16 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 \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m} = \left\langle jk; mq | jk; j' m' \right\rangle \frac{\bra{\alpha' j'} | \hat{T}^{(k)} | \ket{\alpha j}}{\sqrt{2j'+1}}, 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.
For our example of the radiative dipole transition 3d \rightarrow 2p hydrogen, selection rules already reduced the number of matrix elements we needed to calculate from 45 down to 9. The Wigner-Eckart theorem allows us to calculate just 1 matrix element, and rotational symmetry does the rest of the work for us.
Speaking of selection rules, 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.)
16.1 Proof of the Wigner-Eckart theorem
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: \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} However, we can also just evaluate the ladder operators against the angular-momentum eigenstates on either side. This gives us the equation \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}. We can more compactly write the above equation as \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} Although I didn’t focus on it in these notes, it so happens that the Clebsch-Gordan coefficients obey a very similar-looking recursion relation: \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} These equations are basically identical, if we make the identifications (j,m) \rightarrow (j_1, m_1) \\ (k,q) \rightarrow (j_2, m_2) \\ (j', m') \rightarrow (j,m). 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 an extra constant: \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m} = \left\langle jk;mq | jk;j'm' \right\rangle \frac{\bra{\alpha' j'}|\hat{T}^{(k)}|\ket{\alpha j}}{\sqrt{2j'+1}}. 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: \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} |\left\langle jk;mq | jk;j'm' \right\rangle|^2 How do we evaluate the sum? Summing over m and q here is just insertion of a complete set of states: \sum_{m,m',q} |\left\langle jk;mq | jk;j'm' \right\rangle|^2 = \sum_{m,m',q} \left\langle jk;j'm' | jk;mq \right\rangle \left\langle jk;mq | jk;j'm' \right\rangle \\ = \sum_{m'} \left\langle j'm' | j'm' \right\rangle \\ = (2j'+1), 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.
My convention here 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: \hat{J}_{\pm 1}^{(1)} = \mp \frac{1}{\sqrt{2}} \hat{J}_{\pm}, \\ \hat{J}_{0}^{(1)} = \hat{J}_z. 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 \bra{j,m'} \hat{J}_0^{(1)} \ket{j,m} = \left\langle j1;m0 | j1;jm' \right\rangle \frac{\bra{j}|\hat{J}^{(1)}|\ket{j}}{\sqrt{2j+1}}.
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: \bra{j,m'} \hat{J}_0^{(1)} \ket{j,m} = \hbar m\delta_{mm'}. 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.
16.2 Wigner 3j symbols
We define a new object called the Wigner 3j-symbol, which looks like a small 2 \times 3 matrix: \left( \begin{array}{ccc} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{array} \right) The Clebsch-Gordan coefficients are related to this object in the following way: \left\langle j_1 j_2; m_1 m_2 | j_1 j_2; j m \right\rangle = (-1)^{j_1-j_2+m} \sqrt{2j+1} \left( \begin{array}{ccc} j_1 & j_2 & j \\ m_1 & m_2 & -m \end{array} \right). 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 \left( \begin{array}{ccc} j_1 & j_2 & j \\ m_1 & m_2 & m \end{array} \right) = \left( \begin{array}{ccc} j & j_1 & j_2 \\ m & m_1 & m_2 \end{array} \right) = (-1)^{j_1+j_2+j} \left( \begin{array}{ccc} j_2 & j_1 & j \\ m_2 & m_1 & m \end{array} \right). Flipping the signs of all of the m values also gives the same overall sign: \left( \begin{array}{ccc} j_1 & j_2 & j \\ -m_1 & -m_2 & -m \end{array} \right) = (-1)^{j_1+j_2+j} \left( \begin{array}{ccc} j_1 & j_2 & j \\ m_1 & m_2 & m \end{array} \right). 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: \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}} \left( \begin{array}{ccc} l_1 & l_2 & l_3 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{ccc} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \end{array} \right). 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, \left( \begin{array}{ccc} j & 1 & j \\ m & 0 & -m \end{array} \right) = (-1)^{1-j-m} \frac{m}{\sqrt{j(j+1)(2j+1)}}. 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 things at the level of this class, if we encounter a situation where we have to add three angular momenta 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 \bra{j,m'} \hat{J}_0^{(1)} \ket{j,m} = (-1)^{j-1+m} \left( \begin{array}{ccc} j & 1 & j \\ m & 0 & -m \end{array} \right) \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}. Since the left-hand side is just equal to \hbar m, we thus find for the reduced matrix element \bra{j}|\hat{J}^{(1)} |\ket{j} = \hbar (2j+1) \sqrt{j(j+1)}. 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!
16.2.1 Scalar operators and the replacement theorem
Let’s look at some special cases of the Wigner-Eckart theorem now. First, if we let \hat{S} be any scalar operator, then k=q=0, and the Clebsch-Gordan coefficients collapse to Kronecker deltas: \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}}. 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:
Given any two spherical tensors \hat{X}_q^{(k)} and \hat{Z}_q^{(k)} of the same rank k, their matrix elements satisfy the relation
\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} 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.
The replacement theorem 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.
16.3 The projection 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: \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} (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: 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)}. 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: \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} where the conventions for spherical basis map into our standard definitions of the angular momentum ladder operators as \hat{J}_{\pm 1}^{(1)} = \mp \frac{1}{\sqrt{2}} \hat{J}_{\pm}. 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 \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} 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 \bra{\alpha, j,m} \hat{J}{}^2 \ket{\alpha,j,m} = c_j \bra{\alpha j}|\hat{\vec{J}}{}^{(1)}|\ket{\alpha j}. 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}}: \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)}. Thus, we arrive at a formula with all of the reduced matrix elements already replaced:
Given a vector operator in spherical basis \hat{V}_q^{(1)}, its matrix elements are given by the formula
\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}.
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 \bra{\alpha', j, m} \hat{\vec{J}} \cdot \hat{\vec{V}} \ket{\alpha, j, m} = \left\langle j 0; m 0 | j 0; j m \right\rangle \frac{\bra{\alpha' j}|\hat{\vec{J}} \cdot \hat{\vec{V}}|\ket{\alpha j}}{\sqrt{2j+1}}. 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}.
16.4 Example: Lande g-factor
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 \hat{W} = \frac{eB}{2m_e c} (L_z + 2S_z). 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. (Once again, this is borrowing a result from the future that 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 \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). Taking the expectation value within this subspace then gives \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] Now we apply the projection theorem: we have \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]. Similarly for the spin operator we get \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. So the energy correction E_1 = \left\langle \hat{H}_1 \right\rangle for any j is overall proportional to m\hbar; we can write E_1 = g_j \frac{qB}{2m_e c} \hbar m, 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 g_J = \frac{3}{2} + \frac{s(s+1) - l(l+1)}{2j(j+1)}.
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 g_j = \begin{cases} 1 + \frac{1}{2l+1}, & j = l + 1/2; \\ 1 - \frac{1}{2l+1}, & j = l - 1/2. \end{cases} 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.
16.5 Electric multipole expansion
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 U(\vec{r}) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} f_{l,m}(r) Y_l^m(\theta, \phi). 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, \nabla^2 U(r) = 0. However, as we’ve seen before the angular part of the Laplacian is just the orbital angular momentum operator, \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) 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: \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. 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 f_{l,m}(r) = \sqrt{\frac{4\pi}{2l+1}} c_{l,m} r^l. 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, V(\hat{\vec{r}}) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} c_{lm} \hat{Q}_l^m, where the \hat{Q}_l{}^m electric multipole operators are defined in terms of their position space matrix elements by \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}'). 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, as it should. 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 \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).
Note that by construction, the multipole operator \hat{Q}^{(l)} is a rank-l spherical tensor. There exists a similar expansion for the magnetic multipole moments. By the Wigner-Eckart theorem, \bra{\alpha', l_2, m_2} \hat{Q}_m^{(l)} \ket{\alpha, l_1, m_2} = \left\langle l_1, l; m_1, m | l_1, l; l_2, m_2 \right\rangle \frac{\bra{\alpha', l_2}|\hat{Q}^{(l)}|\ket{\alpha, l_1}}{\sqrt{2l_2+1}} If someone tries to tell you about “the dipole moment” or “the quadrupole moment” of a particle, what they really mean is the reduced matrix element here, or something proportional to it. The reduced matrix element can be computed if we know the radial solution for the wavefunction of our particle; a little bit of work with our series expansion above and formulas for integrals of three Y_l^m’s will lead you to the formula \bra{\alpha', l_2}|\hat{Q}^{(l)}|\ket{\alpha, l_1} = q\sqrt{2l_1+1} \left\langle l_1, l; 0, 0 | l_1, l; l_2, 0 \right\rangle \\ \ \times \int_0^{\infty} dr\ r^{l+2} R_{\alpha', l_2}^\star(r) R_{\alpha, l_1}(r). This leads to an interesting selection rule. Remember that the Clebsch-Gordan coefficients are symmetric under flipping the signs of all of the m values, up to an overall minus sign: \left\langle j_1, j_2; -m_1, -m_2 | j_1, j_2; j, -m \right\rangle = (-1)^{j_1+j_2-j} \left\langle j_1, j_2; m_1, m_2 | j_1, j_2; j, m \right\rangle. The C-G coefficient with all of the m numbers equal to zero therefore vanishes if l_1 + l_2 - l is an odd number. (The opposite condition happens to hold for magnetic multipole operators: they vanish if l_1 + l_2 - l is even.) This argument can also be made using the properties of the spherical harmonics under parity, as it turns out.
We also have the usual triangle selection rule, which requires that |l_1 - l_2| \leq l \leq l_1 + l_2. This has interesting implications when we ask about the expectation values of multipole operators, i.e. set l_2 = l_1. Then we must have 0 \leq l \leq 2l_1. In other words, for a state with angular momentum l_1, all electric multipole moments with l > 2l_1 must vanish. Moreover, since 2l_1 - l must now be an even number, we see that l must be even for any non-vanishing electric multipole moment. Writing this out in equation form, the selection rules are: \bra{l_1} \hat{Q}^{(l)} \ket{l_1} = 0\ \textrm{unless}\ l = 0, 2, ..., 2l_1.
Here are a few simple examples of how this can be applied to atomic and nuclear systems:
- For the ground state of the hydrogen atom (l_1=0), all multipole moments (except the charge) must be zero.
- Any orbital angular momentum eigenstate has no electric dipole moment l=1 (up to effects that break parity invariance.)
- The deuteron (a proton and neutron bound together) is observed to have an electric quadrupole moment. This tells us that its ground state is not an l_1=0 eigenstate.