Copyright c 2019 by Robert G.

Littlejohn

Physics 221A
Fall 2019
Notes 19
Irreducible Tensor Operators and the
Wigner-Eckart Theorem†

1. Introduction

The Wigner-Eckart theorem concerns matrix elements of a type that is of frequent occurrence
in all areas of quantum physics, especially in perturbation theory and in the theory of the emis-
sion and absorption of radiation. This theorem allows one to determine very quickly the selection
rules for the matrix element that follow from rotational invariance. In addition, if matrix elements
must be calculated, the Wigner-Eckart theorem frequently offers a way of significantly reducing the
computational effort. We will make quite a few applications of the Wigner-Eckart theorem in this
course.
The Wigner-Eckart theorem is based on an analysis of how operators transform under rotations.
It turns out that operators of a certain type, the irreducible tensor operators, are associated with
angular momentum quantum numbers and have transformation properties similar to those of kets
with the same quantum numbers. An exploitation of these properties leads to the Wigner-Eckart
theorem.

2. Definition of a Rotated Operator

We consider a quantum mechanical system with a ket space upon which rotation operators U (R),
forming a representation of the classical rotation group SO(3), are defined. The representation will
be double-valued if the angular momentum of the system is a half-integer. In these notes we consider
only proper rotations R; improper rotations will be taken up later. The operators U (R) map kets
into new or rotated kets,
|ψ ′ i = U (R)|ψi, (1)

where |ψ ′ i is the rotated ket. We will also write this as

R
|ψi −−→ U (R)|ψi. (2)

In the case of half-integer angular momenta, the mapping above is only determined to within a sign
by the classical rotation R.

† Links to the other sets of notes can be found at:

http://bohr.physics.berkeley.edu/classes/221/1920/221.html.
2 Notes 19: Irreducible Tensor Operators

Now if A is an operator, we define the rotated operator A′ by requiring that the expectation
value of the original operator with respect to the initial state be equal to the expectation value of
the rotated operator with respect to the rotated state, that is,
hψ ′ |A′ |ψ ′ i = hψ|A|ψi, (3)
which is to hold for all initial states |ψi. But this implies
hψ|U (R)† A′ U (R)|ψi = hψ|A|ψi, (4)
or, since |ψi is arbitrary [see Prob. 1.6(b)],
U (R)† A′ U (R) = A. (5)
Solving for A′ , this becomes
A′ = U (R) A U (R)† , (6)
which is our definition of the rotated operator. We will also write this in the form,
R
A −−→ U (R) A U (R)† . (7)
Notice that in the case of half-integer angular momenta the rotated operator is specified by the
SO(3) rotation matrix R alone, since the sign of U (R) cancels and the answer does not depend on
which of the two rotation operators is used on the right hand side. This is unlike the case of rotating
kets, where the sign does matter. Equation (7) defines the action of rotations on operators.

3. Scalar Operators

Now we classify operators by how they transform under rotations. First we define a scalar
operator K to be an operator that is invariant under rotations, that is, that satisfies

U (R) K U (R)† = K,
(8)
for all operators U (R). This terminology is obvious. Notice that it is equivalent to the statement
that a scalar operator commutes with all rotations,
[U (R), K] = 0. (9)
If an operator commutes with all rotations, then it commutes in particular with infinitesimal rota-
tions, and hence with the generators J. See Eq. (12.13). Conversely, if an operator commutes with
J (all three components), then it commutes with any function of J, such as the rotation operators.
Thus another equivalent definition of a scalar operator is one that satisfies

[J, K] = 0.
(10)
The most important example of a scalar operator is the Hamiltonian for an isolated system, not
interacting with any external fields. The consequences of this for the eigenvalues and eigenstates of
the Hamiltonian are discussed in Secs. 7 and 10 below.
 Notes 19: Irreducible Tensor Operators 3

4. Vector Operators

In ordinary vector analysis in three-dimensional Euclidean space, a vector is defined as a collec-

tion of three numbers that have certain transformation properties under rotations. It is not sufficient
just to have a collection of three numbers; they must in addition transform properly. Similarly, in
quantum mechanics, we define a vector operator as a vector of operators (that is, a set of three
operators) with certain transformation properties under rotations.
Our requirement shall be that the expectation value of a vector operator, which is a vector of
ordinary or c-numbers, should transform as a vector in ordinary vector analysis. This means that if
|ψi is a state and |ψ ′ i is the rotated state as in Eq. (1), then

hψ ′ |V|ψ ′ i = Rhψ|V|ψi, (11)

where V is the vector of operators that qualify as a genuine vector operator. In case the notation
in Eq. (11) is not clear, we write the same equation out in components,
X
hψ ′ |Vi |ψ ′ i = Rij hψ|Vj |ψi. (12)
j

Equation (11) or (12) is to hold for all |ψi, so by Eq. (1) they imply (after swapping R and R−1 )

U (R) V U (R)† = R−1 V, (13)

or, in components,
X
U (R) Vi U (R)† = Vj Rji .
j
(14)
We will take Eq. (13) or (14) as the definition of a vector operator.
In the case of a scalar operator, we had one definition (8) involving its properties under conjuga-
tion by rotations, and another (10) involving its commutation relations with the angular momentum
J. The latter is in effect a version of the former, when the rotation is infinitesimal. Similarly, for
vector operators there is a definition equivalent to Eq. (13) or (14) that involves commutation rela-
tions with J. To derive it we let U and R in Eq. (13) have axis-angle form with an angle θ ≪ 1, so
that
i
U (R) = 1 − θn̂ · J, (15)
h̄
and
R = I + θn̂ · J. (16)
See Eqs. (11.22) and (11.32) for the latter. Then the definition (13) becomes
 i   i 
1 − θn̂ · J V 1 + θn̂ · J = (I − θn̂ · J)V, (17)
h̄ h̄
or
[n̂ · J, V] = −ih̄ n̂×V. (18)
4 Notes 19: Irreducible Tensor Operators

Taking the j-th component of this, we have

ni [Ji , Vj ] = −ih̄ ǫjik ni Vk , (19)
or, since n̂ is an arbitrary unit vector,

[Ji , Vj ] = ih̄ ǫijk Vk .

(20)
Any vector operator satisfies this commutation relation with the angular momentum of the system.
The converse is also true; if Eq. (20) is satisfied, then V is a vector operator. This follows since
Eq. (20) implies Eq. (18) which implies Eq. (17), that is, it implies that the definition (13) is satisfied
for infinitesimal rotations. But it is easy to show that if Eq. (13) is true for two rotations R1 and
R2 , then it is true for the product R1 R2 . Therefore, since finite rotations can be built up as the
product of a large number of infinitesimal rotations (that is, as a limit), Eq. (20) implies Eq. (13)
for all rotations. Equations (13) and (20) are equivalent ways of defining a vector operator.
We have now defined scalar and vector operators. Combining them, we can prove various
theorems. For example, if V and W are vector operators, then V · W is a scalar operator, and
V×W is a vector operator. This is of course just as in vector algebra, except that we must remember
that operators do not commute, in general. For example, it is not generally true that V·W = W ·V,
or that V×W = −W×V.
If we wish to show that an operator is a scalar, we can compute its commutation relations with
the angular momentum, as in Eq. (10). However, it may be easier to consider what happens when
the operator is conjugated by rotations. For example, the central force Hamiltonian (16.1) is a scalar
because it is a function of the dot products p · p = p2 and x · x = r2 . See Sec. 16.2.

5. Examples of Vector Operators

Consider a system consisting of a single spinless particle moving in three-dimensional space, for
which the wave functions are ψ(x) and the angular momentum is L = x×p. To see whether x is a
vector operator (we expect it is), we compute the commutation relations with L, finding,
[Li , xj ] = ih̄ ǫijk xk . (21)
According to Eq. (20), this confirms our expectation. Similarly, we find
[Li , pj ] = ih̄ ǫijk pk , (22)
so that p is also a vector operator. Then x×p (see Sec. 4) must also be a vector operator, that is,
we must have
[Li , Lj ] = ih̄ ǫijk Lk . (23)
This last equation is of course just the angular momentum commutation relations, but here with a
new interpretation. More generally, by comparing the adjoint formula (13.89) with the commutation
relations (20), we see that the angular momentum J is always a vector operator.
 Notes 19: Irreducible Tensor Operators 5

6. Tensor Operators

Finally we define a tensor operator as a tensor of operators with certain transformation prop-
erties that we will illustrate in the case of a rank-2 tensor. In this case we have a set of 9 operators
Tij , where i, j = 1, 2, 3, which can be thought of as a 3 × 3 matrix of operators. These are required
to transform under rotations according to
X
U (R) Tij U (R)† = Tkℓ Rki Rℓj , (24)
kℓ

which is a generalization of Eq. (14) for vector operators. As with scalar and vector operators, a
definition equivalent to Eq. (24) may be given that involves the commutation relations of Tij with
the components of angular momentum.
As an example of a tensor operator, let V and W be vector operators, and write

Tij = Vi Wj . (25)

Then Tij is a tensor operator (it is the tensor product of V with W). This is just an example; in
general, a tensor operator cannot be written as the product of two vector operators as in Eq. (25).
Another example of a tensor operator is the quadrupole moment operator. In a system with a
collection of particles with positions xα and charges qα , where α indexes the particles, the quadrupole
moment operator is
X
Qij = qα (3xαi xαj − rα2 δij ). (26)
α

This is obtained from Eq. (15.88) by setting

X
ρ(x) = qα δ(x − xα ). (27)
α

The quadrupole moment operator is especially important in nuclear physics, in which the particles
are the protons in a nucleus with charge q = e. Notice that the first term under the sum (26) is an
operator of the form (25), with V = W = xα .
Tensor operators of other ranks (besides 2) are possible; a scalar is considered a tensor operator
of rank 0, and a vector is considered a tensor of rank 1. In the case of tensors of arbitrary rank, the
transformation law involves one copy of the matrix R−1 = Rt for each index of the tensor.

7. Energy Eigenstates in Isolated Systems

In this section we explore the consequences of rotational invariance for the eigenstates, eigen-
values and degeneracies of a scalar operator. The most important scalar operator in practice is the
Hamiltonian for an isolated system, so for concreteness we will speak of such a Hamiltonian, but the
following analysis applies to any scalar operator.
Let H be the Hamiltonian for an isolated system, and let E be the Hilbert space upon which
it acts. Since H is a scalar it commutes with J, and therefore with the commuting operators J 2
6 Notes 19: Irreducible Tensor Operators

and J3 . Let us denote the simultaneous eigenspaces of J 2 and J3 with quantum numbers j and m
by Sjm , as illustrated in Fig. 13.5. It was shown in Notes 13 that for a given system, j takes on
certain values that must be either integers or half-integers. For example, in central force motion we
have only integer values of j (which is called ℓ in that context), while for the 57 Fe nucleus, which is
discussed in more detail in the next section, we have only half-integer values. For each value of j that
occurs there is a collection of 2j + 1 eigenspaces Sjm of J 2 and J3 , for m = −j, . . . , +j. These spaces
are mapped invertibly into one another by J+ and J− , as illustrated in Fig. 13.5, and if they are
finite-dimensional, then they all have the same dimension. As in Notes 13, we write Nj = dim Sjj ,
which we call the multiplicity of the given j value.
In Notes 13 we constructed a standard angular momentum basis by picking an arbitrary or-
thonormal basis in each stretched space Sjj , with the basis vectors labeled by γ as in Fig. 13.5, where
γ = 1, . . . , Nj . We denote these basis vectors in Sjj by |γjji. Then by applying lowering operators,
we construct an orthonormal basis in each of the other Sjm , for m running down to −j. In this
way we construct a standard angular momentum basis |γjmi on the whole Hilbert space E. In this
construction, it does not matter how the basis |γjji is chosen in Sjj , as long as it is orthonormal.
Now, however, we have a Hamiltonian, and we would like a simultaneous eigenbasis of H, J 2
and J3 . To construct this we restrict H to Sjj for some j (see Sec. 1.23 for the concept of the
restriction of an operator to a subspace, and how it is used in proving that commuting operators
possess a simultaneous eigenbasis). This restricted H is a Hermitian operator on Sjj so it possesses
an eigenbasis on that space.
The spectrum of H on Sjj can be either discrete, continuous, or mixed (in most problems we
will consider in this course it has a continuous spectrum above a threshold energy, and may have
discrete bound states below that). Let us focus on the bound states and assume that H possesses
at least one bound eigenstate |ψi on Sjj with corresponding eigenvalue E. Then |ψi satisfies

J 2 |ψi = j(j + 1)h̄2 |ψi, J3 |ψi = jh̄ |ψi, H|ψi = E|ψi. (28)

Now by applying a lowering operator J− we find

J− H|ψi = HJ− |ψi = EJ− |ψi, (29)

so that J− |ψi is an eigenstate of H, lying in the space Sj,j−1 , with the same eigenvalue E as |ψi ∈ Sjj .
Continuing to apply lowering operators, we generate a set of 2j + 1 linearly independent eigenstates
of H with the same eigenvalue E, that is, E is independent of the quantum number m. These states
span an irreducible, invariant subspace of E.
There may be other irreducible subspaces with the same energy. This can occur in two ways.
It could happen that there is another energy eigenstate in Sjj , linearly independent of |ψi, with the
same energy E. That is, it is possible that E is a degenerate eigenvalue of H restricted to Sjj . In
general, every discrete eigenvalue E of H restricted to Sjj corresponds to an eigenspace, a subspace
of Sjj that may be multidimensional. Choosing an orthonormal basis in this subspace and applying
 Notes 19: Irreducible Tensor Operators 7

lowering operators, we obtain a set of orthogonal, irreducible subspaces of the same value of j, each
with 2j + 1 dimensions and all having the same energy.
It could also happen that there is another bound energy eigenstate, in a different space Sj ′ j ′
for j ′ 6= j, with the same energy E as |ψi ∈ Sjj . This would be a degeneracy of H that crosses
j values. If such a degeneracy exists, then we have at least two irreducible subspaces of the same
energy, one of dimension 2j + 1 and the other of dimension 2j ′ + 1. In other words, degeneracies
can occur either within a given j value or across j values.
These facts that we have accumulated can be summarized by a theorem:

Theorem 1. The discrete energy eigenspaces of an isolated system consist of one or more invari-
ant, irreducible subspaces under rotations, each associated with a definite j value. The different
irreducible subspaces can be chosen to be orthogonal.

Let us look at two examples of how this theorem works out in practice, the first a simple one
with a small number of degrees of freedom that is exactly solvable, and the other a complicated one
with a large number of degrees of freedom, in which all we know about the Hamiltonian is that it is
invariant under rotations.

8. Example: Central Force Motion

For the simple example we take the case of central force motion, for which we use the notation
L, ℓ etc. instead of J, j etc.
In central force motion the stretched subspace Sℓℓ consists of wave functions R(r)Yℓℓ (θ, φ),
where R(r) is any radial wave function. To find the energy eigenstates in this stretched subspace we
solve the radial Schrödinger equation for the given ℓ value, which produces in general a continuous
and a discrete spectrum. We assume there is a discrete spectrum for the given ℓ value and denote the
energy eigenvalues and corresponding radial wave functions by Enℓ and Rnℓ (r), as in Notes 16. By
applying lowering operators to the wave function Rnℓ (r) Yℓℓ (θ, φ), we obtain an irreducible subspace
of degenerate energy eigenfunctions, spanned by {Rnℓ (r) Yℓm (θ, φ), m = +ℓ, . . . , −ℓ}.
Now we consider degeneracies. Is it possible, for a given value of ℓ in a central force problem,
that a bound energy eigenvalue can be degenerate? That is, can there be more than one linearly
independent bound energy eigenstate of a given energy in Sℓℓ ? As discussed in Sec. 16.4, the answer is
no, the boundary conditions on the radial wave functions guarantee that there can be no degeneracy
of this type. In central force problems, we do not have degeneracies within a given ℓ value.
Then is it possible that there is a degeneracy between different values of ℓ? Again, as discussed
in Notes 16, the answer is that in general it is not very likely, since the different radial equations for
different values of ℓ are effectively different Schrödinger equations whose centrifugal potentials are
different.
The fact is that systematic degeneracies require a non-Abelian symmetry. We are already
taking into account the SO(3) symmetry of proper rotations, which explains the degeneracy in the
8 Notes 19: Irreducible Tensor Operators

magnetic quantum number m, so any additional degeneracy will require a larger symmetry group
than SO(3). In the absence of such extra symmetry, degeneracies between different ℓ values can
occur only by “accident,” that is, by fine tuning parameters in a Hamiltonian to force a degeneracy
to happen. This is not likely in most practical situations. Therefore in central force problems we do
not normally expect degeneracies that cross subspaces of different values of ℓ.
As explained in Notes 17, however, the electrostatic model of hydrogen is a notable exception,
due to the symmetry group SO(4) possessed by this model, which is larger than the rotation group
SO(3). The extra symmetry in this model of hydrogen explains why the energy levels En = −1/2n2
(in the right units) are the same across the angular momentum values ℓ = 0, . . . , n − 1. The isotropic
harmonic oscillator in two or more dimensions is another example of a system with extra degeneracy;
such oscillators are approximate models for certain types of molecular vibrations.
For a more complicated example of how Theorem 1 works out in practice we examine some
energy levels of the nucleus 57 Fe, which is important in the Mössbauer effect. We use the opportunity
to digress into some of the interesting physics connected with this effect. We begin with a general
discussion of aspects of the emission and absorption of photons by quantum systems.

9. Emission and Absorpton of Photons

When an atom, nucleus or other quantum system is in an excited state B and emits a photon
while dropping into the ground state A,

B → A + γ, (30)

then in a simple description of the process we say that the energy of the photon is given by

Eγ = EB − EA , (31)

where EB and EA are the energies of the states B and A. If now there is another atom, nucleus or
other system of the same type nearby in its ground state A, then it would appear that that photon
has exactly the right energy to induce the inverse reaction,

A + γ → B, (32)

thereby lifting the second system into the excited state B.

Does this mean, for example, that when an atom in a gas emits a photon that the photon
only travels as far as the nearest neighboring atom before being absorbed again? No, because there
are several effects that complicate the basic picture just presented, modifying the energy Eγ of the
photon so that is it not exactly given by Eq. (31). These include the natural line width of the
transition B → A; Doppler shifts; recoil; and, in the case of nuclei, what are called chemical shifts.
In quantum mechanics the energy of a system is only precisely defined in a process that takes
place over an infinite amount of time. The excited state B of our system is unstable with some
lifetime τ , so its energy EB is only defined to within an uncertainty of order ∆EB = h̄/τ . Assuming
 Notes 19: Irreducible Tensor Operators 9

A is the ground state, it is stable and can exist over an infinite amount of time, so there is no
uncertainty in its energy. Overall, the uncertainty in the energy EB creates uncertainty of order h̄/τ
to the energy of the photon Eγ emitted in the process (30). This can be seen experimentally; if all
other sources of broadening of the spectral line are eliminated, then the energy of photons emitted
in an atomic or nuclear transition does not have a definite value, but rather there is a spread of
order ∆E = h̄/τ about the nominal value EB − EA . This spread is called the natural line width of
the spectral line. The natural line width of spectral lines is examined in some detail in Notes 43.
Similarly, if a photon of energy Eγ encounters a quantum system of the same type at rest in
its ground state A, then if Eγ is roughly within the range ∆E = h̄/τ about the nominal energy
EB − EA it will be able to lift the second system into the excited state B, that is, the inverse reaction
(32) will take place.
On the other hand, if the emitting atom, nucleus or other quantum system is in a state of
motion, then the frequency ωγ = Eγ /h̄ of the emitted photon will be Doppler shifted and may no
longer be within the resonance needed to raise another such system into its excited state. Writing
simply E for the nominal energy EB − EA of the photon, the velocity v needed to shift the photon
out of resonance is given by
v ∆E h̄
= = . (33)
c E Eτ
Whether or not the photon is shifted out of resonance depends on the velocity and other parameters,
but in many practical circumstances one will find that thermal velocities do exceed the value given
by Eq. (33). A similar logic applies in case the receiving system is in a state of motion (or both, as
would be the case of a gas).
Even if the emitting atom or nucleus or other system is at rest, the energy Eγ is not given
exactly by Eq. (31) because of the recoil of the emitting system when the photon is emitted. The
photon has energy E = h̄ω and momentum p = h̄k where ω = c|k|, so by conservation of momentum
the emitting system suffers a recoil and has momentum mv = −h̄k after the photon of frequency ω
is emitted, where m is the mass of the emitting system. Thus Eq. (31) should be replaced by
1
Eγ + mv 2 = EB − EA , (34)
2
where v is the recoil velocity. Some of the available energy goes into kinetic energy of the recoiling
system, and the energy Eγ of the emitted photon is actually less than the nominal value (31). Again,
whether this recoil shift is greater or less than the natural line width depends on the parameters of
the problem.

57
10. The Fe Nucleus and the Mössbauer Effect

The Mössbauer effect involves a transition 57 Fe∗ → 57 Fe + γ between two energy levels of the
57
Fe nucleus, where the simple notation 57 Fe (or A) refers to the ground state and 57 Fe∗ (or B)
refers to an excited state. These states are illustrated in the energy level diagram for the nucleus
10 Notes 19: Irreducible Tensor Operators

given in Fig. 1. The photon emitted has energy 14.4 KeV, and the lifetime of the excited state 57 Fe∗
is τ = 9.8 × 10−8 sec. From these figures we find ∆ω/ω = ∆E/E = 4.7 × 10−13 , where E and ω are
the energy and frequency of the emitted photon. The spread in the energy is very small compared
to the energy. For example, according to Eq. (33), to Doppler shift the photon out of resonance it
would require a velocity of v/c = 4.7 × 10−13 , or v = 0.014 cm/sec.

57
Co

β capture

57
Fe∗∗ ( 52 , −)

γ (122 KeV)
γ (136 KeV)

57
Fe∗ ( 23 , −)
γ (14.4 KeV)
57
Fe ( 21 , −)

Fig. 1. Energy level diagram relevant for the Mössbauer effect in 57 Fe. 57 Fe is the ground state, 57 Fe∗ is an excited
state, and 57 Fe∗∗ is a more highly excited state. Principal transitions via photon emission are shown.

The Mössbauer effect makes use of a source containing iron nuclei in the excited state 57
Fe∗ ,
which emits photons, and a receiver containing iron nuclei in the ground state which may absorb
them by being lifted into the excited state 57 Fe∗ via the reverse reaction. The receiver can be a block
of natural iron, which contains the isotope 57 Fe at the 2% level, behind which a gamma-ray detector
is placed. If the incident photons are within the narrow resonant range of energies, then they will
be absorbed by the block of iron, and the detector will detect nothing. But if they are shifted out
of resonance, the gamma rays will pass through the block of iron and the detector will detect them.
If there is some effect that shifts the frequency of the gamma rays from their nominal energy (for
example, the gravitational red shift in the Pound-Rebka experiment), then a compensating Doppler
shift can be introduced by giving the source some velocity. By measuring the velocity of the source
needed to shift the gamma rays back into resonance, one can measure the shift caused by the effect
in question.
However, plugging in the numbers shows that the recoil shift described by Eq. (34), where m is
the mass of an iron nucleus, is much greater than the natural line width, so the recoil would seem
to spoil the whole idea. But the iron nucleus is not free, rather it is part of a crystal lattice, whose
vibrations are described by a large number of harmonic oscillators, the normal modes of the lattice.
(See the discussion in Sec. 8.2.) Thus the recoil kinetic energy E = (1/2)mv 2 in Eq. (34) is not free
to take on any value, rather it must be some multiple of h̄ω, where ω is the frequency of a normal
mode of the lattice. That is, when the photon is emitted by the nucleus, some number of phonons
 Notes 19: Irreducible Tensor Operators 11

are also emitted into the lattice, representing the recoil energy.

m
M h̄ω

Fig. 2. A one dimensional model of an iron atom coupled to a normal mode of the lattice. The mass of the iron atom
is m, the mass of the crystal is M . The iron atom emits a photon of energy h̄ω and suffers a recoil (an impulse) as a
result.

To model this situation in the simplest possible way, let us imagine an iron atom connected to a
spring, forming a one-dimensional harmonic oscillator, as illustrated in Fig. 2. In the figure m is the
mass of the iron atom while M is the mass of the crystal lattice to which it is coupled. When the
atom emits a photon it suffers an an impulse, that is, a change ∆p in its momentum. The emission
takes place over a short time compared to the frequency of the harmonic oscillator (a normal mode
of the lattice), so the position of the iron atom does not change much during the emission process.
Classically we can model the impulse by the map,

x 7→ x, p 7→ p + ∆p. (35)

To model the effect of the recoil in quantum mechanics, we use the momentum displacement operator
S(b) introduced in Notes 8 [see Eqs. (8.64)–(8.66)]. That is, if |ψi is the state of the oscillator before
the photon is emitted, then the state after the emission is

|ψ ′ i = ei∆px/h̄ |ψi. (36)

The final state can be expanded in energy eigenstates,

X
|ψ ′ i = cn |ni, (37)
n

so that the probability of finding the oscillator in state n after the photon has been emitted is
|cn |2 . In particular, if the initial state of the oscillator |ψi = |ni i is an energy eigenstate, then the
probability |cn |2 is the probability to make a transition ni → n as a result of the recoil. As we say,
n − ni phonons are emitted.
In fact, there is a certain probability that no phonons are emitted at all, that is n = ni and the
oscillator remains in the initial state. Because of quantum mechanics, the recoil energy cannot take
on any value, but rather is quantized, and the value zero is allowed. What makes the 57 Fe nucleus
attractive for the Mössbauer effect is that it has a reasonable probability for this recoilless emission.
Of course there is not only a recoil energy but also a recoil momentum. In the Mössbauer effect,
recoilless emission does not violate conservation of momentum because the entire crystal lattice,
with an effectively infinite mass M (as in the figure) takes up the recoil momentum.
The excited state 57 Fe∗ has only a short lifetime but in practice a population of these excited
57 57
states is maintained in the source as a part of the decay chain of Co, as shown in Fig. 1. Co has
12 Notes 19: Irreducible Tensor Operators

a lifetime of 271 days, which is long enough to make it practical to use it as a source of 57 Fe∗ in a
real experiment. As shown in the figure, 57 Co transforms into an excited state 57 Fe∗∗ by electron
capture, after which 57
Fe∗∗ decays by the emission of a photon into 57
Fe∗ , which is the source of
the photons of interest in the Mössbauer effect.
Mössbauer was awarded the Nobel Prize in 1961 for his discovery of recoilless emission of
gamma ray photons and some of its applications. A notable early application was the Pound-
Rebka experiment, carried out in 1959, in which the Mössbauer effect was used to make the first
measurement of the gravitational red shift. This is the red shift photons experience when climbing
in a gravitational field, in accordance with the 1911 prediction of Einstein. The gravitational red
shift is one of the physical cornerstones of general relativity.

57
11. Energy levels in Fe

To return to the subject of Hamiltonians and their energy levels in isolated systems, let us
draw attention to the three levels 57 Fe, 57 Fe∗ and 57 Fe∗∗ , in Fig. 1. These are energy levels of the
Hamiltonian for the 57 Fe nucleus, and, according to Theorem 1, each must consist of one or more
irreducible subspaces under rotations. In fact, they each consist of precisely one such irreducible
subspace, with a definite j value, which is indicated in the figure ( 12 for the ground state 57 Fe, and
3 5
2 and 2 for the two excited states
57
Fe∗ and 57 Fe∗∗ , respectively). Also indicated are the parities
of these states (all three have odd parity). The parity of energy eigenstates of isolated systems is
discussed in Sec. 20.8.
57
A model for the Hamiltonian of the Fe nucleus views it as a 57-particle system, that is, with
26 protons and 31 neutrons. The Hamiltonian is some function of the positions, momenta and spins
of the particles,
H = H(xα , pα , Sα ), (38)

where α = 1, . . . , 57. In this model the total angular momentum of the system is the sum of the
orbital and spin angular momenta of the nucleons,
57
X
J= xα ×pα + Sα , (39)
α=1

and the “spins” of the various nuclear states shown in Fig. 1 are actually the quantum numbers of
J 2 (that is, we call J the “spin” and use the notation S etc. for it). For example, we say that the
ground state 57
Fe has spin s = 12 .
This model is more or less crude, due to the fact that protons and neutrons are composite
particles, each made up of three quarks, which interact with the quark and gluon fields via the strong
interactions. For our purposes the only thing that matters is that rotations act upon the state space
of the system by means of unitary operators, and that these commute with the Hamiltonian. The
model (38) at least gives us something concrete to think about.
 Notes 19: Irreducible Tensor Operators 13

Each nuclear energy eigenstate consists of a single irreducible subspace under rotations for the
same reasons discussed in connection with central force motion in Sec. 7. That is, extra degeneracy
requires extra symmetry or else an unlikely accident, and neither of these is to be expected in nuclei.
Therefore each energy level is characterized by a unique angular momentum value, as indicated in
the figure.
We can summarize these accumulated facts by stating an addendum to Theorem 1.

Addendum to Theorem 1. With a few exceptions, notably the electrostatic model of hydrogen,
the bound state energy eigenspaces of isolated systems consist of a single invariant, irreducible
subspace under rotations. Thus, the energy eigenvalues are characterized by an angular momentum
quantum number, which is variously denoted ℓ, s, j, etc, depending on the system.

We can now understand why the Hilbert space for spins in magnetic fields consists of a single
irreducible subspace under rotations for a large class of particles, a question that was raised in
Notes 14. For example, if we place the 57 Fe nucleus in a magnetic field that is strong by laboratory
standards, say, 10T, then the energy splitting between the two magnetic substates m = ± 21 will
be of the order of 100 MHz in frequency units, or about 4 × 10−7 eV, or roughly 3 × 10−11 times
smaller than the energy separation from the first excited state 57 Fe∗ . Therefore it is an excellent
approximation to ignore the state 57 Fe∗ and all other excited states of the 57 Fe nucleus, and to treat
the Hilbert space of the nucleus as if it were a single irreducible subspace with s = 21 , that is, the
ground eigenspace. In other words, in the case of nuclei, the 2s + 1-dimensional Hilbert space used
in our study of spins in magnetic fields in Notes 14 is actually a subspace of a larger Hilbert space.
It is, in fact, an energy eigenspace of an isolated system. This in turn explains why the magnetic
moment is proportional to the spin [see Prob. 2(a)].

12. The Spherical Basis

We return to our development of the properties of operators under rotations. We take up the
subject of the spherical basis, which is a basis of unit vectors in ordinary three-dimensional space that
is alternative to the usual Cartesian basis. Initially we just present the definition of the spherical basis
without motivation, and then we show how it can lead to some dramatic simplifications in certain
problems. Then we explain its deeper significance. The spherical basis will play an important role
in the development of later topics concerning operators and their transformation properties.
We denote the usual Cartesian basis by ĉi , i = 1, 2, 3, so that

ĉ1 = x̂, ĉ2 = ŷ, ĉ3 = ẑ. (40)

We have previously denoted this basis by êi , but in these notes we reserve the symbol ê for the
spherical basis.
The spherical basis is defined by
x̂ + iŷ
ê1 = − √ ,
2
14 Notes 19: Irreducible Tensor Operators

ê0 = ẑ,
x̂ − iŷ
ê−1 = √ . (41)
2
This is a complex basis, so vectors with real components with respect to the Cartesian basis have
complex components with respect to the spherical basis. We denote the spherical basis vectors
collectively by êq , q = 1, 0, −1.
The spherical basis vectors have the following properties. First, they are orthonormal, in the
sense that
ê∗q · êq′ = δqq′ . (42)

Next, an arbitrary vector X can be expanded as a linear combination of the vectors ê∗q ,

ê∗q Xq ,
X
X= (43)
q

where the expansion coefficients are

Xq = êq · X. (44)

These equations are equivalent to a resolution of the identity in 3-dimensional space,

ê∗q êq ,
X
I= (45)
q

in which the juxtaposition of the two vectors represents a tensor product or dyad notation.
You may wonder why we expand X as a linear combination of ê∗q , instead of êq . The latter
type of expansion is possible too, that is, any vector Y can be written
X
Y= êq Yq , (46)
q

where
Yq = ê∗q · Y. (47)

These relations correspond to a different resolution of the identity,

êq ê∗q .
X
I= (48)
q

The two types of expansion give the contravariant and covariant components of a vector with respect
to the spherical basis; in this course, however, we will only need the expansion indicated by Eq. (43).

13. An Application of the Spherical Basis

To show some of the utility of the spherical basis, we consider the problem of dipole radiative
transitions in a single-electron atom such as hydrogen or an alkali. It is shown in Notes 41 that the
transition amplitude for the emission of a photon is proportional to matrix elements of the dipole
 Notes 19: Irreducible Tensor Operators 15

operator between the initial and final states. We use an electrostatic, spinless model for the atom,
as in Notes 16, and we consider the transition from initial energy level Enℓ to final level En′ ℓ′ . These
levels are degenerate, since the energy does not depend on the magnetic quantum number m or m′ .
The wave functions have the form,

ψnℓm (r, θ, φ) = Rnℓ (r)Yℓm (Ω), (49)

as in Eq. (16.15).
The dipole operator is proportional to the position operator of the electron, so we must evaluate
matrix elements of the form,
hnℓm|x|n′ ℓ′ m′ i, (50)

where the initial state is on the left and the final one on the right. The position operator x has
three components, and the initial and final levels consist of 2ℓ + 1 and 2ℓ′ + 1 degenerate states,
respectively. Therefore if we wish to evaluate the intensity of a spectral line as it would be observed,
we really have to evaluate 3(2ℓ′ + 1)(2ℓ + 1) matrix elements, for example, 3 × 3 × 5 = 45 in a 3d → 2p
transition. This is actually an exaggeration, as we shall see, because many of the matrix elements
vanish, but there are still many nonvanishing matrix elements to be calculated.
A great simplification can be achieved by expressing the components of x, not with respect to
the Cartesian basis, but with respect to the spherical basis. First we define

xq = êq · x, (51)

exactly as in Eq.(44). Next, by inspecting a table of the Yℓm ’s (see Sec. 15.7), we find that for ℓ = 1
we have
r r 
3 iφ 3 x + iy 
rY11 (θ, φ) = −r sin θe = − √ ,
8π 4π 2
r r
3 3
rY10 (θ, φ) = r cos θ = (z),
4π 4π
r r 
3 −iφ 3 x − iy 
rY1,−1 (θ, φ) = r sin θe = √ , (52)
8π 4π 2

where we have multiplied each Y1m by the radius r. On the right hand side we see the spherical
components xq of the position vector x, as follows from the definitions (41). The results can be
summarized by
r
3
rY1q (θ, φ) = xq , (53)
4π
for q = 1, 0, −1, where q appears explicitly as a magnetic quantum number. This equation reveals
a relationship between vector operators and the angular momentum value ℓ = 1, something we will
have more to say about presently.
16 Notes 19: Irreducible Tensor Operators

Now the matrix elements (50) become a product of a radial integral times an angular integral,
Z ∞
′ ′ ′
hnℓm|xq |n ℓ m i = ∗ (r)rR ′ ′ (r)
r2 dr Rnℓ n ℓ
0
r (54)
4π
Z
× ∗ (θ, φ)Y (θ, φ)Y ′ ′ (θ, φ).
dΩ Yℓm 1q ℓ m
3

We see that all the dependence on the three magnetic quantum numbers (m, q, m′ ) is contained in
the angular part of the integral. Moreover, the angular integral can be evaluated by the three-Yℓm
formula, Eq. (18.67), whereupon it becomes proportional to the Clebsch-Gordan coefficient,

hℓm|ℓ′ 1m′ qi. (55)

The radial integral is independent of the three magnetic quantum numbers (m, q, m′ ), and the trick
we have just used does not help us to evaluate it. But it is only one integral, and after it has
been done, all the other integrals can be evaluated just by computing or looking up Clebsch-Gordan
coefficients.
The selection rule m = q + m′ in the Clebsch-Gordan coefficient (55) means that many of the
integrals vanish, so we have exaggerated the total number of integrals that need to be done. But had
we worked with the Cartesian components xi of x, this selection rule might not have been obvious.
In any case, even with the selection rule, there may still be many nonzero integrals to be done (nine,
in the case 3d → 2p).
The example we have just given of simplifying the calculation of matrix elements for a dipole
transition is really an application of the Wigner-Eckart theorem, which we take up later in these
notes.
The process we have just described is not just a computational trick, rather it has a physical
interpretation. The initial and final states of the atom are eigenstates of L2 and Lz , and the photon
is a particle of spin 1 (see Notes 40). Conservation of angular momentum requires that the angular
momentum of the initial state (the atom, with quantum numbers ℓ and m) should be the same as the
angular momentum of the final state (the atom, with quantum numbers ℓ′ and m′ , plus the photon
with spin 1). Thus, the selection rule m = m′ + q means that q is the z-component of the spin of
the emitted photon, so that the z-component of angular momentum is conserved in the emission
process. As for the selection rule ℓ ∈ {ℓ′ − 1, ℓ′ , ℓ′ + 1}, it means that the amplitude is zero unless the
possible total angular momentum quantum number of the final state, obtained by combining ℓ′ ⊗ 1,
is the total angular momentum quantum number of the initial state. This example shows the effect
of symmetries and conservation laws on the selection rules for matrix elements.
This is only an incomplete accounting of the symmetry principles at work in the matrix element
(50) or (54); as we will see in Notes 20, parity also plays an important role.
 Notes 19: Irreducible Tensor Operators 17

14. Significance of the Spherical Basis

To understand the deeper significance of the spherical basis we examine Table 1. The first
row of this table summarizes the principal results obtained in Notes 13, in which we worked out
the matrix representations of angular momentum and rotation operators. To review those results,
we start with a ket space upon which proper rotations act by means of unitary operators U (R), as
indicated in the second column of the table. We refer only to proper rotations R ∈ SO(3), and
we note that the representation may be double-valued. The rotation operators have generators,
defined by Eq. (12.13), that is, that equation can be taken as the definition of J when the rotation
operators U (R) are given. [Equation (12.11) is equivalent.] The components of J satisfy the usual
commutation relations (12.24) since the operators U (R) form a representation of the rotation group.
Next, since J 2 and Jz commute, we construct their simultaneous eigenbasis, with an extra index γ
to resolve degeneracies. Also, we require states with different m but the same γ and j to be related
by raising and lowering operators. This creates the standard angular momentum basis (SAMB),
indicated in the fourth column. In the last column, we show how the vectors of the standard angular
momentum basis transform under rotations. A basis vector |γjmi, when rotated, produces a linear
combination of other basis vectors for the same values of γ and j but different values of m. This
implies that the space spanned by |γjmi for fixed γ and j, but for m = −j, . . . , +j is invariant under
rotations. This space has dimensionality 2j + 1. It is, in fact, an irreducible invariant space (more
on irreducible subspaces below). One of the results of the analysis of Notes 13 is that the matrices
j
Dm ′ m (U ) are universal matrices, dependent only on the angular momentum commutation relations

and otherwise independent of the nature of the system.

Space Action Ang Mom SAMB Action on SAMB

j
X
Kets |ψi 7→ U |ψi J |γjmi U |γjmi = |γjm′ iDm ′m

m ′
X
3D Space x 7→ Rx iJ êq Rêq = êq′ Dq1′ q
q′
X
Operators A 7→ U AU † ... Tqk U Tqk U † = Tqk′ Dqk′ q
q′

Table 1. The rows of the table indicate different vector spaces upon which rotations act by means of unitary operators.
The first row refers to a ket space (a Hilbert space of a quantum mechanical system), the second to ordinary three-
dimensional space (physical space), and the third to the space of operators. The operators in the third row are the
usual linear operators of quantum mechanics that act on the ket space, for example, the Hamiltonian. The first column
identifies the vector space. The second column shows how rotations R ∈ SO(3) act on the given space. The third column
shows the generators of the rotations, that is, the 3-vector of Hermitian operators that specify infinitesimal rotations.
The fourth column shows the standard angular momentum basis (SAMB), and the last column, the transformation law
of vectors of the standard angular momentum basis under rotations.

At the beginning of Notes 13 we remarked that the analysis of those notes applies to other
spaces besides ket spaces. All that is required is that we have a vector space upon which rotations
18 Notes 19: Irreducible Tensor Operators

act by means of unitary operators. For other vectors spaces the notation may change (we will not
call the vectors kets, for example), but otherwise everything else goes through.
The second row of Table 1 summarizes the case in which the vector space is ordinary three-
dimensional (physical) space. Rotations act on this space by means of the matrices R, which, being
orthogonal, are also unitary (an orthogonal matrix is a special case of a unitary matrix). The action
consists of just rotating vectors in the usual sense, as indicated in the second column.
The generators of rotations in this case must be a vector J of Hermitian operators, that is,
Hermitian matrices, that satisfy
i
U (n̂, θ) = 1 − θn̂ · J, (56)
h̄
when θ is small. Here U really means the same thing as R, since we are speaking of the action
on three-dimensional space, and 1 means the same as the identity matrix I. We will modify this
definition of J slightly by writing J′ = J/h̄, thereby absorbing the h̄ into the definition of J and
making J′ dimensionless. This is appropriate when dealing with ordinary physical space, since it
has no necessary relation to quantum mechanics. (The spherical basis is also useful in classical
mechanics, for example.) Then we will drop the prime, and just remember that in the case of this
space, we will use dimensionless generators. Then we have

U (n̂, θ) = 1 − iθn̂ · J. (57)

But this is equivalent to

R(n̂, θ) = I + θn̂ · J, (58)

as in Eq. (11.32), where the vector of matrices J is defined by Eq. (11.22). These imply

J = iJ, (59)

as indicated in the third column of Table 1. Writing out the matrices Ji explicitly, we have
     
0 0 0 0 0 i 0 −i 0
J1 =  0 0 −i  , J2 =  0 0 0  , J3 =  i 0 0. (60)
0 i 0 −i 0 0 0 0 0

These matrices are indeed Hermitian, and they satisfy the dimensionless commutation relations,

[Ji , Jj ] = iǫijk Jk , (61)

as follows from Eqs. (11.34) and (59).

We can now construct the standard angular momentum basis on three-dimensional space. In
addition to Eq. (60), we need the matrices for J 2 and J± . These are
 
2 0 0
J2 =  0 2 0 (62)
0 0 2
 Notes 19: Irreducible Tensor Operators 19

and  
0 0 ∓1
J± =  0 0 −i  . (63)
±1 i 0
We see that J 2 = 2I, which means that every vector in ordinary space is an eigenvector of J 2 with
eigenvalue j(j + 1) = 2, that is, with j = 1. An irreducible subspace with j = 1 in any vector space
must be 3-dimensional, but in this case the entire space is 3-dimensional, so the entire space consists
of a single irreducible subspace under rotations with j = 1.
The fact that physical space carries the angular momentum value j = 1 is closely related to the
fact that vector operators are irreducible tensor operators of order 1, as explained below. It is also
connected with the fact that the photon, which is represented classically by the vector field A(x)
(the vector potential), is a spin-1 particle.
Since every vector in three-dimensional space is an eigenvector of J 2 , the standard basis consists
of the eigenvectors of J3 , related by raising and lowering operators (this determines the phase
conventions of the vectors, relative to that of the stretched vector). But we can easily check that
the spherical unit vectors (41) are the eigenvectors of J3 , that is,

J3 êq = q êq , q = 0, ±1. (64)

Furthermore, it is easy to check that these vectors are related by raising and lowering operators,
that is,
p
J± êq = (1 ∓ q)(1 ± q + 1) êq±1 , (65)
where J± is given by Eq. (63). Only the overall phase of the spherical basis vectors is not determined
by these relations. The overall phase chosen in the definitions (41) has the nice feature that ê0 = ẑ.
Since the spherical basis is a standard angular momentum basis, its vectors must transform
under rotations according to Eq. (13.85), apart from notation. Written in the notation appropriate
for three-dimensional space, that transformation law becomes
X
Rêq = êq′ Dq1′ q (R). (66)
q′

We need not prove this as an independent result; it is just a special case of Eq. (13.85). This
transformation law is also shown in the final column of Table 1, in order to emphasize its similarity
to related transformation laws on other spaces.
Equation (66) has an interesting consequence, obtained by dotting both sides with ê∗q′ . We use
a round bracket notation for the dot product on the left hand side, and we use the orthogonality
relation (42) on the right hand side, which picks out one term from the sum. We find

ê∗ 1


q′ , Rêq = Dq′ q (R), (67)

which shows that Dq1′ q is just the matrix representing the rotation operator on three-dimensional
space with respect to the spherical basis. The usual rotation matrix contains the matrix elements
20 Notes 19: Irreducible Tensor Operators

with respect to the Cartesian basis, that is,

ĉi , Rĉj = Rij . (68)

See Eq. (11.7). For a given rotation, matrices R and D1 are similar (they differ only by a change of
basis).

15. Reducible and Irreducible Spaces of Operators

In the third row of Table 1 we consider the vector space of operators. The operators in question
are the operators that act on the ket space of our quantum mechanical system, that is, they are
the usual operators of quantum mechanics, for example, the Hamiltonian. Linear operators can
be added and multiplied by scalars, so they form a vector space in the mathematical sense, but of
course they also act on vectors (that is, kets). So the word “vector” is used in two different senses
here. Rotations act on operators according to our definition (6), also shown in the second column of
the table. Thus we have another example of a vector space upon which rotation operators act, and
we can expect that the entire construction of Notes 13 will go through again, apart from notation.
Rather than filling in the rest of the table, however, let us return to the definition of a vector
operator, Eq. (14), and interpret it in a different light. That definition concerns the three components
V1 , V2 and V3 of a vector operator, each of which is an operator itself, and it says that if we rotate
any one of these operators, we obtain a linear combination of the same three operators. Thus, any
linear combination of these three operators is mapped into another such linear combination by any
rotation, or, equivalently, the space of operators spanned by these three operators is invariant under
rotations. Thus we view the three components of V as a set of “basis operators” spanning this space,
which is a 3-dimensional subspace of the space of all operators. (We assume V 6= 0.) A general
element of this subspace of operators is an arbitrary linear combination of the three basis operators,
that is, it has the form
a1 V1 + a2 V2 + a3 V3 = a · V, (69)

a dot product of a vector of numbers a and a vector of operators V.

If a subspace of a vector space is invariant under rotations, then we may ask whether it contains
any smaller invariant subspaces. If not, we say it is irreducible. If so, it can be decomposed into
smaller invariant subspaces, and we say it is reducible. The invariant subspaces of a reducible space
may themselves be reducible or irreducible; if reducible, we decompose them further. We continue
until we have only irreducible subspaces. Thus, every invariant subspace can be decomposed into
irreducible subspaces, which in effect form the building blocks of any invariant subspace.
In the case of a ket space, the subspaces spanned by |γjmi for fixed γ and j but m = −j, . . . , +j
is, in fact, an irreducible subspace. The proof of this will not be important to us, but it is not
hard. What about the three-dimensional space of operators spanned by the components of a vector
operator? It turns out that it, too, is irreducible.
 Notes 19: Irreducible Tensor Operators 21

A simpler example of an irreducible subspace of operators is afforded by any scalar operator K.

If K 6= 0, K can be thought of as a basis operator in a one-dimensional space of operators, in which
the general element is aK, where a is a number (that is, the space contains all multiples of K). Since
K is invariant under rotations [see Eq. (8)], this space is invariant. It is also irreducible, because a
one-dimensional space contains no smaller subspace, so if invariant it is automatically irreducible.
We see that both scalar and vector operators are associated with irreducible subspaces of op-
erators. What about second rank tensor operators Tij ? Such an “operator” is really a tensor of
operators, that is, 9 operators that we can arrange in a 3 × 3 matrix. Assuming these operators are
linearly independent, they span a 9-dimensional subspace of operators that is invariant under rota-
tions, since according to Eq. (24) when we rotate any of these operators we get a linear combination
of the same operators. This space, however, is reducible.
To see this, let us take the example (25) of a tensor operator, that is, Tij = Vi Wj where V
and W are vector operators. This is not the most general form of a tensor operator, but it will
illustrate the points we wish to make. A particular operator in the space of operators spanned by
the components Tij is the trace of Tij , that is,
tr T = T11 + T22 + T33 = V · W. (70)
Being a dot product of two vectors, this is a scalar operator, and is invariant under rotations.
Therefore by itself it spans a 1-dimensional, irreducible subspace of the 9-dimensional space of
operators spanned by the components of Tij . The remaining (orthogonal) 8-dimensional subspace
can be reduced further, for it possesses a 3-dimensional invariant subspace spanned by the operators,
X3 = T12 − T21 = V1 W2 − V2 W1 ,
X1 = T23 − T32 = V2 W3 − V3 W2 ,
X2 = T31 − T13 = V3 W1 − V1 W3 , (71)
or, in other words,
X = V×W. (72)
The components of X form a vector operator, so by themselves they span an irreducible invariant
subspace under rotations. As we see, the components of X contain the antisymmetric part of the
original tensor Tij .
The remaining 5-dimensional subspace is irreducible. It is spanned by operators containing the
symmetric part of the tensor Tij , with the trace removed (or, as we say, the symmetric, traceless
part of Tij ). The following five operators form a basis in this subspace:
S1 = T12 + T21 ,
S2 = T23 + T32 ,
S3 = T31 + T13 ,
S4 = T11 − T22 ,
S5 = T11 + T22 − 2T33 . (73)
22 Notes 19: Irreducible Tensor Operators

The original tensor Tij breaks up in three irreducible subspaces, a 1-dimensional scalar (the
trace), a 3-dimensional vector (the antisymmetric part), and the 5-dimensional symmetric, traceless
part. Notice that these dimensionalities are in accordance with the Clebsch-Gordan decomposition,

1 ⊗ 1 = 0 ⊕ 1 ⊕ 2, (74)

which corresponds to the count of dimensionalities,

3 × 3 = 1 + 3 + 5 = 9. (75)

This Clebsch-Gordan series arises because the vector operators V and W form two ℓ = 1 irreducible
subspaces of operators, and when we form T according to Tij = Vi Wj , we are effectively combining
angular momenta as indicated by Eq. (74). The only difference from our usual practice is that we
are forming products of vector spaces of operators, instead of tensor products of ket spaces.
We have examined this decomposition in the special case Tij = Vi Wj , but the decomposition
itself applies to any second rank tensor Tij . More generally, Cartesian tensors of any rank ≥ 2 are
reducible.
It is possible that a given tensor Tij may have one or more of the three irreducible components
that vanish. The quadrupole moment tensor (26), for example, is already symmetric and traceless,
so its nine components are actually linear combinations of just five independent operators. For
another example, an antisymmetric tensor Tij = −Tji contains only the three-dimensional (vector)
subspace.
For many purposes it is desirable to organize tensors into their irreducible subspaces. This
can be done by going over from the Cartesian to the spherical basis, and then constructing linear
combinations using Clebsch-Gordan coefficients to end up with tensors transforming according to
an irreducible representation of the rotations. We will say more about this process later.

16. Irreducible Tensor Operators

So far we have said nothing about a standard angular momentum basis of operators. The
Cartesian components Vi of a vector operator do form a basis in a 3-dimensional, irreducible subspace
of operators, but they do not transform under rotations as a standard angular momentum basis. We
see this from the definition (14), which shows that if we rotate the basis operators Vi in this subspace,
the coefficients of the linear combinations of the basis operators we obtain are Cartesian components
of the rotation matrix R. When we rotate the basis vectors of a standard angular momentum basis,
the coefficients are components of the D-matrices, as we see in Eq. (13.85). We now define a class
of operators that do transform under rotations as a standard angular momentum basis.
We define an irreducible tensor operator of order k as a set of 2k + 1 operators Tqk , for q =
−k, . . . , +k, that satisfy
X
U Tqk U † = Tqk′ Dqk′ q (U ),
q′
(76)
 Notes 19: Irreducible Tensor Operators 23

for all rotation operators U . We denote the irreducible tensor operator itself by T k , and its 2k + 1
components by Tqk . This definition is really a version of Eq. (13.85), applied to the space of operators.
It means that the components of an irreducible tensor operator are basis operators in a standard
angular momentum basis that spans an irreducible subspace of operators. Thus we place Tqk in the
SAMB column of the third row of Table 1, and the transformation law (76) in the last column. The
three transformation laws in the last column (for three different kinds of spaces) should be compared.
We see that the order k of an irreducible tensor operator behaves like an angular momentum quantum
number j, and q behaves like m.
However, unlike the standard angular momentum basis vectors in ket spaces, irreducible tensor
operators are restricted to integer values of angular momentum quantum number k. The physical
reason for this is that operators, which represent physically observable quantities, must be invariant
under a rotation of 2π; the mathematical reason is that our definition of a rotated operator, given by
Eq. (6), is quadratic U (R), so that the representation of rotations on the vector space of operators
is always a single-valued representation of SO(3).
Let us examine some examples of irreducible tensor operators. A scalar operator K is an
irreducible tensor operator of order 0, that is, it is an example of an irreducible tensor operator T00 .
This follows easily from the fact that K commutes with any rotation operator U , and from the fact
that the j = 0 rotation matrices are simply given by the 1 × 1 matrix (1) [see Eq. (13.68)].
Irreducible tensor operators of order 1 are constructed from vector operators by transforming
from the Cartesian basis to the spherical basis. If we let V be a vector operator as defined by
Eq. (13), and define its spherical components by

Vq = Tq1 = êq · V, (77)

then we have

U (R)Vq U (R)† = êq · (R−1 V) = (Rêq ) · V

X
= Vq′ Dq1′ q (R), (78)
q′

where we use Eq. (66).

The electric quadrupole operator is given as a Cartesian tensor in Eq. (26). This Cartesian
tensor is symmetric and traceless, so it contains only 5 independent components, which span an
irreducible subspace of operators. In fact, this subspace is associated with angular momentum value
k = 2. It is possible to introduce a set of operators Tq2 , q = −2, . . . , +2 that form a standard angular
momentum basis in this space, that is, that form an order 2 irreducible tensor operator. These can
be regarded as the spherical components of the quadrupole moment tensor. We will explore this
subject in more detail later.
24 Notes 19: Irreducible Tensor Operators

17. Commutation Relations of an Irreducible Tensor Operator with J

Above we presented two equivalent definitions of scalar and vector operators, one involving
transformation properties under rotations, and the other involving commutation relations with J. We
will now do the same with irreducible tensor operators. To this end, we substitute the infinitesimal
form (15) of the rotation operator U into both sides of the definition (76).
On the right we will need the D-matrix for an infinitesimal rotation. Since the D-matrix
contains just the matrix elements of U with respect to a standard angular momentum basis [this is
the definition of the D-matrices, see Eq. (13.56)], we require these matrix elements in the case of an
infinitesimal rotation. For θ ≪ 1, Eq. (13.56) becomes
j ′
 i  i
Dm ′ m (n̂, θ) = hjm | 1 − θn̂ · J |jmi = δm′ m − θhjm′ |n̂ · J|jmi. (79)
h̄ h̄
Changing notation (jm′ m) → (kq ′ q) and substituting this and Eq. (15) into the definition (76) of
an irreducible tensor operator, we obtain
 i   i  X  i 
1 − θn̂ · J Tqk 1 + θn̂ · J = Tqk′ δq′ q − θhkq ′ |n̂ · J|kqi , (80)
h̄ h̄ h̄
′ q

or, since n̂ arbitrary unit vector,

X
[J, Tqk ] = Tqk′ hkq ′ |J|kqi. (81)
q′

The operators J on the left- and right-hand sides of Eqs. (80) and (81) are not the same
operators. On the left J is the angular momentum on the same space upon which the operators Tqk
act; in practice this is usually the state space of a quantum system. The J on the right is the angular
momentum operator on a model space in which the matrices Dqk′ q are defined. See the discussion in
Sec. 18.13.
Equation (81) specifies a complete set of commutation relations of the components of J with
the components of an irreducible tensor operator, but it is usually transformed into a different form.
First we take the z-component of both sides and use Jz |kqi = h̄q|kqi, so that

hkq ′ |Jz |kqi = qh̄ δq′ q . (82)

This is Eq. (13.47) with a change of notation. Then Eq. (81) becomes Eq. (89a) below. Next dot
both sides of Eq. (81) with x̂ ± iŷ, and use
p
J± |kqi = (k ∓ q)(k ± q + 1)h̄ |k, q ± 1i, (83)

or
p
hkq ′ |J± |kqi = (k ∓ q)(k ± q + 1)h̄ δq′ ,q±1 . (84)
This is Eq. (13.48b) with a change of notation. Then we obtain Eq. (89b) below. Finally, take the
i-th component of Eq. (81),
X
[Ji , Tqk ] = Tqk′ hkq ′ |Ji |kqi, (85)
q′
 Notes 19: Irreducible Tensor Operators 25

and form the commutator of both sides with Ji ,

X X
[Ji , [Ji , Tqk ]] = [Ji , Tqk′ ]hkq ′ |Ji |kqi = Tqk′′ hkq ′′ |Ji |kq ′ ihkq ′ |Ji |kqi
q′ q′ q′′
X (86)
= Tqk′′ hkq ′′ |Ji2 |kqi,
q′′

where we have used Eq. (81) again to create a double sum. Finally summing both sides over i, we
obtain,
X X
[Ji , [Ji , Tqk ]] = Tqk′′ hkq ′′ |J 2 |kqi. (87)
i q′′

But
hkq ′′ |J 2 |kqi = k(k + 1)h̄2 δq′′ q , (88)
a version of Eq. (13.46), so we obtain Eq. (89c) below.
In summary, an irreducible tensor operator satisfies the following commutation relations with
the components of angular momentum:

[Jz , Tqk ] = h̄q Tqk , (89a)

p k
[J± , Tqk ] = h̄ (k ∓ q)(k ± q + 1) Tq±1 , (89b)
X
[Ji , [Ji , Tqk ]] = h̄2 k(k + 1) Tqk . (89c)
i

We see that forming the commutator with J± plays the role of a raising or lowering operator for the
components of an irreducible tensor operator. As we did with scalar and vector operators, we can
show that these angular momentum commutation relations are equivalent to the definition (76) of
an irreducible tensor operator. This is done by showing that Eqs. (89) are equivalent to Eq. (76) in
the case of infinitesimal rotations, and that if Eq. (76) is true for any two rotations, it is also true
for their product. Thus by building up finite rotations as products of infinitesimal ones we show
the equivalence of Eqs. (76) and (89). Many books take Eqs. (89) as the definition of an irreducible
tensor operator.

18. Statement and Applications of the Wigner-Eckart Theorem

The Wigner-Eckart theorem is not difficult to remember and it is quite easy to use. In this
section we discuss the statement of the theorem and ways of thinking about it and its applications,
before turning to its proof.
The Wigner-Eckart theorem concerns matrix elements of an irreducible tensor operator with
respect to a standard angular momentum basis of kets, something we will write in a general notation
as hγ ′ j ′ m′ |Tqk |γjmi. As an example of such a matrix element, you may think of the dipole matrix
elements hn′ ℓ′ m′ |xq |nℓmi that we examined in Sec. 13. In that case the operator (the position or
dipole operator) is an irreducible tensor operator with k = 1.
26 Notes 19: Irreducible Tensor Operators

The matrix element hγ ′ j ′ m′ |Tqk |γjmi depends on 8 indices, (γ ′ j ′ m′ ; γjm; kq), and in addition
it depends on the specific operator T in question. The Wigner-Eckart theorem concerns the de-
pendence of this matrix element on the three magnetic quantum numbers (m′ mq), and states that
that dependence is captured by a Clebsch-Gordan coefficient. More specifically, the Wigner-Eckart
theorem states that hγ ′ j ′ m′ |Tqk |γjmi is proportional to the Clebsch-Gordan coefficient hj ′ m′ |jkmqi,
with a proportionality factor that is independent of the magnetic quantum numbers. That propor-
tionality factor depends in general on everything else besides the magnetic quantum numbers, that
is, (γ ′ j ′ ; γj; k) and the operator in question. The standard notation for the proportionality factor is
hγ ′ j ′ ||T k ||γji, something that looks like the original matrix element except the magnetic quantum
numbers are omitted and a double bar is used. The quantity hγ ′ j ′ ||T k ||γji is called the reduced
matrix element. With this notation, the Wigner-Eckart theorem states

hγ ′ j ′ m′ |Tqk |γjmi = hγ ′ j ′ ||T k ||γji hj ′ m′ |jkmqi.

(90)

The reduced matrix element can be thought of as depending on the irreducible tensor operator T k
and the two irreducible subspaces (γ ′ j ′ ) and (γj) that it links. Some authors (for example, Sakurai)
√
include a factor of 1/ 2j + 1 on the right hand side of Eq. (90), but here that factor has been
absorbed into the definition of the reduced matrix element. The version (90) is easier to remember
and closer to the basic idea of the theorem.
To remember the Clebsch-Gordan coefficient it helps to suppress the bra hγ ′ j ′ m′ | from the
matrix element and think of the ket Tqk |γjmi, or, more precisely, the (2j + 1)(2k + 1) kets that
are produced by letting m and q vary over their respective ranges. This gives an example of an
operator with certain angular momentum indices multiplying a ket with certain angular momentum
indices. It turns out that such a product of an operator times a ket has much in common with the
product (i.e., the tensor product) of two kets, insofar as the transformation properties of the product
under rotations are concerned. That is, suppose we were multiplying a ket |kqi with the given
angular momentum quantum numbers times another ket |jmi with different angular momentum
quantum numbers. Then we could find the eigenstates of total angular momentum by combining
the constituent angular momenta according to k ⊗ j. Actually, in thinking of kets Tqk |jmi, it is
customary to think of the product of the angular momenta in the reverse order, that is, j ⊗ k. This
is an irritating convention because it makes the Wigner-Eckart theorem harder to remember, but I
suspect it is done this way because in practice k tends to be small and j large.
In any case, thinking of the product of kets, the product

|jmi ⊗ |kqi = |jkmqi (91)

contains various components of total J 2 and Jz , that is, it can be expanded as a linear combination
of eigenstates of total J 2 and J z , with expansion coefficients that are the Clebsch-Gordan coeffi-
cients. The coefficient with total angular momentum j ′ and z-component m′ is the Clebsch-Gordan
coefficient hj ′ m′ |jkmqi, precisely what appears in the Wigner-Eckart theorem (90).
 Notes 19: Irreducible Tensor Operators 27

Probably the most useful application of the Wigner-Eckart theorem is that it allows us to easily
write down selection rules for the given matrix element, based on the selection rules of the Clebsch-
Gordan coefficient occurring in Eq. (90). In general, a selection rule is a rule that tells us when a
matrix element must vanish on account of some symmetry consideration. The Wigner-Eckart the-
orem provides us with all the selection rules that follow from rotational symmetry; a given matrix
element may have other selection rules based on other symmetries (for example, parity). The selec-
tion rules that follow from the Wigner-Eckart theorem are that the matrix element hγj ′ m′ |Tqk |γjmi
vanishes unless m′ = m + q and j ′ takes on one of the values, |j − k|, |j − k| + 1, . . . , j + k.
Furthermore, suppose we actually have to evaluate the matrix elements hγ ′ j ′ m′ |Tqk |γjmi for
all (2k + 1)(2j + 1) possibilities we get by varying q and m. We must do this, for example, in
computing atomic transition rates. (We need not vary m′ independently, since the selection rules
enforce m′ = m+q.) Then the Wigner-Eckart theorem tells us that we actually only have to do one of
these matrix elements (presumably, whichever is the easiest), because if we know the left hand side of
Eq. (90) for one set of magnetic quantum numbers, and if we know the Clebsch-Gordan coefficient on
the right-hand side, then we can determine the proportionality factor, that is, the reduced matrix
element. Then all the other matrix elements for other values of the magnetic quantum numbers
follow by computing (or looking up) Clebsch-Gordan coefficients. This procedure requires that the
first matrix element we calculate be nonzero.
In some other cases, we have analytic formulas for the reduced matrix element. That was
the case of the application in Sec. 13, where the three-Yℓm formula allowed us to compute the
proportionality factor explicitly.

19. The Wigner-Eckart Theorem for Scalar Operators

Let us consider a scalar operator for which k = q = 0, such as the Hamiltonian H for an isolated
system, that is, with T00 = H. In this case the Clebsch-Gordan coefficient is

hj ′ m′ |j0m0i = δj ′ j δm′ m , (92)

so the Wigner-Eckart theorem can be written

hγ ′ j ′ m′ |H|γjmi = Cγj ′ γ δj ′ j δm′ m , (93)

where
Cγj ′ γ = hγ ′ j||H||γji. (94)

We write it this way because Cγj ′ γ can be seen as a set of matrices, labeled by j and indexed by (γ ′ γ).
The size of the j-th matrix is Nj , the multiplicity of the j value in the system under consideration.
In practice the multiplicity is often infinite. The problem of finding the energy eigenvalues of the
system amounts to diagonalizing each of the matrices Cγj ′ γ .
28 Notes 19: Irreducible Tensor Operators

20. Proof of the Wigner-Eckart Theorem

Consider the product of kets |jmi ⊗ |kqi = |jkmqi with the given angular momentum quantum
numbers, and consider the (2j + 1)(2k + 1)-dimensional product space spanned by such kets when we
allow the magnetic quantum numbers m and q to vary over their respective ranges. The eigenstates
|JM i of total J 2 and Jz in this space are given by the Clebsch-Gordan expansion,
X
|JM i = |jkmqihjkmq|JM i. (95)
mq

Moreover, the states |JM i for fixed J and M = −J, . . . , +J form a standard angular momentum
basis in an invariant, irreducible subspace of dimension 2J +1 in the product space. This means that
the basis states |JM i are not only eigenstates of total J 2 and Jz , but they are also linked by raising
and lowering operators. Equivalently, the states |JM i transform as a standard angular momentum
basis under rotations,
X
J
U |JM i = |JM ′ iDM ′ M (U ). (96)
M′

Now consider the (2j + 1)(2k + 1) kets Tqk |γjmi obtained by varying m and q. We construct
linear combinations of these with the same Clebsch-Gordan coefficients as in Eq. (95),
X
|X; JM i = Tqk |γjmihjkmq|JM i, (97)
mq

and define the result to be the ket |X; JM i, as indicated. The indices JM in the ket |X; JM i
indicate that the left-hand side depends on these indices, because the right hand side does; initially
we assume nothing else about this notation. Similarly, X simply stands for everything else the
left-hand side depends on, that is, X is an abbreviation for the indices (γkj).
However, in view of the similarity between Eqs. (95) and (97), we can guess that |X; JM i is
actually an eigenstate of J 2 and Jz with quantum numbers J and M , and that the states |X; JM i
are related by raising and lowering operators. That is, we guess

Jz |X; JM i = M h̄ |X; JM i, (98a)

p
J± |X; JM i = (J ∓ M )(J ± M + 1)h̄ |X; J, M ± 1i, (98b)
J 2 |X; JM i = J(J + 1)h̄2 |X; JM i. (98c)

If true, this is equivalent to the transformation law,

X
J
U |X; JM i = |X; JM ′ i DM ′ M (U ), (99)
M′

exactly as in Eq. (96). Equations (98) and (99) are equivalent because Eq. (98) can be obtained
from Eq. (99) by specializing to infinitesimal rotations, while Eq. (99) can be obtained from Eq. (98)
by building up finite rotations out of infinitesimal ones.
 Notes 19: Irreducible Tensor Operators 29

In Sec. 21 below we will prove that these guesses are correct. For now we merely explore the
consequences. To begin, since |X; JM i is an eigenstate of J 2 and Jz with quantum numbers J and
M , it can be expanded as a linear combination of the standard basis kets |γjmi with the same values
j = J and m = M , but in general all possible values of γ. That is, we have an expansion of the
form,
|γ ′ JM i CγkJMj
X
|X; JM i = ′γ , (100)
γ′

where the indices on the expansion coefficients CγkJMj

′γ simply list all the parameters they can depend
on. These coefficients, do not, however, depend on M , as we show by applying raising or lowering
operators to both sides, and using Eq. (98b). This gives
p
(J ∓ M )(J ± M + 1)h̄ |X; J, M ± 1i

(J ∓ M )(J ± M + 1)h̄ |γ ′ J, M ± 1i CγkJMj

Xp
= ′γ , (101)
γ′

or, after canceling the square roots,

|γ ′ J, M ± 1i CγkJMj
X
|X; J, M ± 1i = ′γ . (102)
γ′

Comparing this to Eq. (100), we see that the expansion coefficients are the same for all M values,
and thus independent of M . We will henceforth write simply CγkJj
′ γ for them.

Now we return to the definition (97) of the kets |X; JM i and use the orthogonality of the
Clebsch-Gordan coefficients (18.50) to solve for the kets Tqk |γjmi. This gives

|γ ′′ JM i CγkJj
X X
Tqk |γjmi = |X; JM ihJM |jkmqi = ′′ γ hJM |jkmqi, (103)
JM γ ′′ JM

where we use Eq. (100), replacing γ ′ with γ ′′ . Now multiplying this by hγ ′ j ′ m′ | and using the
orthonormality of the basis |γjmi, we obtain
′
hγ ′ j ′ m′ |Tqk |γjmi = Cγkj′ γj hj ′ m′ |jkmqi, (104)

which is the Wigner-Eckart theorem (90) if we identify

′
Cγkj′ γj = hγ ′ j ′ ||T k ||γji. (105)

21. Proof of Eq. (99)

To complete the proof of the Wigner-Eckart theorem we must prove Eq. (99), that is, we
must show that the kets |X; JM i transform under rotations like the vectors of a standard angular
momentum basis. To do this we call on the definition of |X; JM i, Eq. (97), and apply U to both
sides,
X
U |X; JM i = U Tqk U † U |γjmihjkmq|JM i. (106)
mq
30 Notes 19: Irreducible Tensor Operators

Next we use the definition of an irreducible tensor operator (76) and the transformation law for
standard basis vectors under rotations, Eq. (13.85), to obtain
j
X
U |X; JM i = Tqk′ |γjm′ i Dm k
′ m (U ) Dq ′ q (U ) hjkmq|JM i. (107)
mq
m′ q ′

We now call on Eq. (18.64) with a change of indices,

j
X
k J′
Dm ′ m (U )Dq ′ q (U ) = hjkm′ q ′ |J ′ M ′ i DM ′ ′′
′ M ′′ (U ) hJ M |jkmqi, (108)
J ′ M ′ M ′′

which expresses the product of D-matrices in Eq. (107) in terms of single D-matrices. When we
substitute Eq. (108) into Eq. (107), the m′ q ′ -sum is doable by the definition (97),
X
Tqk′ |γjm′ ihjkm′ q ′ |J ′ M ′ i = |X; J ′ M ′ i, (109)
m′ q ′

and the mq-sum is doable by the orthogonality of the Clebsch-Gordan coefficients,

X
hJ ′ M ′′ |jkmqihjkmq|JM i = δJ ′ J δM ′′ M . (110)
mq

Altogether, Eq. (107) becomes

X
J′
X
J
U |X; JM i = |X; J ′ M ′ iDM ′ M ′′ (U ) δJ ′ J δM ′′ M = |X; JM ′ iDM ′ M (U ). (111)
J ′ M ′ M ′′ M′

This proves Eq. (99).

Instead of proving Eq. (99), many authors (for example, Sakurai) prove the equivalent set
of statements (98), which involve the actions of the angular momentum operators on the states
|X; JM i. I think the transformation properties under rotations are little easier. In either case, the
rest of the proof is the same.

22. Products of Irreducible Tensor Operators

As we have seen, the idea behind the Wigner-Eckart theorem is that a product of an irreducible
tensor operator Tqk times a ket of the standard basis |γjmi transforms under rotations exactly as the
tensor product of two kets of standard bases with the same quantum numbers, |jmi⊗|kqi. Similarly,
it turns out that the product of two irreducible tensor operators, say, Xqk11 Yqk22 , transforms under
rotations exactly like the tensor product of kets with the same quantum numbers, |k1 q1 i ⊗ |k2 q2 i.
In particular, such a product of operators can be represented as a linear combination of irreducible
tensor operators with order k lying in the range |k1 − k2 |, . . . , k1 + k2 , with coefficients that are
Clebsch-Gordan coefficients. That is, we can write
X
Xqk11 Yqk22 = Tqk hkq|k1 k2 q1 q2 i, (112)
kq
 Notes 19: Irreducible Tensor Operators 31

where the Tqk are new irreducible tensor operators.

To prove this, we first solve for Tqk ,
X
Tqk = Xqk11 Yqk22 hk1 k2 q1 q2 |kqi, (113)
q1 q2

which we must show is an irreducible tensor operator. To do this, we conjugate both sides of this
with a rotation operator U and use the fact that X and Y are irreducible tensor operators,
X
U Tqk U † = U Xqk11 U † U Yqk22 U † hk1 k2 q1 q2 |kqi
q1 q2
X
= Xqk′1 Yqk′ 2 Dqk′1q1 (U )Dqk′2q2 (U ) hk1 k2 q1 q2 |kqi. (114)
1 2 1 2
q1 q2
q1′ q2′

Next we use Eq. (18.64) with a change of symbols,

X
Dqk′1q1 (U )Dqk′2q2 (U ) = K
hk1 k2 q1′ q2′ |KQ′ iDQ′ Q (U )hKQ|k1 k2 q1 q2 i, (115)
1 2
KQQ′

which we substitute into Eq. (114). Then the q1′ q2′ -sum is doable in terms of the expression (113)
for Tqk ,
X
Xqk′1 Yqk′ 2 hk1 k2 q1′ q2′ |KQ′ i = TQK′ , (116)
1 2
q1′ q2′

and the q1 q2 -sum is doable by the orthogonality of the Clebsch-Gordan coefficients,

X
hKQ|k1 k2 q1 q2 ihk1 k2 q1 q2 |kqi = δKk δQq . (117)
q1 q2

Then Eq. (114) becomes

X X
U Tqk U † = TQK′ DQ
K
′ Q δKk δQq = Tqk′ Dqk′ q (U ). (118)
KQQ′ q′

This shows that Tqk is an irreducible tensor operator, as claimed.

As an application, two vector operators V and W, may be converted into k = 1 irreducible
tensor operators Vq and Wq by going over to the spherical basis. From these we can construct
k = 0, 1, 2 irreducible tensor operators according to
X
Tqk = Vq1 Wq2 h11q1 q2 |kqi. (119)
q1 q2

This will yield the same decomposition of a second rank tensor discussed in Sec. 15, where we found
a scalar (k = 0), a vector (k = 1), and a symmetric, traceless tensor (k = 2).
32 Notes 19: Irreducible Tensor Operators

Problems

1. This will help you understand irreducible tensor operators better. Let E be a ket space for some
system of interest, and let A be the space of linear operators that act on E. For example, the
ordinary Hamiltonian is contained in A, as are the components of the angular momentum J, the
rotation operators U (R), etc. The space A is a vector space in its own right, just like E; operators
can be added, multiplied by complex scalars, etc. Furthermore, we may be interested in certain
subspaces of A, such as the 3-dimensional space of operators spanned by the components Vx , Vy , Vz
of a vector operator V.
Now let S be the space of linear operators that act on A. We call an element of S a “super”
operator because it acts on ordinary operators; ordinary operators in A act on kets in E. We will
denote super-operators with a hat, to distinguish them from ordinary operators. (This terminology
has nothing to do with supersymmetry.)
Given an ordinary operator A ∈ A, it is possible to associate it in several different ways with a
super-operator. For example, we can define a super operator ÂL , which acts by left multiplication:

ÂL X = AX, (120)

where X is an arbitrary ordinary operator. This equation obviously defines a linear super-operator,
that is, ÂL (X + Y ) = ÂL X + ÂL Y , etc. Similarly, we can define a super-operator associated with
A by means of right multiplication, or by means of the forming of the commutator, as follows:

ÂR X = XA,
(121)
ÂC X = [A, X].

There are still other ways of associating an ordinary operator with a super-operator. Let R be a
classical rotation, and let U (R) be a representation of the rotations acting on the ket space E. Thus,
the operators U (R) belong to the space A. Now associate such a rotation operator U (R) in A with
a super-operator Û (R) in S, defined by

Û (R)X = U (R) X U (R)† . (122)

Again, Û (R) is obviously a linear super-operator.

(a) Show that Û (R) forms a representation of the rotations, that is, that

Û (R1 )Û (R2 ) = Û (R1 R2 ). (123)

This is easy.
Now let U (R) be infinitesimal as in Eq. (15), and let

i
Û (R) = 1 − θn̂ · Ĵ. (124)
h̄
 Notes 19: Irreducible Tensor Operators 33

(Here the hat on n̂ denotes a unit vector, while that on Ĵ denotes a super-operator.) Express the
super-operator Ĵ in terms of ordinary operators. Write Eqs. (89) in super-operator notation. Work
out the commutation relations of the super-operators Ĵ.

(b) Now write out nine equations, specifying the action of the three super-operators Jˆi on the the
basis operators Vj . Write the answers as linear combinations of the Vj ’s. Then write out six more
equations, specifying the action of the super raising and lowering operators, Jˆ± , on the three Vj .
Now find the operator A that is annihilated by Jˆ+ . Do this by writing out the unknown operator
as a linear combination of the Vj ’s, in the form

A = ax Vx + ay Vy + az Vz , (125)

and then solving for the coefficients ai . Show that this operator is an eigenoperator of Jˆz with
eigenvalue +h̄. In view of these facts, the operator A must be a “stretched” operator for k = 1;
henceforth write T11 for it. This operator will have an arbitrary, complex multiplicative constant,
call it c. Now apply Jˆ− , and generate T01 and T−1
1
. Choose the constant c to make T01 look as simple
as possible. Then write
Tq1 = êq · V, (126)
and thereby “discover” the spherical basis.

2. This problem concerns quadrupole moments and spins. It provides some background for prob-
lem 3.

(a) In the case of a nucleus, the spin Hilbert space Espin = span{|smi, m = −s, . . . , +s} is actually
the ground state of the nucleus. It is customary to denote the angular momentum j of the ground
state by s. This state is (2s+ 1)-fold degenerate. The nuclear spin operator S is really the restriction
of the total angular momentum of the nucleus J to this subspace of the (much larger) nuclear Hilbert
space.
Let Akq and Bqk be two irreducible tensor operators on Espin . As explained in these notes, when
we say “irreducible tensor operator” we are really talking about the collection of 2k + 1 operators
obtained by setting q = −k, . . . , +k. Use the Wigner-Eckart theorem to explain why any two such
operators of the same order k are proportional to one another. This need not be a long answer.
Thus, all scalars are proportional to a standard scalar (1 is convenient), and all vector operators
(for example, the magnetic moment µ) are proportional to a standard vector (S is convenient), etc.
For a given s, what is the maximum value of k? What is the maximum order of an irreducible
tensor operator that can exist on space Espin for a proton (nucleus of ordinary hydrogen)? A deuteron
(heavy hydrogen)? An alpha particle (nucleus of helium)? These rules limit the electric and magnetic
multipole moments that a nucleus is allowed to have, as is discussed more fully in Notes 26.

(b) Let A and B be two vector operators (on any Hilbert space, not necessarily Espin ), with spherical
components Aq , Bq , as in Eq. (77). As explained in the notes, Aq and Bq are k = 1 irreducible
34 Notes 19: Irreducible Tensor Operators

tensor operators. As explained in Sec. 22, it is possible to construct irreducible tensor operators Tqk
for k = 0, 1, 2 out of the nine operators, {Aq Bq′ , q, q ′ = −1, 0, 1}. Write out the three operators T00 ,
T11 and T22 in terms of the Cartesian products Ai Bj . Just look up the Clebsch-Gordan coefficients.
There are nine operators in T00 , Tq1 and Tq2 , but I’m only asking you to compute these three to save
you some work.
Show that T00 is proportional to A · B, that T11 is proportional to a spherical component of
A×B, and that T22 can be written in terms of the components of the symmetric and traceless part
of the Cartesian tensor Ai Bj , which is
1 1
(Ai Bj + Aj Bi ) − (A · B)δij . (127)
2 3
(c ) In classical electrostatics, the quadrupole moment tensor Qij of a charge distribution ρ(x) is
defined by Z
Qij = d3 x ρ(x)[3xi xj − r2 δij ], (128)

where x is measured relative to some origin inside the charge distribution. The quadrupole mo-
ment tensor is a symmetric, traceless tensor. The quadrupole energy of interaction of the charge
distribution with an external electric field E = −∇φ is
1X ∂ 2 φ(0)
Equad = Qij . (129)
6 ij ∂xi ∂xj
This energy must be added to the monopole and dipole energies, plus the higher multipole energies.
In the case of a nucleus, we choose the origin to be the center of mass, whereupon the dipole
moment and dipole energy vanish. The monopole energy is just the usual Coulomb energy qφ(0),
where q is the total charge of the nucleus. Thus, the quadrupole term is the first nonvanishing
correction. However, the energy must be understood in the quantum mechanical sense.
Let {xα , α = 1, . . . , Z} be the position operators for the protons in a nucleus. The neutrons are
neutral, and do not contribute to the electrostatic energy. The electric quadrupole moment operator
for the nucleus is defined by
X
Qij = e (3xαi xαj − rα2 δij ), (130)
α
where e is the charge of a single proton. In an external electric field, the nuclear Hamiltonian
contains a term Hquad , exactly in the form of Eq. (129), but now interpreted as an operator.
The operator Qij , being symmetric and traceless, constitutes the Cartesian specification of a
k = 2 irreducible tensor operator, that you could turn into standard form Tq2 , q = −2, . . . , +2 using
the method of part (b) if you wanted to. We’ll stay with the Cartesian form here, however. When
the operator Qij is restricted to the ground state (really a manifold of 2s + 1 degenerate states), it
remains a k = 2 irreducible tensor operator. According to part (a), it must be proportional to some
standard irreducible tensor operator, for which 3Si Sj − S 2 δij is convenient. That is, we must be
able to write
Qij = a(3Si Sj − S 2 δij ), (131)
 Notes 19: Irreducible Tensor Operators 35

for some constant a.

It is customary in nuclear physics to denote the “quadrupole moment” of the nucleus by the
real number Q, defined by
Q = hss|Q33 |ssi, (132)
where |ssi is the stretched state. Don’t confuse Qij , a tensor of operators, with Q, a single number.
The book, Modern Quantum Mechanics by J. J. Sakurai gives the interaction energy of a nucleus
in an external electric field as
eQ h ∂ 2 φ   ∂2φ   ∂2φ  i
2 2
Hint = S x + S y + S2 , (133)
2s(s − 1)h̄2 ∂x2 ∂y 2 ∂z 2 z

where φ is the electrostatic potential for the external field satisfying the Laplace equation ∇2 φ = 0
and where the coordinate axes are chosen so that the off-diagonal elements of ∂ 2 φ/∂xi ∂xj vanish.
Here φ and its derivatives are evaluated at the center of mass of the nucleus and φ satisfies the
Laplace equation rather than the Poisson equation because the sources of the external electric field
are outside the nucleus.
Express the quantity a in Eq. (131) in terms of Q, and derive a version of Eq. (133). This
equation, copied out of the book, has an error in it; correct it.

3. This is Sakurai, problem 3.29, p. 247; or Sakurai and Napolitano, problem 3.33, p. 261.
A spin- 23 nucleus situated at the origin is subjected to an external inhomogeneous electric field.
The basic electric quadrupole interaction is given by Eq. (133) (but corrected), where as above φ
satisfies the Laplace equation and the off-diagonal components ∂ 2 φ/∂xi ∂xj vanish. Show that the
interaction energy can be written

A(3Sz2 − S 2 ) + B(S+
2 2
+ S− ), (134)

and express A and B in terms of the nonvanishing second derivatives of φ, evaluated at the origin.
Determine the energy eigenkets (in terms of |mi, where m = ± 23 , ± 21 ) and the corresponding energy
eigenvalues. Is there any degeneracy?

4. Some questions on the Mössbauer effect and the Pound-Rebka experiment.

The gravitational red shift is a prediction of general relativity, but the basic physics behind it
can be understood in elementary terms. In the Pound-Rebka experiment, photons were launched
upward from the ground, just outside a building at Harvard University. These photons were then
received by a detector just outside the window of an upper floor of the building, approximately
20 meters above the ground.
Suppose at the very moment a photon is released at the ground, a physicist with a detector
in hand jumps out of the upper storey window. As the photon is climbing upward in the earth’s
gravitational field, suffering a red shift in the process, the physicist is falling, accelerating downward.
At a certain point the detector captures the photon and the physicist measures its frequency. This
36 Notes 19: Irreducible Tensor Operators

frequency is blue shifted compared to what it would be if the physicist had just held the detector
out the window, instead of jumping, because of the Doppler shift due to the downward motion of
the detector.
It turns out that the gravitational red shift and the blue shift due to the falling detector (cum
physicist) exactly cancel. Thus, the physicist finds a measured frequency of the photon exactly
equal to the frequency it had when emitted at ground level. You can use this fact to calculate the
gravitational red shift as seen by a detector that is just held out the window, not falling.
This situation is similar to that illustrated in the “shoot the monkey” demonstration used in
elementary physics classes, except that instead of an arrow shot at the monkey particles of light are
used. The basic physical reasoning used here is close to that employed by Einstein in his 1911 paper
which first predicted the gravitational red shift.

(a) If an atom is free (not a part of a crystal lattice or otherwise bound to anything else), then it
suffers some recoil on emitting a photon, which produces a shift ∆ω in the frequency of the emitted
photon. In the case of the 14.4 KeV photon emitted by the 57 Fe nucleus, calculate the fractional
shift ∆ω/ω due to this recoil and compare to the natural line width.

(b) If the 57 Fe atom is free-floating in a gas at 300 K, calculate the average ∆ω/ω due to the Doppler
shift due to the thermal motion of the atom.

(c) Calculate the ∆ω/ω for the gravitational red shift of the same photon climbing (as in the Pound-
Rebka experiment) about 20 meters in the earth’s gravitational field, and compare to the ∆ω/ω due
to the natural line width. You will see that the experiment was a delicate one that required careful
measurement and attention to systematic errors.