chapter 13. angular momentum: general theory

13. Angular Momentum Theory, February 10, 2014 1

Chapter 13. Angular Momentum: General Theory

Section 13.1. Definition and Basic Properties of the Angular Momentum

§1 Introduction

§2 The definition of the angular momentum operator

§3 L2 commutes with Lx, Ly, and Lz

§4 The eigenstates of L2 and Lz

§5 L2 and Lz commute with other operators

§6 A more convenient notation

Section 13.2. The Eigenvalues of L2 and Lz

§7 A summary of our task

§8 The ladder operators

§9 Find expressions for L+|Lz; j,m〉 and L−|Lz; j,m〉§10 The constants C+(j,m) and C−(j,m)

§11 The matrix elements of L+, L−, Lx, Ly in the basis set |j,m〉§12 Determining j and m

§13 For a given j, m has a highest and a lowest value

§14 The values of m and j

§15 Summary

Section 13.3. Eigenvalue Problems for Angular Momentum Operators

§16 Introduction

§17 The matrix-and-vector representation for angular momentum problems

§18 The case j = 1

§19 The eigenvalues and eigenvectors of Lx for j = 1

§20 The energy of the electron in the hydrogen atom exposed to a magnetic field


§21 Physical consequences

§22 Using a “dumb” choice of axis

Appendix 13.1. A Compact Form of the Commutation Relations

Appendix 13.2. Commutator Algebra

c©2014 Horia Metiu

Section 13.1. Definition and Basic Properties of the Angular Mo-

mentum

§ 1 Introduction. There are many reasons why you should study angular

momentum in quantum mechanics. It describes rotational motion, which

is essential for understanding microwave and infrared spectra of molecules.

Moreover, the spin of a particle is an angular momentum. A particle whose

spin differs from zero has a magnetic moment which means that we can

change its energy by acting on it with an external magnetic field. NMR and

ESR spectroscopies are based on this property. You cannot understand these

techniques without a through understanding of the properties of angular

momentum. Finally, the total angular momentum is conserved during a

collision between two molecules. This conservation places limitations on the

reaction probability and the energies of the products after reaction.

In this chapter we define angular momentum through the commutation

relations between the operators representing its projections on the coordinate

axes. These commutation relations allow us to determine the eigenstates of

the angular momentum operator and to derive all matrix elements needed in

calculations. The present chapter derives the general theory of angular mo-


mentum. The subsequent ones apply this knowledge to derive the properties

of spin and of the orbital angular momentum.

§ 2 The definition of the angular momentum operator. In quantum me-

chanics, we encounter two kinds of angular momenta. The orbital angular

momentum is similar to the angular momentum in classical mechanics, which

is defined by

~L = ~r × ~p (1)

where ~r is the position of a particle and ~p is its momentum. To define

the operator ~L that represents the orbital angular momentum in quantum

mechanics, we replace, in Eq. 1, ~r and ~p with the appropriate operators ~r

and ~p. This means that

~L = ~r × ~p (2)

In addition to the orbital angular momentum, the particles studied in

quantum mechanics have an angular momentum called spin. This is not

caused by the motion of the particle (e.g. by its rotation); it is instead an

intrinsic property of the particle similar to charge or mass. This angular

momentum cannot be defined by Eq. 2.

Born, Heisenberg, and Jordan1 noted that the angular momentum defined

by Eq. 2 satisfies the commutation relations

[Lx, Ly] = ihLz (3)

1M. Born, W. Heisenberg, and P. Jordan, Zur Quantenmechanik II, Zeitschrift

fur Physik 35 (8–9), 557-615 (published August 1926; received November 16, 1925).

http://link.springer.com/article/10.1007/BF01379806. English translation in: B. L. van

der Waerden, editor, Sources of Quantum Mechanics (Dover Publications, 1968, ISBN

0-486-61881-1).


[Ly, Lz] = ihLx (4)

[Lz, Lx] = ihLy (5)

The angular momentum is a vector and the operators Lx, Ly and Lz are

the components of this vector on a Cartesian coordinate system. They rec-

ommended that Eqs. 3–5 be used as a definition: a vector operator whose

components satisfy Eqs. 3–5 represents an angular momentum. This defini-

tion is valid for both the orbital angular momentum and the spin. A compact

way of writing these expressions, using the Levi-Civita tensor, is explained

in Appendix 13.1. That notation is helpful when programming angular-

momentum calculations in Mathematica. It also allows a more compact, but

harder to decipher, notation.

In this chapter we will use often a variety of expressions involving com-

mutators; Appendix 13.2 reviews several useful relationships involving them.

§ 3 L2 commutes with Lx, Ly, and Lz. We know from classical mechanics

that the rotational energy is proportional to the square of the angular mo-

mentum vector. Since this is an important quantity, we are naturally led to

examine the operator

L2 = L2x + L2

y + L2z (6)

I show below that this operator commutes with Lx, Ly, and Lz. We already

know that the components of the angular momentum do not commute with

each other and therefore they cannot have joint eigenvalues. Physically this

means that there is no state in which two of the projections of the angular

momentum have well defined values. However, there are states that are eigen-

states of the angular momentum squared and one of its projections (e.g. Lz).


This means that one can create in the laboratory states in which a measure-

ment of either the square of the angular momentum or of its projection on a

(Cartesian) coordinate axis will result in unique, well defined values (in other

words the probability that these values are determined in a measurement is

equal to 1).

Showing that L2 commutes with Lx, and Ly and Lz, is a matter of pa-

tiently using the commutator properties. We start with (use Eq. 6)

[Lz, L2] = [Lz, L

2

x] + [Lz, L2

y] + [Lz, L2

z] (7)

Any operator commutes with a power of itself, so

[Lz, L2

z] = 0 (8)

Next I use the relationship, given in Appendix 13.2 (Eq. 162),

[A, B2] = [A, B]B + B[A, B],

which gives

[Lz, L2

x] = [Lz, Lx]Lx + Lx[Lz, Lx] (9)

Using Eq. 5 in Eq. 9, to obtain

[Lz, L2

x] = ihLyLx + ihLxLy = ih(

LyLx + LxLy

)

(10)

You can show similarly that

[Lz, L2

y] = −ih(

LyLx + LxLy

)

(11)

Using Eqs. 8, 10, and 11 in Eq. 7 gives

[L2, Lz] = 0 (12)


In the same way, you can show that

[L2, Lx] = 0 (13)

and

[L2, Ly] = 0 (14)

Common sense tells us that if Eq. 12 is true, then Eqs. 13 and 14 must also

be true, because there is no physical agent that differentiates between the

OZ, and OX, and OY directions, unless one acts with a vectorial external

field on the system.

§ 4 The eigenstates of L2 and Lz. Since L2 commutes with Lz, there exists

a set of eigenfunctions of L2 that are also eigenfunctions of Lz. Because Lz

does not commute with Lx or with Ly, a joint eigenstate of L2 and Lz cannot

be an eigenstate of Lx or of Ly.

Let us examine in more detail what this means. Since L2 and Lz commute,

there exists a set of kets |Lz;α, β〉 that satisfy the eigenvalue equations for

those two operators:

L2|Lz;α, β〉 = α|Lz;α, β〉 (15)

Lz|Lz;α, β〉 = β|Lz;α, β〉 (16)

The symbol Lz inside the ket |Lz;α, β〉 reminds me that this is an eigenket of

Lz, not of Lx or Ly. If the system is in the state |Lz;α, β〉, we are guaranteed

to get the value α if we measure L2 and the value β if we measure Lz. α

is related to the rotational energy. β is the value that the projection of the

angular momentum on the Z-axis can take, which is related to the orientation

of the vector ~L.


But L2 also commutes with Lx, so there must exist a set of kets |Lx;α, β〉such that

L2|Lx;α, β〉 = α|Lx;α, β〉 (17)

Lx|Lx;α, β〉 = β|Lx;α, β〉 (18)

As long as no electric or magnetic field acts on the system there is no physical

difference between the OX and the OZ axes. Therefore, the values that β

can take when the system is in the state |Lx;α, β〉 are the same as the values

it can take when the system is in |Lz;α, β〉.In what follows, we concentrate on the joint eigenstates of L2 and Lz. At

this point there is no reason to prefer Lz over Lx or Ly, but we must pick

one.

Suppose that we have performed an experiment that places the system

in the state |Lz;α, β〉. We can measure L2 and get the value α, or we can

measure Lz and get the value β. However, if we measure Lx, when the system

is in state |Lz;α, β〉, the probability that Lx has the value β ′ and L2 has the

value α′ is

|〈Lx;α′, β ′ |Lz;α, β〉|2

When the system is in the state |Lz;α, β〉, we know L2 and Lz with certainty

but we only know the probability that Lx takes a given value. For someone

familiar with quantum mechanics, this is not surprising. Lz and Lx do not

commute so the eigenstate |Lz;α, β〉 cannot be an eigenstate of Lx. In ad-

dition, because Lz and Lx do not commute, the uncertainties in their values

are connected by the Robinson inequality.

These results are incomprehensible within classical physics, where there


is no physical law to prevent us from knowing with certainty all three pro-

jections of any vector once we know the state of the system (i.e. the values

of the position and of the momentum).

§ 5 L2 and Lz commute with other operators. One difficulty in grasping

the physics of angular momentum comes from the fact that in many con-

crete problems, Lx, Ly, and Lz are not the only operators describing the

system. For example, we might examine two particles interacting through a

central force. This means that the potential energy V (r) depends only on

the distance r between the particles and is independent of their orientation

in space. You know two examples of such systems: the hydrogen atom and

the diatomic molecule.

The eigenvalue problem for the Hamiltonian of the hydrogen atom in

spherical coordinates is (see Metiu, Quantum Mechanics, p. 206)

− h2

2µ

1

r2

∂

∂r

(

r2∂ψ

∂r

)

+L2

2µr2ψ + V (r)ψ = Eψ (19)

We see that besides the angular momentum squared L2, the equation contains

the potential energy V (r) and the radial kinetic energy (the first term in

Eq. 19). You have also learned that the Hamiltonian H of the hydrogen

atom commutes with L2 and with Lz. This means that H, L2, and Lz have

common eigenstates, denoted by |n, j,m〉. In the state |n, j,m〉, the energy

of the atom is (see Metiu, p. 300)

En = − µe2z2

2(4πε0)2h2

1

n2, (20)

the square of the angular momentum is

h2j(j + 1), (21)


and the projection of the angular momentum on the z-axis is

hm (22)

In other words, we have

H|n, j,m〉 = − µe2z2

2(4πε0)2h2

1

n2|n, j,m〉 (23)

L2|n, j,m〉 = h2j(j + 1)|n, j,m〉 (24)

Lz|n, j,m〉 = hm|n, j,m〉 (25)

Here j indicates the eigenvalue of L2, and m indicates the eigenvalue of Lz.

We learn from this example that the eigenstates of L2 and Lz may also be

eigenstates of other observables that commute with L2 and Lz (in this case,

the energy). These eigenstates are labeled by additional quantum numbers

(n, in this case); they are |n, j,m〉, not just |j,m〉.A similar situation appears in the case of the diatomic molecules. If we

make the rigid-rotor approximation and the harmonic approximation, the

eigenvalue equation for the energy, Eq. 19, becomes (see Metiu, Chapter 16)

− h2

2µ

1

r2

∂

∂r

(

r2∂ψ

∂r

)

+L2

2µr20

ψ +1

2k(r − r0)

2ψ = Eψ (26)

The eigenstates are |ψ〉 = |v, j,m〉, in which the molecule has the vibrational

energy

Ev = hω(

v +1

2

)

(27)

and the values of L2 and Lz given by Eqs. 21 and 22. Eqs. 24 and 25 hold

for this example also.

In general, when L2 and Lz commute with other operators, their eigen-

states must be eigenstates of those operators. When we need to make this


explicit, these eigenstates will be labeled |a, j,m〉, where a tells us the values

that the quantities represented by these other operators take, when the sys-

tem is in a pure state |a, j,m〉 and a measurement of those other quantities

is made.

§ 6 A more convenient notation. I now change notation by writing

α ≡ h2j(j + 1) (28)

β ≡ hm (29)

|Lz;α, β〉 ≡ |Lz; j,m〉 (30)

At this point pretend that you have never heard of angular momentum and

that these equations are a change of variable to be justified because they

simplify the algebra that follows. In this state of blissful ignorance you do

not know the values of α and β and therefore you do not know the values of

j and m. In what follows, I will derive the possible values of j and m (i.e.

the eigenvalues of L2 and Lz).

In the new notation, the eigenvalue equations Eqs. 15 and 16 become

L2|Lz; j,m〉 = h2j(j + 1)|Lz; j,m〉 (31)

and

Lz|Lz; j,m〉 = hm|Lz; j,m〉 (32)

Setting α = h2j(j + 1) needs further examination. We must have

α ≥ 0 (33)

because α is a value that L2 can take. Our change of notation connects j to

α through the equation α = h2j(j + 1) which has two solutions for j:


j1 = −1

2

(

1 +√

1 + 4α/h2

)

(34)

and

j2 =1

2

(

−1 +√

1 + 4α/h2

)

(35)

Because α/h2 ≥ 0, both roots j1 and j2 are real, j1 is negative, and j2 is non-

negative. Remember that α is a physical observable while j is an arbitrary

mathematical quantity introduced for convenience. You can verify that the

negative values of j give the same values of α as the non-negative ones. We

can choose either for indexing the kets, and we choose

j ≥ 0 (36)

Note also that m must be a real number (hm is a value of Lz). Because

Lz is equally likely to be positive or negative, we expect m to be able to

take both positive and negative values. Moreover, we guess that if m is an

allowed value, then −m is also allowed since the projection of the angular

momentum vector on the OZ axis can be oriented along OZ (m > 0) or

opposite to it (m < 0), and in a field-free region of space these two states are

equally probable.

Section 13.2. The Eigenvalues of L2 and Lz

§ 7 A summary of our task. To set up the theory of angular momentum,

we need to do the following.

1. Since most calculations in quantum mechanics are done by representing

operators by matrices, we need to use the kets |Lz; j,m〉 as a basis set


and find out how to calculate the matrix elements 〈Lz; j,m | O |Lz; j′, m′〉

for any operator O that is a function of Lx and/or Ly.

2. We also need to find the values of j and m allowed by the eigenvalue

equations Eqs. 31–32.

To reach these objectives, we only have the commutation relations Eqs. 3–

5 and our wits. The strategy is simple but tedious. We know how L2 and

Lz act on |Lz; j,m〉. Therefore we must use the commutation relations to

express Lx and Ly in terms of L2 and Lz. Determining the allowed values of

j and m is more subtle and cannot be reduced to a few sentences.

§ 8 The ladder operators. In vector calculus, one often describes a vector

vx, vy, vz through its spherical components vx + ivy, vx − ivy, vz. It might

therefore be useful to take a look at the operators

L+ ≡ Lx + iLy (37)

and

L− ≡ Lx − iLy (38)

along with Lz and L2. If we find out how L+ and L− act on |Lz; j,m〉, then

it is easy to use

Lx =L+ + L−

2(39)

and

Ly =L+ − L−

2i(40)

to find out how Lx and Ly act on |Lz; j,m〉. Working with L+ and L− will

prove fruitful.


Our calculations rely on the commutation relations and I derive here a

number of relations involving commutators that we need. The commutators

of L+ and L− with L2 and Lz are easily derived from the definitions, Eqs. 37

and 38, and the original commutation relations, Eqs. 3–5. They are

[Lz, L+] = hL+ (41)

and

[Lz, L−] = −hL− (42)

We also have

[L2, L+] = [L2, L−] = 0 (43)

because L2 commutes with both Lx and Ly, hence with both L+ and L−.

To carry out our program of determining what happens when L+ and L−

act on |Lz; j,m〉, we need to find expressions that contain L+ and/or L− in

the left-hand side and only Lz and L2 in the right-hand side. I derive below

several such expressions. One of them is

L−L+ = (Lx − iLy)(Lx + iLy) (used Eqs. 37 and 38)

= L2

x + L2

y + i(LxLy − LyLx)

= L2

x + L2

y + i[Lx, Ly]

= L2

x + L2

y + i2hLz (used Eq. 3)

= L2 − L2

z − hLz (used L2 = L2

x + L2

y + L2

z) (44)

This is the kind of equation we seek because it expresses L+ and L− in terms

of L2 and Lz. This is good because I know how the latter two operators act

on | Lz; j,m〉. Similarly

L+L− = L2 − L2

z + hLz (45)


We can add Eqs. 44 and 45 to obtain

L+L− + L−L+

2= L2 − L2

z (46)

Using the fact that the operators Lx, Ly, and Lz represent observables and

are therefore Hermitian (so that L†x = Lx and (iLx)

† = −iLx, etc.), we have

L†+ = L− (47)

and

L†− = L+ (48)

I can rewrite Eq. 46, with the help of Eqs. 47 and 48, as

L†−L− + L†

+L+

2= L2 − L2

z (49)

These manipulations were made to ensure that the right-hand side contains

operators whose action on |Lz; j,m〉 is known (see Eqs. 31 and 32).

§ 9 Find expressions for L+|Lz; j,m〉 and L−|Lz; j,m〉. For convenience,

I will drop the marker Lz from the eigenkets |Lz; j,m〉 and reinstate it only

when there is some possibility of confusion.

The fact that L2 and L+ commute leads to

L2L+|j,m〉 = L+L2|j,m〉 = h2j(j + 1)L+|j,m〉 (50)

This equation tells us that if |j,m〉 is an eigenstate of L2 with the eigenvalue

h2j(j+1) then L+|j,m〉 is also an eigenstate of L2, with the same eigenvalue

h2j(j + 1).

Since [Lz, L+] = hL+ (Eq. 41), we have

(

LzL+ − L+Lz

)

|j,m〉 = hL+|j,m〉 (51)


Use Lz|j,m〉 = hm|j,m〉 in the left-hand side and rearrange the terms to

obtain:

Lz

(

L+|j,m〉)

= h(m+ 1)(

L+|j,m〉)

(52)

Eq. 52 tells us that L+|j,m〉 is either zero, which is uninteresting, or it is an

eigenstate of Lz with the eigenvalue h(m+ 1). This means that

L+|j,m〉 = C+(j,m)|j,m+ 1〉 (53)

where C+(j,m) is a number that may or may not depend on j and m. It is

prudent to assume that it does.

Note that in Eq. 53, L+ acts on |j,m〉 and does not affect the value of j

but changes m. This is in agreement with Eq. 50, which says that L+|j,m〉is an eigenfunction of L2 with the eigenvalue h2j(j + 1).

A similar argument starts from [Lz, L−] = −hL− (Eq. 42) and concludes

that

L−|j,m〉 = C−(j,m)|j,m− 1〉 (54)

where C−(j,m) is another, unknown constant. Also, L−|j,m〉 is an eigen-

function of L2 corresponding to the eigenvalue h2j(j + 1).

The operators L+ and L− are called ladder operators; L+ is called a raising

operator and L−, a lowering operator. When L+ acts on |j,m〉, it increases

m by 1 and multiplies the result by C+(j,m); when L− acts on |j,m〉, it

decreases m by 1 and multiplies the result by C−(j,m). The operators L+

and L− do not affect j in |j,m〉, because both L+ and L− commute with L2

and therefore they have common eigenstates with L2. In other words, they

must convert a state in which L2 has the value h2j(j + 1) into a state in

which L2 has the same value h2j(j + 1).


§ 10 The constants C+(j,m) and C−(j,m). To determine C+(j,m) or

C−(j,m), we need to derive first a few helpful results. Consider

|λ〉 = A|x〉 (55)

where |x〉 is an arbitrary ket and A is an arbitrary operator. We have

〈λ |λ〉 = 〈Ax | Ax〉 = 〈x | A†A |x〉 (56)

I used here the property 〈Ax | z〉 = 〈x | A†z〉 of adjoint operators (|x〉 and |z〉are arbitrary kets).

In addition, if

A|x〉 = a|y〉 (57)

where a is a number, then

〈λ |λ〉 = 〈ay | ay〉 = a∗a〈y | y〉 (58)

Let us use Eqs. 56 and 58 to determine C+. To do this I apply Eq. 56 to

|λ〉 ≡ L+|j,m〉 (59)

and obtain

〈λ |λ〉 = 〈j,m | L†+L+ | j,m〉 (60)

We also have (see Eq. 53)

|λ〉 = L+|j,m〉 = C+|j,m+ 1〉 (61)

Use this to calculate

〈λ |λ〉 = C∗+C+〈j,m+ 1 | j,m+ 1〉 = C∗

+C+ (62)


I used the fact that 〈j,m + 1 | j,m + 1〉 = 1. The two expressions for 〈λ |λ〉must be equal and therefore

〈j,m | L†+L+ | j,m〉 = C∗

+C+ (63)

I am very pleased with this equation, for two reasons. First, I obtained

a formula for C+. Second, this equation contains L⊥+L+, which, according to

Eq. 44 is (use L⊥+ = L−)

L−L+ = L2 − L2

z − hLz

This expresses L−L+ in terms of L2 and Lz. I know how L2 and Lz act

on |j,m〉 and therefore I can readily calculate the matrix element in Eq. 63.

This gives

〈j,m | L−L+ | j,m〉 = h2j(j + 1) − h2m2 − hm (64)

Eq. 63 determines the absolute value of C+ but not its phase. Since all

physical properties of a ket are independent of a phase factor in front of it,

I can take in C+ any phase I want. I choose it so that

C+(j,m) =√

〈j,m | L†+L+ | j,m〉 (65)

Now use Eq. 44 in Eq. 65:

C+(j,m) =

√

〈j,m |(

L2 − L2z − hLz

)

| j,m〉 (66)

Using the eigenvalue equations for L2 and Lz (Eqs. 31–32), we rewrite Eq. 66

as

C+(j,m) = h√

j(j + 1) −m(m+ 1) (67)

(we used 〈j,m | j,m〉 = 1).


Finally, putting C+(j,m) given by Eq. 67 into Eq. 53 gives

L+|j,m〉 = h√

j(j + 1) −m(m+ 1) |j,m+ 1〉 (68)

Note that L+|j,m〉 is an eigenstate of L2 corresponding to the eigenvalue

hj(j + 1) and an eigenstate of Lz corresponding to the eigenvalue (m+ 1)h.

However, |j,m〉 is not an eigenstate of L+. It should not be, because L+

does not commute with L2 or with Lz.

To calculate C−(j,m), we follow the reasoning used to find C+(j,m). This

leads to

C−(j,m) = h√

j(j + 1) −m(m− 1) (69)

and

L−|j,m〉 = h√

j(j + 1) −m(m− 1) |j,m− 1〉 (70)

We can now calculate (use Eqs. 67, 70, 39)

Lx|j,m〉 =1

2(L+ + L−)|j,m〉

=1

2[C+(j,m)|j,m+ 1〉 + C−(j,m)|j,m− 1〉] (71)

Similarly (use Eqs. 67, 70, 40)

Ly|j,m〉 =1

2i[C+(j,m)|j,m+ 1〉 − C−(j,m)|j,m− 1〉] (72)

§ 11 The matrix elements of L+, L−, Lx, Ly in the basis set |j,m〉. Now

that we know how L+ and L− act on |j,m〉 (Eqs. 68 and 70), we can calculate

the matrix elements

〈j′, m′ | L+ | j,m〉 = h√

j(j + 1) −m(m+ 1) 〈j′, m′ | j,m+ 1〉

= h√

j(j + 1) −m(m+ 1) δj′jδm′,m+1 (73)


The only non-zero matrix elements are those for whichm′ = m+1 and j′ = j.

Therefore only

〈j,m + 1 | L+ | j,m〉 = h√

j(j + 1) −m(m+ 1) (74)

differs from zero.

The matrix elements of L− are

〈j′, m′ | L− | j,m〉 = h√

j(j + 1) −m(m− 1) δj′jδm′,m−1 (75)

and only

〈j,m− 1 | L− | j,m〉 = h√

j(j + 1) −m(m− 1) (76)

differs from zero.

Since (see Eq. 39)

Lx =L+ + L−

2

we have

〈j′, m′ | Lx | j,m〉 =1

2(C+(j,m) δj′jδm′,m+1 + C−(j,m) δj′jδm′,m−1) (77)

Only the matrix elements

〈j,m + 1 | Lx | j,m〉 =1

2C+(j,m) (78)

and

〈j,m− 1 | Lx | j,m〉 =1

2C−(j,m) (79)

differ from zero.

Using (see Eq. 40)

Ly =L+ − L−

2i


with Eqs. 73 and 75 leads to

〈j′, m′ | Ly | j,m〉 =1

2i(C+(j,m) δj′jδm′,m+1 − C−(j,m) δj′jδm′,m−1) (80)

Only the matrix elements

〈j,m+ 1 | Ly | j,m〉 =1

2iC+(j,m) (81)

and

〈j,m − 1 | Ly | j,m〉 = − 1

2iC−(j,m) (82)

differ from zero.

This covers all the matrix elements used in applications. However, we

still don’t know what values j and m can take, since we do not know the

eigenvalues of L2 and Lz. Next, we start working to find them.

§ 12 Determining j and m. We have decided to use the eigenstates |j,m〉 of

L2 and Lz as a basis set and have managed to calculate the matrix elements

of Lx, Ly, L+, and L−. The remaining task is to determine the allowed values

of j and m. We will find that j can only be integer or half-integer and that

m in | Lx; j,m〉 can only take integer or half-integer values between −j and

j. For example, if j = 1 then m can only be one of −1, 0, 1; if j = 1

2then m

can only be one of −1

2, +1

2.

§ 13 For a given j, m has a highest and a lowest value. A classical inter-

pretation of angular momentum tells us that if L2 is fixed then Lz must have

an upper and a lower bound. Indeed the projection of a vector on an axis

can at most be equal to the length ` of the vector: the projection has the

upper bound ` and the lower bound −`. Since quantum mechanics is not


classical mechanics, it will be reassuring to prove that, for a given j, m has

an upper and a lower bound.

Recall that (Eqs. 46 and 49)

L+L− + L−L+

2= L2 − L2

z (83)

andL†−L− + L†

+L+

2= L2 − L2

z (84)

From properties of the inner product, we know that for any ket |x〉 and

any operator A, we have 〈x | A†A |x〉 ≥ 0 (with equality only when A|x〉 = 0).

Taking the matrix element of the operator in Eq. 84 with the non-zero ket

|j,m〉 gives

〈j,m | L†−L− | j,m〉 + 〈j,m | L†

+L+ | j,m〉 = 2h2(j(j + 1) −m2) (85)

Since the left-hand side cannot be negative, we conclude that

j(j + 1) −m2 ≥ 0 (86)

This means that if j is fixed then m has both a lower bound and an upper

bound. We don’t know what the bounds are but we know that they exist, so

let us denote them by m` and mu.

This gives us two new equations to play with:

L+|j,mu〉 = 0 (87)

L−|j,m`〉 = 0 (88)

To derive them, observe that, according to Eq. 53, L+|j,mu〉 = C+(j,mu)|j,mu+

1〉. Since no value of m can be larger than mu, the state |j,mu + 1〉 can-

not exist and the ket is zero (which proves Eq. 87). One arrives at Eq. 88

similarly.


Now let us act with L− on Eq. 87 and with L+ on Eq. 88, to obtain

L−L+|j,mu〉 = 0 (89)

L+L−|j,m`〉 = 0 (90)

Why do I do that? Because I know that (Eq. 44)

L−L+ = L2 − L2

z − hLz;

the right-hand side of this equation contains only operators that act on |j,m〉according to known rules, which means that I can calculate

0 = L−L+|j,mu〉 = (L2 − L2

z − hLz)|j,mu〉

= (h2j(j + 1) − h2m2

u − h2mu)|j,mu〉 (91)

This implies that

j(j + 1) −m2

u −mu = 0 (92)

Similarly, using (see Eq. 45)

L+L− = L2 − L2

z + hLz

in Eq. 90 gives

j(j + 1) −m2

` +m` = 0 (93)

Solving Eq. 93 for j(j + 1) and using the result in Eq. 92 gives

mu(mu + 1) = m`(m` − 1) (94)

This equation has two solutions

mu = −m` (95)


and

mu = m` − 1 (96)

The second one is excluded by the fact that mu ≥ m` by definition.

We see therefore that mu = −m`. Symmetry requires the possible pro-

jections of L along OZ to be equal to and of opposite sign to the possible

projections in the direction opposite to OZ. Eq. 95 agrees with this require-

ment.

§ 14 The values of m and j. We have learned a lot but we still don’t know

which values j and m can take. Here we settle this question.

Let us consider a ket |j,m0〉 where m0 is an allowed value of m for the

given j. Acting on that ket with L+ repeatedly, we obtain

L+|j,m0〉 = C+(j,m0)|j,m0 + 1〉,L2

+|j,m0〉 = C+(j,m0)C+(j,m0 + 1)|j,m0 + 2〉,etc.

After nu steps, this procedure must generate a ket proportional to |j,mu〉. If

it did not, then we could apply L+ indefinitely and generate kets in which

m > mu; this contradicts the fact that m has a largest value, namely mu.

Therefore we must have

mu = m0 + nu (97)

where nu is a non-negative integer. Similarly, applying L− to |j,m0〉 some n`

times produces a ket proportional to |j,m`〉, so that

m` = m0 − n` (98)

with n` a non-negative integer. From Eqs. 97 and 98, it follows that

mu −m` = (m0 + nu) − (m0 − n`) = nu + n` ≡ n (99)


where n is a non-negative integer. Since we already know that mu = −m`

(Eq. 95), we conclude that

2mu = n (100)

We also know that (see Eq. 92)

j(j + 1) = m2

u +mu = mu(mu + 1) (101)

This has two solutions

j = mu (102)

and

j = −mu − 1 (103)

Eq. 103 is not compatible with the facts that mu ≥ 0 and j ≥ 0, so we

can eliminate it as irrelevant. We also know that mu = −m`, which used in

Eq. 102 gives

m` = −j (104)

Now combining mu = n/2 (Eq. 100) with Eq. 102 gives

j =n

2(105)

We have obtained a startling result: j can be either a non-negative integer

(if n is even) or a half-integer (if n is odd). In addition, the largest value of

m is j (Eq. 102) and the smallest value of m is −j (Eq. 104). Therefore m

can take the 2j + 1 values −j, −j + 1,. . . , j − 1, j.

The theory of orbital angular momentum (see Metiu, Chapter 14, p. 220ff),

based on the equation ~L = ~r× ~p, tells us that half-integer values of j are not

allowed for orbital angular momentum. However, by using the commutation


relations to define the angular momentum, we have found that j can be ei-

ther integer or half-integer. We know of no additional condition that we can

use to conclude that half-integers are not allowed in general. Therefore the

postulate that the have a half-integer intrinsic angular momentum (the spin)

is in harmony with conclusions derived from the commutation relations.

§ 15 Summary. This lengthy derivation led us to the following results. L2

and Lz have common eigenstates |Lz; j,m〉 for which

L2|Lz; j,m〉 = h2j(j + 1) |Lz; j,m〉 (106)

and

Lz|Lz; j,m〉 = hm |Lz; j,m〉. (107)

j can be either a non-negative integer or a positive half-integer (that is, of

the form integer/2). m can take 2j + 1 discrete values

m = −j,−j + 1,−j + 2, . . . , j − 2, j − 1, j (108)

For example, if j = 3

2then m can take 2 × 3

2+ 1 = 4 values, namely

m = −3

2, −3

2+ 1,

3

2− 1,

3

2= −3

2, −1

2,

1

2,

3

2

If j = 3 then m takes the 2 × 3 + 1 = 7 values −3,−2, 1, 0, 1, 2, 3.

We have also found that

Lx|Lz; j,m〉 =1

2[C+(j,m)|Lz; j,m+ 1〉 + C−(j,m)|Lz; j,m− 1〉] (109)

and

Ly|Lz; j,m〉 =1

2i[C+(j,m)|Lz; j,m+ 1〉 − C−(j,m)|Lz; j,m− 1〉] (110)


with

C+(j,m) = h√

j(j + 1) −m(m+ 1) (111)

and

C−(j,m) = h√

j(j + 1) −m(m− 1). (112)

Note that |j,m〉 = 0 if m 6∈ [−j, j].The matrix elements needed in computations are

〈Lz; j′, m′ | Lz | Lz; j,m〉 = δjj′δmm′hm (113)

〈Lz; j′, m′ | L2 | Lz; j,m〉 = δjj′δmm′h2j(j + 1) (114)

〈Lz; j′, m′ | Lx | Lz; j,m〉 = δjj′

[

C+(j,m)δm′,m+1 + C−(j,m)δm′,m−1

2

]

(115)

〈Lz ; j′, m′ | Ly | Lz; j,m〉 = δjj′

[

C+(j,m)δm′,m+1 − C−(j,m)δm′,m−1

2i

]

(116)

Note that all these matrix elements are zero if j 6= j′. This means that

in the basis set |Lz; j,m〉, the matrices of Lx and Ly consist of smaller

matrices strung along the diagonal; there are no non-zero matrix elements

〈j,m | Oj′, m′〉 in which j 6= j′ if O = Lx, Ly, L+, or L− or any function of

them.

Section 13.3. Eigenvalue Problems for Angular Momentum Oper-

ators

§ 16 Introduction. We have seen several times that kets are very useful for

theoretical analysis, but practical calculations often use the representation of

kets by vectors and of operators by matrices. In this section I review briefly

the general theory and then apply it to the case of j = 3 as an example.


which in vector-and-matrix notation is written as

Oψ = oψ (123)

or

O11 · · · O1N

......

ON1 · · · ONN

c1...

cN

= o

c1...

cN

(124)

The solution of Eq. 124 provides N eigenvectors

ψ(α) ≡ c1(α), . . . , cN (α), α = 1, . . . , N (125)

and N eigenvalues o(α), α = 1, . . . , N .

From these we can calculate the N eigenkets |ψ(1)〉, . . . , |ψ(N)〉 by using

|ψ(α)〉 =N∑

k=1

ck(α)|ηk〉, α = 1, 2, . . . , N (126)

§ 17 The matrix-and-vector representation for angular momentum problems.

It is natural to use for angular momentum calculations the basis set |j,m〉.In this basis set, the matrix elements of Lz, L

2, Lx, and Ly are given by

Eqs. 113–116. When looking at these equations, we notice that all matrix

elements in which j 6= j′ are zero. The matrices of the operators L2 and Lz


are diagonal and those of Lx and Ly are “block diagonal”:

0

0 •• 0 •

• 0

0 •• 0 •

• 0 •• 0 •

• 0. . .

All matrix elements other than the dots are also zero.

We have arranged the rows and the columns in the following order:

• j = 0, m = 0

• j = 1, m = −1

• j = 1, m = 0

• j = 1, m = +1

• etc.

The state with j = 0 generates a 1× 1 block, the states with j = 1 generate

a 3 × 3 block, . . . ; in general, the states having an arbitrary j generate a

(2j + 1) × (2j + 1) block. This break-up into blocks makes this basis set

particularly simple to use in calculations. Physically it signifies that states


with different values of j are not coupled by any of the operators L2, Lz, Lx,

and Ly.

Because of this “block” property, we can treat physical problems involving

a specific j within a reduced space generated by the basis set

|j,−j〉, |j,−j + 1〉, . . . , |j, j − 1〉, |j, j〉 (127)

§ 18 The case j = 1. Let us solve some problems for j = 1. The basis set

is

|1,−1〉, |1, 0〉, |1, 1〉 (128)

Since we know that j = 1, the first index is superfluous and we use instead

the notation | − 1〉, |0〉, |1〉. The matrix representing an operator O in this

subspace is

O =

〈−1 | O | − 1〉〈−1 | O | 0〉〈−1 | O | 1〉〈0 | O | − 1〉〈0 | O | 0〉〈0 | O | 1〉〈1 | O | − 1〉〈1 | O | 0〉〈1 | O | 1〉

(129)

The look of a matrix depends on the order of the kets in the basis set; the

physical conclusions derived by using the matrix do not. However, vectors

and matrices must be derived with the same basis set in order for us to use

ordinary matrix and vector algebra.

Eqs. 113 and 114 tell us that the matrix elements 〈1, m | L2 | 1, m′〉 and

〈1, m | Lz | 1, m′〉 are zero if m 6= m′. Therefore the corresponding matrices

are diagonal:

L2 = h2(1)(1 + 1)

1 0 0

0 1 0

0 0 1

(130)


and

Lz =

−h 0 0

0 0 0

0 0 h

(131)

No surprise here. The basis set consists of eigenstates |j,m〉 (with j = 1) of

L2 and Lz and therefore the matrices of these operators, in that basis set,

are diagonal and the diagonal elements are the eigenvalues of the operator.

The matrices corresponding to Lx and Ly are more interesting. Let us

look at Lx, and use Eq. 115 together with Eqs. 111 and 112. I calculate, as

an example, 〈−1 | Lx | − 1〉. We have

〈m′ = −1 | Lx |m = −1〉 = 0 (132)

because (see Eq. 115)

δm′,m+1 = δ−1,−1+1 = δ−1,0 = 0

and

δm′,m−1 = δ−1,−1−1 = δ−1,−2 = 0

Another example is (use Eq. 115)

〈m′ = −1 | Lx |m = 0〉 =C+(1, 0)δ−1,0+1 + C−(1, 0)δ−1,0−1

2=C−(1, 0)

2(133)

with

C−(1, 0) = h√

1(1 + 1) − 0(0 − 1) = h√

2 (134)

Together these give

〈−1 | Lx | 0〉 =h√2

(135)


Patiently calculating all the matrix elements leads to2

Lx =

0 h√2

0

h√2

0 h√2

0 h√2

0

(136)

These matrix elements were calculated3 in WorkBook13.Angular Momentum.

The matrix must be Hermitian because Lx represents an observable, and

therefore we need only calculate the matrix elements in the lower-left triangle;

the ones in the upper-right triangle are then obtained from 〈m′ | O |m〉 =

〈m | O |m′〉∗. In addition, the Kronecker deltas in Eq. 115 tell us that the

diagonal elements (that is, those with m′ = m) are zero and that in the

lower-left triangle, only 〈0 | Lx | − 1〉 and 〈1 | Lx | 0〉 are nonzero.

Exercise 1 Show that for j = 1

Ly =h√2

0 −i 0

i 0 −i

0 i 0

(137)

Is this matrix Hermitian?

2The top row is m′ = −1, the middle row is m′ = 0, and the bottom row is m′ = 1;

the left column is m = −1, the middle column is m = 0, and the right column is m = 1.

In all entries j = 1.3Before Mathematica was invented these calculations were done by hand and were so

tedious that many a student developed a sturdy dislike for quantum theory of angular

momentum. I have friends who avoided any research on subjects in which angular mo-

mentum was important. Nowadays the proofs are still tedious but at least the calculations

are easier.


Exercise 2 Verify that the matrices representing the operators satisfy the

commutation relations

LxLy − LyLx = ihLz

LyLz − LzLy = ihLx

LzLx − LxLz = ihLy

and that

L2 = L2

x + L2

y + L2

z

§ 19 The eigenvalues and eigenvectors of Lx for j = 1. Now that we

have a matrix representation of Lx, we can calculate the eigenvalues and

eigenvectors of this operator. We need to know them because they appear in

some problems in spectroscopy. I solved the eigenvalue problem in Cell 6 of

the Mathematica file WorkBook13.Angular momentum. The results are (Cell

6 of WorkBook 13):

eigenvalue eigenvector

−h 1

2, − 1√

2, 1

2

0 − 1√2, 0, 1√

2

h 1

2, 1√

2, 1

2

The eigenvectors are normalized. I remind you that in this basis set the

eigenvectors of Lz are 1, 0, 0, 0, 1, 0, and 0, 0, 1.We are pleased to see that Lx has the same eigenvalues as Lz. Since

there is no force on the system that breaks spherical symmetry, there is no


preferred direction in space. The possible values of the projection of the

vector ~L on the x-axis must be the same as those of the projection on the

z-axis.

Exercise 3 Show that Ly has the same eigenvalues as Lz.

The eigenstates of Lx in the basis set |Lz; 1, m〉 are (see WorkBook13, Cell

6)

eigenvalue eigenket

−h |Lx; 1,−1〉 =1

2|Lz; 1,−1〉 − 1√

2|Lz; 1, 0〉 +

1

2|Lz; 1, 1〉 (138)

0 |Lx; 1, 0〉 = − 1√2|Lz; 1,−1〉 +

1√2|Lz; 1, 1〉 (139)

h |Lx; 1, 1〉 =1

2|Lz; 1,−1〉 +

1√2|Lz; 1, 0〉 +

1

2|Lz; 1, 1〉 (140)

I have had to augment the notation with a new index: |Lz; j,m〉 are eigenkets

of Lz , and |Lx; j,m〉 are those of Lx.

If we manage to place the system in the state |Lx; 1,−1〉, a measure-

ment of Lx is guaranteed to give the result −h. However, if we measure Lz,

when the system is in the state |Lx; 1,−1〉, we obtain the value −h with the

probability

P−1 ≡ |〈Lz ; 1,−1 |Lx; 1,−1〉|2

=

[

1

2〈Lz; 1,−1 |Lz ; 1,−1〉 − 1√

2〈Lz ; 1,−1 |Lz ; 1, 0〉 +

1

2〈Lz; 1,−1 |Lz ; 1, 1〉

]2

=1

4(141)


To get this, I used the orthonormality condition 〈Lz; 1, m |Lz ; 1, m′〉 = δmm′.

Similarly, the probability P0 that Lz is zero is given by

P0 ≡ |〈Lz; 1, 0 |Lx; 1,−1〉|2 =1

2(142)

and the probability P1 that Lz is 1 by

P1 ≡ |〈Lz; 1, 1 |Lx; 1,−1〉|2 =1

4(143)

P−1 + P0 + P1 = 1, which is a must.

This result is consistent with the fact that Lz does not commute with

Lx and therefore there is no state in which the value of both is known with

certainty.

We can also calculate the average value of Lz,

〈Lz〉 ≡ 〈Lx; 1,−1 | Lz |Lx; 1,−1〉,

when the system is in the state |Lx; 1,−1〉. Using Eq. 138 in the right-hand

side (for |Lx; 1,−1〉) gives

〈Lx; 1,−1 | Lz |Lx; 1,−1〉 = 〈Lx; 1,−1 | Lz

(

1

2

)

|Lz ; 1,−1〉

− 〈Lx; 1,−1 | Lz

(

1√2

)

|Lz ; 1, 0〉

+ 〈Lx; 1,−1 | Lz

(

1

2

)

|Lz ; 1, 1〉

Using Eq. 138 again, as well as 〈Lz ; j,m′ | Lz |Lz; j,m〉 = δmm′hm, leads to

〈Lz〉 =(

1

2

)

(−h)(

1

2

)

−(

1√2

)

(0) +(

1

2

)

(h)(

1

2

)

= 0 (144)

This is the same as

〈Lz〉 =1∑

m=−1

hmPm = (−h)14

+ (0)1

2+ (h)

1

4= 0, (145)


as it should be. This result is physically reasonable. Since no forces break

spherical symmetry, there is no bias to prefer positive values of Lz over

negative values. The average is zero because the value −h is as probable as

h.

§ 20 The energy of the electron in the hydrogen atom exposed to a magnetic

field. When we discuss the hydrogen atom, I will show that when the atom

is exposed to a magnetic field ~B, the energy of its electron changes. The

Hamiltonian is

H = H0 −e

2me

~L · ~B (146)

where me is the mass of the electron, e is the proton charge, and H0 is the

Hamiltonian of the atom in the absence of the field.

I am ignoring other terms in the Hamiltonian since I only want to illus-

trate how we handle this kind of problem. The question is: what happens to

the energy of the electron when it is exposed to the magnetic field? I answer

first the question for the case when I was smart enough to pick the z-axis

along ~B. In this case ~B = 0, 0, B and ~L · ~B = Lz. The Hamiltonian is

H = H0 −e

2me

Lz B,

where B is the magnitude of the field ~B. Since Lz and H0 commute, they

have joint eigenstates, denoted by |n, j,m〉, where n gives the energy En in

the absence of the field.

I want to know what happens to the energy of the states |2, j,m〉 when

the field is on. To answer this question, I will use these states as a basis set.

Therefore I have

H|n, j,m〉 = H0|n, j,m〉 − eB

2me

Lz|n, j,m〉


=(

En − eB

2me

m)

|n, j,m〉 (147)

We see that |n, j,m〉 are eigenstates of the Hamiltonian defined by Eq. 146.

The energy spectrum is given by

En − eB

2me

m (148)

As noted above, En is the energy of the atom in the absence of the field:

En = −Ry

n2, (149)

where Ry is the Rydberg constant. For the case n = 2, in the absence of the

field, the atom has four degenerate states: |2, 0, 0〉, |2, 1,−1〉, |2, 1, 0〉, and

|2, 1, 1〉. If the field is on, the energy of these states changes to that given by

Eq. 148 and depends on the quantum number m. The states are

state |n, j,m〉 energy change

|2, 0, 0〉 −Ry/4 unchanged

|2, 1,−1〉 −Ry/4 + (eB/2me)h increased

|2, 1, 0〉 −Ry/4 unchanged

|2, 1, 1〉 −Ry/4 − (eB/2me)h decreased

The energy-level diagram changes, when the field is turned on, from having

four states of the same energy E2 to having three distinct energy levels,

E2 + eBh/2me, E2, and E2 − eBh/2me. The level having the energy E2 is

doubly degenerate: the states |2, 0, 0〉 and |2, 1, 0〉 have energy E2.

§ 21 Physical consequences. In the absence of the field, an atom excited in

any one of the levels of energy E2 will emit a photon of frequency

(E2 − E1)/h =[

−Ry

4−(

−Ry

1

)]

/h =Ry

h

(

4 − 1

4

)

(150)


When the atom is in a magnetic field, the E2 levels corresponding to j = 1

splits into three levels as shown in Fig. 1. In the absence of the magnetic

field, the atom excited in a state with n = 2 emits a photon of frequency

Ω2. When the field is present, three emission frequencies, Ω1, Ω2, and Ω3

are possible. Not all of them will be observed because of selection rules that

render the rate of some transitions equal to zero.4

E2,1

E2,0

E2,-1

E1,0

|2,1,1⟩

|2,1,0⟩ and |2,2,0⟩

|2,1,-1⟩

|1,0,0⟩

W1 W3W2

Figure 1: The energy levels of a hydrogen atom in a magnetic field for n = 1

and n = 2

4The spectrum of the hydrogen atom in a magnetic field will be discussed in detail in

a future chapter. For a complete discussion we need to take into account electron spin,

proton spin, and some relativistic effects, which add terms to the Hamiltonian in addition

to the ones used here.


§ 22 Using a “dumb” choice of axis. We decided to use as a basis set the

eigenstates |Lz;n, j,m〉 of Lz. Then we wisely decided that the z-axis coin-

cides with the vector ~B, so that the interaction energy between the electron

and the field became (e/2me)BLz. The Hamiltonian H = H0 − (e/2me)BLz

commutes with Lz and with L2 and has |Lz;n, j,m〉 as eigenstates. This

makes the matrix of H, in the chosen basis, diagonal. The eigenvalue prob-

lem is then trivial.

What would happen if we were not so clever and decided to take the

x-axis along ~B? We have ~B = B, 0, 0 and the Hamiltonian is

H = H0 − (e/2me)BLx (151)

The basis set is still |Lz;n, j,m〉. H no longer commutes with Lz and the

basis-set kets are no longer eigenstates of H. The matrix of H, in the basis

set |Lz;n, j,m〉, is no longer diagonal.

The matrix elements of the Hamiltonian are given by

〈Lz;n, j,m | H |Lz;n, j,m′〉 = −Ry

n2− eB

2me

〈Lz ;n, j,m | Lx |Lz;n, j,m′〉

(152)

For n = 2, we have j = 0 and m = 0, or j = 1 and m = −1, 0, 1. We have

listed the equations giving the matrix elements 〈Lz ; 2, j,m | Lx |Lz ; 2, j,m′〉 in

§15 (see Eqs. 115, 111, 112). The matrixH corresponding to the Hamiltonian

in Eq. 151 is (see WorkBook 13)


|2, 0, 0〉 |2, 1,−1〉 |2, 1, 0〉 |2, 1, 1〉

|2, 0, 0〉|2, 1,−1〉|2, 1, 0〉|2, 1, 1〉

−Ry/4 0 0 0

0 −Ry/4 − eB2me

h/√

2 0

0 − eB2me

h/√

2 −Ry/4 − eB2me

h/√

2

0 0 − eB2me

h/√

2 −Ry/4

The eigenvalues of this matrix were calculated in Cell 7 of WorkBook 13.

They are −Ry/4, −Ry/4, −Ry/4 − eB2me

h, and −Ry/4 + eB2me

h. They are

exactly the same as the values obtained when we took the z-axis along ~B.

They must be, because the coordinate system and its axes are only in our

heads. The atom does not know about them. The energies are measurable

quantities and if they depend on our choice of axes, the theory would be

erroneous. Note however that the eigenvectors of H with ~B taken along OX

differ from the eigenvectors of H with ~B taken along OZ. This is all right.

The eigenvectors are not measurable quantities. However, any observable

calculated with these eigenvectors is the same whether ~B is oriented along

OX or along OZ.

You see here a conspicuous feature of quantum mechanics. The mathe-

matics depends on the choice of basis set and coordinate axes, but the physics

does not.

Appendix 13.1. A Compact Form of the Commutation Relations

It is sometimes useful to write the commutation relations Eqs. 3–5 in a

more compact form. To do this we write the components of the vector ~L as


L1, L2, and L3, with L1 ≡ Lx, L2 ≡ Ly, and L3 ≡ Lz. Then we can write

[Lα, Lj] = ih3∑

k=1

εαjkLk, α = 1, 2, 3 (153)

Here εαjk is the Levi-Civita tensor. Since this symbol appears often in

physics, it is worth explaining it in detail. To do so, we need to intro-

duce the concept of permutation. We start with an ordered list of symbols,

such as 1, 2, 3, or A, B, C or m, α, 3. A permutation is an operation

that changes the order of the symbols in the list. For example, suppose the

permutation Π1 acts on 1, 2, 3 to produce 1, 3, 2. We can write this as

Π11, 2, 3 = 1, 3, 2. Π1 has nothing to do with the numbers 2 and 3; it

simply exchanges the second and third objects in the list. That is, it takes

the object in position 2 and places it in position 3, and it takes the object in

position 3 and places it in position 2. Because of this,

Π1A,B,C = A,C,B and Π1m, α, 3 = m, 3, α (154)

A transposition is a permutation that interchanges the position of two

elements in the list. Π1 is a transposition, but

Π21, 2, 3 = 2, 3, 1 (155)

is not. Any permutation can be written as a succession of transpositions.

For example, we can reach 2, 3, 1 from 1, 2, 3 by two transpositions:

1, 2, 3 → 2, 1, 3 → 2, 3, 1 (156)

This decomposition of a permutation into successive transpositions is not

unique. The permutation Π2 in Eq. 155 is also equivalent to the following


sequence of transpositions:

1, 2, 3 → 1, 3, 2 → 3, 1, 2 → 3, 2, 1 → 2, 3, 1 (157)

This takes four transpositions, while Eq. 156 took only two, but the outcome

is the same Π2.

One can prove the following theorem: the number of transpositions through

which a given permutation is expressed is either odd or even. In other words,

Π2 might be expressed using two or four or six transpositions, but never using

one, or three, or five. A permutation that can be decomposed into an even

number of transpositions has the signature +1 and is called ‘even’; one that

can be decomposed into an odd number of transpositions has the signature

−1 and is called ‘odd’.

Mathematica has a function Permutations[list] that generates all per-

mutations of the objects in the list. For example, the lists generated by

Permutations[1,2,3] are

Permutations Signature

1,2,3 +1

1,3,2 −1

2,1,3 −1

2,3,1 +1

3,1,2 +1

3,2,1 −1

The Mathematica function Signature[a] gives the signature of the permu-

tation that creates the list a from the ordered version of the list a.


We can now return to the Levi-Civita symbol. It is defined by

εijk =

signature[i, j, k] if i 6= j and i 6= k and j 6= k

0 otherwise(158)

Thus ε112 = 0, ε123 = 1, ε213 = −1, etc.

In many books, Eq. 153 is written as

[Lα, Lj] = ihεαjkLk

with the understanding that repeated indices are summed over. This is called

Einstein’s convention.

Exercise 4 Show that the cross-product ~w = ~v×~u of vector algebra can be

written as

wi =3∑

j=1

3∑

k=1

εijkvjuk, i = 1, 2, 3

Appendix 13.2. Commutator Algebra

This chapter relies heavily on commutators so I collect here some infor-

mation about them. These relationships are useful in many areas of quantum

mechanics.

From the definition of the commutator [A, B] = AB − BA, it is obvious

that

[A, B] = −[B, A] (159)

and

[A, B + C] = [A, B] + [A, C] (160)


It is also easy to prove that

[f(A), A] = 0 (161)

if A is the operator of an observable and f is an arbitrary function.

Not so obvious, but still easy to prove, is

[A, BC] = [A, B]C + B[A, C] (162)

Indeed, we have (from the definition of a commutator)

[A, BC ] = ABC − BCA (163)

Since (from the definition) AB = [A, B] + BA, we can rewrite Eq. 163 as

[A, BC] = [A, B]C + BAC − BCA,

from which it follows that

[A, BC] = [A, B]C + B(AC − CA)

= [A, B]C + B[A, C]

This is what we wanted to prove.

If we take C = B in Eq. 162, we have

[A, B2] = [A, B]B + B[A, B] (164)

Then taking C = B2 in Eq. 162:

[A, B3] = [A, BB2] = [A, B]B2 + B[A, B2] (use Eq. 162)

= [A, B]B2 + B([A, B]B + B[A, B]) (use Eq. 162)

= [A, B]B2 + B[A, B]B + B2[A, B]


Repeating this procedure, we obtain

[A, Bn] =n−1∑

s=0

Bs[A, B]Bn−s−1 (165)

For fun, let’s apply this to A = x and B = p, for which we know that

[x, p] = ihI (166)

where I is the unit operator. We have

[x, pn] =n−1∑

s=0

ps[x, p]pn−s−1 =n−1∑

s=0

ps ihI pn−s−1

= ihpn−1 + p(ih)pn−2 + · · · + pn−1(ih)

= ih(npn−1) = ih∂pn

∂p

Exercise 5 Show that if

f(p) =∞∑

n=0

fnpn

where fn, n ≥ 0, are numbers, then

[x, f(p)] = ih∂f(p)

∂p

Exercise 6 Show that

[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0

chapter 13. angular momentum: general theory

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