Lecture Notes No. 10
1
Orbital angular momentum
1.1
The algebra of angular momentum
In classical mechanics you might recall that the angular momentum for a
particle in motion is given by
~ = ~r × p~
L
where “×” refers to the cross product between the two vectors. In quantum
mechanics, the corresponding operator is found by using the operators for
position and momentum. Hence the three components of the orbital angular
momentum operator are
Lx = ŷ p̂z − ẑ p̂y ,
Ly = ẑ p̂x − x̂ p̂z ,
Lz = x̂ p̂y − ŷ p̂x .
We include the term “orbital” because later on we will learn about a different
type of angular momentum called spin. For the rest of this lecture we will
mainly drop the term orbital, but keep in mind that the angular momentum
discussed in this section is always orbital.
~
Consider the commutator between the first two components of L,
[Lx , Ly ] = [ŷ p̂z − ẑ p̂y , ẑ p̂x − x̂ p̂z ] = ŷ[p̂z , ẑ] p̂x + p̂y [ẑ , p̂z ] x̂
= −i~ ŷ p̂x + i~ x̂ p̂y = i~ Lz
(1)
By cycling through the components x → y → z → x and x → z → y → x
we can also show that
[Ly , Lz ] = i~ Lx ,
[Lz , Lx ] = i~ Ly .
(2)
Hence we see that the different components of angular momentum do not
commute with each other, meaning that quantum states cannot be simultaneous eigenstates of two different components of angular momentum (unless
the eigenvalue of all angular momentum components is 0). Next consider the
~ 2 = L2x + L2y + L2z
commutator of Lz with L
~ 2 ] = [Lz , L2 + L2 ] = [Lz , Lx ] Lx + Lx [Lz , Lx ] + [Lz , Ly ] Ly + Ly [Lz , Ly ]
[Lz , L
x
y
= i~ Ly Lx + i~ Lx Ly − i~ Lx Ly − i~ Ly Lx = 0
By symmetry we also have
~ 2] = 0 ,
[Lx , L
~ 2] = 0 .
[Ly , L
~2
Hence, we can choose our eigenstates to be simultaneous eigenstates of L
~ The convention is to choose Lz for the component.
and one component of L.
1
1.2
Quantization of angular momentum
Let us now define the two operators L± ≡ Lx ± iLy . These operators are not
Hermitian, instead L− = (L+ )† . Using the commutation relations in (1) and
(2), we can readily derive the very useful commutation relations
[Lz , L± ] = [Lz , Lx ± iLy ] = i~Ly ± ~Lx = ±~L± ,
as well as
[L+ , L− ] = [Lx + iLy , Lx − iLy ] = −i[Lx , Ly ] + i[Ly , Lx ] = −2i(i~)Lz = 2~Lz
~ 2 , which we
Suppose we have a state |ψi that is an eigenstate of Lz and L
~ 2 . Let the eigenvalues be ν and
know is possible since Lz commutes with L
λ respectively:
~ 2 |ψi = λ|ψi
Lz |ψi = ν|ψi
L
Now consider the new states L+ |ψi and L− |ψi. Acting with Lz on these
states we find
Lz L+ |ψi = L+ Lz |ψi + [Lz , L+ ]|ψi = νL+ |ψi + ~L+ |ψi = (ν + ~)L+ |ψi
Lz L− |ψi = L− Lz |ψi + [Lz , L− ]|ψi = νL− |ψi − ~L− |ψi = (ν − ~)L− |ψi
~ 2 on these states we get
If we act with L
~ 2 L+ |ψi = L+ L
~ 2 |ψi + [L
~ 2 , L+ ]|ψi = λL+ |ψi + 0 = λL+ |ψi
L
~ 2 L− |ψi = L− L
~ 2 |ψi + [L
~ 2 , L− ]|ψi = λL− |ψi + 0 = λL− |ψi
L
Hence, L+ creates a new state which raises Lz by one unit of ~ but has the
~ 2 eigenvalue as |ψi, while L− creates a new state which lowers Lz by
same L
~ 2 eigenvalue as |ψi. For this reason,
one unit of ~ and also has the same L
L+ and L− are called “raising” and “lowering” operators respectively. But
as in the case of the harmonic oscillator there could be a potential problem.
~ 2 for the state (L+ )n |ψi must satisfy
The expectation value of L
~ 2 (L+ )n |ψi = hψ|(L− )n (L2 +L2 +L2 )(L+ )n |ψi ≥ hψ|(L− )n L2 (L+ )n |ψi
hψ|(L− )n L
x
y
z
z
since L2x + L2y is positive definite. But we also have that
~ 2 (L+ )n |ψi = λhψ|(L− )n (L+ )n |ψi
hψ|(L− )n L
hψ|(L+ )n L2z (L+ )n |ψi = (ν + n~)2 hψ|(L− )n (L+ )n |ψi .
Hence, consistency requires either λ ≥ (ν + n~)2 , or hψ|(L− )n (L+ )n |ψi =
0. This last possibility is true only if (L+ )n |ψi = 0. Eventually, for large
2
enough n, (ν + n~)2 > λ, so there must exist a positive integer n such that
(L+ )n |ψi = 0. Likewise there must be another positive integer q such that
(L− )q |ψi = 0 otherwise we will run into an inconsistency at the other end.
Let us suppose we chose |ψi such that L+ |ψi = 0. Mathematicians call
such a state a “highest weight state”. Since L+ and L− shift the eigenvalue
of Lz by units of ~, let us write the Lz eigenvalue for the highest weight state
as ν = ` ~, where ` is a dimensionless number. Hence Lz acting on this state
is
Lz |ψi = ` ~ |ψi .
A convenient notation is to write this state as
|ψi ≡ |`, `i ,
which is a special case of the states |`, mi, where the first value indicates the
highest value for Lz while the second indicates the actual value for Lz ,
Lz |`, mi = m~|`, mi .
(3)
Since |ψi is highest weight, we have that m = `. We can get to the more
general states |`, mi by acting ` − m times with L− on |`, `i.
~ 2 can be written as
Now L
~ 2 = 1 (L+ L− + L− L+ ) + L2z = 1 ([L+ , L− ] + 2L− L+ ) + L2z
L
2
2
2
= L− L+ + ~Lz + Lz .
(4)
We can also write this as
~ 2 = 1 (L+ L− + L− L+ ) + L2 = 1 ([L− , L+ ] + 2L+ L− ) + L2
L
z
z
2
2
= L+ L− − ~Lz + L2z .
(5)
~ 2 acting on this highest weight state is
Hence L
~ 2 |`, `i = (L− L+ + ~Lz + L2 )|`, `i = 0 + (`~2 + `2 ~2 )|`, `i
L
z
2
= `(` + 1)~ |`, `i .
(6)
There must exist an integer q where (L− )q |`, `i = 0. To find q in terms
of `, we have that
0 = h`, `|(L+ )q (L− )q |`, `i = h`, `|(L+ )q−1 L+ L− (L− )q−1 |`, `i
~ 2 − L2 + ~Lz )(L− )q−1 |`, `i
= h`, `|(L+ )q−1 (L
z
q−1
= h`, `|(L+ ) (`(` + 1)~2 − (` − q + 1)2 ~2 + (` − q + 1)~2 (L− )q−1 |`, `i
= (2` + 1 − q)q~2 h`, `|(L+ )q−1 (L− )q−1 |`, `i
3
Consistency then requires that q = 2` + 1. Since q is an integer, we see that
` must be integer or half-integer. We also see that the state with the lowest
value of Lz is (L− )2` |`, `i:
Lz (L− )2` |`, `i = (` − 2`)~(L− )2` |`, `i = −`~(L− )2` |`, `i .
Therefore, up to a normalization factor this is the state |`, −`i which has
the lowest value of m, m = −`. We will later show that m must be an
integer to have a sensible wave-function. Hence, ` must also be an integer.
The numbers ` and m are called the orbital angular momentum quantum
numbers.
We now determine the normalization factors in the transformations between the different eigenstates of Lz . We define the states |`, mi to be normalized:
h`, m|`, mi = 1 .
(7)
If we now act with L− on this state we get L− |`, mi = C|`, m − 1i where C
is to be determined. To find C consider the inner product
~ 2 − L2z + ~Lz )|`, mi = ~2 (`(` + 1) − m(m − 1))
h`, m|L+ L− |`, mi = h`, m|(L
But this is also
h`, m|L+ L− |`, mi = h`, m − 1|C ∗ C|`, m − 1i = |C|2 ,
and so
p
C = ~ `(` + 1) − m(m − 1)
By a similar argument we can show that
p
L+ |`, mi = ~ `(` + 1) − m(m + 1) |`, m + 1i .
(8)
(9)
These factors are known as Clebsch-Gordon coefficients.
2
Angular momentum and spherical harmonics
Let us now see how orbital angular momentum is related to the spherical
harmonics discussed in the number 9 notes. We first show that the Lz =
∂
−i~ ∂φ
:
−i~
∂
∂x ∂
∂y ∂
= −i~
− i~
←− chain rule
∂φ
∂φ ∂x
∂φ ∂y
∂
∂
= −r sin θ sin φ −i~
+ r sin θ cos φ −i~
∂x
∂y
= −ŷ p̂x + x̂ p̂y = Lz
4
If we then act with Lz on Y`m (θ, φ) we have
r
∂
2` + 1 imφ |m|
Lz Y`m (θ, φ) = −i~
e P` (cos θ) = m~ Y`m (θ, φ).
∂φ
4π
Hence, Y`m (θ, φ) is an eigenfunction of Lz with eigenvalue m~. And so we
learn that the m is the Lz quantum number.
To get L+ and L− in terms of angular derivatives is a little messy and we
refer the interested reader to appendix A.1 for the details. Their expressions
are
∂
∂
±iφ
+ i cot θ
±
L± = ~ e
∂θ
∂φ
Notice that the angular momentum operators only contain angular derivatives; there is no r dependence whatsoever.
~ 2 in terms of derivatives is given by1
Finally, L
∂
1 ∂2
1 ∂
2
2
~
sin θ
+
L = −~
.
sin θ ∂θ
∂θ sin2 θ ∂φ2
This combination of derivatives is the same that appears in the angular part
of the Laplacian. We further recall that Y`m (θ, φ) are eigenfunctions of this
operator. Therefore,
~ 2 Y`m (θ, φ) = ~2 `(` + 1)Y`m (θ, φ)
L
~ 2 . We
and thus we see that ` is the same quantum number we found with L
can now also see that since all components of angular momentum only involve
angular derivatives and that the angular part of the Laplacian is proportional
~ 2 , then for a radially symmetric potential
to L
~ ] = 0.
[H, L
~ 2 and
Hence we can have states that are simultaneously eigenstates of H, L
Lz .
A
Appendix
A.1
Derivation of L+ and L−
Here we derive L+ and L− in terms of angular derivatives. We first write
L± = ŷ p̂z − ẑ p̂y ± i ẑ p̂x ∓ i x̂ p̂z
= i (∓x̂ − iŷ) p̂z + i ẑ (±p̂x + ip̂y )
1
See appendix A.2 for the details.
5
The individual terms in this expression are
(∓x̂ − iŷ) = r sin θ(∓ cos φ − i sin φ) = ∓r sin θ e±iφ
ẑ = r cos θ
∂
∂r ∂
∂θ ∂
∂
sin θ ∂
p̂z = −i~
= −i~
+
= −i~ cos θ −
∂z
∂z ∂r ∂z ∂θ
∂r
r ∂θ
∂r ∂
∂θ ∂
∂φ ∂
∂r ∂
∂θ ∂
∂φ ∂
±p̂x + ip̂y = −i~ ∓
∓
∓
+i
+i
+i
∂x ∂r ∂x ∂θ ∂x ∂φ
∂y ∂r
∂y ∂θ
∂y ∂φ
∂
cos θ cos φ ∂
sin φ ∂
= −i~ ± sin θ cos φ ±
∓
∂r
r
∂θ r sin θ ∂φ
cos θ sin φ ∂
cos φ ∂
∂
+i
+isin θ sin φ + i
∂r
r
∂θ
r sin θ ∂φ
∂
cos θ ∂
1 ∂
= ∓i~ e±iφ sin θ +
±i
∂r
r ∂θ
r sin θ ∂φ
In evaluating the partial derivatives we used that
p
x2 + y 2
2
2
2
2
,
r =x +y +z ,
tan θ =
z
tan φ =
x
.
y
Putting this all together, we are left with
∂
∂
±iφ
L± = ~ e
+ i cot θ
±
∂θ
∂φ
A.2
~ 2 in terms of angular derivatives
L
~ 2 as
We write L
~ 2 = L2 + L2 + L2 = 1 (L+ L− + L− L+ ) + L2
L
x
y
z
z
2
2
~
∂
∂
∂
∂
iφ
−iφ
e
+ i cot θ
e
−
+ i cot θ
=
2
∂θ
∂φ
∂θ
∂φ
∂
∂
∂
∂
∂2
−iφ
iφ
+e
−
+ i cot θ
e
+ i cot θ
− ~2 2
∂θ
∂φ
∂θ
∂φ
∂φ
2
2
1 ∂
∂
∂
1 ∂
∂
1 ∂2
2
2
= −~
+ cot θ
+
= −~
sin θ
+
∂θ2
∂θ sin2 θ ∂φ2
sin θ ∂θ
∂θ sin2 θ ∂φ2
where the middle term in the left hand side of the last line comes from
∂
i cot θ ∂φ
acting on the e−iφ and e+iφ terms.
6
A.3
Uncertainty of the angular momementum vector
Consider the highest weight angular momentum state |`, `i, which is an eigenstate of Lz , but not of Lx or Ly . This means there is some uncertainty in Lx
and Ly . Using our result for generalized Heisenberg uncertainty, we have
1
~
`
σLx σLy ≥ h`, `|[Lx , Ly ]|`, `i = h`, `|Lz |`, `i = ~2 .
2i
2
2
In fact, we can actually argue that the uncertainty is an equality for this
state: The expectation values for Lx and Ly are
hLy i = h`, `| 2i1 (L+ − L− )|`, `i = 0 ,
hLx i = h`, `| 21 (L+ + L− )|`, `i = 0,
while the expectation values for L2x and L2y are
`
hL2x i = h`, `| 14 (L+ + L− )2 |`, `i = h`, `| 14 L+ L− |`, `i = h`, `| ~2 Lz |`, `i = ~2 ,
2
`
hL2y i = −h`, `| 41 (L+ − L− )2 |`, `i = h`, `| 41 L+ L− |`, `i = h`, `| ~2 Lz |`, `i = ~2 ,
2
giving
σLx
q
p
= hL2x i − hLx i2 = σLy = hL2y i − hLy i2 = ~
r
`
.
2
We also know that
~ 2 − L2 )|`, `i = ~2 ` |`, `i ,
(L2x + L2y )|`, `i = (L
z
and so there is no uncertainty in the combined length squared of these two
transverse components. Since (σLx )2 +(σLx )2 equals the eigenvalue of L2x +L2y ,
~ in the x−y plane is completely unspecified.
this means that the direction of L
If we look at the figure, it means that
the angular momentum vector is pointing somewhere along the dashed ring, but
where on the ring is completely uncertain.
The uncertainty in the angular direction of
L
L
z
the angular momentum vector is the polar
~ which is
angle of L,
p 2
Lx + L2y
1
θ = arctan
= arctan √ .
Lz
`
7
Let me stress that a measurement will not collapse the wave-function to one
where the angle is known. The point is that one cannot measure the angle
without uncertainty. This is because to know the angle one must know all
the components, but this is not possible since the corresponding operators
don’t commute with each other.
8