15 Spherical tensors and selection rules

Our treatment of addition of angular momentum so far has been somewhat general, although we saw an example for energy spittings in 2p hydrogen where the language of total angular momentum and Clebsch-Gordan coefficients made things simpler to describe.

But the real power of the formalism we’ve developed is only visible once we consider not just energy eigenvalues, but more complicated matrix elements between angular momentum states. The language of spherical tensors and the Wigner-Eckart theorem will greatly simplify such calculations through the magic of decomposition into block-diagonal objects using representations of angular momentum.

This is, by far, the most technical subject we will deal with this semester. As a result, it needs a good motivating example to remind us that the technical matter is worth struggling through!

15.0.1 Motivating example: dipole radiative transition

Once again, let’s consider a hydrogenic atom, and go back to ignoring the effects of spin: the states of the orbiting electron can thus be labeled in the usual way \ket{nlm}. Although these quantum numbers are conserved for the system in isolation, the electron can undergo a radiative transition, in which a photon is emitted and the state of the electron can change.

We aren’t ready to describe photons and photon emission in full detail yet, but for present purposes it’s enough to know that photon emission can be put through a form of multipole expansion, and the leading effect is given by the dipole transition matrix element, which is proportional to the position \hat{\vec{r}} of the the electron. In other words, the dipole transition probability amplitude from state \ket{nlm} to state \ket{n'l'm'} is proportional to \bra{n'l'm'} \hat{\vec{r}} \ket{nlm}. There are a lot of possible transitions to consider for any given initial state! Even for fixed \ket{n'l'm'} this is actually three matrix elements, for each of the components of \hat{\vec{r}}. If we want to consider a transition from nl to n'l', there is also a significant amount of degeneracy from the magnetic quantum numbers, so that the total number of matrix elements in such a transition is 3(2l'+1)(2l+1). For a transition 3d \rightarrow 2p for example, from l=2 down to l=1 (see the appendix for a refresher on hydrogen energy levels and labels), this gives 45 matrix elements in total.

To evaluate these matrix elements, we insert complete sets of position states, using the fact that we already know our wavefunctions for the energy eigenstates take the form \left\langle \vec{r} | nlm \right\rangle = R_{nl}(r) Y_l^m(\theta, \phi). This means that the matrix element becomes an integral, \bra{n'l'm'} \hat{\vec{r}} \ket{nlm} = \int_0^\infty dr\ r^3 R_{n'l'}^\star(r) R_{nl}(r) \int d\Omega (Y_{l'}^{m'})^\star(\theta, \phi) Y_l^m(\theta, \phi) \left(\frac{\vec{r}}{r}\right) where I’ve rewritten \vec{r} = r (\vec{r}/r), since the latter quantity in spherical coordinates only depends on the spherical angles, allowing us to split up the integration as shown above.

Now, this is sufficient to solve the problem; we can just open up Mathematica and start doing integrals. However, if we do this, we will find out after the fact that many of our integrals just give zero; for example, evaluating the matrix elements for 3d \rightarrow 2p and for the \hat{x} direction, 9 out of 15 angular integrals vanish. There is not, in what we have written so far, any obvious reason for this to be true; the one thing we could try to look at is parity, but as we saw parity of spherical harmonics only depends on the value of l, so it doesn’t tell us why certain \ket{2m'} \rightarrow \ket{1m} transitions would vanish. The feeling that we are missing something only increases if we look at other transitions: if you try to calculate the 4f \rightarrow 1s transition matrix element with \hat{\vec{r}} in Mathematica, you will find that every single angular integral is zero - the transition is completely forbidden!

Of course, we expect that many of these matrix elements are related by rotational symmetry. Unfortunately, while we know that the \ket{nlm} are eigenstates of angular momentum, the operator \hat{\vec{r}} in the middle is not manifestly rotation-invariant if we expand out in Cartesian components, which we know mix together under arbitrary rotations. Let’s try to understand the problem a bit more concretely.

15.1 Rotation and vector operators; selection rules

Let’s zoom out a bit and consider an arbitrary vector-valued operator \hat{\vec{V}}. We know that if we take an expectation value \left\langle \hat{\vec{V}} \right\rangle = \bra{\psi} \hat{\vec{V}} \ket{\psi} in some quantum state, the result is a classical vector, which means that it transforms under rotation of our system as (writing it out in Cartesian components) \left\langle \hat{V}_i \right\rangle \rightarrow \sum_j R_{ij} \left\langle \hat{V}_j \right\rangle, where R_{ij} is a standard 3x3 rotation matrix. However, we also have a quantum description of rotations: rotating our system maps our state \ket{\psi} \rightarrow \hat{\mathcal{D}}(R) \ket{\psi}. So we also have the relation \left\langle \hat{V}_i \right\rangle \rightarrow \left\langle \hat{\mathcal{D}}^{-1}(R) \hat{V}_i \hat{\mathcal{D}}(R) \right\rangle. Since all of the above is true for an arbitrary quantum state, we have an identity between sets of operators, \hat{\mathcal{D}}^{-1}(R) \hat{V}_i \hat{\mathcal{D}}(R) = \sum_j R_{ij} \hat{V}_j.

Being careful about rotation and vectors

As a word of warning, the above is only true for vectors \hat{\vec{V}} that are “part of our system”, so that they are indeed expected to rotate along with our states. But there are often physical systems where external vectors are present. An example would be an external magnetic field, \vec{B}. Such a vector does not transform under rotations, since if it’s outside the system, we expect our physical results to change when we change the orientation of our experiment relative to the external \vec{B}-field.

At the level we’re currently doing things, external vectors will typically not be operators, since we’re treating the electromagnetic fields classically. But later on when we worry about quantum effects and electromagnetism, it will still be true that vector operators don’t rotate if they correspond to external fields, so be careful!

If we consider the infinitesmal version of this relation, rotating by \phi = \epsilon then on the left-hand side we have (1 + \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar}) \hat{V}_i (1 - \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar} ) = \sum_j R_{ij}(\epsilon) \hat{V}_j \\ \Rightarrow \hat{V}_i - \frac{i\epsilon}{\hbar} [\hat{V}_i, \hat{\vec{J}} \cdot \vec{n} ] = \sum_j R_{ij}(\epsilon) \hat{V}_j. We can also expand out the right-hand side, since the matrix R_{ij} simplifies for rotation by an infinitesmal angle about a given axis; the result is, referring to Merzbacher, \sum_j R_{ij}(\epsilon) \hat{V}_j \approx \hat{V}_i - \epsilon (\vec{n} \times \hat{\vec{V}})_i. = \hat{V}_i - \epsilon (\epsilon_{ijk} n_j \hat{V}_k). The \hat{V}_i term cancels on both sides, and by comparing the \mathcal{O}(\epsilon) terms, we find that the commutator with \hat{\vec{J}} gives a cross product with the normal vector, which in terms of the Cartesian components implies the commutation identity [\hat{V}_i, \hat{J}_j] = i\hbar \epsilon_{ijk} \hat{V}_k.

(If you don’t want to mess around with cross products, you can arrive at the same result just by setting \vec{n} equal to the coordinate axis directions in turn and plugging in the corresponding rotation matrices.)

This looks promising to help us to understand why some of the matrix elements are zero when we’re working with angular momentum matrix elements, since it relates \hat{\vec{V}} to a commutator that involves the angular momentum operator. For example, we know from this equation that [\hat{V}_z, \hat{J}_z] = 0, which we can apply as follows: \bra{j'm'} [\hat{V}_z, \hat{J}_z] \ket{jm} = 0 \\ \hbar (m - m') \bra{j'm'} \hat{V}_z \ket{jm} = 0 applying \hat{J}_z to the eigenstates on the left and right. So we see immediately that the matrix elements of \hat{V}_z vanish unless m = m'. This is an example of a selection rule; a condition that matrix elements have to meet in order to be non-zero. For the example of a 3d \rightarrow 2p radiative transition above, this means that out of 15 possible matrix elements for \hat{z}, only three will be non-zero. Selection rules are extremely powerful since they can tell us important physics without having to calculate anything, and they save us a lot of effort when we do calculate!

What about the other two matrix elements? Looking again at the commutator with \hat{J}_z (singled out since it is what we’re using to define our basis), we have \bra{j'm'} [\hat{V}_x, \hat{J}_z] \ket{jm} = -i\hbar \bra{j'm'} \hat{V}_y \ket{jm} \\ \bra{j'm'} [\hat{V}_y, \hat{J}_z] \ket{jm} = i\hbar \bra{j'm'} \hat{V}_x \ket{jm}.

This is a little less straightforward since we’re now connecting matrix elements of different components together, but we can see that linear combinations of these two equations will relate the same matrix element to itself:

\bra{j'm'} [\hat{V}_x \pm i\hat{V}_y, \hat{J}_z] \ket{jm} = \hbar \bra{j'm'} (-i \hat{V}_y \mp \hat{V}_x) \ket{jm} or applying the \hat{J}_z operators on the left again, \bra{j'm'} (\hat{V}_x \pm i\hat{V}_y) \ket{jm} \hbar(m - m') = \mp \hbar \bra{j'm'} (\hat{V}_x \pm i\hat{V}_y) \ket{jm} \\ \Rightarrow \hbar (m - m' \pm 1) \bra{j'm'} (\hat{V}_x \pm i \hat{V}_y) \ket{jm} = 0. So we’ve found our other two selection rules for a vector operator: the combinations \hat{V}_{x} \pm i\hat{V}_y only connect two angular-momentum eigenstates that satisfy m' = m \pm 1, respectively.

Going back to check our 3d \rightarrow 2p result again, I said that in Mathematica I found that only 6 out of 15 matrix elements for \hat{x} were non-zero. This matches what we would expect from the selection rule; three of the matrix elements have m' = m + 1 and the other three must be m' = m - 1.

15.2 Spherical basis

The three combinations of components of \hat{\vec{V}} which map to themselves under commutation with \hat{J}_z (and thus, under rotations around the z-axis) represent a change of basis from Cartesian basis to what is known as the spherical basis, written \hat{V}_q^{(1)}: \hat{V}_1^{(1)} = -\frac{1}{\sqrt{2}} (\hat{V}_x + i \hat{V}_y), \\ \hat{V}_0^{(1)} = \hat{V}_z, \\ \hat{V}_{-1}^{(1)} = \frac{1}{\sqrt{2}} (\hat{V}_x - i\hat{V}_y). Although it didn’t matter for the selection rules, here we divide the combinations by \sqrt{2} to maintain overall normalization; there is also a minus sign on the +1 component, which ensures the basis is orthonormal. By convention, the subscript runs through q=\{-1, 0, +1\} in this basis; the superscript is part of a larger notation for “spherical tensors” that we’ll explain fully later, but for now it helps us keep track of whether we’re working in spherical basis or Cartesian basis.

It’s also useful to write out the new basis that we’re changing to explicitly: the basis vectors are \begin{align*} \hat{e}_1 &= -\frac{1}{\sqrt{2}} \left( \hat{x} + i\hat{y}\right) \\ \hat{e}_0 &= \hat{z} \\ \hat{e}_{-1} &= \frac{1}{\sqrt{2}} \left( \hat{x} - i\hat{y} \right). \end{align*} (I’m using the hats now to denote unit vectors, although you can take these as operator definitions too.) The complex nature of this basis makes it obvious that this is designed to work in quantum mechanics, although if you’ve studied the formulation of classical electromagnetism using complex numbers, you may recognize the \hat{e}_{\pm 1} vectors as the unit vectors describing left and right circular polarization of a light wave.

We can check that these vectors indeed define an orthonormal basis, as long as we’re careful to keep track of complex conjugation: \hat{e}_q^\star \hat{e}_{q'} = \delta_{qq'} We can, of course, expand any arbitrary spatial vector in terms of the spherical basis components, \vec{X} = \sum_{q} \hat{e}_q^\star X_q^{(1)} where X_q^{(1)} = \hat{e}_q \cdot \vec{X}. We could have also expanded in the vectors \hat{e}_q instead of the conjugates \hat{e}_q^\star; for our present purposes this won’t change anything as long as we’re consistent, so we’ll use the convention above.

Now, it probably won’t surprise you to learn that this is not an arbitrary choice of basis; it is intricately related to rotational symmetry. If we write out the spherical harmonics with a factor of r added, we notice the following: r Y_1^1(\theta, \phi) = -r \sqrt{\frac{3}{8\pi}} \sin \theta e^{i\phi} = \sqrt{\frac{3}{4\pi}} \left( - \frac{x + iy}{\sqrt{2}} \right) \\ r Y_1^0(\theta, \phi) = r \sqrt{\frac{3}{4\pi}} \cos \theta = \sqrt{\frac{3}{4\pi}} z, \\ r Y_1^{-1}(\theta, \phi) = r \sqrt{\frac{3}{8\pi}} \sin \theta e^{-i\phi} = \sqrt{\frac{3}{4\pi}} \left( \frac{x-iy}{\sqrt{2}} \right). Thus, we see that the components of \vec{r} in spherical basis are proportional to the spherical harmonics: writing r_q = \hat{e}_q \cdot \vec{\hat{r}}, we have r_q^{(1)} = \sqrt{\frac{4\pi}{3}} r Y_1^q(\theta, \phi). Notice that the index q of the spherical basis vectors is now acting exactly like the magnetic quantum number m for an l=1 angular momentum operator. (The fact that this is specifically the l=1 spherical harmonics is related to the way we’ve defined our notation with the superscript (1).)

In fact, it makes perfect sense in retrospect that specifically the l=1 spherical harmonics are what appear here. These spherical harmonics transform under the 3-dimensional representation of the rotation group SO(3). Vectors also have three components, and they also transform in the same 3-dimensional representation. All we’ve done with our change of basis is make it so that the specific choice of representation now matches, i.e. in spherical basis the rotation matrix that we apply to a vector is the very same rotation matrix we apply to the spherical harmonics in the \ket{1m} eigenbasis.

Because of the way the spherical harmonics appear in expanding \vec{r} in spherical basis, returning to our concrete dipole transition example, we can divide up the integral for transition matrix elements further. Our result from above splits further as follows: \bra{n'l'm'} \hat{r}_q \ket{nlm} = \int_0^{\infty} dr\ r^3 R_{n'l'}^\star(r) R_{nl}(r) \sqrt{\frac{4\pi}{3}} \\ \times \int d\Omega (Y_{l'}^{m'})^\star(\theta, \phi) Y_1^q (\theta, \phi) Y_l^m(\theta, \phi). This is nice, because it explicitly isolates the fact that the angular integral is exactly what determines whether our matrix element will be non-zero or not. However, integration over three spherical harmonics is still not an obviously nice thing to do, and there is no hint of the selection rules that we already found above in this expression yet.

The way to reconcile them is that there is an algebraic way to simplify the angular integration, which will cause our selection rules to reappear; this is worth a look at, since it highlights some nice algebraic techniques for dealing with problems where spherical harmonics appear in general.

15.3 Algebraic identities for spherical harmonics

We’ve now developed two somewhat different-looking approaches to dealing with angular momentum: the algebraic approach, involving ladder operators and Clebsch-Gordan coefficients, and the coordinate-space approach, in which solving the angular differential equation led us to the spherical harmonics. Of course, angular momentum is angular momentum, so these are really two descriptions of the same physics. In fact, as we stated before the only difference between the two is an (angular) position ket: Y_l^m(\theta, \phi) = \left\langle \theta, \phi | l,m \right\rangle. We can also write the position bra as \bra{\vec{n}}, where \vec{n} is a unit vector pointing in the direction given by the spherical angles (and a radial distance, but I’m neglecting that and just focusing on angular dependence.) Thinking of it in this way, it’s clear that we can write this position ket as a rotation of the z-axis unit vector: \ket{\vec{n}} = \hat{\mathcal{D}}(\phi, \theta) \ket{\hat{z}}. where \hat{\mathcal{D}} is a rotation operator describing the rotation from the z-axis to \vec{n}. Now, we insert a complete set of states: \ket{\vec{n}} = \sum_{l'} \sum_{m'} \hat{\mathcal{D}}(\phi, \theta) \ket{l',m'} \left\langle l',m' | \hat{z} \right\rangle. If we take the inner product with a particular eigenstate \bra{l,m}, then we have: \left\langle l,m | \vec{n} \right\rangle = \sum_{l'} \sum_{m'} \bra{l,m} \hat{\mathcal{D}}(\phi, \theta) \ket{l',m'} \left\langle l',m' | \hat{z} \right\rangle.

The matrix element containing the rotation \hat{\mathcal{D}} is the Wigner D-matrix, which we encountered briefly before when discussing addition of angular momentum: once again, it is defined as \mathcal{D}_{m'm}^{(j)}(R) = \bra{j,m'} \exp \left( \frac{-i (\hat{\vec{J}} \cdot \vec{n}) \phi}{\hbar} \right) \ket{j,m}. Notice that in particular, the D-matrix preseves the orbital quantum number j, only mixing states with the same j and different m (this gives a block-diagonal form that corresponds to a direct sum in addition of angular momentum, remember.) Because of this structure, the sum over l' for the spherical harmonics above collapses, and we have \left\langle l,m | \vec{n} \right\rangle = \sum_{m'} \mathcal{D}_{m'm}^{(l)}(R) \left\langle l,m' | \hat{z} \right\rangle.

Now, we work backwards by noticing that the two inner products on both sides are just spherical harmonics. In particular, the one on the right-hand side is a spherical harmonic evaluated at \theta = 0. This is always zero unless m'=0, because we can write m' \left\langle l,m' | \hat{z} \right\rangle = \bra{l,m'} \hat{L}_z \ket{\hat{z}} = \bra{l,m'} (\hat{x} \hat{p}_y - \hat{y} \hat{p}_x) \ket{\hat{z}} = 0 so either the harmonic is zero or m'=0. With m'=0 the \phi-dependence in the spherical harmonics drops out, so we have: \left\langle l,m' | \hat{z} \right\rangle = (Y_l^{m'})^\star (\theta=0, \phi) = \sqrt{\frac{2l+1}{4\pi}} P_l (\cos 0) \delta_{m'0} = \sqrt{\frac{2l+1}{4\pi}} \delta_{m'0}. This collapses the sum and leaves just the \mathcal{D}-matrix with m'=0: dividing through, we find the identity \mathcal{D}_{m0}^{(l)}(\phi, \theta) = \sqrt{\frac{4\pi}{2l+1}} (Y_l^m)^\star (\theta, \phi). This is, clearly, a very useful identity for considering the rotation of m=0 eigenstates.

15.3.1 Combination identity for spherical harmonics

We can learn some much more interesting formulas by considering addition of angular momentum along with this sort of exercise. Given two angular momentum operators \hat{\vec{J}}_1 and \hat{\vec{J}}_2, we know that rotations act independently on the two product-basis states. So if we’re interested in the amplitude of a rotated eigenstate, we can write it as a single Wigner D-matrix or as a pair of D-matrices: \bra{j_1, j_2; m_1', m_2'} \hat{\mathcal{D}}(R) \ket{j_1, j_2; m_1, m_2} = \bra{j_1 m_1'} \hat{\mathcal{D}}_1(R) \ket{j_1 m_1} \bra{j_2 m_2'} \hat{\mathcal{D}}_2(R) \ket{j_2 m_2} \\ = \mathcal{D}_{m_1'm_1}^{(j_1)} \mathcal{D}_{m_2'm_2}^{(j_2)}. This is easy to write, but not so easy to use. In particular, if we try to write the product on the right as a single matrix acting on the product states, it will not be block-diagonal; the matrix will be reducible. Fortunately, the change of basis to an irreducible (block-diagonal) matrix is one we’ve already studied: the appropriate basis is the total angular momentum basis. To see this, let’s insert some complete sets of total-\hat{\vec{J}} states in the left-hand side above: \bra{j_1, j_2; m_1', m_2'} \hat{\mathcal{D}}(R) \ket{j_1, j_2; m_1, m_2} = \sum_{j,m,m'} C_{m_1'm_2'}^{jm'} \bra{j,m'} \hat{\mathcal{D}}(R) \ket{j,m} C_{m_1m_2}^{jm} \\ = \sum_{j,m,m'} C_{m_1'm_2'}^{jm'} C_{m_1m_2}^{jm} \mathcal{D}_{mm'}^{(j)} where I’m using shorthand to write the Clebsch-Gordan coefficients, C_{m_1 m_2}^{jm} \equiv \left\langle j_1 j_2; m_1 m_2 | j_1 j_2; j m \right\rangle and I’m ignoring complex conjugates since the C-G coefficients are all real. The above result is completely general, and expresses the product-basis D-matrix elements in terms of the total basis ones. However, this can also be combined with the other identity we just derived to prove a very useful formula.

Let’s take j_1 = l_1 and j_2 = l_2 to both be orbital angular momenta. If we then let m_1 = m_2 = 0, the original product-basis formula for this matrix element becomes \mathcal{D}_{m_1'0}^{(l_1)} \mathcal{D}_{m_2'0}^{(l_2)} = \frac{4\pi}{\sqrt{(2l_1+1)(2l_2+1)}} (Y_{l_1}^{m_1'} (\theta, \phi) Y_{l_2}^{m_2'} (\theta, \phi))^{\star} The choice of m_1 and m_2 forces C_{m_1 m_2}^{jm} above to vanish unless m=0 as well, which lets us rewrite the other expression with a single \mathcal{D}-matrix: \bra{j_1, j_2; m_1', m_2'} \hat{\mathcal{D}}(R) \ket{j_1, j_2; 0, 0} = \sum_{l',m'} C_{m_1'm_2'}^{l'm'} C_{00}^{l'0} \sqrt{\frac{4\pi}{2l'+1}} (Y_{l'}^{m'})^\star (\theta,\phi) This is a very interesting identity: we’ve showed that the product of two spherical harmonics can be written in terms of a sum over individual harmonics. Dropping the superfluous primes, we have Y_{l_1}^{m_1} (\theta, \phi) Y_{l_2}^{m_2} (\theta, \phi) = \frac{\sqrt{(2l_1+1)(2l_2+1)}}{4\pi} \\ \times \sum_{l',m'} \left\langle l_1 l_2; m_1 m_2 | l_1 l_2; l' m' \right\rangle \left\langle l_1 l_2; 0 0 | l_1 l_2; l' 0 \right\rangle \sqrt{\frac{4\pi}{2l'+1}} Y_{l'}^{m'} (\theta, \phi). This is a relation of special functions, but mathematically nothing operationally has changed from above. All we are doing here is rewriting a reducible product of two states (two spherical harmonics) as a sum over irreducible basis states (single spherical harmonics.)

The most powerful application of this derivation appears if we multiply both sides by a third spherical harmonic (Y_l^m)^\star(\theta, \phi), and then integrate over the solid angle. This allows us to use the orthogonality of the spherical harmonics, \int d\Omega (Y_{l'}^{m'})^\star (\theta, \phi) Y_l^m(\theta, \phi) = \delta_{l,l'} \delta_{m,m'}. Using this identity the integral under the sum on the right vanishes unless m=m' and l=l', and thus: \int d\Omega (Y_l^m)^\star (\theta, \phi) Y_{l_1}^{m_1}(\theta, \phi) Y_{l_2}^{m_2}(\theta, \phi) \\ = \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2l+1)}} \left\langle l_1 l_2; 0 0 | l_1 l_2; l 0 \right\rangle \left\langle l_1 l_2; m_1 m_2 | l_1 l_2; l m \right\rangle. This is a beautiful formula, and really shows off the power of our Hilbert-space formulation of quantum mechanics. The integral on the left-hand side is a complicated, calculus-based expression, incidentally the sort of expression that appears atomic and nuclear spectroscopy calculations. On the other hand, the right-hand side expression is a simple product of algebraic quantities, the Clebsch-Gordan coefficients. So we’ve traded an intimidating integral for some simple algebra or the use of a table of C-G coefficients. This is the Hilbert-space approach to quantum mechanics in a nutshell!

Finally, back to our radiative transition problem, where we had used the spherical basis to write the result as \bra{n'l'm'} \hat{r}_q \ket{nlm} = \int_0^{\infty} dr\ r^3 R_{n'l'}^\star(r) R_{nl}(r) \sqrt{\frac{4\pi}{3}} \int d\Omega (Y_{l'}^{m'})^\star(\theta, \phi) Y_1^q (\theta, \phi) Y_l^m(\theta, \phi).

Now we know that this three-Y integral is simply proportional to a Clebsch-Gordan coefficient, \int d\Omega (...) \sim \left\langle l1; mq | l1;l'm' \right\rangle. This lets us recover a pair of selection rules, due to the properties of the C-G coefficients. The first is that m' = q+m, reproducing what we had already found for the dipole transition using spherical basis. But the C-G coefficients let us go further, because they also vanish unless the two orbital quantum numbers are within 1 of each other, i.e. there is a second selection rule that is now evident, |l-1| \leq l' \leq l+1. So not only do we not have to do any angular integrals, but we see immediately that most of the matrix elements we would naively write down are automatically zero. In the case of our example transition 3d \rightarrow 2p, we have l=2, l'=1, so the latter selection rule is satisfied. But we can now see that, for example, the 4f \rightarrow 2p transition, or 3d \rightarrow 1s, are completely forbidden through the dipole operator.

Incidentally, this is an example of an application of the Wigner-Eckart theorem, which we’ll derive shortly; the Wigner-Eckart theorem allows a very general and powerful splitting of matrix elements between angular momentum eigenstates, into algebraic angular parts and a smaller number of “reduced” matrix elements.

15.4 Spherical tensors

The simplification of our calculation for the dipole transition in terms of spherical harmonics is actually much more general than it might seem! The key insight comes from representation theory: the \hat{\vec{r}} vector transforms under the j=1 representation of the rotation group (like any vector operator.) The fact that the result ended up proportional to an \ell=1 spherical harmonic was no accident, because by construction the spherical harmonic Y_l^m transforms under rotations according to its \ell value.

What does this look like in position space? Let’s write down the general formula: we start with Y_l^m(\vec{n}) = \left\langle \vec{n} | lm \right\rangle. Under a general rotation, the direction ket rotates as \ket{\vec{n}} \rightarrow \hat{\mathcal{D}}(R) \ket{\vec{n}} = \ket{\vec{n}'}. We can then write out the new spherical harmonic in terms of the ones oriented in the old direction: Y_l^m(\vec{n}') = \bra{\vec{n}} \hat{\mathcal{D}}(R)^{\dagger} \ket{lm} \\ = \sum_{l',m'} \left\langle \vec{n} | l'm' \right\rangle \bra{l'm'} \hat{\mathcal{D}}(R)^{\dagger} \ket{lm} \\ = \sum_{m'} Y_l^{m'}(\vec{n}) (\mathcal{D}_{mm'}^{(l)}(R))^\star. We take this transformation rule as a definition of a spherical tensor operator: a spherical tensor operator of rank k, written \hat{T}_q^{(k)}, transforms under rotations as \hat{\mathcal{D}}^\dagger(R) \hat{T}_q^{(k)} \hat{\mathcal{D}}(R) = \sum_{q'=-k}^{k} \mathcal{D}_{qq'}^{(k)}{}^\star(R) T_{q'}^{(k)}. In other words, spherical tensor operators are constructed such that under rotation, their components transform according to the corresponding Wigner D-matrix. As we already stumbled into, a vector operator in spherical basis is precisely a spherical tensor operator of rank 1 (which explains the funny notation we used before.)

If we’re handed a particular tensor, how can we tell if it is a spherical tensor or not? Inferring the finite rotation properties aren’t always easy, but the infinitesmal form can be quite useful. If we subtitute the infinitesmal rotation \mathcal{D}(R) = \left(1 - \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar} \right) then the spherical tensor \hat{T}_q^{(k)} must satisfy \left(1 + \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar} \right) \hat{T}_q^{(k)} \left(1 - \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar} \right) \\ = \sum_{q'=-k}^{k} \hat{T}_{q'}^{(k)} \bra{kq'} \left(1 + \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar} \right)\ket{kq}. The first term on the right just gives T_q^{(k)}, canceling the same term on the left. From the remaining order-\epsilon terms, we end up with a commutator: [\hat{\vec{J}} \cdot \vec{n}, \hat{T}_q^{(k)}] = \sum_{q'} T_{q'}^{(k)} \bra{kq'} \hat{\vec{J}} \cdot \vec{n} \ket{kq}. Having the commutator for an arbitrary direction \vec{n} isn’t very convenient, but we can just plug in the cardinal directions; having all of the commutation relations for x,y,z is equivalent to this equation. The z-equation is particularly simple: [\hat{J}_z, \hat{T}_q^{(k)}] = \hbar q \hat{T}_q^{(k)}, while for x and y, it’s most convenient to write in terms of raising and lowering operators: [\hat{J}_{\pm}, \hat{T}_q^{(k)}] = \hbar \sqrt{(k \mp q)(k \pm q + 1)} T_{q \pm 1}^{(k)}. These two commutation relations can be taken to prove the m-selection rule: for any two angular momentum eigenstates \ket{\alpha, j, m} and \ket{\alpha', j', m'}, where \alpha represents all of the non-angular momentum quantum numbers, we see that \bra{\alpha', j', m'} [\hat{J}_z, \hat{T}_q^{(k)}] \ket{\alpha, j, m} = \hbar q \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m}. But since we now have \hat{J}_z eigenstates on both sides, we can get rid of the commutator, finding that [(m'-m - q) \hbar] \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m} = 0 so that the matrix element vanishes unless m' = q + m. The matrix elements must also satisfy the triangular selection rule, |j-k| \leq j' \leq j+k. This isn’t as easy to prove on its own, but it follows directly from the Wigner-Eckart theorem, which we will prove shortly. Spherical tensors give us the power of selection rules for any physical system, not just those which can be expressed using spherical harmonics.

Example: commutation of a vector dot product

The commutation relations allow us to check whether a particular object that we’ve been handed is a spherical tensor of a given rank or not. Let’s try the example of the vector dot product, which we expect to be invariant under rotation - so we should find that it is a rank-0 tensor. The commutator with \hat{J}_z is [\hat{J}_z, \hat{\vec{U}} \cdot \hat{\vec{V}}] = \sum_i [\hat{J}_z, \hat{U}_i \hat{V}_i] \\ = \sum_i [\hat{J}_z, \hat{U}_i] \hat{V}_i + \hat{U}_i [\hat{J}_z, \hat{V}_i] \\ = \sum_{i,k} \left[ (-i \hbar \epsilon_{izk} \hat{U}_k) \hat{V}_i + \hat{U}_i (-i \hbar \epsilon_{izk} \hat{V}_k) \right] Since by convention \epsilon_{xyz} = 1, we see that the non-vanishing components with z in the middle are \epsilon_{xzy} = -1 and \epsilon_{yzx} = +1. Thus, the only surviving terms in the sum above are -i\hbar \left[\hat{U}_x \hat{V}_y - \hat{U}_y \hat{V}_x + \hat{U}_y \hat{V}_x - \hat{U}_x \hat{V}_y \right] so [\hat{J}_z, \hat{U} \cdot \hat{V}] = 0. The argument for the other two commutators with \hat{J}_x and \hat{J}_y follows in exactly the same way; it doesn’t matter whether there’s a z in the middle or one of the other directions. So indeed, \hat{\vec{U}} \cdot \hat{\vec{V}} is a scalar.

By the way, we didn’t have to expand out the explicit components of \epsilon_{ijk} above; there’s a faster way to see the general result. We can write [\hat{J}_j, \hat{\vec{U}} \cdot \hat{\vec{V}}] = -i\hbar \sum_{i,k} \left[ \epsilon_{ijk} \hat{U}_k \hat{V}_i + \epsilon_{ijk} \hat{U}_i \hat{V}_k \right] But since we’re summing over both i and k, we’re free to swap the two labels with each other; this gives us a minus sign due to the total antisymmetry of \epsilon_{ijk}. If we swap one of the two terms above, we find [\hat{J}_j, \hat{\vec{U}} \cdot \hat{\vec{V}}] = -i \hbar \sum_{i,k} \epsilon_{ijk} \left[ - \hat{U}_i \hat{V}_k + \hat{U}_i \hat{V}_k \right] = 0.

One of the problems with spherical tensors is that the discussion tends to become very formal and impractical right out of the gate (as in Sakurai.) Let’s try to fix that by looking concretely at Cartesian tensors of ranks 0, 1, and 2, and explicitly seeing how they break down into spherical tensors. (Tensors of higher rank than 2 are rarely encountered anyway.)

We don’t have much to say about rank-0 spherical tensors, i.e. scalars. In fact, a rank-0 Cartesian tensor is always a rank-0 spherical tensor; neither object transforms under a rotation. Let’s start our serious study at rank 1.

15.4.1 Rank 1 tensors

A rank-1 Cartesian tensor is just a vector operator, for which we know the commutation relations with angular momentum already: [\hat{V}_i, \hat{J}_j] = i\hbar \epsilon_{ijk} \hat{V}_k.

From these commutation relations, we derived that the spherical tensor components of an arbitrary vector are just the spherical basis components we wrote down above: \hat{V}_1^{(1)} = -\frac{\hat{V}_x + i\hat{V}_y}{\sqrt{2}} \\ \hat{V}_0^{(1)} = \hat{V}_z \\ \hat{V}_{-1}^{(1)} = \frac{\hat{V}_x - i\hat{V}_y}{\sqrt{2}} We also record the inverse relations, which are sometimes helpful when we’re starting a Cartesian expression: \hat{V}_x = \frac{1}{\sqrt{2}} (\hat{V}_{-1}^{(1)} - \hat{V}_{1}^{(1)}) \\ \hat{V}_y = \frac{i}{\sqrt{2}} (\hat{V}_{-1}^{(1)} + \hat{V}_{1}^{(1)}) \\ \hat{V}_z = \hat{V}_0^{(1)}

Exercise - deriving the spherical basis

Using the defining commutation relation above, show that you recover exactly the spherical basis expansion given above, including the normalization.

Answer:

If we let j=z, then we find [\hat{J}_z, \hat{V}_x] = i\hbar \hat{V}_y \\ [\hat{J}_z, \hat{V}_y] = -i\hbar \hat{V}_x \\ [\hat{J}_z, \hat{V}_z] = 0. Combining the first two commutators, we see that [\hat{J}_z, \hat{V}_x \pm i \hat{V}_y] = i\hbar \hat{V}_y \pm \hbar \hat{V}_x = \pm \hbar (\hat{V}_x \pm i \hat{V}_y) So these combinations of \hat{V}_x and \hat{V}_y, along with \hat{V}_z on its own, will satisfy the defining relation for a rank-1 spherical tensor. The remaining commutator is needed to fix the relative normalization. [\hat{J}_{\pm}, \hat{V}_z] = -[\hat{V}_z, \hat{J}_x] \mp i [\hat{V}_z, \hat{J}_y] \\ = -i \hbar \epsilon_{zxy} \hat{V}_y \pm \hbar \epsilon_{zyx} \hat{V}_x \\ = \mp \hbar (\hat{V}_x \pm i \hat{V}_y) and comparing to the formula for [\hat{J}_{\pm}, \hat{T}_q^{(k)}] above, we find the extra factor of -\sqrt{2} that yields exactly the spherical-basis expansion we wrote down above.

15.4.2 Rank 2 tensors

This is where things start to get complicated, because a rank-2 Cartesian tensor is generally reducible in terms of rotations, which means we will have to write it as a combination of spherical tensors of different ranks. We consider the example of a dyadic tensor operator, which we can think of simply as a direct or outer product of a pair of vector operators: U_i V_j has 9 components. To see how this reducible tensor splits into spherical tensors, we rewrite it as follows: U_i V_j = \frac{1}{3} (\vec{U} \cdot \vec{V}) \delta_{ij} + \frac{1}{2} (U_i V_j - U_j V_i) + \left( \frac{U_i V_j + U_j V_i}{2} - \frac{1}{3} (\vec{U} \cdot \vec{V}) \delta_{ij} \right). We’ve proved that the first term is a scalar already in the example above; the presence of the identity matrix \delta_{ij} doesn’t change anything. It’s possible to use the commutation relations in the same way to show that the second term is a rank-1 spherical tensor, and the final term is rank 2, but there are a lot of components to check (3 and then 5), and it’s rather laborious.

Instead, I’ll argue that any rank-2 Cartesian tensor can be decomposed in the following way: T_{ij} = E \delta_{ij} + A_{ij} + S_{ij} where A_{ij} is a totally antisymmetric tensor, and S_{ij} is totally symmetric but has zero trace, i.e. \sum_i S_{ii} = 0. In fact, we can construct everything explicitly: E = \frac{1}{3} \sum_i T_{ii} \\ A_{ij} = \frac{1}{2} (T_{ij} - T_{ji}) \\ S_{ij} = \frac{1}{2}(T_{ij} + T_{ji}) - \frac{1}{3} \delta_{ij} \sum_k T_{kk} Note that in three dimensions, we have 3 distinct non-zero components of A_{ij} and 5 of S_{ij}, so in all we recover the 1+3+5=9 entries in an arbitrary 3x3 Cartesian tensor.

Let’s see how things transform under rotations, starting with the scalar. For any rank-2 tensor, we get one rotation matrix for each index, so: E \delta_{ij} \rightarrow R_{im} R_{jn} E \delta_{mn} = R_{im} (R^T)_{ni} E = (R^T R)_{ij} E. But recall now that rotation matrices are orthogonal matrices, which means that they satisfy R^T = R^{-1}. So the rotated term is just E times the identity, or E \delta_{ij} \rightarrow E \delta_{ij}. As expected, this is a rank-0 tensor; it doesn’t rotate at all.

On to the antisymmetric piece: A_{ij} \rightarrow R_{im} R_{jn} A_{mn} Notice that this is still an antisymmetric tensor in terms of i and j: A_{ji} \rightarrow R_{jm} R_{in} A_{mn} = -R_{jn} R_{im} A_{mn}. It is left as an exercise to prove that this object transforms as a rank-1 tensor; the proof is too much of a detour from our goal right now. (If you’re trying to prove this, start by writing A_{ij} = \epsilon_{ijk} B_k; you will need to invoke the fact that \det(R) = 1.)

Similarly, the symmetric traceless part transforms as S_{ij} \rightarrow R_{im} R_{jn} S_{mn}, which is still symmetric; furthermore, we see that S_{ii} \rightarrow R_{im} R_{in} S_{mn} = \delta_{mn} S_{mn} = S_{mm}, so the trace remains zero. This is the point of the decomposition, of course; in terms of the Cartesian components a generic rotation will mix all 9 components together, whereas with this separation the effects of rotation are block-diagonal.

Because of the way these components break apart, this gives us a natural way to construct E, A, S for any two-index Cartesian tensor T_{ij}:

Calculate the trace to find E = T_{ii};
Antisymmetrize to find A_{ij} = \frac{1}{2}(T_{ij} - T_{ji});
Symmetrize and remove the trace to find E_{ij} = \frac{1}{2} (T_{ij} + T_{ji}) - \frac{1}{3} E \delta_{ij}.

Since this decomposition is completely generic, it’s useful to have expressions for what the spherical tensors look like in terms of the original tensor. Here they are, stated without proof: given the Cartesian rank-2 tensor T_{ij} = E \delta_{ij} + A_{ij} + S_{ij}, its spherical tensor components are T_0^{(0)} = E \\ T_0^{(1)} = A_{xy} \\ T_{\pm 1}^{(1)} = \mp \frac{1}{\sqrt{2}} (A_{yz} \pm i A_{zx}) \\ T_0^{(2)} = \sqrt{\frac{3}{2}} S_{zz} \\ T_{\pm 1}^{(2)} = \mp (S_{zx} \pm i S_{zy}) \\ T_{\pm 2}^{(2)} = \frac{1}{2} (S_{xx} - S_{yy} \pm 2i S_{xy}).

15.4.3 Example: dyadic matrix element

Let’s see an application of how we might put these expressions to use in practice. Suppose that we are given the potential function V(x,y,z) = V_0 \hat{x} \hat{y}. We can rewrite xy as \hat{\vec{r}}_x \hat{\vec{r}}_y, so we recognize it as a component of the dyadic tensor \hat{\vec{r}}_i \hat{\vec{r}}_j. In fact, the dyad of two position vectors is very common, so it’s worth working out in generality. Notice that since both vectors are the same, the antisymmetric piece vanishes automatically, A_{xy} = 0 and therefore T_q^{(1)} = 0. For the symmetric parts, we have E = \frac{1}{3} (x^2 + y^2 + z^2) = \frac{1}{3} r^2 \\ S_{ij} = r_i r_j - \frac{1}{3} \delta_{ij} r^2 This allows us to expand out the rank-2 spherical tensor components in coordinates: T_0^{(2)} = \frac{1}{\sqrt{6}} (-x^2 - y^2 + 2z^2) \\ T_1^{(2)} = -z(x+iy) \\ T_2^{(2)} = \frac{1}{2} (x+iy)^2 Interestingly, these once again look very similar to spherical harmonics; in fact, we can write T_q^{(2)} = r^2 \sqrt{\frac{8\pi}{15}} Y_2^q(\theta, \phi). The spherical harmonics Y_l^m are themselves spherical tensors of rank l, and since they were constructed from (part of) position eigenkets it’s no surprise that they show up when writing out tensors involving the position vector.

Now, back to our application. From the above formulas, it’s easy to see that xy = \frac{i}{2} (T_{-2}^{(2)} - T_{2}^{(2)}). Without having to do anything else, we can immediately apply the selection rules above: for any matrix elements \bra{\alpha', l', m'} V \ket{\alpha, l, m}, we immediately see that |\Delta l| \leq 2 and \Delta m = \pm 2.

It also happens that |\Delta l| = 1 is forbidden not by a selection rule, but by parity. Clearly, the operator \hat{x} \hat{y} is symmetric under parity, which flips the sign of both x and y. On the other hand, based on the definition of the spherical harmonics, it’s straightforward to show that under parity, Y_l^m(\theta, \phi) \rightarrow (-1)^l Y_l^m(\theta, \phi), and therefore \hat{P} \ket{\alpha, l, m} = (-1)^l \ket{\alpha, l, m}. If we were studying an operator which was odd under parity, then the even-l transition matrix elements would vanish instead. But xy is even, so we have |\Delta l| = 0 or 2.

15.5 Combination of spherical tensors

The combination of spherical tensors to form another spherical tensor is often a very useful technique. In fact, for an object like the dyadic tensor where we’re combining two rank-1 spherical tensors, it’s a straightforward way to derive the components in terms of \hat{U}_i and \hat{V}_i.

In fact, we already know how to do this: the rules for combination of spherical tensors are exactly the same as those for addition of angular momentum, and the coefficients are just the Clebsch-Gordan coefficients! If X_{q_1}^{(k_1)} and Z_{q_2}^{(k_2)} are irreducible spherical tensors of rank k_1 and k_2, then T_q^{(k)} = \sum_{q_1, q_2} \left\langle k_1 k_2; q_1 q_2 | k_1 k_2; k q \right\rangle X_{q_1}^{(k_1)} Z_{q_2}^{(k_2)} is an irreducible spherical tensor of rank k. The proof of this statement is straightforward; we just look at how both sides transform under rotation, and end up with products of Clebsch-Gordan coefficients to simplify on the right-hand side. I won’t go through the proof here since it’s messy, but it’s on page 251 of Sakurai if you’re interested.

It’s similarly possible to derive an inverse formula, X_{q_1}^{(k_1)} Z_{q_2}^{(k_2)} = \sum_{k=|k_1-k_2|}^{k_1+k_2} \sum_{q=-k}^{k} \left\langle k_1 k_2; kq | k_1 k_2; q_1 q_2 \right\rangle T_q^{(k)}

These formulas give us explicit ways to construct higher-rank spherical tensors from lower-rank ones, or to decompose a Carteisan tensor into spherical components. For example, to write out the rank-2 spherical tensor components of the dyadic tensor \hat{U}_i \hat{V}_j, all we have to do is look up a table of Clebsch-Gordan coefficients for the addition of two j=1 states to give j=2, and we’ll find the following: \hat{T}_{\pm 2}^{(2)} = \hat{U}_{\pm 1} \hat{V}_{\pm 1} \\ \hat{T}_{\pm 1}^{(2)} = \frac{1}{\sqrt{2}} (\hat{U}_{\pm 1} \hat{V}_0 + \hat{U}_0 \hat{V}_{\pm 1}) \\ \hat{T}_{\pm 0}^{(2)} = \frac{1}{\sqrt{6}} (\hat{U}_{1} \hat{V}_{-1} + 2 \hat{U}_0 \hat{V}_0 + \hat{U}_{-1} \hat{V}_1).

Due to the appearance of Clebsch-Gordan coefficients, the combination of spherical tensors to get other tensors follows the usual set of selection rules, which we can understand in terms of representations of the rotation group. For example, the decomposition of the dyadic tensor can be written using plethysm in the form \mathbf{1} \otimes \mathbf{1} = \mathbf{2} \oplus \mathbf{1} \oplus \mathbf{0}. 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, \hat{T}_{ijk} = \hat{U}_i \hat{V}_j \hat{W}_k. How will this decompose into spherical tensors? Well, \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} 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.