# The Wigner-Eckart theorem

Last time, we continued our discussion of spherical tensor operators $$\hat{T}_q^{(k)}$$, which are operators that transform under rotations in a nice way, i.e. like spherical harmonics (which is to say, like angular momentum eigenstates.) The more rigorous definition is that a spherical tensor of rank $$k$$ transforms in the $$(2k+1)$$-dimensional irreducible representation of the rotation group.

We ended with a formula for decomposing products of spherical tensors,

\begin{aligned} \hat{X}_{q_1}^{(k_1)} \hat{Z}_{q_2}^{(k_2)} = \sum_{k=|k_1-k_2|}^{k_1+k_2} \sum_{q=-k}^{k} \sprod{k_1 k_2; kq}{k_1 k_2; q_1 q_2} \hat{T}_q^{(k)} \end{aligned}

This is an explicit, algebraic version of the more abstract decomposition we've seen before of a direct product into a direct sum. For example, we talked about the dyadic tensor $$\hat{T}_{ij} \equiv \hat{U}_i \hat{V}_j$$ before, which using the formula above takes the form

\begin{aligned} \hat{U}_r^{(1)} \hat{V}_s^{(1)} = \sum_{k=0}^{2} \sum_{q=-k}^k \sprod{11;kq}{11;rs} \hat{T}_q^{(k)}. \end{aligned}

The matrix elements appearing here are Clebsch-Gordan coefficients, and they follow the usual set of selection rules: we have $$0 \leq k \leq 2$$ and then $$q = r+s$$ (which forces $$q$$ and thus $$k$$ to be integer valued.) In terms of representations of the rotation group, we wrote this decomposition before in the form

\begin{aligned} \mathbf{1} \otimes \mathbf{1} = \mathbf{2} \oplus \mathbf{1} \oplus \mathbf{0}. \end{aligned}

The nice thing about our more concrete formula is that we have the coefficients of the decomposition as well. If you go open up a table of Clebsch-Gordan coefficients, you should be able to reproduce the formulas we found last time for the dyadic tensor components.

Tensors of rank higher than two are rarely encountered, but if you do run into one, now you have the formalism to decompose it into spherical tensor components. The trick is to remember that the direct product, like ordinary multiplication, is distributive over (direct) sums. For example, suppose we were to construct a three-index tensor from vectors,

\begin{aligned} \hat{T}_{ijk} = \hat{U}_i \hat{V}_j \hat{W}_k. \end{aligned}

How will this decompose into spherical tensors? Well,

\begin{aligned} \mathbf{1} \otimes \mathbf{1} \otimes \mathbf{1} = \mathbf{1} \otimes (\mathbf{2} \oplus \mathbf{1} \oplus \mathbf{0}) \\ = (\mathbf{3} \oplus \mathbf{2} \oplus \mathbf{1}) \oplus (\mathbf{2} \oplus \mathbf{1} \oplus \mathbf{0}) \oplus \mathbf{1} \\ = \mathbf{3} \oplus \mathbf{2}_2 \oplus \mathbf{1}_3 \oplus \mathbf{0} \end{aligned}

where the subscript tells us how many distinct spherical tensors there will be of a given rank. As always, we should check the number of states: for a three-index tensor formed in this way we expect to find 27 unique components, and on the right we have $$7+2(5)+3(3)+1 = 27$$, so everything checks out. If we actually want to calculate matrix elements of $$\hat{T}_{ijk}$$, we still need the coefficients of this decomposition; those we can obtain using the above formulas and Clebsch-Gordan tables.

### The Wigner-Eckart theorem

I'm going to begin by stating the Wigner-Eckart theorem; then I'll explain what it means, and then we'll prove it. Given a spherical tensor operator $$\hat{T}_q^{(k)}$$, its matrix elements with respect to eigenstates of angular momentum satisfy the relation

\begin{aligned} \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m} = \sprod{jk; mq}{jk; j' m'} \frac{\bra{\alpha' j'} | \hat{T}^{(k)} | \ket{\alpha j}}{\sqrt{2j'+1}}, \end{aligned}

where $$\bra{\alpha' j'} | \hat{T}^{(k)} | \ket{\alpha j}$$ is known as the reduced matrix element, and is independent of the magnetic quantum numbers $$m,m',q$$. Once again, the Clebsch-Gordan coefficients have showed up on the right-hand side here. Exactly what the reduced matrix element is varies depending on what operator we're evaluating, which is why we don't have a precise formula for it; we'll see some examples shortly.

What does the Wigner-Eckart theorem mean? It's simply a statement of rotational symmetry: angular-momentum eigenstates $$\ket{j,m}$$ with the same $$j$$ but different $$m$$ are related to each other by rotations. Thus, if we calculate one matrix element involving $$\ket{j,m}$$ and $$\ket{j',m'}$$, we get all of the other ones by applying rotations (which simply takes the form of Clebsch-Gordan coefficients in the formula.) In fact, this is generally how we compute the reduced matrix element in practice: we calculate the left-hand side for one choice of magnetic quantum numbers, and then use the formula to get all the other choices for free.

The Wigner-Eckart theorem immediately implies our two selection rules $$m' = m+q$$ and $$|j-k| \leq j' \leq j+k$$ for spherical tensor operators without having to worry about integrals over spherical harmonics, simply due to the Clebsch-Gordan coefficients. (Of course, a matrix element which passes the selection rules is not guaranteed to be non-zero; other properties of $$\hat{T}^{(k)}$$ can give further restrictions, like parity in our example from last time.)

Let's go through the proof of the theorem. We focus on the evaluation of the commutator of $$\hat{T}_q^{(k)}$$ with the angular momentum ladder operators:

\begin{aligned} \bra{\alpha', j', m'} [\hat{J}_{\pm}, \hat{T}_q^{(k)}] \ket{\alpha, j, m} = \hbar \sqrt{(k \mp q)(k \pm q + 1)} \bra{\alpha', j', m'} \hat{T}_{q \pm 1}^{(k)} \ket{\alpha, j, m} \end{aligned}

However, we can also just evaluate the ladder operators against the angular-momentum eigenstates on either side. This gives us the equation

\begin{aligned} \sqrt{(j' \pm m')(j' \mp m' + 1)} \bra{\alpha', j', m' \mp 1} \hat{T}_q^{(k)} \ket{\alpha, j, m} = \\ \sqrt{(j \mp m)(j \pm m + 1)} \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m \pm 1} + \\ \sqrt{(k \mp q)(k \pm q + 1)} \bra{\alpha', j', m'} \hat{T}_{q \pm 1}^{(k)} \ket{\alpha, j, m}. \end{aligned}

This equation should look familiar; it looks very similar to the recursion relation for the Clebsch-Gordan coefficients themselves. In fact, we can more compactly write the above equation as

\begin{aligned} \sqrt{(j' \pm m')(j' \mp m' + 1)} X_{m' \mp 1, m}^{q,k} = \\ \sqrt{(j \mp m)(j \pm m + 1)} X_{m',m \pm 1}^{q, k} + \sqrt{(k \mp q)(k \pm q + 1)} X_{m',m}^{q \pm 1, k} \end{aligned}

whereas the recursion relation for the Clebsch-Gordan coefficients was

\begin{aligned} \sqrt{(j \pm m)(j \mp m + 1)} C_{m_1, m_2}^{j, m \mp 1} = \\ \sqrt{(j_1 \mp m_1)(j_1 \pm m_1 + 1)} C_{m_1 \pm 1, m_2}^{j,m} + \sqrt{(j_2 \mp m_2)(j_2 \pm m_2 + 1)} C_{m_1, m_2 \mp 1}^{j,m} \end{aligned}

These equations are basically identical, if we make the identifications

\begin{aligned} (j,m) \rightarrow (j_1, m_1) \\ (k,q) \rightarrow (j_2, m_2) \\ (j', m') \rightarrow (j,m). \end{aligned}

This does not imply that our spherical-tensor matrix elements $$X_{m',m}^{q,k}$$ are equal to the Clebsch-Gordan coefficients; we can multiply the $$X$$'s by an overall constant factor, and they will still satisfy this linear equation. The normalization of the Clebsch-Gordan coefficients was fixed by orthonormality of the angular-momentum eigenstates that they connect together, but since now we have matrix elements of an unknown operator $$\hat{T}_q^{(k)}$$, we have no such normalization condition.

So our matrix elements $$X_{m',m}^{q,k}$$ must be proportional to the Clebsch-Gordan coefficients we would obtain for addition of angular momenta $$j$$ and $$k$$ to obtain $$j'$$. The recursion relation we have above relates matrix elements with different magnetic quantum numbers $$m,m',q$$, but not with different $$j,j',k$$, so we see that the unknown normalization constant can depend on the latter numbers. We write the normalization as the reduced matrix element times a proportionality constant:

\begin{aligned} \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m} = \sprod{jk;mq}{jk;j'm'} \frac{\bra{\alpha' j'}|\hat{T}^{(k)}|\ket{\alpha j}}{\sqrt{2j'+1}}. \end{aligned}

The constant $$\sqrt{2j'+1}$$ is a normalization constant, the use of which isn't immediately obvious. To see why we include it, let's take the squared amplitude and then sum over magnetic quantum numbers on both sides:

\begin{aligned} \sum_{m,m',q} |\bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m}|^2 = \frac{1}{2j'+1} |\bra{\alpha', j'}|\hat{T}^{(k)}|\ket{\alpha,j}|^2 \sum_{m,m',q} |\sprod{jk;mq}{jk;j'm'}|^2 \end{aligned}

How do we evaluate the sum? Summing over $$m$$ and $$q$$ here is just insertion of a complete set of states:

\begin{aligned} \sum_{m,m',q} |\sprod{jk;mq}{jk;j'm'}|^2 = \sum_{m,m',q} \sprod{jk;j'm'}{jk;mq} \sprod{jk;mq}{jk;j'm'} \\ = \sum_{m'} \sprod{j'm'}{j'm'} \\ = (2j'+1), \end{aligned}

i.e. the number of distinct $$m'$$ states. So our normalization ensures that the square of the reduced matrix element is just the sum of the squares of the full matrix elements at each magnetic quantum number $$m',m,q$$.

Incidentally, my convention is slightly different from Sakurai's: he divides by $$(2j+1)$$ and not $$(2j'+1)$$. Unfortunately he never explains this choice, so I'm not sure if it's a typo or if there is some other motivation for using that normalization. Many books will not include an extra normalization factor at all; this is all a matter of conventions, so whatever you're doing just make sure you're consistent!

Let's do an example of finding a reduced matrix element before we move on. A particularly simple operator to consider is the angular momentum operator $$\hat{\vec{J}}$$ (we'll also need this reduced matrix element shortly!) We already know that we can write the spherical tensor components in the following way:

\begin{aligned} \hat{J}_{\pm 1}^{(1)} = \mp \frac{1}{\sqrt{2}} \hat{J}_{\pm}, \\ \hat{J}_{0}^{(1)} = \hat{J}_z. \end{aligned}

Since the reduced matrix element is $$q$$-independent, we can obtain it just by studying the $$q=0$$ component here. We can also ignore the $$\alpha$$ quantum numbers, since $$j$$ and $$m$$ are the only quantum numbers that $$\hat{\vec{J}}$$ will act on. Finally, we must set $$j'=j$$, since we already know that matrix elements between different $$j$$ values will vanish. The Wigner-Eckart theorem thus gives us

\begin{aligned} \bra{j,m'} \hat{J}_0^{(1)} \ket{j,m} = \sprod{j1;m0}{j1;jm'} \frac{\bra{j}|\hat{J}^{(1)}|\ket{j}}{\sqrt{2j+1}}. \end{aligned}

This is not useful yet, because there are unknown quantities on both sides of the equation! Fortunately, the left-hand side is easy to evaluate, remembering that $$\hat{J}_0^{(1)} = \hat{J}_z$$:

\begin{aligned} \bra{j,m'} \hat{J}_0^{(1)} \ket{j,m} = \hbar m\delta_{mm'}. \end{aligned}

Now we just need a Clebsch-Gordan coefficient, but not one we'll easily find in a table since we're adding $$j=1$$ to a second arbitrary $$j$$. This is a great place to see some general formulas that will let us quickly find the answer. To do that, I'm going to introduce an alternate notation that you should be exposed to.

We define a new object called the Wigner $$3j$$-symbol, which looks like a small $$2 \times 3$$ matrix:

\begin{aligned} \threej{j_1}{m_1}{j_2}{m_2}{j_3}{m_3} \end{aligned}

The Clebsch-Gordan coefficients are related to this object in the following way:

\begin{aligned} \sprod{j_1 j_2; m_1 m_2}{j_1 j_2; j m} = (-1)^{j_1-j_2+m} \sqrt{2j+1} \threej{j_1}{m_1}{j_2}{m_2}{j}{-m}. \end{aligned}

There are two good reasons to define the $$3j$$-symbols. First, they are basically the Clebsch-Gordan coefficients with nicer symmetry properties. Any cyclic permutation of the columns of a $$3j$$-symbol is equal, and any other permutation of the columns picks up a factor of $$(-1)^{j_1+j_2+j}$$, for example

\begin{aligned} \threej{j_1}{m_1}{j_2}{m_2}{j}{m} = \threej{j}{m}{j_1}{m_1}{j_2}{m_2} = (-1)^{j_1+j_2+j} \threej{j_2}{m_2}{j_1}{m_1}{j}{m}. \end{aligned}

Flipping the signs of all of the $$m$$ values also gives the same overall sign:

\begin{aligned} \threej{j_1}{-m_1}{j_2}{-m_2}{j}{-m} = (-1)^{j_1+j_2+j} \threej{j_1}{m_1}{j_2}{m_2}{j}{m}. \end{aligned}

Some of the other formulas we've derived look much nicer in terms of the $$3j$$-symbols. In particular, our formula for integration over three spherical harmonics takes a much more symmetric-looking form:

\begin{aligned} \int d\Omega Y_{l_1}^{m_1}(\theta, \phi) Y_{l_2}^{m_2}(\theta, \phi) Y_{l_3}^{m_3}(\theta, \phi) = \sqrt{\frac{(2l_1+1)(2l_2+1)(2l_3+1)}{4\pi}} \threej{l_1}{0}{l_2}{0}{l_3}{0} \threej{l_1}{m_1}{l_2}{m_2}{l_3}{m_3}. \end{aligned}

The $$3j$$-symbols satisfy selection rules, just like the Clebsch-Gordan coefficients; however, because of the notational rearrangement the rules look slightly different. For example, in the above integral the right-hand side vanishes unless $$m_1 + m_2 = -m_3$$, and the three $$l$$ values must all satisfy the triangle inequality with the other two.)

The second reason is that there is a standard formula, the Racah formula, for the value of an arbitrary $$3j$$-symbol. I won't reproduce the Racah formula here, since it's rather messy (although it's the kind of thing which is quite well-suited to implementation in a computer program), but it allows us to derive relatively compact formulas for certain special cases, like the one we're dealing with here. In particular, we can derive the special-case formula we need,

\begin{aligned} \threej{j}{m}{1}{0}{j}{-m} = (-1)^{1-j-m} \frac{m}{\sqrt{j(j+1)(2j+1)}}. \end{aligned}

Before we go back to our derivation, I'll just note in passing that there exist generalizations of the $$3j$$ symbols, such as the $$6j$$-symbols for addition of three angular momenta, and the $$9j$$-symbols for addition of four angular momenta. For our purposes in this class, if we encounter a situation where we have to add three angular momenta (like the homework) it's easier to just add two of them first, and then add the third to the combination.

Finally, back to our reduced matrix element derivation. Replacing our Clebsch-Gordan coefficient with a $$3j$$ symbol, and using the fact that $$m'=m$$, we have

\begin{aligned} \bra{j,m'} \hat{J}_0^{(1)} \ket{j,m} = (-1)^{j-1+m} \threej{j}{m}{1}{0}{j}{-m} \frac{\bra{j}|\hat{J}^{(1)}|\ket{j}}{\sqrt{2j+1}} \\ = \frac{1}{2j+1} \frac{m}{\sqrt{j(j+1)}} \bra{j}|\hat{J}^{(1)}|\ket{j}. \end{aligned}

Since the left-hand side is just equal to $$\hbar m$$, we thus find for the reduced matrix element

\begin{aligned} \bra{j}|\hat{J}^{(1)} |\ket{j} = \hbar (2j+1) \sqrt{j(j+1)}. \end{aligned}

I'll stop here, but it would be good practice to use this to evaluate some matrix elements with $$q = \pm 1$$, and compare them to what you get using the ladder operators $$\hat{J}_{\pm}$$ - they had better agree!