*Michael Fowler,
1/13/08*

Physics is full of vectors: _{} and so on. Classically, a (three-dimensional) vector is
defined by its properties under rotation: the three components corresponding to
the Cartesian *x*, *y*, and *z* axes transform as

_{},

with the usual rotation matrix, for example

_{}

for rotation about the *z*-axis. (We’ll use _{} interchangeably.)

A *tensor* is a
generalization of a such a vector to an object with more than one suffix, such
as, for example, _{}(having 9 and 27 components respectively in three dimensions)
with the requirement that these components mix among themselves under rotation
by each individual suffix following the vector rule, for example

_{}

where *R* is the
same rotation matrix that transforms a vector.
Tensors written in this way are called *Cartesian tensors* (since the suffixes refer to Cartesian
axes). The number of suffixes is the *rank* of the Cartesian tensor, a rank *n* tensor has of course 3* ^{n}* components.

Tensors are common in physics: they are essential in describing stress, distortion and flow in solids and liquids. Tensor forces play an important role in the dynamics of the deuteron, and in fact tensors arise for any charge distribution more complicated than a dipole. Going to four dimensions, and generalizing from rotations to Lorentz transformations, Maxwell’s equations are most naturally expressed in tensor form, and tensors are central to General Relativity.

To get back to non-relativistic physics, since the defining
property of a tensor is its behavior under rotations, spherical polar
coordinates are sometimes a more natural basis than Cartesian coordinates. In fact, in that basis tensors (called *spherical tensors*) have rotational
properties closely related to those of angular momentum eigenstates, as will
become clear in the following sections.

As a preliminary to discussing general tensors in quantum mechanics, we briefly review the rotation operator and quantum vector operators. (A full treatment is given in my 751 lecture.)

Recall that the rotation operator turning a ket through an
angle _{} (the vector direction
denotes the axis of rotation, its magnitude the angle turned through) is

_{}

Since _{} commutes with the
total angular momentum squared _{} we can restrict our
attention to a given *total *angular
momentum *j*, having as usual an
orthonormal basis set _{}, or _{} for short, with 2*j* + 1 components, a general ket _{} in this space is then:

_{}.

Rotating this ket,

_{}

Putting in a complete set of states, and using the standard notation for matrix elements of the rotation operator,

_{}

_{} is standard notation
(see the earlier lecture).

So the ket rotation transformation is

_{}

with the usual matrix-multiplication rules.

Now suppose we apply the rotation operator to one of the *basis
kets* _{}, what is the result?

_{}

Note the *reversal*
of *m*, *m*¢ compared with the operation on the set of component
coefficients of the general ket.

(You may be thinking: wait a minute, _{} *is* a ket in the space—it can be written _{} with _{}, so we could use the previous rule _{} to get _{}. Reassuringly, this
leads to the same result we just found.)

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:

_{}

which will yield the same matrix elements, so the same physics.

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:

_{}

from which

_{}

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 *J _{z}*.

A *quantum mechanical
vector operator *_{} 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,

_{}.

To see what this implies, it is easiest to look at a simple
case. For an infinitesimal rotation
about the *z*-axis,

_{}

the vector transforms

_{}

The unitary Hilbert space operator *U* corresponding to
this rotation _{} so

_{}

The requirement that the two transformations above, the
infinitesimal classical rotation generated by _{} and the infinitesimal
unitary transformation _{}, are in fact the same thing yields the commutation relations
of a vector operator with angular momentum:

_{}

From this result and its cyclic equivalents, the components
of *any* vector operator _{}must satisfy:

_{}.

*Exercise*: verify that the components of _{} do in fact satisfy
these commutation relations.

(*Note*: 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 _{}. For kets _{} and _{} to go to _{} and _{} respectively under a
rotation *U*, *T* itself must transform as _{} (recall _{}). The point is that
this is a Schrödinger rather than a Heisenberg-type transformation: we’re
rotating the kets, not the operators.)

From the definition given earlier, under rotation the elements of a rank two Cartesian tensor transform as::

_{}

where *R _{ij}*
is the rotation matrix for a vector.

It is illuminating to consider a particular example of a
second-rank tensor, _{}, where _{} and _{} are ordinary
three-dimensional vectors.

The problem with this tensor is that it is *reducible*, using the word in the same sense as in our
discussion of group representations is discussing addition of angular
momenta. That is to say, combinations
of the elements can be arranged in sets such that rotations operate only within
these sets. This is made evident by
writing:

_{}

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 _{}, 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.

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!

This is of course no coincidence: as we shall make more
explicit below, a three-dimensional vector is mathematically isomorphic to a
quantum spin one, the tensor we have written is therefore a direct product of
two spins one, so, exactly as we argues in discussing addition of angular
momenta, it will be a reducible representation of the rotation group, and will
be a sum of representations corresponding to the possible total angular momenta
from adding two spins one, that is, *j *= 0, 1, 2.

As discussed earlier, the matrix elements of the rotation operator

_{}

within a definite *j* subspace are written

_{}

so under rotation operator a basis state _{} transforms as:

_{}

The essential point is that these irreducible subgroups into which Cartesian tensors 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 2*k* + 1
operators _{} such that under
rotation they transform among themselves with exactly the same matrix of
coefficients as that for the 2*j* + 1
angular momentum eigenkets _{} for *k* = *j*, that is,

_{}.

To see the properties of these spherical tensors, it is
useful to evaluate the above equation for infinitesimal rotations, for which _{}

Specifically, consider an infinitesimal rotation _{} (Strictly speaking,
this is not a real rotation, but the formalism doesn’t care, and the result we
derive can be confirmed by rotation about the *x* and *y* directions and adding
appropriate terms.)

The equation is

_{}

and equating terms linear in _{}

_{}

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.

The *j* = 1
angular momentum eigenkets are just the familiar spherical harmonics

_{}

The rotation operator will transform (*x*, *y*, *z*)
as an ordinary vector in three-space, and this is evidently equivalent to

_{}

It follows that the spherical representation of a three
vector _{}has the form:

_{}

In line with spherical tensor notation, the components _{}are denoted _{}

By definition, an irreducible tensor operator _{} transforms under
rotation like an angular momentum eigenket _{}. Therefore, rotating
the ket _{},

_{}.

The product of the two *D* matrices appearing is
precisely the set of coefficients to rotate *the direct product of eigenkets*
_{} where _{} is the angular
momentum 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 see that _{} acting on _{} generates a state
having total angular momentum the sum of (*k*, *q*) and (*j*, *m*).

To link up (more or less) with Shankar’s notation: our
direct product state _{} is the same as _{} in the notation _{} 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 _{} where this denotes a
state of total angular momentum *j _{tot}*,

This is the standard Clebsch-Gordan sum:

_{}

The summed terms give a unit operator within this (2*j*_{1}
+ 1)(2*j*_{2} +1) dimensional space, the term _{} is a Clebsch-Gordan
coefficient. The only nonzero
coefficients have *m _{tot}* =

Translating into our _{} notation, and cleaning
up,

_{}

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 *m _{tot}* =

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, so the matrix element of a tensor between two states will
have the form

_{},

where the *j*’s and *m*’s denote the usual angular
momentum eigenstates and the *a*’s are *nonangular*
quantum numbers, such as those for radial states.

The basic point of the Wigner–Eckart theorem is that* the angular dependence
of these matrix elements can be factored out, and 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 *a*’s),
is traditionally written as a bracket with double lines, that is,

_{}

The denominator is the conventional normalization of the double-bar matrix element. The proof is given in, for example, Sakurai (page 239) and is not that difficult. The basic strategy is to put the defining identities

_{}

between _{} bras and kets, then
get rid of the _{}by having them operate on the bra or ket. This generates a series of linear equations
for _{} 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 Shankar’s problem 15.3.3 (actually 15.3.4 in the new Second Edition, which looks identical to the old Second Edition, so watch out!)**: 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 a,
and also it commutes with *J *^{2},
so cannot change the total angular momentum.

So, in the Wigner-Eckart equation, replace _{} on the left-hand side
by _{}, which is just *J _{z}*. 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*.