Last time, we ended partway through considering the dipole transition matrix element for a hydrogenic atom, and we found (by changing to a spherical basis) that the result is proportional to an integral over three spherical harmonics:
\[ \begin{aligned} \bra{n'l'm'} \hat{r}_q \ket{nlm} \sim \int d\Omega (Y_{l'}^{m'})^\star(\theta, \phi) Y_1^q (\theta, \phi) Y_l^m(\theta, \phi). \end{aligned} \]
We know what the spherical harmonics are, so we can certainly just open Mathematica and do the integral; but for the specific example of a \( 3d \rightarrow 2p \) transition we brought up, there are 45 different integrals to do. The good news is that we can actually do the integral algebraically!
To see how, we were considering applying rotations to position eigenkets of the form \( \ket{\vec{n}} \), where \( \vec{n} \) is a unit vector. We stopped with the expression:
\[ \begin{aligned} \sprod{l,m}{\vec{n}} = \sum_{m'} \bra{l,m} \hat{\mathcal{D}}(\phi, \theta) \ket{l,m'} \sprod{l,m'}{\hat{z}}. \end{aligned} \]
The matrix element containing the rotation \( \hat{\mathcal{D}} \) is the Wigner D-matrix, which we encountered last time; if the spherical harmonics are position-space descriptions of the angular momentum eigenstates, then the D-matrix is the position-space description of the rotation operator itself. The other 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
\[ \begin{aligned} m' \sprod{l,m'}{\hat{z}} = \bra{l,m'} \hat{L}_z \ket{\hat{z}} = \bra{l,m'} (\hat{x} \hat{p}_y - \hat{y} \hat{p}_x) \ket{\hat{z}} = 0 \end{aligned} \]
so either the harmonic is zero or \( m'=0 \). We don't know what \( \phi \) is, but for \( m=0 \) the \( \phi \)-dependence in the spherical harmonics drops out:
\[ \begin{aligned} \sprod{l,m'}{\hat{z}} = (Y_l^{m'})^\star (\theta=0, \phi) = \sqrt{\frac{2l+1}{4\pi}} P_l (\cos 0) \delta_{m'0} = \sqrt{\frac{2l+1}{4\pi}} \delta_{m'0}. \end{aligned} \]
So the \( m' \) sum collapses too, leaving just the \( \mathcal{D} \)-matrix with \( m'=0 \): dividing through, we find the identity
\[ \begin{aligned} \mathcal{D}_{m0}^{(l)}(\phi, \theta, 0) = \sqrt{\frac{4\pi}{2l+1}} (Y_l^m)^\star (\theta, \phi). \end{aligned} \]
This is, clearly, a very useful identity for considering the rotation of \( m=0 \) eigenstates (as you saw on the homework.)
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:
\[ \begin{aligned} \bra{j_1, j_2; m_1', m_2'} \hat{\mathcal{D}}(R) \ket{j_1, j_2; m_1, m_2} = \bra{j_1 m_1'} \hat{\mathcal{D}}_1(R) \ket{j_1 m_1} \bra{j_2 m_2'} \hat{\mathcal{D}}_2(R) \ket{j_2 m_2} \\ = \mathcal{D}_{m_1'm_1}^{(j_1)} \mathcal{D}_{m_2'm_2}^{(j_2)}. \end{aligned} \]
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:
\[ \begin{aligned} \bra{j_1, j_2; m_1', m_2'} \hat{\mathcal{D}}(R) \ket{j_1, j_2; m_1, m_2} = \sum_{j,m,m'} C_{m_1'm_2'}^{jm'} \bra{j,m'} \hat{\mathcal{D}}(R) \ket{j,m} C_{m_1m_2}^{jm} \\ = \sum_{j,m,m'} C_{m_1'm_2'}^{jm'} C_{m_1m_2}^{jm} \mathcal{D}_{mm'}^{(j)} \end{aligned} \]
where I'm using shorthand to write the Clebsch-Gordan coefficients,
\[ \begin{aligned} C_{m_1 m_2}^{jm} \equiv \sprod{j_1 j_2; m_1 m_2}{j_1 j_2; j m} \end{aligned} \]
and I'm ignoring complex conjugates that should be here 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
\[ \begin{aligned} \mathcal{D}_{m_1'0}^{(l_1)} \mathcal{D}_{m_2'0}^{(l_2)} = \frac{4\pi}{\sqrt{(2l_1+1)(2l_2+1)}} (Y_{l_1}^{m_1'} (\theta, \phi) Y_{l_2}^{m_2'} (\theta, \phi))^{\star} \end{aligned} \]
The choice of \( m_1 \) and \( m2 \) forces \( C{m_1 m_2}^{jm} \) above to vanish unless \( m=0 \) as well, which lets us rewrite that expression:
\[ \begin{aligned} \bra{j_1, j_2; m_1', m_2'} \hat{\mathcal{D}}(R) \ket{j_1, j_2; 0, 0} = \sum_{l',m'} C_{m_1'm_2'}^{l'm'} C_{00}^{l'0} \sqrt{\frac{4\pi}{2l'+1}} (Y_{l'}^{m'})^\star (\theta,\phi) \end{aligned} \]
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
\[ \begin{aligned} Y_{l_1}^{m_1} (\theta, \phi) Y_{l_2}^{m_2} (\theta, \phi) = \frac{\sqrt{(2l_1+1)(2l_2+1)}}{4\pi} \\ \times \sum_{l',m'} \sprod{l_1 l_2; m_1 m_2}{l_1 l_2; l' m'} \sprod{l_1 l_2; 0 0}{l_1 l_2; l' 0} \sqrt{\frac{4\pi}{2l'+1}} Y_{l'}^{m'} (\theta, \phi). \end{aligned} \]
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 \( (Y_lm)\star(\theta, \phi) \), and then integrate over the solid angle. This allows us to use the orthogonality of the spherical harmonics,
\[ \begin{aligned} \int d\Omega (Y_{l'}^{m'})^\star (\theta, \phi) Y_l^m(\theta, \phi) = \delta_{l,l'} \delta_{m,m'}. \end{aligned} \]
Using this identity the integral under the sum on the right vanishes unless \( m=m' \) and \( l=l' \), and thus:
\[ \begin{aligned} \int d\Omega (Y_l^m)^\star (\theta, \phi) Y_{l_1}^{m_1}(\theta, \phi) Y_{l_2}^{m_2}(\theta, \phi) = \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2l+1)}} \sprod{l_1 l_2; 0 0}{l_1 l_2; l 0} \sprod{l_1 l_2; m_1 m_2}{l_1 l_2; l m}. \end{aligned} \]
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 were stuck at the result
\[ \begin{aligned} \bra{n'l'm'} \hat{r}_q \ket{nlm} = \int_0^{\infty} dr\ r^3 R_{n'l'}^\star(r) R_{nl}(r) \sqrt{\frac{4\pi}{3}} \int d\Omega (Y_{l'}^{m'})^\star(\theta, \phi) Y_1^q (\theta, \phi) Y_l^m(\theta, \phi). \end{aligned} \]
Now we know that this three-Y integral is simply proportional to a Clebsch-Gordan coefficient,
\[ \begin{aligned} \int d\Omega (...) \sim \sprod{l1; mq}{l1;l'm'}. \end{aligned} \]
This gives us a pair of selection rules, due to the properties of the C-G coefficients: we must have
\[ \begin{aligned} |l-1| \leq l' \leq l+1, \\ m' = q+m. \end{aligned} \]
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 \( l=2 \), \( l'=1 \), out of 45 possible combinations only 9 survive the selection rule. 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.
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
\[ \begin{aligned} Y_l^m(\vec{n}) = \sprod{\vec{n}}{lm}. \end{aligned} \]
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:
\[ \begin{aligned} Y_l^m(\vec{n}') = \bra{\vec{n}} \hat{\mathcal{D}}(R)^{-1} \ket{lm} \\ = \sum_{l',m'} \sprod{\vec{n}}{l'm'} \bra{l'm'} \hat{\mathcal{D}}(R)^{-1} \ket{lm} \\ = \sum_{m'} Y_l^{m'}(\vec{n}) (\mathcal{D}_{mm'}^{(l)}(R))^\star. \end{aligned} \]
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
\[ \begin{aligned} \hat{\mathcal{D}}^\dagger(R) \hat{T}_q^{(k)} \hat{\mathcal{D}}(R) = \sum_{q=-k}^{k} D_{qq'}^{(k)}{}^\star(R) T_{q'}^{(k)}. \end{aligned} \]
The appeal of writing things in terms of spherical tensor operators is now clear; the appearance of the Wigner D-matrix in the transformation rule ensures that the rotation of a spherical tensor preserves the value of \( k \), so that the rotation is block-diagonal (in fact, irreducible) in terms of the spherical tensor decomposition.
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
\[ \begin{aligned} \mathcal{D}(R) = \left(1 - \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar} \right) \end{aligned} \]
then the spherical tensor \( \hat{T}_q^{(k)} \) must satisfy
\[ \begin{aligned} \left(1 + \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar} \right) \hat{T}_q^{(k)} \left(1 - \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar} \right) \\ = \sum_{q'=-k}^{k} \hat{T}_{q'}^{(k)} \bra{kq'} \left(1 + \frac{i (\hat{\vec{J}} \cdot \vec{n}) \epsilon}{\hbar} \right)\ket{kq}. \end{aligned} \]
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:
\[ \begin{aligned} [\hat{\vec{J}} \cdot \vec{n}, \hat{T}_q^{(k)}] = \sum_{q'} T_{q'}^{(k)} \bra{kq'} \hat{\vec{J}} \cdot \vec{n} \ket{kq}. \end{aligned} \]
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:
\[ \begin{aligned} [\hat{J}_z, \hat{T}_q^{(k)}] = \hbar q \hat{T}_q^{(k)}, \end{aligned} \]
while for \( x \) and \( y \), it's most convenient to write in terms of raising and lowering operators:
\[ \begin{aligned} [\hat{J}_{\pm}, \hat{T}_q^{(k)}] = \hbar \sqrt{(k \mp q)(k \pm q + 1)} T_{q \pm 1}^{(k)}. \end{aligned} \]
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
\[ \begin{aligned} \bra{\alpha', j', m'} [\hat{J}_z, \hat{T}_q^{(k)}] \ket{\alpha, j, m} = \hbar q \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m}. \end{aligned} \]
But since we now have \( \hat{J}_z \) eigenstates on both sides, we can get rid of the commutator, finding that
\[ \begin{aligned} [(m'-m - q) \hbar] \bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m} = 0 \end{aligned} \]
so that the matrix element vanishes unless
\[ \begin{aligned} m' = q + m. \end{aligned} \]
The matrix elements must also satisfy the triangular selection rule,
\[ \begin{aligned} |j-k| \leq j' \leq j+k. \end{aligned} \]
This isn't as easy to prove on its own, but it follows directly from the Wigner-Eckart theorem, which we will prove shortly. As we saw last time, for the electric dipole operator being able to write everything in terms of spherical harmonics gave us selection rules and massively simplified the result. Spherical tensors give us the power of selection rules for any physical system, not just those which can be expressed using spherical harmonics.
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 I claimed last time was a scalar, i.e. a rank-0 spherical tensor. The commutator with \( \hat{J}_z \) is
\[ \begin{aligned} [\hat{J}_z, \hat{\vec{U}} \cdot \hat{\vec{V}}] = \sum_i [\hat{J}_z, \hat{U}_i \hat{V}_i] \\ = \sum_i [\hat{J}_z, \hat{U}_i] \hat{V}_i + \hat{U}_i [\hat{J}_z, \hat{V}_i] \\ = \sum_{i,k} \left[ (-i \hbar \epsilon_{izk} \hat{U}_k) \hat{V}_i + \hat{U}_i (-i \hbar \epsilon_{izk} \hat{V}_k) \right] \end{aligned} \]
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
\[ \begin{aligned} -i\hbar \left[\hat{U}_x \hat{V}_y - \hat{U}_y \hat{V}_x + \hat{U}_y \hat{V}_x - \hat{U}_x \hat{V}_y \right] \end{aligned} \]
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
\[ \begin{aligned} [\hat{J}_j, \hat{\vec{U}} \cdot \hat{\vec{V}}] = -i\hbar \sum_{i,k} \left[ \epsilon_{ijk} \hat{U}_k \hat{V}_i + \epsilon_{ijk} \hat{U}_i \hat{V}_k \right] \end{aligned} \]
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
\[ \begin{aligned} [\hat{J}_j, \hat{\vec{U}} \cdot \hat{\vec{V}}] = -i \hbar \sum_{i,k} \epsilon_{ijk} \left[ - \hat{U}_i \hat{V}_k + \hat{U}_i \hat{V}_k \right] = 0. \end{aligned} \]
The algebra obviously starts to become rather imposing as we go to higher orders, particularly once we're looking for commutators that don't vanish anymore, but in principle we can carry out the decomposition of an object like our dyadic \( \hat{U}_i \hat{V}_j \) just by way of the commutation relations. Fortunately, there are more practical approaches we can use, making use of the Clebsch-Gordan coefficients.
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 - next time.