We're about to define a lot of extra machinery, with regard to vector operators and then the more generalized tensor operators. Before we get into it, I'll give you a little motivation. Vector operators, and especially tensor operators, can have many different components; if we're interested in their matrix elements, we often have to calculate a large number of possibilities. However, treating the components of these operators as independent is too naive, particularly in the case when we have a system with rotational invariance.
Our goal in all of the following will be to derive the Wigner-Eckart theorem, a powerful result which greatly simplifies the calculation of matrix elements in the presence of rotational symmetry. It's another example of a repeating theme we've seen, that we can use our angular momentum formalism to nicely separate certain aspects of the angular and radial behavior of a quantum system.
So far, we've made repeated references to "vector operators", things like \( \hat{\vec{x}} \) and \( \hat{\vec{L}} \). The definition of these objects is straightforward; we can typically create them just by promoting the corresponding classical operator. If we measure \( \hat{\vec{x}} \), i.e. if we measure all components of the position \( x,y,z \) simultaneously, then the system is put into a position eigenstate and we recover a more familiar-looking vector.
At this point I'm going to switch to index notation, instead of arrows: so \( \hat{V}_i \) is a vector operator, and the latin index \( i \) runs from 1 to 3 (or \( x \) to \( z \).) For a classical vector, the transformation under a rotation is given by the rotation matrix,
\[ \begin{aligned} V_i \rightarrow \sum_j R_{ij} V_j. \end{aligned} \]
Is there a similar transformation rule for vector operators? So far we've dealt with rotation by considering its action on the state kets,
\[ \begin{aligned} \ket{\alpha} \rightarrow \hat{\mathcal{D}}(R) \ket{\alpha}. \end{aligned} \]
This means that the expectation value of any operator transforms under rotation as
\[ \begin{aligned} \ev{\hat{O}} = \bra{\alpha} \hat{O} \ket{\alpha} \rightarrow \bra{\alpha} \hat{\mathcal{D}}^\dagger(R) \hat{O} \hat{\mathcal{D}}(R) \ket{\alpha} \end{aligned} \]
Now, it's completely general to adopt a Heisenberg-picture-like approach here, and decide that we're going to let the rotation act on the operators and leave the states unchanged. This is, in fact, a much more convenient approach for dealing with vector operators. For the particular case of a vector operator, we expect that the expectation value itself should transform classically under a rotation:
\[ \begin{aligned} \ev{\hat{V}_i} \rightarrow \sum_j R_{ij} \ev{\hat{V}_j}. \end{aligned} \]
This ensures that the expectation value itself behaves like a classical vector, so we will properly get back the classical limit. Comparing to the expectation value above, since we require this relation to hold for any state \( \ket{\alpha} \), we arrive at the operator identity
\[ \begin{aligned} \hat{\mathcal{D}}^{\dagger}(R) \hat{V}_i \hat{\mathcal{D}}(R) = \sum_{j} R_{ij} \hat{V}_j. \end{aligned} \]
As always, we can learn more by considering what this implies for an infinitesmal rotation, generated by some angular momentum operator:
\[ \begin{aligned} \hat{\mathcal{D}}(R) = 1 - \frac{i\epsilon}{\hbar} (\hat{\vec{J}} \cdot \vec{n}). \end{aligned} \]
This gives us for the condition above
\[ \begin{aligned} \hat{V}_i - \frac{i\epsilon}{\hbar} [\hat{V}_i, \hat{\vec{J}} \cdot \vec{n}] = \sum_j R_{ij}(\vec{n}, \epsilon) \hat{V}_j. \end{aligned} \]
Evaluating the classical rotation matrix for an arbitrary \( \vec{n} \) is complicated, but we know that in quantum mechanics we prefer to work in terms of rotations around the fixed coordinate axes anyway. Recall that the infinitesmal rotation matrix about the \( z \)-axis is given by
\[ \begin{aligned} R(\vec{z}, \epsilon) = \left(\begin{array}{ccc} 1& -\epsilon & 0 \\ \epsilon & 1 & 0 \\ 0 & 0 & 1 \end{array} \right). \end{aligned} \]
In tensor notation, this becomes
\[ \begin{aligned} R_{ij}(\hat{z}, \epsilon) = \delta_{ij} - \epsilon \epsilon_{ijz} \end{aligned} \]
In fact, the same formula holds for rotation about coordinate axis \( k \), replacing the last term with \( \epsilon_{ijk} \). In other words, if \( \vec{n}_k \) labels one of the axes, then
\[ \begin{aligned} \hat{V}_i - \frac{i\epsilon}{\hbar} [\hat{V}_i, \hat{J}_k] = \sum_j (\delta_{ij} - \epsilon \epsilon_{ijk}) \hat{V}_j, \end{aligned} \]
or canceling the first terms,
\[ \begin{aligned} [\hat{V}_i, \hat{J}_k] = i \hbar \epsilon_{ikj} \hat{V}_j. \end{aligned} \]
This commutation relation can be taken as a definition for a vector operator. You'll recognize this as the angular-momentum commutation relation, which isn't too surprising since \( \hat{\vec{J}} \) itself is also a vector operator.
A much easier definition is that of a scalar operator. Scalars are objects which don't transform at all under rotation, and so if \( \hat{K} \) is a scalar operator, we require that
\[ \begin{aligned} \hat{\mathcal{D}}^{\dagger}(R) \hat{K} \hat{\mathcal{D}}(R) = \hat{K}. \end{aligned} \]
This implies a much simpler commutation relation with angular momentum, namely
\[ \begin{aligned} [\hat{K}, \hat{J}_i] = 0. \end{aligned} \]
We can use these definitions to construct more complicated operators, and study how they transform. For example, it's straightforward to show that if \( \hat{U}_i \) and \( \hat{V}_i \) are two vector operators, then the dot product \( \hat{\vec{U}} \cdot \hat{\vec{V}} \) is a scalar operator, while \( \hat{\vec{U}} \times \hat{\vec{V}} \) is a vector operator.
These are expected results from manipulation of ordinary vectors, but don't forget that these are operators and don't commute! Both \( \hat{\vec{U}} \cdot \hat{\vec{V}} \) and \( \hat{\vec{V}} \cdot \hat{\vec{U}} \) are guaranteed to be scalar operators, but they're not necessarily the same scalar operator.
We can generalize further to spatial tensors, objects which carry more than one index. In classical mechanics, the moment of inertia tensor is probably the most familiar example. In the index notation, the transformation properties of tensors are given by applying one rotation matrix per index, for example
\[ \begin{aligned} T_{ijk} \rightarrow \sum_{i',j',k'} R_{ii'} R_{jj'} R_{kk'} T_{i'j'k'}. \end{aligned} \]
The number of indices on a tensor is called the rank; our example here has rank three. We call a tensor written like this a Cartesian tensor, because we're using the Cartesian coordinates \( x,y,z \) to label its components. Note that scalars are just tensors of rank 0, and vectors are rank-1 tensors.
Although a Cartesian tensor is easy to write down, it's often very hard to work with, particularly when dealing with rotations. Just like our addition of angular momentum example above, the rotation of Cartesian tensors is reducible; we can split it into different sub-objects which rotate in different ways. We can see this by looking at one of the simplest possible tensors, the dyadic, which is a fancy word for a tensor made by sticking two vectors together:
\[ \begin{aligned} T_{ij} = U_i V_j. \end{aligned} \]
This can be rewritten by gathering certain terms together:
\[ \begin{aligned} U_i V_j = \frac{1}{3} (\vec{U} \cdot \vec{V}) \delta_{ij} + \frac{1}{2} (U_i V_j - U_j V_i) + \left( \frac{U_i V_j + U_j V_i}{2} - \frac{1}{3} (\vec{U} \cdot \vec{V}) \delta_{ij} \right). \end{aligned} \]
It's clear that these expressions are equal. But now we've managed to identify parts that transform differently under rotations. The first term is proportional to the dot product, which is a scalar; it doesn't transform at all under rotation. The second term is just the cross product, which rotates as a single (axial) vector. The leftover pieces are another tensor, specifically a symmetric tensor with trace zero; this happens to be precisely the 5-dimensional object which transforms irreducibly under the rotation group. The number of independent terms in each is 1 + 3 + 5, so we still have 9 terms in total.
This decomposition looks suspiciously like the decomposition of the sum of two \( j=1 \) angular momenta into objects with total \( j=0,1,2 \). This is, in fact, exactly what we have just done without knowing it. The pieces which transform uniformly under rotations that we have identified are examples of spherical tensors.
There is nothing special about our choice of the dyadic construction for this tensor; any two-index Cartesian tensor can be decomposed into a scalar, a vector, and a symmetric two-component tensor. You can see from how the Cartesian tensor rotates that we always treat it as a \( 1 \otimes 1 \) with respect to angular momentum \( l \), and the decomposition \( 2 \oplus 1 \oplus 0 \) follows from that.
Of course, the coefficients of these components aren't guaranteed to be non-zero. Clearly if we take a two-index tensor which is already symmetric and traceless, then the scalar and vector parts will vanish. In our previous example of the inertia tensor \( I_{ij} \), the tensor is symmetric but not traceless; its trace (the sum of the diagonal moments) is exactly the \( j=0 \) component, and is easily verified to be invariant under rotations. For an antisymmetric two-index tensor \( T_{ij} = -T_{ji} \), only the vector component is non-zero (a simple example would be the cross product.)
Our solution to having reducible products of rotation matrices for angular momentum eigenstates was a change of basis; in the \( \ket{j m} \) basis, the rotation matrix was block-diagonal and irreducible. Similarly, a change of basis away from Cartesian coordinates will be very helpful here. We introduce a new, complex basis known as the spherical basis:
\[ \begin{aligned} \hat{e}_1 = -\frac{1}{\sqrt{2}} \left( \hat{x} + i\hat{y}\right) \\ \hat{e}_0 = \hat{z} \\ \hat{e}_{-1} = \frac{1}{\sqrt{2}} \left( \hat{x} - i\hat{y} \right). \end{aligned} \]
(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:
\[ \begin{aligned} \hat{e}_q^\star \hat{e}_{q'} = \delta_{qq'} \end{aligned} \]
We can, of course, expand any arbitrary spatial vector in terms of the spherical basis components,
\[ \begin{aligned} \vec{X} = \sum_{q} \hat{e}_q^\star X_q \end{aligned} \]
where
\[ \begin{aligned} X_q = \hat{e}_q \cdot \vec{X}. \end{aligned} \]
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.
To gain a proper appreciation for the spherical basis, let's see a brief example. 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 transition probability amplitude from state \( \ket{nlm} \) to state \( \ket{n'l'm'} \) is proportional to
\[ \begin{aligned} \bra{n'l'm'} \hat{\vec{r}} \ket{nlm}. \end{aligned} \]
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 \( \hat{\vec{r}} \). Furthermore, ignoring the fine structure the energy labels are degenerate in \( m \), so if we want to study a transition from \( nl \) to \( n'l' \), there are in total \( 3(2l'+1)(2l+1) \) matrix elements to compute. For a transition \( 3d \rightarrow 2p \) for example, this gives 45 matrix elements in total.
Fortunately, the spherical basis will greatly reduce the amount of calculation we have to do, by exploiting the rotational symmetry of our system. Let's expand in that basis, with coefficients
\[ \begin{aligned} r_q = \hat{e}_q \cdot \vec{r}. \end{aligned} \]
In terms of the \( x,y,z \) components of \( \hat{\vec{r}} \), this gives the components we wrote above. But there's another, not completely obvious way to obtain the same combinations of Cartesian components, and it comes from the spherical harmonics. Notice that
\[ \begin{aligned} r Y_1^1(\theta, \phi) = -r \sqrt{\frac{3}{8\pi}} \sin \theta e^{i\phi} = \sqrt{\frac{3}{4\pi}} \left( - \frac{x + iy}{\sqrt{2}} \right) \\ r Y_1^0(\theta, \phi) = r \sqrt{\frac{3}{4\pi}} \cos \theta = \sqrt{\frac{3}{4\pi}} z, \\ r Y_1^{-1}(\theta, \phi) = r \sqrt{\frac{3}{8\pi}} \sin \theta e^{-i\phi} = \sqrt{\frac{3}{4\pi}} \left( \frac{x-iy}{\sqrt{2}} \right). \end{aligned} \]
Thus, we see that the components of \( \vec{r} \) in spherical basis are proportional to the spherical harmonics,
\[ \begin{aligned} r Y_1^q(\theta, \phi) = \sqrt{\frac{3}{4\pi}} r_q. \end{aligned} \]
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.
This relationship to the spherical harmonics makes the matrix elements particularly easy to evaluate: we now have
\[ \begin{aligned} \bra{n'l'm'} \hat{r}_q \ket{nlm} = \int_0^{\infty} dr\ r^3 R_{n'l'}^\star(r) R_{nl}(r) \sqrt{\frac{4\pi}{3}} \int d\Omega (Y_{l'}^{m'})^\star(\theta, \phi) Y_1^q (\theta, \phi) Y_l^m(\theta, \phi). \end{aligned} \]
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:
\[ \begin{aligned} Y_l^m(\theta, \phi) = \sprod{\theta, \phi}{l,m}. \end{aligned} \]
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. 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:
\[ \begin{aligned} \ket{\vec{n}} = \hat{\mathcal{D}}(\phi, \theta) \ket{\hat{z}}. \end{aligned} \]
where \( \hat{\mathcal{D}} \) is a rotation operator describing the rotation from \( \hat{z} \) to \( \vec{n} \). Now, we insert a complete set of states:
\[ \begin{aligned} \ket{\vec{n}} = \sum_{l'} \sum_{m'} \hat{\mathcal{D}}(\phi, \theta) \ket{l',m'} \sprod{l',m'}{\hat{z}}. \end{aligned} \]
If we take the inner product with a particular eigenstate \( \bra{l,m} \), then the sum over \( l' \) on the right collapses:
\[ \begin{aligned} \sprod{l,m}{\vec{n}} = \sum_{m'} \bra{l,m} \hat{\mathcal{D}}(\phi, \theta) \ket{l,m'} \sprod{l,m'}{\hat{z}}. \end{aligned} \]
Next time: we'll finish this and come back to radiative transitions.