Rotations, spherical tensor operators and the Wigner-Eckart Theorem
From A Web-Based Quantum Mechanics Course with In-Class Tutorials by
Marianne Breinig The University of Tennessee, Department of Physics and Astronomy
http://electron6.phys.utk.edu/qm2/ ; this module is:
http://electron6.phys.utk.edu/qm2/modules/m4/wigner.htm
1. Adding two angular momenta
Vectors in the subspace E(j1,j2) may be expanded in terms of
or
basis vectors. The basis vectors of one basis can be written as
linear combinations of basis vectors of the other basis.
.
The expansion coefficients
are called the
Clebsch-Gordan coefficients. Sometimes the Clebsch-Gordan coefficients are
written in terms of the Wigner 3-j symbols. We have
Properties of the Wigner 3-j symbols:
We can permute the columns of the 3-j symbol. An even permutation does not
alter its value. An odd permutation multiplies the initial value by
.
Moreover,
.
This implies
or
Adding more than two angular momenta
When adding the angular momenta j1, j2, and j3, many different states of the
system may correspond to the same value of j and m.
Let
Then
The Racah W coefficients and 6-j symbols transform from one scheme of
adding three angular momenta to another.
The 9-j symbols transform from one scheme of adding four angular momenta
to another.
Links:
Racah coefficients, 6-j symbols, 9-j symbols
3-j, 6-j, and 9-j symbol calculators (1)
3-j, 6-j, and 9-j symbol calculators (2)
We will now establish a connection between the Clebsch-Gordan coefficients
and the matrix elements of operators that transform like the spherical
harmonics under rotation.
The rotation matrices
Angular momentum and rotations are closely linked. Let us, for a moment,
return to the rotation matrices. An arbitrary rotation can be accomplished in
three steps, known as Euler rotations, and it is therefore characterized by
three angles, known as Euler angles. We may proceed as follows.
First rotate the body ccw about the z-axis by an angle . Then rotate ccw
about the y’-axis by an angle . Finally rotate ccw about the z’-axis by an
angle .
The rotation matrix is R()=Rz’()Ry’()Rz(). (Note: In quantum mechanics
it is advantageous to perform the second rotation about the y’-axis. In
classical mechanics the second rotation is often performed about the x’-axis.)
Rz’() and Ry’() are rotations about body-fixed axis. We want to express these
rotations in terms of rotations about space-fixed axis. We may write
Ry’()=Rz()Ry()Rz-1(). Similarly, Rz’()=Ry’()Rz()Ry’-1(). We can now rewrite
R() as
R()=Rz’()Ry’()Rz().
=Ry’()Rz()Ry’-1()Ry’()Rz()
=Ry’()Rz()Rz()
=Rz()Ry()Rz-1()Rz()Rz()
=Rz()Ry()Rz(). Rotations about the same axis commute.)
The rotation operator U(R()) therefore may be written as
U(R())=U(Rz())U(Ry())U(Rz()).
It only involves rotations about space-fixed axis. We already know the rotation
operators for rotations about space-fixed axes. (See notes.)
For a spin ½ particle U(R())=U(Rz())U(Ry())U(Rz()) becomes
using
.
(See notes.)
The matrix elements of the second rotation are purely real. This is a result of
choosing the y’-axis for our second Euler rotation.
For a system with angular momentum J the matrix elements of the rotation
operator in the subspace E(j) in the {|j,m>} basis for a rotation R(n,) are
Uj
m',m(R)=<j,m'|exp(-iJn/)|j,m>.
In E matrix elements between states with different values j vanish. In E the
matrix of the rotation operator is block diagonal. Uj(R) is referred to as the
(2j+1)-dimensional irreducible representation of the rotation operator U(R).
The Uj(R) form a group.
The matrix elements of the rotation operator,
R() are
for the rotation
The only part which mixes different m values is the middle rotation. We can
define a new matrix dj with elements
,
which contains the nontrivial part of the rotation matrix Uj() . The matrix
elements
coefficients.
are also listed on the standard table of Clebsch-Gordan
We can establish a connection between the rotation operator and the spherical
harmonics by constructing the state vector
(This state vector is an eigenvector of the
position operator.)
Let
and
Then
and
(Matrix elements between states with different values l vanish.)
We have
We therefore have
or
This is the connection between the rotation operator and the spherical
harmonics. When m = 0 then
Transformation properties of the spherical harmonics
How do the spherical harmonics transform under rotation?
Let
Then
But
, therefore
The function
can also be viewed as an operator in coordinate space. (It
could, for example, be the operator of an experiment that measures an
emission pattern and projects it onto a known distribution.) The coordinate
representation of this operator operating on the state vector |> is
(r),
just as the coordinate representation of the operator X operating on the state
vector |> is x(r).
Rotating the observable
we obtain
. (See notes.)
[U(R)|r>=|Rr>, U(R)(r)=(R-1r), UT(R)(r)=(Rr),
UT(R)Ylm(n)(r)=Ylm(Rn)(Rr),
UT(R)Ylm(n)U(R)(r)=UT(R)Ylm(n)(R-1r)=Ylm(Rn)(r)=Ylm(n')(r)]
We may therefore write
We define as an irreducible tensor operator of rank k any set of 2k+1
quantities that transform like the spherical harmonics Ykq under rotation.
Tkq is an irreducible or spherical tensor of rank k with 2k+1 components if
or, equivalently
or
Considering the infinitesimal form of this expression we have
or, multiplying out the terms,
This yields
.
The two commutation relations can also be taken as a definition of a
spherical tensor of rank k.
Examples of tensor operators:
A scalar operator is an irreducible tensor of rank 0. It consist of one
component and commutes with all components of the angular momentum J.
It therefore fulfills the commutation relations that define a spherical tensor
of rank 0.
The spherical components of a vector operator A are defined as
Using our definition of a vector operator
we can show that
these components satisfy the commutation relations that define a spherical
tensor of rank 1. For example,
etc.
A vector operator therefore is a spherical tensor operator of rank 1.
A Cartesian tensor
is defined through
Let A and B be two vectors, then Tij=AiBj are the elements of a Cartesian
tensor. Cartesian tensors are reducible, they can be decomposed into
objects that transform differently under rotation. For Tij=AiBj we have
is a scalar,
represents three independent
components which transform like a vector.
represents the components of a traceless,
symmetric tensor with 5 independent components. Tij has been decomposed
into three objects that transform like the spherical harmonics with l = 0, 1,
and 2 respectively. It has been decomposed into irreducible spherical
tensors.
The Wigner-Eckart theorem
Why do we care about tensor operators?
The matrix elements of tensor operators with respect to angular momentum
eigenstates satisfy
The double bar matrix element is independent of m, m’, and q. This is the
Wigner-Eckart theorem.
(In the above equation (2j'+1)-1/2 is factored out of the double bar matrix
element. This factoring is not unique. Different books may factor out a
different factor.)
The matrix element
is written as the product of two
factors. The first factor is a Clebsch-Gordan coefficient which depends only on
the way the system is oriented with respect to the z-axis. It does not depend
on the physical nature of the particular tensor operator. The second term
depends on the physical nature of the operator and the system, but it does not
depend on the magnetic quantum numbers m, m’, and q. To evaluate
for the various combinations of m, m’, and q we need to
only evaluate one matrix element. The others are related via the ClebschGordan coefficients.
Certain selection rules for tensor operators follow directly from the selection
rule of the Clebsch-Gordan coefficients. The matrix element
is zero unless m’=m+q and
Proof of the Wigner-Eckart theorem:
Using the commutation relations that define a tensor operator we have
yields
or
Similarly we evaluate the matrix elements of
We find the same recursion relations for the
as we found
for the Clebsch-Gordan coefficients with j1=j, j2=k, and j=j’, m1=m, m2=q, and
m=m’ . Therefore the m behavior of the
must be the same
as that of the Clebsch-Gordan coefficients. We write
Tensor Operators
Review of the Rotation Operator
Recall that the matrix elements of the rotation operator
  
U R
e

i  J
within a definite j subspace are written
  
Dm jm R 
 j, m e

i  J
j, m .
Rotating a Ket
By a ket, we mean here an element of the j subspace expressed as a sum over the basis
kets j, m (which we’ll just write as m here) with a set of coefficients  ,
m
 
j

m  j
m
m
and we are finding how that set of coefficients transforms under rotation. (Any such ket
will stay within the j subspace under rotation, because the rotation operator commutes
2
with J .)

 
Rotating the ket by  to    m m ,
  e

i  J

   m m m e

i  J
m
m , m
j
  Dm m  m m .
m , m
So the ket rotation transformation is
 m    Dm jm  m , or    D
m, m
with the usual matrix-multiplication rules.
Rotating a Basis Vector
j, m
Now we look at something slightly different: suppose we apply the rotation operator to
j, m , what is the result?
one of the basis vectors
e

i  J
j , m   j , m j , m e
m

i  J
j , m   j , m Dm jm  R  .
m
Notice the reversal of m, m compared with the operation on the set of component
coefficients of the ket!
Rotating an Operator
Just as in the Schrödinger versus Heisenberg formulations, we can either apply the
rotation operator to the kets and leave the operators alone, or we can leave the kets alone,
and rotate the operators:
Ae
i  J
Ae

i  J
 U † AU
which will yield the same matrix elements, so the same physics.
Confusingly, there is a slightly different situation in which we need to rotate an operator,
and it gives an opposite result. Suppose an operator T acts on a ket  to give the ket
T  . For kets  and T  to go to U  and UT  respectively under a rotation
†
†
1
U, T itself must transform as T  UTU (recall U  U ).
Scalar Operators
A scalar operator is an operator which is invariant under rotations, for example the
Hamiltonian of a particle in a spherically symmetric potential. (There are many less
trivial examples of scalar operators, such as the dot product of two vector operators, as in
a spin-orbit coupling.)
The transformation of an operator under an infinitesimal rotation is given by:
S  U † ( R) SU ( R) with U ( R)  1 
i J  nˆ
from which
 i J  nˆ 
S S
, S.


It follows that a scalar operator S, which does not change at all, must commute with all
the components of the angular momentum operator, and hence must have a common set
of eigenkets with, say, J 2 and Jz.
Vector Operators
(Shankar 15.3, Sakurai 3.10)
We have of course been using vector operators all along, for example x , L, S and so on.
A vector operator has three components, corresponding to the Cartesian x, y, and z axes.
Classically, such a three-dimensional vector transforms under rotation as
Vi   Rij V j ,
with the usual rotation matrix, for example
 cos

Rz ( )   sin 
 0

 sin 
cos
0
0

0
1 
for rotation about the z-axis.
A quantum mechanical vector operator V is defined by requiring that the expectation
values of its three components in any state transform like the components of a classical
vector under rotation. It follows from this that the operator itself must transform
vectorially,
Vi  U †  R VU
 R    RijV j
i
To see what this implies, it is easiest to look at a simple case. For an infinitesimal
rotation about the z-axis,
 1  0 


Rz ( )    1 0 
 0 0 1


the vector transforms
 Vx 
 
 Vy 
V 
 z
 1 

  1
0 0

0  Vx 
 
0  Vy 

1 
 Vz 
 Vx  Vy 


  Vy  Vx 
 V

z


The unitary Hilbert space operator U corresponding to this rotation
so
U †VU
 1  i J z /
i
 Vi 
i
Vi 1  i J z / 
 J z , Vi .
U  Rz      1 
i J z
,
The requirement that the two transformations above, the infinitesimal classical rotation
†
 R  , are in
i
generated by Rz ( ) and the infinitesimal unitary transformation U  R VU
fact the same thing yields the commutation relations of a vector operator with angular
momentum:
i  J z ,Vx    Vy
i  J z ,Vy    Vx .
From this result and its cyclic equivalents, the components of any vector operator V must
satisfy:
Vi , J j   i ijk Vk
.
Cartesian Tensor Operators
The standard definition of a Cartesian tensor is that each of its suffixes transforms under
rotation as do the components of an ordinary three-dimensional vector. Specifically, the
nine components of a second rank tensor transform as:
Tij  Tij   Rii R jjTij .
A particular example of a second-rank tensor is
three-dimensional vectors.
Tij  U iV j
, where U and V are ordinary
The problem with this tensor is that it is reducible—combinations of the elements can be
arranged in groups such that rotations operate only within these groups. This is made
evident by writing:
U iV j 
U iV j  U jVi    U iV j  U jVi  U V   .
U V
 ij 

ij 
3
2
2
3


The first term, the dot product of the two vectors, is clearly a scalar under rotation, the
second term, which is an antisymmetric tensor has three independent components which
are the vector components of the vector product U V , and the third term is a symmetric
traceless tensor, which has five independent components. Altogether, then, there are 1 +
3 + 5 = 9 components, as required.
Spherical Tensors
Notice the numbers of elements of these irreducible subgroups: 1, 3, 5. These are exactly
the numbers of elements of angular momenta representations for j = 0, 1, 2! The first
term is of course trivial: the scalar by definition is not affected by rotation, and neither is
an j = 0 state.
To deal with the second and third terms, we introduce tensor operators having three and
five components, such that under rotation these sets of components transform among
themselves just as do the sets of eigenkets of angular momentum in the j = 1 and j = 2
representations respectively.
Recall that the matrix elements of the rotation operator
    e
U R

i  J
within a definite j subspace are written
j
mm
D
 R   
j, m e

i  J
j, m
so under rotation operator a basis state j, m transforms as:
e

i  J
j , m   j , m j , m e
m

i  J
j , m   j , m Dm jm  R  .
m
Notice the reversal of m, m compared with the usual operation of a matrix on a vector:
this is because we are rotating the basis here, rather than rotating relative to the basis.
The basic idea is that these irreducible subgroups into which Cartesian tensors apparently
decompose under rotation (generalizing from our one example) form a more natural basis
set of tensors for problems with rotational symmetries.
Definition: We define a spherical tensor of rank k as a set of 2k + 1 components
Tkq , q  k , k  1, , k such that under rotation they transform like a set of angular
momentum eigenkets, that is,
U  R  TkqU †  R    Dq kqTkq
q
.
To see the properties of these spherical tensors, it is useful to evaluate the above equation
D k    k , q I  i  J / k , q
for infinitesimal rotations, for which qq  
, to find
 J  , Tkq     k q  k  q  1 Tkq 1
 J z , Tkq   qTkq .
Sakurai observes that this set of commutation relations could be taken as the definition of
the spherical tensors.
Notational note: we have followed Shankar here in having the rank k as a subscript, the
“magnetic” quantum number q as a superscript, the same convention used for the
spherical harmonics (but not for the D matrices!) Sakurai, Baym and others have the
rank above, usually in parentheses, and the magnetic number below. Fortunately, all use
k for rank and q for magnetic quantum number.
A Spherical Vector
The j = 1 angular momentum eigenkets are just the familiar spherical harmonics
Y10 
3 z
, Y11 
4 r
3 x  iy
.
4 2r
The rotation operator will transform (x, y, z) as an ordinary vector in three-space, and this
is evidently equivalent to
j  1, m   j  1, m Dm jm  R 
m
It follows that the spherical representation of a three vector
T11 
Vx  iVy
2
V , V , V  has the form:
x
y
z
 V11 , T10  Vz  V10 .
In line with spherical tensor notation, the components
T , T
1
1
0
1
, T11 
q
are denoted T1 .
Matrix Elements of Tensor Operators Between Angular Momentum
Eigenkets
q
By definition, an irreducible tensor operator Tk transforms under rotation like an angular
q
momentum eigenket k , q . Therefore, rotating the ket Tk j, m ,
UTkq j, m  UTkqU 1U j , m   Dq kqTkq  Dm jm j , m
q
m
.
The product of the two D matrices appearing is precisely the set of coefficients to rotate
k , q  j, m where k , q is the angular momentum
the direct product of eigenkets
eigenket having j = k, m = q.
We have met this direct product of two angular momentum eigenkets before: this is just a
system having two angular momenta, such as orbital plus spin angular momenta. So we
q
j, m generates a state having total angular momentum the sum of
see that Tk acting on
(k, q) and (j, m).
To link up (more or less) with Shankar’s notation: our direct product state k , q  j, m
is the same as k , q; j, m in the notation j1 , m1; j2 , m2 for a product state of two angular
momenta (possibly including spins). Such a state can be written as a sum over states of
the form jtot , mtot ; j1 , j2 where this denotes a state of total angular momentum j , ztot
direction component mtot, made up of two spins having total angular momentum j1, j2
respectively.
This is the standard Clebsch-Gordan sum:
j1 , m1 ; j2 , m2 
j1  j2
jtot


jtot  j1  j2 mtot  jtot
jtot , mtot ; j1 , j2
jtot , mtot ; j1 , j2 j1 , m1 ; j2 , m2 .
The summed terms give a unit operator within this (2j1 + 1)(2j2 +1) dimensional space,
the term jtot , mtot ; j1 , j2 j1 , m1; j2 , m2 is a Clebsch-Gordan coefficient. The only nonzero
coefficients have mtot = m1 + m2, and jtot restricted as noted, so for given m1, m2 we just
set mtot = m1 + m2, we don’t sum over mtot, and the sum over jtot begins at |mtot|.
Translating into our k , q  j, m notation, and cleaning up,
k , q; j, m 
k j

jtot  q  m
jtot , q  m; k , j
jtot , q  m; k , j k , q; j , m .
We are now able to evaluate the angular component of the matrix element of a spherical
tensor operator between angular momentum eigenkets: we see that it will only be nonzero
for mtot = m1 + m2, and jtot at least |mtot|.
The Wigner-Eckart Theorem
At this point, we must bear in mind that these tensor operators are not necessarily just
functions of angle. For example, the position operator is a spherical vector multiplied by
the radial variable r, and kets specifying atomic eigenstates will include radial quantum
numbers as well as angular momentum.
Therefore, the matrix element of a tensor between two states will look like
 2 , j2 , m2 Tkq 1 , j1 , m1 , where the j’s and m’s denote the usual angular momentum
eigenstates and the ’s are nonangular quantum numbers, such as radial states.
The basic point of the Wigner–Eckart theorem is that the angular dependence of these
matrix elements can be factored out, and that it is given by the Clebsch-Gordan
coefficients. Having factored it out, the remaining dependence, which is only on the total
angular momentum in each of the kets, not the relative orientation, and of course on the
’s, is traditionally written as a bracket with double lines, that is,
 2 , j2 , m2 Tkq 1 , j1 , m1   2 , j2 Tk 1 , j1  j2 , m2 k , q; j1 , m1 .
The proof is given in, for example, Sakurai (page 239) and is not that difficult. The basic
strategy is to put the defining identities
 J  , Tkq     k q  k  q  1 Tkq 1
 J z , Tkq   qTkq
between  , j, m bras and kets, then get rid of the J  and J z by having them operate on
q
the bra or ket. This generates a series of linear equations for  2 , j2 , m2 Tk 1 , j1 , m1
matrix elements with m variables differing by one, and in fact this set of linear equations
is identical to the set that generates the Clebsch-Gordan coefficients, so we must
conclude that these spherical tensor matrix elements, ranging over possible m and j
values, are exactly proportional to the Clebsch-Gordan coefficients—and that is the
theorem.
A Few Hints for 15.3.3: that first matrix element comes from adding a spin j to a spin 1,
writing the usual maximum m state, applying the lowering operator to both sides to get
the total angular momentum j + 1, m = j state, then finding the same m state orthogonal to
that, which corresponds to total angular momentum j (instead of j + 1).
For the operator J, the Wigner-Eckart matrix element simplifies because J cannot affect
, and also it commutes with J2, so cannot change the total angular momentum.
q
0
So, in the Wigner-Eckart equation, replace Tk on the left-hand side by J1 , which is just
Jz. The result of (1) should follow.
(2) First note that a scalar operator cannot change m. Since c is independent of A we can
take A = J to find c.