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
13. Angular Momentum Theory, February 10, 2014 2
§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-
13. Angular Momentum Theory, February 10, 2014 3
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).
13. Angular Momentum Theory, February 10, 2014 4
[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).
13. Angular Momentum Theory, February 10, 2014 5
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)
13. Angular Momentum Theory, February 10, 2014 6
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.
13. Angular Momentum Theory, February 10, 2014 7
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
13. Angular Momentum Theory, February 10, 2014 8
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)
13. Angular Momentum Theory, February 10, 2014 9
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
13. Angular Momentum Theory, February 10, 2014 10
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:
13. Angular Momentum Theory, February 10, 2014 11
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
13. Angular Momentum Theory, February 10, 2014 12
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.
13. Angular Momentum Theory, February 10, 2014 13
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)
13. Angular Momentum Theory, February 10, 2014 14
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)
13. Angular Momentum Theory, February 10, 2014 15
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).
13. Angular Momentum Theory, February 10, 2014 16
§ 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)
13. Angular Momentum Theory, February 10, 2014 17
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).
13. Angular Momentum Theory, February 10, 2014 18
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)
13. Angular Momentum Theory, February 10, 2014 19
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
13. Angular Momentum Theory, February 10, 2014 20
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
13. Angular Momentum Theory, February 10, 2014 21
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.
13. Angular Momentum Theory, February 10, 2014 22
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)
13. Angular Momentum Theory, February 10, 2014 23
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)
13. Angular Momentum Theory, February 10, 2014 24
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
13. Angular Momentum Theory, February 10, 2014 25
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)
13. Angular Momentum Theory, February 10, 2014 26
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.
13. Angular Momentum Theory, February 10, 2014 27
Let us assume that on physical grounds we have decided that a certain
physical system can be described by a ket space generated by the orthonormal
basis set |η1〉, |η2〉, . . . , |ηN 〉. An arbitrary ket |ψ〉 is then represented as
|ψ〉 =N∑
i=1
|ηi〉 〈ηi |ψ〉 ≡N∑
i=1
|ηi〉 ci (117)
and an arbitrary operator O by
O =N∑
i=1
N∑
k=1
|ηi〉 〈ηi | O | ηk〉 〈ηk| ≡N∑
i=1
N∑
k=1
|ηi〉Oik〈ηk| (118)
Since we are assumed to know |ηi〉, i = 1, 2, . . . , N , knowing the vector
ψ = c1, c2, . . . , cN (119)
is equivalent to knowing |ψ〉 (because of Eq. 117). Eq. 118 tells us that we
know how to operate with O if we know the matrix elements
Oik ≡ 〈ηi | O | ηk〉 (120)
The basis set kets |ηi〉 are represented by the vectors
η1 = 1, 0, . . . , 0...
ηN = 0, 0, . . . , 1
(121)
and therefore
ψ =N∑
i=1
ciηi (122)
The eigenvalue problem O|ψ〉 = o|ψ〉 is represented by the eigenvalue problem
for the matrix O∑
k
Oikck = o ci,
13. Angular Momentum Theory, February 10, 2014 28
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
13. Angular Momentum Theory, February 10, 2014 29
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
13. Angular Momentum Theory, February 10, 2014 30
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)
13. Angular Momentum Theory, February 10, 2014 31
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)
13. Angular Momentum Theory, February 10, 2014 32
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.
13. Angular Momentum Theory, February 10, 2014 33
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
13. Angular Momentum Theory, February 10, 2014 34
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)
13. Angular Momentum Theory, February 10, 2014 35
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)
13. Angular Momentum Theory, February 10, 2014 36
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〉
13. Angular Momentum Theory, February 10, 2014 37
=(
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)
13. Angular Momentum Theory, February 10, 2014 38
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.
13. Angular Momentum Theory, February 10, 2014 39
§ 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)
13. Angular Momentum Theory, February 10, 2014 40
|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
13. Angular Momentum Theory, February 10, 2014 41
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
13. Angular Momentum Theory, February 10, 2014 42
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.
13. Angular Momentum Theory, February 10, 2014 43
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)
13. Angular Momentum Theory, February 10, 2014 44
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]
13. Angular Momentum Theory, February 10, 2014 45
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
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