ) = −h̄L̂+ψ + L̂z (L̂+ψ)
(33)
(34)
(35)
Rearrange to get
L̂z (L̂+ψ) = (b + h̄)(L̂+ψ),
(36)
which =⇒ L̂+ψ is eigenfctn of L̂z with eigenvalue b + h̄. Label simult.
e’fctns of L̂2 and L̂z by ψab, then
L̂+ψab = ψa,b+h̄
L̂−ψab = ψa,b−h̄
(37)
(38)
in general.
? So we begin to get the picture—L̂± move us up and down the ladder of
L̂z quantum numbers. Recalling the SHO, need to find out where the top
and bottom of the ladder are!
7
Know b2 ≤ a from above, so sequence b, b±h̄, b±2h̄ · · · must terminate
above and below, i.e. there exist bmax and bmin for each choice of a:
L̂+ψa,bmax = 0
L̂−ψa,bmin = 0
,
,
L̂z ψa,bmax = bmaxψa,bmax
L̂z ψa,bmin = bminψa,bmin
(39)
(40)
Now apply L̂2 to ψa,bmax to find a. Convenient to have form
L̂2
= L̂2x + L̂2y + L̂2z = L̂−L̂+ + i[L̂y , L̂x] + L2z
=⇒ L̂2 = L̂−L̂+ + L̂2z + h̄L̂z
(41)
(42)
and similarly
L̂2 = L̂+L̂− + L̂2z − h̄L̂z
(43)
L̂2ψa,bmax = aψa,bmax = (b2max + h̄bmax)ψa,bmax
(44)
Therefore using (42)
and same argument applied to ψa,bmin using (43) gives
L̂2ψa,bmin = aψa,bmin = (b2min − h̄bmin)ψa,bmin
(45)
a = b2max + h̄bmax
a = b2min − h̄bmin
(46)
(47)
Summarize:
Difference of (46) and (47) is
0 = (bmax − bmin)(bmax + bmin) + h̄(bmax + bmin)
(48)
which has soln. only when bmax = −bmin.
Now recall b is eigenvalue of L̂z , showed b = mh̄, m = 0, ±1, ±2 · · · so
must have
bmax − bmin = nh̄
8
(49)
for some integer n > 0. (48) and (49) together now say
bmax =
nh̄
≡ `h̄
2
(50)
where I’ve defined 2` to be the integer n in (49). In terms of this `, plug
back into (46) and find
a = h̄2`(` + 1).
(51)
Allowed values of quantum numbers which follow from
commutation relations:
1 3
L̂2 : h̄2`(` + 1), ` = 0, , 1, , 2, · · ·
2 2
L̂z : mh̄, m = −`, m = −`, −` + 1, · · · ` − 1, `
(52)
(53)
But note for orbital angular momentum specifically we showed m =
integer, so only integer values of ` are allowed. Half-integer values will be
encountered when we discuss spin.
9