Last time, we introduced the idea of a spherical tensor. A spherical tensor of rank \( k \) transforms under rotations in the same way that a spherical harmonic with \( \ell=k \) would, i.e. it satisfies the relation
\[ \begin{aligned} \hat{\mathcal{D}}^\dagger(R) \hat{T}_q^{(k)} \hat{\mathcal{D}}(R) = \sum_{q=-k}^{k} D_{qq'}^{(k)}{}^\star(R) T_{q'}^{(k)}. \end{aligned} \]
or equivalently the algebraic relations
\[ \begin{aligned} [\hat{J}_z, \hat{T}_q^{(k)}] = \hbar q \hat{T}_q^{(k)}, \\ [\hat{J}_{\pm}, \hat{T}_q^{(k)}] = \hbar \sqrt{(k \mp q)(k \pm q + 1)} T_{q \pm 1}^{(k)}. \end{aligned} \]
We found that this implies specific commutation relations for \( \hat{T}_q^{(k)} \) with the total angular momentum operator \( \hat{\vec{J}} \), which in turn implies that matrix elements of any spherical tensor obey corresponding selection rules.
Dealing with matrix elements of angular-momentum eigenstates will be greatly simplified by use of spherical tensor operators, as we'll see shortly. But first, let's run through some explicit examples to try to get a better sense for what a spherical tensor operator is. We talked about the trivial rank-0 (scalar) case last time, so let's move on to rank 1.
A rank-1 Cartesian tensor is just a vector operator, for which we know the commutation relations with angular momentum already:
\[ \begin{aligned} [\hat{V}_i, \hat{J}_j] = i\hbar \epsilon_{ijk} \hat{V}_k. \end{aligned} \]
What are the spherical tensor components \( \hat{V}_q^{(1)} \) of this vector operator? If we let \( j=z \), then we find
\[ \begin{aligned} [\hat{J}_z, \hat{V}_x] = i\hbar \hat{V}_y \\ [\hat{J}_z, \hat{V}_y] = -i\hbar \hat{V}_x \\ [\hat{J}_z, \hat{V}_z] = 0. \end{aligned} \]
Combining the first two commutators, we see that
\[ \begin{aligned} [\hat{J}_z, \hat{V}_x \pm i \hat{V}_y] = i\hbar \hat{V}_y \pm \hbar \hat{V}_x = \pm \hbar (\hat{V}_x \pm i \hat{V}_y) \end{aligned} \]
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.
\[ \begin{aligned} [\hat{J}_{\pm}, \hat{V}_z] = -[\hat{V}_z, \hat{J}_x] \mp i [\hat{V}_z, \hat{J}_y] \\ = -i \hbar \epsilon_{zxy} \hat{V}_y \pm \hbar \epsilon_{zyx} \hat{V}_x \\ = \mp \hbar (\hat{V}_x \pm i \hat{V}_y) \end{aligned} \]
and comparing to the formula for \( [\hat{J}_{\pm}, \hat{T}_q^{(k)}] \) above, we find an extra factor of \( -\sqrt{2} \). In the end, we have that the spherical tensor components of an arbitrary vector are
\[ \begin{aligned} \hat{V}_1^{(1)} = -\frac{\hat{V}_x + i\hat{V}_y}{\sqrt{2}} \\ \hat{V}_0^{(1)} = \hat{V}_z \\ \hat{V}_{-1}^{(1)} = \frac{\hat{V}_x - i\hat{V}_y}{\sqrt{2}} \end{aligned} \]
These are exactly the projections onto the spherical basis vectors that we defined last time; now you can see where they come from. We also record the inverse relations, which are sometimes helpful when we're starting a Cartesian expression:
\[ \begin{aligned} \hat{V}_x = \frac{1}{\sqrt{2}} (\hat{V}_{-1}^{(1)} - \hat{V}_{1}^{(1)}) \\ \hat{V}_y = \frac{i}{\sqrt{2}} (\hat{V}_{-1}^{(1)} + \hat{V}_{1}^{(1)}) \\ \hat{V}_z = \hat{V}_0^{(1)} \end{aligned} \]
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.
Our rewriting of the dyadic tensor:
\[ \begin{aligned} U_i V_j = \frac{1}{3} (\vec{U} \cdot \vec{V}) \delta_{ij} + \frac{1}{2} (U_i V_j - U_j V_i) + \left( \frac{U_i V_j + U_j V_i}{2} - \frac{1}{3} (\vec{U} \cdot \vec{V}) \delta_{ij} \right). \end{aligned} \]
We've proved that the first term is a scalar. 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:
\[ \begin{aligned} T_{ij} = E \delta_{ij} + A_{ij} + S_{ij} \end{aligned} \]
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, I can construct everything explicitly:
\[ \begin{aligned} E = \frac{1}{3} \sum_i T_{ii} \\ A_{ij} = \frac{1}{2} (T_{ij} - T_{ji}) \\ S_{ij} = \frac{1}{2}(T_{ij} + T_{ji}) - \frac{1}{3} \delta_{ij} \sum_k T_{kk} \end{aligned} \]
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:
\[ \begin{aligned} E \delta_{ij} \rightarrow R_{im} R_{jn} E \delta_{mn} = R_{im} (R^T)_{ni} E = (R^T R)_{ij} E. \end{aligned} \]
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
\[ \begin{aligned} E \delta_{ij} \rightarrow E \delta_{ij}. \end{aligned} \]
As expected, this is a rank-0 tensor; it doesn't rotate at all.
On to the antisymmetric piece:
\[ \begin{aligned} A_{ij} \rightarrow R_{im} R_{jn} A_{mn} \end{aligned} \]
Notice that this is still an antisymmetric tensor in terms of \( i \) and \( j \):
\[ \begin{aligned} A_{ji} \rightarrow R_{jm} R_{in} A_{mn} = -R_{jn} R_{im} A_{mn}. \end{aligned} \]
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
\[ \begin{aligned} S_{ij} \rightarrow R_{im} R_{jn} S_{mn}, \end{aligned} \]
which is still symmetric; furthermore, we see that
\[ \begin{aligned} S_{ii} \rightarrow R_{im} R_{in} S_{mn} = \delta_{mn} S_{mn} = S_{mm}, \end{aligned} \]
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.
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
\[ \begin{aligned} T_{ij} = E \delta_{ij} + A_{ij} + S_{ij}, \end{aligned} \]
its spherical tensor components are
\[ \begin{aligned} T_0^{(0)} = E \\ T_0^{(1)} = A_{xy} \\ T_{\pm 1}^{(1)} = \mp \frac{1}{\sqrt{2}} (A_{yz} \pm i A_{zx}) \\ T_0^{(2)} = \sqrt{\frac{3}{2}} S_{zz} \\ T_{\pm 1}^{(2)} = \mp (S_{zx} \pm i S_{zy}) \\ T_{\pm 2}^{(2)} = \frac{1}{2} (S_{xx} - S_{yy} \pm 2i S_{xy}). \end{aligned} \]
Let's see an application of how we might put these expressions to use in practice. Suppose that we are given the potential function
\[ \begin{aligned} V(x,y,z) = V_0 \hat{x} \hat{y}. \end{aligned} \]
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
\[ \begin{aligned} E = \frac{1}{3} (x^2 + y^2 + z^2) = \frac{1}{3} r^2 \\ S_{ij} = r_i r_j - \frac{1}{3} \delta_{ij} r^2 \end{aligned} \]
This allows us to expand out the rank-2 spherical tensor components in coordinates:
\[ \begin{aligned} T_0^{(2)} = \frac{1}{\sqrt{6}} (-x^2 - y^2 + 2z^2) \\ T_1^{(2)} = -z(x+iy) \\ T_2^{(2)} = \frac{1}{2} (x+iy)^2 \end{aligned} \]
Interestingly, these once again look very similar to spherical harmonics; in fact, we can write
\[ \begin{aligned} T_q^{(2)} = r^2 \sqrt{\frac{8\pi}{15}} Y_2^q(\theta, \phi) \end{aligned} \]
Again, the spherical harmonics \( Y_l^m \) are always spherical tensors of rank \( l \), but the converse isn't always true.
Now, back to our application. From the above formulas, it's easy to see that
\[ \begin{aligned} xy = \frac{i}{2} (T_{-2}^{(2)} - T_{2}^{(2)}). \end{aligned} \]
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,
\[ \begin{aligned} Y_l^m(\theta, \phi) \rightarrow (-1)^l Y_l^m(\theta, \phi), \end{aligned} \]
and therefore
\[ \begin{aligned} \hat{P} \ket{\alpha, l, m} = (-1)^l \ket{\alpha, l, m}. \end{aligned} \]
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.
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
\[ \begin{aligned} T_q^{(k)} = \sum_{q_1, q_2} \sprod{k_1 k_2; q_1 q_2}{k_1 k_2; k q} X_{q_1}^{(k_1)} Z_{q_2}^{(k_2)} \end{aligned} \]
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 it on the board since it's messy, but the proof is on page 251 of Sakurai if you're interested.
It's similarly possible to derive an inverse formula,
\[ \begin{aligned} X_{q_1}^{(k_1)} Z_{q_2}^{(k_2)} = \sum_{k=|k_1-k_2|}^{k_1+k_2} \sum_{q=-k}^{k} \sprod{k_1 k_2; kq}{k_1 k_2; q_1 q_2} T_q^{(k)} \end{aligned} \]
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:
\[ \begin{aligned} \hat{T}_{\pm 2}^{(2)} = \hat{U}_{\pm 1} \hat{V}_{\pm 1} \\ \hat{T}_{\pm 1}^{(2)} = \frac{1}{\sqrt{2}} (\hat{U}_{\pm 1} \hat{V}_0 + \hat{U}_0 \hat{V}_{\pm 1}) \\ \hat{T}_{\pm 0}^{(2)} = \frac{1}{\sqrt{6}} (\hat{U}_{1} \hat{V}_{-1} + 2 \hat{U}_0 \hat{V}_0 + \hat{U}_{-1} \hat{V}_1). \end{aligned} \]
Next time: finally the Wigner-Eckart theorem!