Appendix B — Compact review of fall semester

This may or may not actually stay in these written notes; but for now, it’s a convenient location for this. Some review of previous material was requested by my students prior to the fall 2024 final exam. Maybe this will become a FAQ eventually!

B.1 Wigner-Eckart, revisited

Here is the Wigner-Eckart theorem again:

\bra{\alpha', j', m'} \hat{T}_q^{(k)} \ket{\alpha, j, m} = \left\langle jk; mq | jk; j' m' \right\rangle \frac{\bra{\alpha' j'} | \hat{T}^{(k)} | \ket{\alpha j}}{\sqrt{2j'+1}},

This formula can be puzzling, on its own - the notation is pretty dense. What is the “reduced matrix element” on the right (the thing with double bars), and what is the purpose of this theorem?

Here are a couple of key points to remember.

  1. Don’t worry about the reduced matrix element. The reduced ME is not really something that we usually care about, on its own. The real physics content of the W-E theorem is rotational symmetry, specifically the fact that if we change the values of m, m', q on the left-hand side, we can find the answer easily using Clebsch-Gordan coefficients.

We saw two other results that are basically just rewriting the W-E theorem: the replacement theorem, and the projection theorem. But there’s another useful rewriting that I didn’t do before, which doesn’t have a name (and the notation gets a little dense, but bear with me): \bra{\alpha', j', m'_2} \hat{T}_{q_2}^{(k)} \ket{\alpha, j, m_2} = \frac{\left\langle jk; m_2 q_2 | jk; j' m'_2 \right\rangle}{\left\langle jk; m_1 q_1 | jk; j' m'_1 \right\rangle} \bra{\alpha', j', m'_1} \hat{T}_{q_1}^{(k)} \ket{\alpha, j, m_1}.

In words: all matrix elements that only differ by changing the magnetic quantum numbers m, m', q are related to each other by Clebsch-Gordan coefficients.

  1. The role of \hat{T}_q^{(k)} is to ensure angular momentum is conserved. As I’ve stressed, W-E is a key result that exploits rotational symmetry. But as we’ve also observed in the past, a good symmetry always corresponds to a conserved quantity, and good rotational symmetry is equivalent to conservation of angular momentum. One of the simple consequences of this is that quantum states with different angular momenta have zero overlap, for example (writing \ket{jm}): \left\langle 10 | 00 \right\rangle = 0.

This is pretty trivial - all of the \ket{jm} states are part of an orthonormal basis, so of course their inner products are all zero. But we could have an operator \hat{O} that gives a valid transition matrix element between them: \bra{10} \hat{O} \ket{00} \neq 0. This is a transition because one way you can interpret it is that the operator \hat{O} acts on the state \ket{00}, and the result has some non-zero component of \ket{10}. As a result, we have a non-zero probability to find the system in state \ket{10} once \hat{O} acts on the initial state \ket{00}.

Now, the W-E theorem tells us that there are specific operators that are allowed to give such a transition. Let’s see how that works: switching to spherical basis, we have \bra{10} \hat{T}_q^{(k)} \ket{00} = \left\langle 0k;0q | 0k;10 \right\rangle (...) where I’m just writing the Clebsch-Gordan because we don’t care about the rest - the question is, is this C-G coefficient automatically zero or not? Unpacking the notation, this C-G appears in the problem of adding angular momentum 0 to angular momentum k, to get final angular momentum 1. We remember the two selection rules are: m' = q + m \\ |j-k| \leq j' \leq j+k or in this specific case, q = 0 \\ j' = k.

In other words, the only kind of operator that can give a non-zero transition matrix element for these states is a spherical tensor of the form \hat{T}_0^{(1)}. (This could be the position operator in which case \hat{T}_0^{(1)} = \hat{z}, but it could be another kind of vector operator too.)

That’s the mathematical content, now let’s think about what this means in terms of the physics. The initial state \ket{00} has no angular momentum at all, but the final state \ket{10} has one unit of angular momentum. So conservation of angular momentum demands that the operator in the transition matrix element must carry one unit of angular momentum! This is a very general statement, in fact; let’s think about a totally general transition mediated by a vector operator, \bra{j'm'} \hat{T}_q^{(1)} \ket{jm} \sim \left\langle j1;mq | j1;j'm' \right\rangle. The triangle selection rule tells us simply that |j-1| \leq j' \leq j+1 - vector operators carry one unit of angular momentum, so they can only mediate transitions between initial and final states that differ by (no more than) one unit of angular momentum. More generally, spherical operators carry k units of angular momentum, allowing larger jumps in j, with the W-E theorem always telling us (through the Clebsch-Gordans) what is allowed so that angular momentum is always conserved.

B.1.1 Multipole transitions and hydrogen, revisited

I already have another appendix on hydrogen, but here let’s quickly review the specific context of transition matrix elements, following on from our Wigner-Eckart discussion. As a brief reminder of the minimum info we need for this discussion, hydrogen (and hydrogen-like atom) energy levels carry the quantum numbers \ket{nlm}, neglecting spin effects, where n is the “principal quantum number”, labeling the energy level E_n (up to small corrections that we’ll mostly talk about next semester.) The angular momentum is also related to the principal quantum number by the condition l < n, which means that the ground state n=1 has no angular momentum (l=0 only), n=2 can have l=0 or 1, and so on.

As above, all of these states form an orthonormal basis so they don’t normally overlap: \left\langle n'l'm' | nlm \right\rangle = 0 unless all quantum numbers match. Of course, in the real world transitions do happen, with emission of light. We haven’t developed the full machinery to describe these transitions yet, but we saw before that one possibility is the presence of an external electric field which can induce the transitions, which can be described using a multipole expansion, \bra{\vec{r}'} \hat{Q}_m^{(l)} \ket{\vec{r}} = q \sqrt{\frac{4\pi}{2l+1}} r^l Y_l^m(\theta, \phi) \delta(\vec{r} - \vec{r}'), (written in spherical tensor basis), the simplest of which is the dipole operator \vec{d} = q\vec{r}, where q would be the charge of the electron for the case of describing transitions in hydrogen.

One possibility is that there is a response of the hydrogen atom to the external field that doesn’t cause the state to change, in which case the “transition” ME has the same state on both sides: \bra{n,l_1,m_1} \hat{Q}_m^{(l)} \ket{n,l_1,m_1}. We discussed this in detail in the lecture chapter on Wigner-Eckart, but I’ll remind you of the key features. First, from the W-E theorem and the triangle selection rule, we have that 0 \leq l \leq 2l_1, so no multipole moments with l > 2l_1 are allowed; for example, even the dipole moment l=1 vanishes for hydrogen in the ground state, n = l_1 = 0.

The other more subtle observation is that some extra matrix elements vanish due to parity; I gave a fancy argument using symmetry properties of C-G coefficients in the notes, but there is also a simpler argument. For a “moment” (one of these MEs that has the same state on both sides), the action of parity on the state vectors doesn’t matter since any factor of (-1) that could appear is squared; so, only the action on the operator itself matters. For the dipole operator \vec{d}, like any other vector it is odd under parity. But this means that \bra{n,l_1,m_1} \hat{\vec{d}} \ket{n,l_1,m_1} \rightarrow \bra{n,l_1,m_1} (-\hat{\vec{d}}) \ket{n,l_1,m_1}. So if parity is a good symmetry, all electric dipole moments vanish for the eigenstates \ket{nlm} of hydrogen. The same is true for any odd l (e.g. the octupole moment Q_m^{(3)}.) In general, parity acts as \hat{P} \ket{nlm} = (-1)^l \ket{nlm} \\ \hat{P}^\dagger \hat{T}_q^{(k)} \hat{P} = (-1)^{k+\pi_T} \hat{T}_q^{(k)} where \pi_T = \{0,1\} is the intrinsic parity of the operator, which depends on what operator it is. For example, the position operator \hat{\vec{r}} has positive intrinsic parity \pi = 0, so it picks up a minus sign under parity (since it is a spherical tensor with k=1.)

This lets us get to the question of what transitions can be mediated by the dipole term of an external electric field: now we have from W-E \bra{n_2,l_2,m_1} q\hat{\vec{r}} \ket{n_1,l_1,m_1} \sim \left\langle l_1,1;m_1,q | l_1,1;l_2,m_2 \right\rangle which, from the triangle inequality, tells us that |l_1 - 1| \leq l_2 \leq l_1 + 1. What about parity? Under a parity reflection, we know that \vec{r} gives a minus sign, so the overall matrix element goes as \langle ... \rangle \rightarrow (-1)^{l_1+l_2+1} \langle ... \rangle. This vanishes if it goes to minus itself, which means l_1 + l_2 has to be odd. So the final result is that l_2 = l_1 \pm 1 (matching what we just argued above, that the electric dipole moment is zero, so l has to change with a dipole operator in the middle!)

B.2 Other equations, briefly

The above section received multiple requests for review, but here I include very brief discussion on a few other equations/concepts that I was asked to highlight.

  • Interaction picture: I was asked about this, but I’ll just reassure you here that this won’t appear on an exam. It’s an important concept to know about, and will be even more useful with perturbation theory next semester, but it’s complex enough (with two types of time evolution at once) that it’s not really practical to test on a sit-down exam.

  • Baker-Hausdorff lemma: This is a useful little formula that I presented, which I’ll repeat here: if \hat{A} and \hat{G} are Hermitian operators, then e^{i\hat{G} \lambda} \hat{A} e^{-i\hat{G} \lambda} = \hat{A} + i \lambda [\hat{G}, \hat{A}] - \frac{\lambda^2}{2} [\hat{G}, [\hat{G}, \hat{A}] ] + ... \\ \ + \left( \frac{i^n \lambda^n}{n!} \right) [\hat{G}, [\hat{G}, [\hat{G}, ...[\hat{G}, \hat{A}] ] ]...] + ...

This is sort of a corollary of the Baker-Campbell-Hausdorff formula, which states that if \hat{X} and \hat{Y} are two operators, the product of their exponentials is e^{\hat{X}} e^{\hat{Y}} = e^{\hat{Z}}, where \hat{Z} = \hat{X} + \hat{Y} + \frac{1}{2} [\hat{X}, \hat{Y}] + \frac{1}{12} [\hat{X}, [\hat{X}, \hat{Y}]] - \frac{1}{12} [\hat{Y}, [\hat{X}, \hat{Y}]] + ...

For ordinary numbers that commute with each other, this is just the usual formula that e^{x} e^{y} = e^{x+y}, but when we have operators that don’t commute we end up with an infinite number of extra terms containing commutators. (To see why this is happening, try series expanding e^{\hat{X}} e^{\hat{Y}} and then e^{\hat{X} + \hat{Y}}, and you’ll quickly notice that the terms match in contents but not in the order of the operators.)

These are cumbersome infinite series that are useful for certain formal applications, but otherwise are only really used in two circumstances:

  1. The series truncates itself: for example, if the operators are Pauli matrices, taking multiple commutators will quickly give back the identity matrix, and the infinite series can collapse. We saw examples of this when talking about time evolution in two-state systems; another example was in the SHO, see chapter 7 where the Baker-Hausdorff lemma initially appeared.

  2. We are studying an approximation, and there is a small parameter that lets us truncate the series. In the Baker-Hausdorff example, if \lambda is small enough we can just keep the first commutator and throw the rest out to get an answer up to \mathcal{O}(\lambda^2).

For most purposes (especially on an exam), it may be worth remembering the first term of the formula at most; often one can just work with power series directly instead of using these formulas (like we did in the two-state system.)

  • Probability currents: The principle of conservation of probability is encoded in the continuity equation, \frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0, where \rho = |\psi|^2 is the local probability density and the probability current \vec{j} is \vec{j} = \frac{\hbar}{m} {\rm Im}(\psi^\star \nabla \psi). As we discussed, the latter quantity is sensitive to the phase of the wavefunction. For simple things like plane waves, the probability current is just zero.

This is a useful formal result: conservation of probability is important (it’s roughly the statement that nothing is leaving or entering our physical system.) In practice, the main use that we saw was that imposing this equation can give you a quick way to match across boundaries in wave mechanics problems.

  • Brillouin zone: This came up in the context of studying finite translation symmetry. I’m out of time to cover it in detail here, but the short version is that if we have a lattice of points, there is a maximum possible momentum that comes from study of translation symmetry, leading to the condition (for a lattice with spacing a) of -\frac{\pi}{a} \leq k \leq \frac{\pi}{a}.

This is the Brillouin zone: the space of all possible momentum values allowed in the system. The cutoff on momentum leads to a cutoff on allowed energies as well.

  • Special functions in wavefunction solutions: In particular, the ones that show up in three dimensions. Here’s a lightning-quick review:

For a spherically symmetric potential, the solution breaks apart into spherical harmonics times radial functions, \psi_{Elm}(r) = R_{El}(r) Y_l^m(\theta, \phi).

The free solution (zero or constant potential) is given in terms of the dimensionless \rho = kr, where k is the usual wave number, \hbar k = \sqrt{2m(E-V_0)}. The solutions are spherical Bessel functions j_l(\rho) and spherical Neumann functions n_l(\rho). These functions behave as powers of \rho at short distances, j_l(\rho) \sim \rho^l, \\ n_l(\rho) \sim \rho^{-(l+1)}. Because of the divergent behavior, the n_l(\rho) (for any l) are not allowed as part of the solution for any region that includes the origin, r=0.

At large distances \rho \rightarrow \infty, these functions behave like modified \sin and \cos, with oscillating behavior. Continuing the analogy, we can construct the spherical equivalent of complex exponential functions by taking the linear combinations h_l^{(1)} = j_l(\rho) + in_l(\rho), \\ h_l^{(2)} = j_l(\rho) - in_l(\rho), which are the spherical Hankel functions. Asymptotically, up to some extra spherical factors, they behave roughly as h_l^{(1)} \sim e^{ikr}, \\ h_l^{(2)} \sim e^{-ikr}. If we have a bound-state solution, then k at large distances is imaginary (k \rightarrow i\kappa), which leads to real-exponential behavior h_l^{(1)} \sim e^{-\kappa r} and h_l^{(2)} \sim e^{+\kappa r} - for such solutions, clearly only the Hankel functions of the first kind are allowed (the h_l^{(2)} blow up at infinity.)