Last time, we wrote down the Wigner-Eckart theorem:
\[ \begin{aligned} \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m} = \sprod{jk; mq}{jk; j' m'} \frac{\bra{\alpha' j'} | \hat{T}^{(k)} | \ket{\alpha j}}{\sqrt{2j'+1}}, \end{aligned} \]
Let's look at some special cases of the Wigner-Eckart theorem now. First, if we let \( \hat{S} \) by any scalar operator, then \( k=q=0 \), and the Clebsch-Gordan coefficients collapse to Kronecker deltas:
\[ \begin{aligned} \bra{\alpha', j', m'} \hat{S} \ket{\alpha, j, m} = \delta_{jj'} \delta_{mm'} \frac{\bra{\alpha' j'}|\hat{S}|\ket{\alpha j}}{\sqrt{2j'+1}}. \end{aligned} \]
Unsurprisingly, we find that scalar operators can never connect eigenstates with different angular momentum quantum numbers. It's easy to rearrange this formula to calculate the reduced matrix element, but since there is no \( j \) or \( m \) dependence anywhere there's not really a good reason to do so.
There is one interesting thing to point out here. One of our main motivations for working with spherical tensors was to end up with objects which are block-diagonal in the space of angular momentum eigenstates \( \ket{j,m} \). If we think of the above evaluation of \( \hat{S} \) as a matrix, we see that the form of the \( j \)-block is very simple; the matrix is proportional to the identity. If we had another scalar operator \( \hat{S}' \), the same would be true; on the subspace given by a particular \( j \), the matrices representing \( \hat{S} \) and \( \hat{S}' \) are proportional to each other.
This seems rather trivial here, but the Wigner-Eckart theorem actually implies a much more general result. If we have any two spherical tensors \( \hat{X}_q^{(k)} \) and \( \hat{Z}_q^{(k)} \) with the same rank \( k \), then we can use the Wigner-Eckart theorem twice, cancel out the identical Clebsch-Gordan coefficients, and find the result
\[ \begin{aligned} \bra{\alpha', j', m'} \hat{X}_q^{(k)} \ket{\alpha, j, m} = \frac{\bra{\alpha' j'}|\hat{X}^{(k)}|\ket{\alpha j}}{\bra{\beta' j'}|\hat{Z}^{(k)}|\ket{\beta j}} \bra{\beta', j', m'} \hat{Z}_q^{(k)} \ket{\beta, j, m} \end{aligned} \]
In other words, the matrices representing any two spherical tensors of matching rank on any subspace (labelled by \( j \) and \( j' \)) are proportional to one another. This corollary of the Wigner-Eckart theorem is known as the replacement theorem. It can come in handy when we want to replace the matrix elements of one operator, which are difficult to calculate, with those of another which are easier. Note that we can even use the replacement theorem when the non-rotational quantum numbers are different, since all the dependence on those quantum numbers is captured by the reduced matrix element.
Note that we won't always be able to calculate all of the matrix elements of \( \hat{X}_q^{(k)} \) with one particular chosen replacement operator, \( \hat{Z}_q^{(k)} \). (We are guaranteed that the matrices on a subspace are proportional to one another, but it's possible for the constant of proportionality to be zero.) For example, a very convenient choice for replacement is a spherical tensor built from the angular momentum vector \( \hat{\vec{J}} \) itself, since its matrix elements are particularly simple between \( \ket{j m} \) states. However, the angular momentum operator only connects states with equal \( j \), so we can't use this replacement to find any matrix elements of another spherical tensor with \( \Delta j \neq 0 \).
Despite these drawbacks, looking at matrix elements of \( \hat{\vec{J}} \) will allow us to derive one more very useful corollary from the Wigner-Eckart theorem.
Let's consider the case of an arbitrary vector operator \( \hat{\vec{V}} \). In general, finding the reduced matrix element will be a difficult exercise. Fortunately, we can use the replacement theorem to derive a useful, general formula for certain matrix elements of vector operators, without having to compute the reduced matrix element itself each time. We can relate the matrix elements of any vector operator to the elements of the angular momentum operator:
\[ \begin{aligned} \bra{\alpha', j, m'} \hat{V}_q^{(1)} \ket{\alpha, j, m} = \frac{\bra{\alpha' j}|\hat{V}^{(1)}|\ket{\alpha j}}{\bra{\alpha j}|\hat{J}^{(1)}|\ket{\alpha j}} \bra{\alpha, j, m'} \hat{J}_q^{(1)} \ket{\alpha, j, m} \end{aligned} \]
(Note that this only works for matrix elements of \( \hat{\vec{V}} \) between eigenstates with the same \( j \).) To determine the reduced matrix elements, we start by evaluating the matrix elements of the dot product \( \hat{\vec{J}} \cdot \hat{\vec{V}} \). First, notice that the dot product in terms of spherical basis vectors looks noticeably different:
\[ \begin{aligned} U_q^{(1)} \cdot V_q^{(1)} = U_0^{(1)} V_0^{(1)} - U_{1}^{(1)} V_{-1}^{(1)} - U_{-1}^{(1)} V_1^{(1)}. \end{aligned} \]
You can verify that this reduces to the normal dot product if we go back to Cartesian coordinates. Now, let's choose \( m'=m \) and evaluate:
\[ \begin{aligned} \bra{\alpha', j,m} \hat{\vec{J}} \cdot \hat{\vec{V}} \ket{\alpha, j, m} = \bra{\alpha', j, m} (\hat{J}_0 \hat{V}_0 - \hat{J}_1 \hat{V}_{-1} - \hat{J}_{-1} \hat{V}_1) \ket{\alpha, j,m } \\ = m \hbar \bra{\alpha', j, m} \hat{V}_0^{(1)} \ket{\alpha, j, m} + \frac{\hbar}{\sqrt{2}} \sqrt{(j+m)(j-m+1)} \bra{\alpha', j, m-1} \hat{V}_{-1}^{(1)} \ket{\alpha, j, m} - \\ \frac{\hbar}{2} \sqrt{(j-m)(j+m+1)} \bra{\alpha', j, m+1} \hat{V}_{1}^{(1)} \ket{\alpha, j, m} \end{aligned} \]
where you'll recall that \(\hat{J}_{\pm 1}^{(1)} = \mp \hat{J}_{\pm} / \sqrt{2}\). The actual coefficients of all of these terms aren't so imporatnt, actually; what really matters is that all three terms are equal to some function of \( j \) and \( m \) times some matrix element of \( \hat{V}_q^{(1)} \). But by the Wigner-Eckart theorem, all three of these matrix elements are themselves equal to a function of \( j \) and \( m \) times the reduced matrix element for \( \hat{V} \). Thus, we can collect everything together and rewrite
\[ \begin{aligned} \bra{\alpha', j,m} \hat{\vec{J}} \cdot \hat{\vec{V}} \ket{\alpha, j, m} = c_{jm} \bra{\alpha' j}|\hat{\vec{V}}{}^{(1)}|\ket{\alpha j} \end{aligned} \]
Furthermore, we know that the \( c_{jm} \) can't actually be a function of \( m \) at all, since \(\hat{\vec{J}} \cdot \hat{\vec{V}}\) is a scalar operator; so we can rewrite the constants as simply \(c_j\). To finish the derivation, we note that the \(c_{j}\) don't depend on our choice of \( \alpha' \) or \( \hat{\vec{V}} \) either, so if we choose \(\hat{\vec{V}} = \hat{\vec{J}}\) and let \( \alpha' = \alpha \), we find that
\[ \begin{aligned} \bra{\alpha, j,m} \hat{J}{}^2 \ket{\alpha,j,m} = c_j \bra{\alpha j}|\hat{\vec{J}}{}^{(1)}|\ket{\alpha j}. \end{aligned} \]
We can calculate \( c_j \) from this, but all we really wanted was the ratio between the reduced matrix elements of \( \hat{\vec{J}} \) and \( \hat{\vec{V}} \):
\[ \begin{aligned} \frac{\bra{\alpha', j, m}| \hat{\vec{V}}{}^{(1)} |\ket{\alpha, j,m}}{\bra{\alpha, j, m}| \hat{\vec{J}}{}^{(1)} |\ket{\alpha, j, m}} = \frac{\bra{\alpha', j, m} \hat{\vec{J}} \cdot \hat{\vec{V}} \ket{\alpha, j, m}}{\hbar^2 j(j+1)}. \end{aligned} \]
Thus, we arrive at the projection theorem, a formula with all of the reduced matrix elements already replaced:
\[ \begin{aligned} \bra{\alpha', j, m'} \hat{V}_q^{(1)} \ket{\alpha, j, m} = \frac{\bra{\alpha', j, m} \hat{\vec{J}} \cdot \hat{\vec{V}} \ket{\alpha, j, m}}{\hbar^2 j(j+1)} \bra{jm'} \hat{J}_q^{(1)} \ket{jm}. \end{aligned} \]
Notice that, following Sakurai, I'm writing the \( m \) labels on the \( \hat{\vec{J}} \cdot \hat{\vec{V}} \) eigenstates. However, we know that there can be no \( m \)-dependence in that matrix element, since the dot product is a scalar: if we apply the Wigner-Eckart theorem again to the dot product itself, then we have
\[ \begin{aligned} \bra{\alpha', j, m} \hat{\vec{J}} \cdot \hat{\vec{V}} \ket{\alpha, j, m} = \sprod{j 0; m 0}{j 0; j m} \frac{\bra{\alpha' j}|\hat{\vec{J}} \cdot \hat{\vec{V}}|\ket{\alpha j}}{\sqrt{2j+1}}. \end{aligned} \]
That Clebsch-Gordan coefficient is always 1, so as promised the \( m \)-dependence vanishes completely. If we try to write it as a reduced matrix element, we pick up some extra normalization factors, so it's better to leave the \( m \) label on the state. Just remember it won't matter (as we'll see in our explicit example shortly.)
A brief note on the physical interpretation of the projection theorem is in order. The name comes from the fact that to find arbitrary matrix elements of \( \hat{\vec{V}} \), we only need to consider the dot product \( \hat{\vec{J}} \cdot \hat{\vec{V}} \), which is precisely the projection of \( \hat{\vec{V}} \) onto the total angular momentum vector. Since \( \hat{\vec{J}} \) is the generator of rotations, this projection is exactly the part of \( \hat{\vec{V}} \) which is rotationally invariant. The leftover part aside from this projection is not rotationally invariant, and therefore will give zero if we try to take an expectation value in an eigenstate of rotation \( \ket{j,m} \).
Let's start seeing some practical applications of all this machinery, beginning with an example we studied before in the context of 2p hydrogen, namely the splitting of energy levels due to an applied external magnetic field. The Hamiltonian picks up a contribution of the form
\[ \begin{aligned} \hat{W} = \frac{eB}{2m_e c} (L_z + 2S_z). \end{aligned} \]
We calculated the effect of this contribution for the example of the \( 2p \) orbital of a hydrogen atom explicitly; now we're prepared for a completely general treatment. This term breaks the rotational symmetry of the electron, so that the states \( \ket{j m} \) are no longer energy eigenstates. However, as long as \( B \) is small we can ignore this effect, and just work with the \( B=0 \) eigenstates. (This will be justified rigorously when we come to perturbation theory.)
This operator is not proportional to \( \hat{J}_z \), but thanks to the projection theorem, we know that its matrix elements in a subspace given by fixed \( j \) are proportional to the matrix elements of \( \hat{J}_z \). Inside this subspace, we also have fixed \( l \) and \( s \) quantum numbers. This allows us to easily evaluate the reduced matrix elements of dot products: we can rewrite
\[ \begin{aligned} \hat{\vec{L}} \cdot \hat{\vec{J}} = \hat{\vec{L}} \cdot (\hat{\vec{L}} + \hat{\vec{S}}) \\ = \hat{\vec{L}}{}^2 + \frac{1}{2} (\hat{\vec{J}}{}^2 - \hat{\vec{L}}{}^2 - \hat{\vec{S}}{}^2) \\ \hat{\vec{S}} \cdot \hat{\vec{J}} = \hat{\vec{S}}{}^2 + \frac{1}{2} (\hat{\vec{J}}{}^2 - \hat{\vec{L}}{}^2 - \hat{\vec{S}}{}^2). \end{aligned} \]
Taking the expectation value within this subspace then gives
\[ \begin{aligned} \bra{j,m}\hat{\vec{L}} \cdot \hat{\vec{J}}\ket{j,m} = \frac{\hbar^2}{2} \left[ j(j+1) + l(l+1) - s(s+1) \right] \\ \bra{j,m}\hat{\vec{S}} \cdot \hat{\vec{J}}\ket{j,m} = \frac{\hbar^2}{2} \left[ j(j+1) + s(s+1) - l(l+1) \right] \end{aligned} \]
Now we apply the projection theorem: we have
\[ \begin{aligned} \bra{j,m} \hat{L}_z \ket{j,m} = \bra{j,m} \hat{L}_0^{(1)} \ket{j,m} = \frac{\bra{j,m} \hat{\vec{J}} \cdot \hat{\vec{L}} \ket{j,m}}{\hbar^2 j(j+1)} \bra{j,m} \hat{J}_0^{(1)} \ket{j,m} \\ = \frac{\hbar m}{2j(j+1)} \left[ j(j+1) + l(l+1) - s(s+1) \right]. \end{aligned} \]
Similarly for the spin operator we get
\[ \begin{aligned} \bra{j,m} \hat{S}_z \ket{j,m} = \frac{1}{2j(j+1)} [j(j+1) + s(s+1) - l(l+1)] m \hbar. \end{aligned} \]
So the energy correction \( E_1 = \ev{\hat{H}_1} \) for any \( j \) is overall proportional to \( m\hbar \); we can write
\[ \begin{aligned} E_1 = g_j \frac{qB}{2m_e c} \hbar m, \end{aligned} \]
where \( g_j \), also known as the Lande \( g \)-factor, contains all of the information on the various angular momentum quantum numbers. Putting together the pieces from above, we find
\[ \begin{aligned} g_J = \frac{3}{2} + \frac{s(s+1) - l(l+1)}{2j(j+1)}. \end{aligned} \]
If we have simply \( s=1/2 \), for example in a hydrogenic atom, then the expression for \( g \) will simplify substantially: we must have \( j = l \pm 1/2 \) and so
\[ \begin{aligned} g_j = \begin{cases} 1 + \frac{1}{2l+1}, & j = l + 1/2; \\ 1 - \frac{1}{2l+1}, & j = l - 1/2. \end{cases} \end{aligned} \]
As we saw before for just the \( 2p \) orbital, the magnetic field completely splits apart the energies of the \( \ket{nljm} \) eigenstates. This small splitting of the \( \ket{nljm} \) energy eigenvalues is known as the Zeeman effect.
In the theory of classical electromagnetism, if we are interested in studying a particle with charge \( q \) moving in an external electric potential \( V(\vec{r}) \), an often-useful technique is to expand the potential energy function out in terms of spherical harmonics: if \( V(\vec{r}) = qU(\vec{r}) \), then we can always write
\[ \begin{aligned} U(\vec{r}) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} f_{l,m}(r) Y_l^m(\theta, \phi). \end{aligned} \]
This is known as the multipole expansion. We can carry out the same expansion in quantum mechanics, finding the same expression with the coordinates promoted to operators in the usual way.
Based on the assumption that we are studying an external electric field, we can specialize a bit more. Given that all of the charges generating the potential are outside the region where our particle is moving, we know that the potential must satisfy the Laplace equation,
\[ \begin{aligned} \nabla^2 U(r) = 0. \end{aligned} \]
However, as we've seen before the angular part of the Laplacian is just the orbital angular momentum operator,
\[ \begin{aligned} \nabla^2 U(r) = \frac{1}{r} \frac{\partial^2}{\partial r^2} (rU(r)) - \frac{\hat{\vec{L}}{}^2}{\hbar^2 r^2} U(r) \end{aligned} \]
Applying this to the power series above, the angular momentum operator just acts on the spherical harmonics to give back \( \hbar^2 l(l+1) \), so that we end up with a very simple equation for the radial wavefunctions:
\[ \begin{aligned} \frac{1}{r} \frac{\partial^2}{\partial r^2} (r f_{l,m}(r)) - \frac{l(l+1)}{r^2} f_{l,m}(r) = 0. \end{aligned} \]
We've solved this equation before; the solutions are \( r^l \) and \( r^{-(l+1)} \), and as usual we're interested in working near the origin (since our electric field is being generated from "outside" the region of interest), and thus we have
\[ \begin{aligned} f_{l,m}(r) = \sqrt{\frac{4\pi}{2l+1}} c_{l,m} r^l. \end{aligned} \]
Here the \( c_{l,m} \) are unknown coefficients, and we've added some extra normalization factors for later convenience. Putting everything together, we see that we can define the potential in terms of a power series of operators,
\[ \begin{aligned} V(\hat{\vec{r}}) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} c_{lm} \hat{Q}_l^m, \end{aligned} \]
where the \( \hat{Q}_l^m \) electric multipole operators are defined in terms of their position space matrix elements by
\[ \begin{aligned} \bra{\vec{r}'} \hat{Q}_l{}^m \ket{\vec{r}} = q \sqrt{\frac{4\pi}{2l+1}} r^l Y_l^m(\theta, \phi) \delta(\vec{r} - \vec{r}'). \end{aligned} \]
Note that since \( Y_0^0 \) is just the constant \( 1/\sqrt{4\pi} \), the \( l=0 \) multipole operator reduces to just the charge \( q \) of our quantum particle. Moving on to \( l=1 \), you can easily convince yourself that the three \( \hat{Q}_1^m \) operators are just the components of a vector, specifically the vector \( \vec{d} = q \vec{r} \), i.e. the dipole operator.
At \( l=2 \), we find the quadrupole operator, which is now a rank-2 spherical tensor: its components are
\[ \begin{aligned} \hat{Q}_2^{\pm 2} = \frac{\sqrt{6}}{4} q (x \pm iy)^2 \\ \hat{Q}_2^{\pm 1} = \mp \frac{\sqrt{6}}{2} q z(x \pm iy) \\ \hat{Q}_2^{0} = \frac{1}{2} q (3z^2 - r^2). \end{aligned} \]
Next time, we'll apply Wigner-Eckart to the multiple expansion, and then on to hydrogen fine structure.