20. The Wigner-Eckart Theorem

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 emission and absorption of radiation. This theorem allows one to determine 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, a study of which was initiated in 19. Transformation of Operators 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.

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.

We denote the usual Cartesian basis by ${\hat{\mathbf{c}}}_i$, $i=1,2,3$, so that

We have previously denoted this basis by ${\hat{\mathbf{e}}}_i$, but in these notes we reserve the symbol ${\hat{\mathbf{e}}}$ for the spherical basis.

The spherical basis is defined by

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 ${\hat{\mathbf{e}}}_q$, $q=1,0,-1$.

The spherical basis vectors have the following properties. First, they are orthonormal, in the sense that

Next, an arbitrary vector $\Xvec$ can be expanded as a linear combination of the vectors ${\hat{\mathbf{e}}}^\cc_q$,

where the expansion coefficients are

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

in which the juxtaposition of the two vectors is dyad notation for representing the tensor product.

You may wonder why we expand $\Xvec$ as a linear combination of ${\hat{\mathbf{e}}}_q^\cc$, instead of ${\hat{\mathbf{e}}}_q$. The latter type of expansion is possible too, that is, any vector $\Yvec$ can be written

where

These relations correspond to a different resolution of the identity,

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 (6).

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 42. Emission and Absorption of Radiation{} that the transition amplitude for the emission of a photon is proportional to matrix elements of the dipole operator between the initial and final states. We use an electrostatic, spinless model for the atom, as in 16. Central Force Motion, and we consider the transition from initial energy level $E_{n\ell}$ to final level $E_{n'\ell'}$. These levels are degenerate, since the energy does not depend on the magnetic quantum number $m$ or $m'$. The wave functions have the form,

as in (15).

The dipole operator is proportional to the position operator of the electron, so we must evaluate matrix elements of the form,

where the initial state is on the left and the final one on the right. The position operator $\xvec$ has three components, and the initial and final levels consist of $2\ell+1$ and $2\ell'+1$ 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 $3(2\ell'+1)(2\ell+1)$ matrix elements, for example, $3\times 3 \times 5=45$ in a $3d \to 2p$ transition. This is because the energy of the photon is $E_\gamma=E_{n\ell}-E_{n'\ell'}$, which is independent of the initial and final $m$ and $m'$ quantum numbers. This count 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 $\xvec$, not with respect to the Cartesian basis, but with respect to the spherical basis. First we define

exactly as in Eq.(7). Next, by inspecting a table of the $Y_{\ell m}$'s (see Table of $Y_{\ell m}$'s), we find that for $\ell=1$ we have

where we have multiplied each $Y_{1m}$ by the radius $r$. On the right hand side we see the spherical components $x_q$ of the position vector $\xvec$, as follows from the definitions (4). The results can be summarized by

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 $\ell=1$, something we will have more to say about presently.

Now the matrix elements (13) become a product of a radial integral times an angular integral,

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-$Y_{\ell m}$ formula, (83), whereupon it becomes proportional to the Clebsch-Gordan coefficient,

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 (20) 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 $x_i$ of $\xvec$, 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 $3d \to 2p$).

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 is the main topic of 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 $L^2$ and $L_z$, and the photon is a particle of spin 1 (see 41. The Quantized Electromagnetic Field). Conservation of angular momentum requires that the angular momentum of the initial state (the atom, with quantum numbers $\ell$ and $m$) should be the same as the angular momentum of the final state (the atom, with quantum numbers $\ell'$ 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 $\ell\in\{\ell'-1,\ell',\ell'+1\}$, it means that the amplitude is zero unless the possible total angular momentum quantum number of the final state, obtained by combining $\ell'\otimes 1$, 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 (13) or (19); as we will see in 21. Parity, parity also plays an important role.

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 13. Representations of the Angular Momentum Operators and Rotations, in which we worked out the matrix representations of angular momentum and rotation operators. To review those results, we start with a ket space (first column) upon which proper rotations act by means of unitary operators $U(\Rmat)$ (second column). We refer only to proper rotations $\Rmat \in SO(3)$, and we note that the representation may be double-valued.

The rotation operators have generators, defined by (13), that is, that equation can be taken as the definition of $\Jvec$ when the rotation operators $U(\Rmat)$ are given. [(11) is equivalent.] The generators appear in the third column of the table. The components of $\Jvec$ satisfy the usual commutation relations (24) since the operators $U(\Rmat)$ form a representation of the rotation group.

Next, since $J^2$ and $J_z$ commute, we construct their simultaneous eigenbasis, with an extra index $\gamma$ to resolve degeneracies. Also, we require states with different $m$ but the same $\gamma$ 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. This table entry is essentially (60). A basis vector $\ket{\gamma jm}$, when rotated, produces a linear combination of other basis vectors for the same values of $\gamma$ and $j$ but different values of $m$. This implies that the space spanned by $\ket{\gamma jm}$ for fixed $\gamma$ and $j$, but for $m=-j,\ldots,+j$ is invariant under rotations. This space has dimensionality $2j+1$. It is, in fact, an irreducible invariant space.

Note that the $D$-matrices that appear in the transformation law in the last column are universal matrices, dependent only on the angular momentum commutation relations and phase conventions, but independent of the nature of the system. See the discussion in Seec. Universal Coefficients and Matrices.

Now we turn to the other rows of the table. At the beginning of 13. Representations of the Angular Momentum Operators and Rotations\ 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 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 $\Rmat$, 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 $\Jvec$ of Hermitian operators, that is, Hermitian matrices, that satisfy

when $\theta$ is small. Here $U$ really means the same thing as $\Rmat$, since we are speaking of the action on three-dimensional space, and 1 means the same as the identity matrix $\Imat$. We will modify this definition of $\Jvec$ slightly by writing $\Jvec' = \Jvec/\hbar$, thereby absorbing the $\hbar$ into the definition of $\Jvec$ and making $\Jvec'$ 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

But by a change of notation this is the same as

as in (32), where the vector of matrices $\Jmatvec$ is defined by (22). These imply

as indicated in the third column of Table 1. Writing out the matrices $J_i$ explicitly, we have

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

as follows from (24) and (34).

We can now construct the standard angular momentum basis on three-dimensional space. In addition to (25), we need the matrices for $J^2$ and $J_\pm$. These are

and

We see that $J^2=2\Imat$, 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 $\Avec(\xvec)$ (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 $J_3$, 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 (4) are the eigenvectors of $J_3$, that is,

Furthermore, it is easy to check that these vectors are related by raising and lowering operators, that is,

where $J_\pm$ is given by (28). Only the overall phase of the spherical basis vectors is not determined by these relations. The overall phase chosen in the definitions (4) has the nice feature that ${\hat{\mathbf{e}}}_0={\hat{\mathbf{z}}}$.

Since the spherical basis is a standard angular momentum basis, its vectors must transform under rotations according to (91), apart from notation. Written in the notation appropriate for three-dimensional space, that transformation law becomes

We need not prove this as an independent result; it is just a special case of (91), in a different notation. 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.

(31) has an interesting consequence, obtained by dotting both sides with ${\hat{\mathbf{e}}}^\cc_{q'}$. We use a round bracket notation for the dot product on the left hand side, and we use the orthogonality relation (5) on the right hand side, which picks out one term from the sum. We find

which shows that $D^1_{q'q}$ 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 with respect to the Cartesian basis, that is,

See (7). For a given rotation, matrices $\Rmat$ and $D^1(\Rmat)$ are similar (they differ only by a change of basis).

In the third row of the table the vector space upon which rotations act is the space of operators, that is, the usual linear operators in quantum mechanics that map the ket space into itself. The action of rotations on operators is indicated in the second column, which is the same as the definition of the rotated operator presented by (7). The third column, for the generators of rotations, is left blank. It is explored in Prob.~\prbrwigeck-1. For the remaining two columns of the third row of Table 1, we turn the definition of a new class of operators.

Irreducible Tensor Operators¶

We define an irreducible tensor operator of order $k$ as a set of $2k+1$ operators $T^k_q$, for $q=-k,\ldots,+k$, that satisfy

for all rotation operators $U$. We denote the irreducible tensor operator itself by $T^k$, and its $2k+1$ components by $T^k_q$. This definition is really a version of (91), 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 $T^k_q$ in the SAMB column of the third row of Table 1, and the transformation law (34) 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 $m$.

The index $k$ of an irreducible tensor operator of a physical observable is restricted to integer values, $k=0,1,\ldots$. This is unlike the case of the standard angular momentum basis states in a ket space, in which the index $j$ can take on either integer or half-integer values. The physical reason for this is that operators that represent physically observable quantities must be invariant under a rotation of $2\pi$. The mathematical reason is that our definition of a rotated operator, given by (6), is quadratic $U(\Rmat)$, so that the representation of rotations on the vector space of operators is always a single-valued representation of $SO(3)$.

Let us look at 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 $T^0_0$. 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 $1\times1$ matrix $(1)$ [see (70)].

Irreducible tensor operators of order 1 are constructed from vector operators by transforming from the Cartesian basis to the spherical basis. If we let $\Vvec$ be a vector operator as defined by (14), and define its spherical components by

then we have

where we use (31) and (74).

The electric quadrupole operator is given as a Cartesian tensor in (29). 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 $T^2_q$, $q=-2,\ldots,+2$ 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. This subject is explored in more detail in Prob.~\prbrwigeck-2.

Commutation Relations of an Irreducible Tensor Operator with $\Jvec$¶

Above we presented two equivalent definitions of scalar and vector operators, one involving transformation properties under rotations, and the other involving commutation relations with $\Jvec$. We will now do the same with irreducible tensor operators. To this end, we substitute the infinitesimal form (16) of the rotation operator $U$ into both sides of the definition (34).

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 (58)], we require these matrix elements in the case of an infinitesimal rotation. For $\theta\ll 1$, (58) becomes

Changing notation $(jm'm) \to (kq'q)$ and substituting this and (16) into the definition (34) of an irreducible tensor operator, we obtain

or, since ${\hat{\mathbf{n}}}$ arbitrary unit vector,

The operators $\Jvec$ on the left- and right-hand sides of (39) and (40) are not the same operators. On the left $\Jvec$ is the angular momentum on the same space upon which the operators $T^k_q$ act; in practice this is usually the state space of a quantum system. The $\Jvec$ on the right is the angular momentum operator on a model space in which the matrices $D^k_{q'q}$ are defined. See the discussion in Universal Coefficients and Matrices.

(40) specifies a complete set of commutation relations of the components of $\Jvec$ 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 $J_z \ket{kq} = \hbar q \ket{kq}$, so that

This is (48) with a change of notation. Then (40) becomes (48) below. Next dot both sides of (40) with ${\hat{\mathbf{x}}}\pm i{\hat{\mathbf{y}}}$, and use

or

This is (50) with a change of notation. Then we obtain (49) below. Finally, take the $i$-th component of (40),

and form the commutator of both sides with $J_i$,

where we have used (40) again to create a double sum. Finally summing both sides over $i$, we obtain,

But

a version of (47), so we obtain (50) below.

In summary, an irreducible tensor operator satisfies the following commutation relations with the components of angular momentum:

We see that forming the commutator with $J_\pm$ 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 (34) of an irreducible tensor operator. This is done by showing that (48) are equivalent to (34) in the case of infinitesimal rotations, and that if (34) 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 (34) and (48). Many books take (48) as the definition of an irreducible tensor operator.

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 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 $\matrixelement {\gamma'j'm'} {T^k_q} {\gamma jm}$. As an example of such a matrix element, you may think of the dipole matrix elements $\matrixelement {n'\ell'm'} {x_q} {n\ell m}$ that we examined in An Application of the Spherical Basis. In that case the operator (the position or dipole operator) is an irreducible tensor operator with $k=1$.

The matrix element $\matrixelement {\gamma'j'm'} {T^k_q} {\gamma jm}$ depends on 8 indices, $(\gamma'j'm';\gamma jm; kq)$, and in addition it depends on the specific operator $T$ in question. The Wigner-Eckart theorem concerns the dependence 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 $\matrixelement {\gamma'j'm'} {T^k_q} {\gamma jm}$ is proportional to the Clebsch-Gordan coefficient $\braket{j'm'}{jkmq}$, with a proportionality factor that is independent of the magnetic quantum numbers. That proportionality factor depends in general on everything else besides the magnetic quantum numbers, that is, $(\gamma'j';\gamma j;k)$ and the operator in question. The standard notation for the proportionality factor is $\reducedme{\gamma'j'}{T^k}{\gamma j}$, something that looks like the original matrix element except the magnetic quantum numbers are omitted and a double bar is used. The quantity $\reducedme{\gamma'j'}{T^k}{\gamma j}$ is called the reduced matrix element. With this notation, the Wigner-Eckart theorem states

The reduced matrix element can be thought of as depending on the irreducible tensor operator $T^k$ and the two irreducible subspaces $(\gamma'j')$ and $(\gamma j)$ that it links. Some authors (for example, Sakurai) include a factor of $1/\sqrt{2j+1}$ on the right hand side of (51), but here that factor has been absorbed into the definition of the reduced matrix element. The version (51) 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 $\bra{\gamma'j'm'}$ from the matrix element and think of the ket $T^k_q \ket{\gamma jm}$, 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 $\ket{kq}$ with the given angular momentum quantum numbers times another ket $\ket{jm}$ with different angular momentum quantum numbers. Then we could find the eigenstates of total angular momentum by combining the constituent angular momenta according to $k\otimes j$. Actually, in thinking of kets $T^k_q \ket{jm}$, it is customary to think of the product of the angular momenta in the reverse order, that is, $j\otimes k$. 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

contains various components of total $J^2$ and $J_z$, 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 coefficients. The coefficient with total angular momentum $j'$ and $z$-component $m'$ is the Clebsch-Gordan coefficient $\braket{j'm'}{jkmq}$, precisely what appears in the Wigner-Eckart theorem (51).

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 (51). 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 theorem 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 selection rules that follow from the Wigner-Eckart theorem are that the matrix element $\matrixelement{\gamma j'm'}{T^k_q}{\gamma jm}$ vanishes unless $m'=m+q$ and $j'$ takes on one of the values, $|j-k|, |j-k|+1, \ldots, j+k$.

Furthermore, suppose we actually have to evaluate the matrix \ \fi\fi $\matrixelement{\gamma' j'm'}{T^k_q}{\gamma jm}$ 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 (51) 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 An Application of the Spherical Basis, where the three-$Y_{\ell m}$ formula allowed us to compute the proportionality factor explicitly.

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 $T^0_0=H$. In this case the Clebsch-Gordan coefficient is

so the Wigner-Eckart theorem can be written

where

The matrix elements of the Hamiltonian of an isolated system in a standard angular momentum basis are diagonal in both $j$ and $m$, and moreover are independent of $m$.

If we wish to find the eigenvalues of the Hamiltonian we can diagonalize the matrices $C^j_{\gamma'\gamma}$, where $j$ labels the matrix and $\gamma,\gamma'=1,\ldots,N_j$, where $N_j$ is the multiplicity of the given $j$ value.

Recursion Relations for Matrix Elements¶

Many books rationalize the Wigner-Eckart theorem by showing that matrix elements of the form $\matrixelement{\gamma'j'm'}{T^k_q}{\gamma jm}$ satisfy the same recursion relations as the Clebsch-Gordan coefficients $\braket{j'm'}{jkmq}$, and by arguing that that fact implies that the two are proportional. The conclusion is correct, but the ususal presentations fall far short of actually proving it. Nevertheless, it is worthwhile outlining the usual argument.

We obtain a selection rule for the matrix element $\matrixelement{\gamma'j'm'}{T^k_q}{\gamma jm}$ by sandwiching $\bra{\gamma'j'm'}$ and $\ket{\gamma jm}$ around the commutation relation (48). This gives

so either $m'=m+q$ or $\matrixelement{\gamma'j'm'}{T^k_q}{\gamma jm}=0$.

Similarly, we obtain recursion relations by sandwiching $\bra{\gamma'j'm'}$ and $\ket{\gamma jm}$ around the commutation relations (49). Noting that

we obtain

which may be compared to the recursion relation (63) for the Clebsch-Gordan coefficients. Similarly, we find

which may be compared to (64).

The rest of the proof is based on an argument from linear algebra that proceeds as follows. Let $\Mmat$ be an $n\times n$ matrix and $x$ an $n$-dimensional vector. The equation $\Mmat x=0$ has a set of solutions $\{x\}$ that constitute a vector space, called the kernel of $\Mmat$. If $\det\Mmat\ne0$, the kernel is the trivial space $\{0\}$, but if $\det\Mmat=0$ then the dimension of the kernel is $\ge1$.

If the kernel is 1-dimensional, then any solution $x$ is proportional to any nonzero solution $x_0$, that is, $x=c x_0$. In the application to the proof of the Wigner-Eckart theorem, the matrix $\Mmat$ is the set of coefficients of the recursion relations, the unknown $x$ is the collection of matrix elements $\matrixelement{\gamma'j'm'}{T^k_q}{\gamma jm}$, $x_0$ is the set of Clebsch-Gordan coefficients $\braket{j'm'}{jkmq}$, and $c$ is the reduced matrix element.

Since filling in the details of this approach is somewhat tedious, we present an alternative proof of the Wigner-Eckart theorem, one based on the observation that if a set of vectors indexed by $jm$ values transforms under rotations as a standard angular momentum basis, then it is a standard angular momentum basis.

Proof of the Wigner-Eckart Theorem¶

Consider the product of kets $\ket{jm}\otimes \ket{kq} = \ket{jkmq}$ 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 $\ket{JM}$ of total $J^2$ and $J_z$ in this space are given by the Clebsch-Gordan expansion,

Moreover, the states $\ket{JM}$ for fixed $J$ and $M=-J,\ldots,+J$ 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 $\ket{JM}$ are not only eigenstates of total $J^2$ and $J_z$, but they are also linked by raising and lowering operators. Equivalently, the states $\ket{JM}$ transform as a standard angular momentum basis under rotations,

Now consider the $(2j+1)(2k+1)$ kets $T^k_q \ket{\gamma jm}$ obtained by varying $m$ and $q$. We construct linear combinations of these with the same Clebsch-Gordan coefficients as in (60),

and define the result to be the ket $\ket{X;JM}$, as indicated. The indices $JM$ in the ket $\ket{X;JM}$ 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 $(\gamma kj)$.

However, in view of the similarity between (60) and (62), we can guess that $\ket{X;JM}$ is actually an eigenstate of $J^2$ and $J_z$ with quantum numbers $J$ and $M$, and that the states $\ket{X;JM}$ are related by raising and lowering operators. That is, we guess

If true, this is equivalent to the transformation law,

exactly as in (61). Equations~(63) and (66) are equivalent because (63) can be obtained from (66) by specializing to infinitesimal rotations, while (66) can be obtained from (63) by building up finite rotations out of infinitesimal ones.

In Proof of (66) below we will prove that these guesses are correct. For now we merely explore the consequences. To begin, since $\ket{X;JM}$ is an eigenstate of $J^2$ and $J_z$ with quantum numbers $J$ and $M$, it can be expanded as a linear combination of the standard basis kets $\ket{\gamma j m}$ with the same values $j=J$ and $m=M$, but in general all possible values of $\gamma$. That is, we have an expansion of the form,

where the indices on the expansion coefficients $C^{kJMj}_{\gamma'\gamma}$ 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 (64). This gives

or, after canceling the square roots,

Comparing this to (67), we see that the expansion coefficients are the same for all $M$ values, and thus independent of $M$. We will henceforth write simply $C^{kJj}_{\gamma'\gamma}$ for them.

Now we return to the definition (62) of the kets $\ket{X;JM}$ and use the orthogonality of the Clebsch-Gordan coefficients (54) to solve for the kets $T^k_q\ket{\gamma jm}$. This gives