QM Lecture 8: Angular Momentum

**By: Tao Steven Zheng (郑涛)**

## **Angular Momentum**

In classical mechanics, the angular momentum

L
→

{\displaystyle \vec L}

 is defined as

L
→

=

r
→

×

p
→

{\displaystyle \vec L = \vec r \times \vec p}

where

r
→

=
(
x
,
y
,
z
)

{\displaystyle \vec{r}=(x, y, z)}

 is the displacement vector, and

p
→

=
(

p
x

,

p
y

,

p
z

)

{\displaystyle \vec {p} = (p\_x, p\_y, p\_z)}

 the linear momentum vector. When the angular momentum is expressed in its components in Cartesian coordinates (by computing the cross product explicitly), we get

L
→

=

i

j

k

x

y

z

p
x

p
y

p
z

=
(
y

p
z

−
z

p
y

)

i

(
z

p
x

−
x

p
z

)

j

(
x

p
y

−
y

p
x

)

k

{\displaystyle \begin{align}
\vec{L} = \begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
 x & y & z \\
 p\_x & p\_y & p\_z \end{vmatrix} = (yp\_z - zp\_y)\mathbf{i} + (zp\_x - xp\_z)\mathbf{j} + (xp\_y - yp\_x)\mathbf{k} \\[5 pt]
\end{align} }

For the sake of convenience, let us express each component of

L
→

=
(

L
x

,

L
y

,

L
z

)

{\displaystyle \vec{L} = (L\_x, L\_y, L\_z) }

 as

L
x

=
y

p
z

−
z

p
y

L
y

=
z

p
x

−
x

p
z

L
z

=
x

p
y

−
y

p
x

{\displaystyle \begin{align}
L\_x &= yp\_z - zp\_y \\[5 pt]
L\_y &= zp\_x - xp\_z \\[5 pt]
L\_z &= xp\_y - yp\_x \\[5 pt]
\end{align} }

The quantum mechanical analogon of angular momentum is the **angular momentum operator**. Using the coordinate representation of the momentum operator, we can write out the components of the angular momentum operator by making the substitutions

p
^

x

=
−
i
ℏ

∂

∂
x

p
^

y

=
−
i
ℏ

∂

∂
y

p
^

z

=
−
i
ℏ

∂

∂
z

{\displaystyle \hat{p}\_x = -i\hbar \frac{\partial}{\partial x} \qquad
\hat{p}\_y = -i\hbar \frac{\partial}{\partial y} \qquad
\hat{p}\_z = -i\hbar \frac{\partial}{\partial z} \qquad }

Subsequently, the components of the angular momentum operator are

L
^

x

=
−
i
ℏ

(

y

∂

∂
z

−
z

∂

∂
y

)

L
^

y

=
−
i
ℏ

(

z

∂

∂
x

−
x

∂

∂
z

)

L
^

z

=
−
i
ℏ

(

x

∂

∂
y

−
y

∂

∂
x

)

{\displaystyle \begin{align}
\hat{L}\_x &= -i\hbar \left(y\frac{\partial}{\partial z} - z \frac{\partial}{\partial y}\right) \\[5 pt]
\hat{L}\_y &= -i\hbar \left(z\frac{\partial}{\partial x} - x \frac{\partial}{\partial z}\right) \\[5 pt]
\hat{L}\_z &= -i\hbar \left(x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x}\right) \\[5 pt]
\end{align} }

One should note that there are three angular momentum operators in quantum mechanics: the orbital angular momentum (

L
^

{\displaystyle \hat{L} }

), the spin angular momentum (**spin** for short,

S
^

{\displaystyle \hat{S}}

), and the total angular momentum (

J
^

=

L
^

S
^

{\displaystyle \hat{J} = \hat{L} + \hat{S} }

). Note that the total angular momentum is always conserved due to **Noether's theorem**.

## **Commutation Relations**

Each pair of angular momentum operators are non-commutative. This can be demonstrated by simple calculation using the commutator. We will drop the hat-notation convention in this derivation for simplicity. Begin by expressing each angular momentum operator in terms of the components of position and linear momentum. Doing this for

L
y

{\displaystyle L\_y}

 first and using the commutator property

[
A
,
B
C
# ]
[
A
,
B
]
[
A
,
C
]

{\displaystyle [A, B + C] = [A, B] + [A, C] }

 gives

[

L
x

,

L
y

]

=
[

L
x

,
z

p
x

−
x

p
z

]

=
[

L
x

,
z

p
x

]
−
[

L
x

,
x

p
z

]

{\displaystyle \begin{align}
\left[L\_x, L\_y \right] &= [L\_x, z p\_x - x p\_z] \\[5 pt]
&= [L\_x, z p\_x] - [L\_x, x p\_z]
\end{align} }

Since

[
A
,
B
C
# ]
[
A
,
B
]
C
B
[
A
,
C
]

{\displaystyle [A, BC] = [A, B]C + B[A, C]}

, we can expand each of the terms to get a pair of commutators:

[

L
x

,
z

p
x

]

=
[

L
x

,
z
]

p
x

z
[

L
x

,

p
x

]

[

L
x

,
z

p
z

]

=
[

L
x

,
x
]

p
z

x
[

L
x

,

p
z

]

{\displaystyle \begin{align}
\left[L\_x, z p\_x \right] &= [L\_x, z] p\_x + z [L\_x, p\_x] \\[5 pt]
\left[L\_x, z p\_z \right] &= [L\_x, x] p\_z + x [L\_x, p\_z]
\end{align} }

Next, use the definition of angular momentum to calculate each commutator on the right hand side of each equation above.

[

L
x

,
z

]

=
[
y

p
z

−
z

p
y

,
z
]

=
[
y

p
z

,
z
]
−
[
z

p
y

,
z
]

=
y
[

p
z

,
z
]
[
y
,
z
]

p
z

−
z
[

p
y

,
z
]
−
[
z
,
z
]

p
y

=
y
[

p
z

,
z
]

=
−
i
ℏ
y

{\displaystyle \begin{align}
\left[L\_x, z \right] &= [y p\_z - z p\_y, z] \\[5 pt]
&= [y p\_z, z] - [z p\_y, z] \\[5 pt]
&= y[p\_z,z] + [y, z]p\_z - z[p\_y, z] - [z, z]p\_y \\[5 pt]
&= y[p\_z, z] \\[5 pt]
&= -i\hbar y
\end{align} }

[

L
x

,

p
x

]

=
[
y

p
z

−
z

p
y

,

p
x

]

=
[
y

p
z

,

p
x

]
−
[
z

p
y

,

p
x

]

=
y
[

p
z

,

p
x

]
[
y
,

p
x

]

p
z

−
z
[

p
y

,

p
x

]
−
[
z
,

p
x

]

p
y

=
0

{\displaystyle \begin{align}
\left[L\_x, p\_x\right] &= [y p\_z - z p\_y, p\_x] \\[5 pt]
&= [y p\_z, p\_x] - [z p\_y, p\_x] \\[5 pt]
&= y[p\_z,p\_x] + [y, p\_x]p\_z - z[p\_y, p\_x]- [z, p\_x]p\_y \\[5 pt]
&= 0 \\[5 pt]
\end{align}
 }

[

L
x

,
x

]

=
[
y

p
z

−
z

p
y

,
x
]

=
[
y

p
z

,
x
]
−
[
z

p
y

,
x
]

=
y
[

p
z

,
x
]
[
y
,
x
]

p
z

−
z
[

p
y

,
x
]
−
[
z
,
x
]

p
y

=
0

{\displaystyle \begin{align}
\left[L\_x, x\right] &= [y p\_z - z p\_y, x] \\[5 pt]
&= [y p\_z, x] - [z p\_y, x] \\[5 pt]
&= y[p\_z,x] + [y, x]p\_z - z[p\_y, x] - [z, x]p\_y \\[5 pt]

&= 0 \\[5 pt]
\end{align}
 }

[

L
x

,

p
z

]

=
[
y

p
z

−
z

p
y

,

p
z

]

=
[
y

p
z

,

p
z

]
−
[
z

p
y

,

p
z

]

=
y
[

p
z

,

p
z

]
[
y
,

p
z

]

p
z

−
z
[

p
y

,

p
z

]
−
[
z
,

p
z

]

p
y

=
−
[
z
,

p
z

]

p
y

=
−
i
ℏ

p
y

{\displaystyle \begin{align}
\left[L\_x, p\_z \right] &= [y p\_z - z p\_y, p\_z] \\[5 pt]
&= [y p\_z, p\_z] - [z p\_y, p\_z] \\[5 pt]
&= y[p\_z, p\_z] + [y, p\_z]p\_z - z[p\_y, p\_z] - [z, p\_z] p\_y \\[5 pt]
&= -[z, p\_z] p\_y \\[5 pt]
&= -i\hbar p\_y
\end{align} }

Assembling all the above parts together, we get the final result:

[

L
x

,

L
y

]

=
[

L
x

,
z
]

p
x

z
[

L
x

,

p
x

]
−
[

L
x

,
x
]

p
z

−
x
[

L
x

,

p
z

]

=
−
i
ℏ
y

p
x

i
ℏ
x

p
y

=
i
ℏ
(
x

p
y

−
y

p
x

)

=
i
ℏ

L
z

{\displaystyle \begin{align}
\left[L\_x, L\_y \right] &= [L\_x, z ] p\_x + z [L\_x, p\_x] - [L\_x, x] p\_z - x [L\_x, p\_z] \\[5 pt]
&= -i\hbar y p\_x + i\hbar x p\_y \\[5 pt]
&= i\hbar (x p\_y - y p\_x) \\[5 pt]
&= i\hbar L\_z
\end{align}

 }

By going through the cyclic permutations of the indices

(
x
→
y
,
y
→
z
,
z
→
x
)

{\displaystyle (x \to y, y \to z, z \to x) }

, we will discover that the components of angular momentum satisfy the following commutation relations:

[

L
^

x

,

L
^

y

# ]
i
ℏ

L
^

z

[

L
^

y

,

L
^

z

# ]
i
ℏ

L
^

x

[

L
^

z

,

L
^

x

# ]
i
ℏ

L
^

y

{\displaystyle [\hat{L}\_x, \hat{L}\_y] = i\hbar \hat{L}\_z \qquad
[\hat{L}\_y, \hat{L}\_z] = i\hbar \hat{L}\_x \qquad
[\hat{L}\_z, \hat{L}\_x] = i\hbar \hat{L}\_y \qquad }

The fact that the components of angular momentum do not commute indicates that we can only specify one component at a time. This also shows that the components of angular momentum cannot share the same eigenvectors. Although we cannot ascertain all of the individual components of angular momentum simultaneously, there is more information available about the state given by the square of the angular momentum operator:

L
^

2

=

L
^

x
2

L
^

y
2

L
^

z
2

{\displaystyle \hat{L}^2 = {\hat{L}^2\_x} + {\hat{L}^2\_y} + {\hat{L}^2\_z}}

. One interesting and important property to prove is that the square of the angular momentum operator

L
^

2

{\displaystyle \hat{L}^2}

 commutes with every component of angular momentum

(

L
^

x

,

L
^

y

,

L
^

z

)

{\displaystyle (\hat{L}\_x, \hat{L}\_y, \hat{L}\_z)}

.

In addition, there are two more notable operators related to angular momentum called the **raising operator** and **lowering operator**. The raising operator is defined as

L
^

=

L
^

x

i

L
^

y

{\displaystyle \hat{L}\_+ = \hat{L}\_x + i\hat{L}\_y }

 and the lowering operator is

L
^

−

=

L
^

x

−
i

L
^

y

{\displaystyle \hat{L}\_- = \hat{L}\_x - i\hat{L}\_y }

. The raising and lowering operators are useful for deriving the eigenvalues of angular momentum operators:

L
^

2

l
,
m
⟩

=

ℏ
2

l
(
l
1
)

l
,
m
⟩

L
^

z

l
,
m
⟩

=
ℏ
m

l
,
m
⟩

{\displaystyle \begin{align}
\hat{L}^2 |l,m \rangle &= \hbar^2 l(l + 1)|l,m\rangle \\[5 pt]

\hat{L}\_z |l,m\rangle &= \hbar m |l,m \rangle

\end{align} }

and

L
^

l
,
m
⟩

=
ℏ

l
(
l
1
)
−
m
(
m
1
)

l
,
m
1
⟩

L
^

−

l
,
m
⟩

=
ℏ

l
(
l
1
)
−
m
(
m
−
1
)

l
,
m
−
1
⟩

{\displaystyle \begin{align}
\hat{L}\_+ |l,m \rangle &= \hbar \sqrt{l(l + 1)-m(m+1)}|l,m+1\rangle \\[5 pt]
\hat{L}\_- |l,m \rangle &= \hbar \sqrt{l(l + 1)-m(m-1)}|l,m-1\rangle
\end{align} }

where

l

{\displaystyle l}

 is called **azimuthal quantum number**, and

m

{\displaystyle m}

 (usually denoted

m
l

{\displaystyle m\_l}

) is called the **magnetic quantum number**. For a more detailed discussion on the eigenvalue problem for angular momentum, click here.

## **Uncertainty Relations**

Using the generalized uncertainty principle for two operators

A
^

{\displaystyle \hat{A}}

 and

B
^

{\displaystyle \hat{B} }

σ

A

2

σ

B

2

≥

(

1

2
i

⟨
[

A
^

,

B
^

]
⟩

)

2

{\displaystyle {\sigma}\_{A}^{2}{\sigma}\_{B}^{2} \ge {\left(\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\right)}^{2}}

we can calculate the uncertainty relations for the components of angular momentum using the commutators. For example, for the

L
^

x

{\displaystyle \hat{L}\_x}

 and

L
^

y

{\displaystyle \hat{L}\_y}

σ

L
x

2

σ

L
y

2

≥

(

1

2
i

⟨
[

L
^

x

,

L
^

y

]
⟩

)

2

.

{\displaystyle {\sigma}\_{L\_x}^{2}{\sigma}\_{L\_y}^{2} \ge {\left(\frac{1}{2i}\langle[\hat{L}\_x,\hat{L}\_y]\rangle\right)}^{2}.}

Simply substitute the commutator relation

[

L
^

x

,

L
^

y

# ]
i
ℏ

L
^

z

{\displaystyle [\hat{L}\_x,\hat{L}\_y] = i\hbar \hat{L}\_z}

 into the uncertainty principle to get

σ

L
x

2

σ

L
y

2

≥

(

ℏ
2

⟨

L
^

z

⟩

)

2

{\displaystyle {\sigma}\_{L\_x}^{2}{\sigma}\_{L\_y}^{2} \ge {\left(\frac{\hbar}{2}\langle\hat{L}\_z\rangle\right)}^{2}}

or

σ

L
x

σ

L
y

≥

ℏ
2

⟨

L
^

z

⟩
.

{\displaystyle {\sigma}\_{L\_x}{\sigma}\_{L\_y} \ge \frac{\hbar}{2}\langle\hat{L}\_z\rangle.}

## **Problems**

### **Problem 1: Angular Momentum Operator Relations**

Prove the following relations for the angular momentum operator

L
^

=
(

L
^

x

,

L
^

y

,

L
^

z

)

{\displaystyle \hat{L} = (\hat{L}\_x, \hat{L}\_y, \hat{L}\_z)}

:

(
1
)

[

L
2

^

,

L
^

# ]
0

(
2
)

L
^

×

L
^

=
i
ℏ

L
^

{\displaystyle \begin{align}
&(1) \; [\hat{L^2},\hat{L}] = 0 \\[5 pt]
&(2) \; \hat{L} \times \hat{L} = i\hbar \hat{L}
\end{align}}

### **Problem 2: Ladder Operators**

The raising operator and lowering operator for angular momentum are defined as

L
^

=

L
^

x

i

L
^

y

{\displaystyle \hat{L}\_+ = \hat{L}\_x + i\hat{L}\_y }

 and

L
^

−

=

L
^

x

−
i

L
^

y

{\displaystyle \hat{L}\_- = \hat{L}\_x - i\hat{L}\_y }

 respectively. Show that

L
^

2

=

L
^

z
2

1
2

(

L
^

L
^

−

L
^

−

L
^

)

{\displaystyle \hat{L}^2 = \hat{L}^2\_z + \frac{1}{2}(\hat{L}\_+\hat{L}\_- + \hat{L}\_-\hat{L}\_+)}

.

### **Problem 3:** **Commutation Relations of Ladder Operators**

The raising operator and lowering operator for angular momentum are defined as

L
^

=

L
^

x

i

L
^

y

{\displaystyle \hat{L}\_+ = \hat{L}\_x + i\hat{L}\_y }

 and

L
^

−

=

L
^

x

−
i

L
^

y

{\displaystyle \hat{L}\_- = \hat{L}\_x - i\hat{L}\_y }

 respectively. Prove the following relations:

(
1
)

[

L
^

2

,

L
^

# ]
[

L
^

2

,

L
^

−

# ]
0

(
2
)

[

L
^

z

,

L
^

# ]
ℏ

L
^

(
3
)

[

L
^

z

,

L
^

−

# ]
−
ℏ

L
^

−

(
4
)

[

L
^

,

L
^

−

# ]
2
ℏ

L
^

z

{\displaystyle \begin{align}
&(1) \; [\hat{L}^2,\hat{L}\_+] = [\hat{L}^2,\hat{L}\_-] = 0 \\[5 pt]
&(2) \; [\hat{L}\_z,\hat{L}\_+] = \hbar \hat{L}\_+ \\[5 pt]
&(3) \; [\hat{L}\_z,\hat{L}\_-] = -\hbar \hat{L}\_- \\[5 pt]
&(4) \; [\hat{L}\_+,\hat{L}\_-] = 2\hbar \hat{L}\_z \\
\end{align}}