Appendix C — Algebraic identities for spherical harmonics

These notes in various places contain 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. We’ve gone back and forth between these descriptions; in this appendix, we’ll exploit the connection to find some useful identities for the spherical harmonics.

Translation between the spherical harmonics and the angular momentum eigenstates is readily done with 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.

C.0.1 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!

C.0.2 Recovering the vector selection rules

Finally, let’s use this approach to reiterate some key results from the main text on selection rules for vector transition matrix elements. Explicitly, we can look at the electric dipole transition operator, where we had used the spherical basis to write the necessary matrix element 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. We also know that the C-G coefficients vanish unless the two orbital quantum numbers are within 1 of each other, i.e. we can also immediately read off the second selection rule, |l-1| \leq l' \leq l+1. The selection rules save us from the trouble of even trying to compute most of the angular integrals; the remaining non-vanishing ones can be easily found algebraically, using a C-G table.