Last time, we ended studying orbital angular momentum and gave some formulas in coordinate space for the operators \( \hat{L}{x,y} \). These were a little messy, and in fact switching to the ladder operators \( L{\pm} = L_x \pm i L_y \) gives us the slightly nicer formula
\[ \begin{aligned} \bra{x,y,z} \hat{L}_{\pm} \ket{\psi} = -i \hbar e^{\pm i \phi} \left( \pm i \frac{\partial}{\partial \theta} - \cot \theta \frac{\partial}{\partial \phi} \right) \sprod{x,y,z}{\psi}. \end{aligned} \]
The last operator to consider is the squared total angular momentum, \( \hat{\vec{L}}^2 \). We can construct that from what we've already done:
\[ \begin{aligned} \hat{\vec{L}}{}^2 = \hat{L}_z^2 + \frac{1}{2} (\hat{L}_+ \hat{L}_- + \hat{L}_- \hat{L}_+), \end{aligned} \]
which with a bit of algebra gives
\[ \begin{aligned} \bra{x,y,z} \hat{\vec{L}}{}^2 \ket{\psi} = -\hbar^2 \left[ \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial}{\partial \theta} \right) \right] \sprod{x,y,z}{\psi}. \end{aligned} \]
This is precisely the angular part of the Laplacian operator \( \nabla^2 \) in spherical coordinates! In fact, if we expand out the full three-dimensional kinetic energy term in spherical coordinates, we find that
\[ \begin{aligned} \frac{\hat{\vec{p}}{}^2}{2m} = \frac{-\hbar^2}{2m} \nabla^2 = \frac{-\hbar^2}{2mr^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r}\right) + \frac{\hat{\vec{L}}{}^2}{2mr^2}. \end{aligned} \]
We could have started with the classical Hamiltonian in spherical coordinates and identified the angular momentum from there as well, leading to the same expression.
This is a good point to introduce the idea of a complete set of commuting observables (or CSCO.) The importance of a set of commuting observables, i.e. \( \hat{A}, \hat{B}, \hat{C}, ... \) such that
\[ \begin{aligned} [\hat{A}, \hat{B}] = [\hat{A}, \hat{C}] = [\hat{B}, \hat{C}] = 0, \end{aligned} \]
is that we can label eigenstates simultaneously with each of the eigenvalues \( \ket{a, b, c...} \). A complete set of commuting observables occurs when we can uniquely label a complete basis of our Hilbert space with the eigenvalues in our set. In other words, using the eigenkets of our CSCO operators, we can uniquely specify any state at all in our Hilbert space as a list of eigenvalues (known in this context as the quantum numbers of a given eigenstate.)
Another way to state the significance of this last point is that if we measure every observable in our CSCO, we expect the system to collapse to a single, unique state which is a simultaneous eigenvector of all of the observables \( \ket{a,b,c...} \). Thus, the CSCO represents the maximum set of things we can know simultaneously about our system. The ideal situation occurs when the Hamiltonian itself is part of our CSCO, since this makes the time evolution especially simple: any state in the CSCO just picks up a phase according to its energy.
It's important to note that the choice of a CSCO is, in general, not unique. As a trivial example, for the simple two-state system we can use \( \hat{S}_x \) to label eigenstates just as well as \( \hat{S}_z \). Moreover, we saw that we can construct the operator \( \hat{\vec{S}}^2 \) even in the two-state system, so \( {\hat{\vec{S}}^2, \hat{S_z}} \) could be a CSCO. However, we don't gain any new information from \( \hat{\vec{S}}^2 \) in this case, since it is proportional to the identity. An even more pathological example comes from the simple harmonic oscillator, where we can construct the projection operators
\[ \begin{aligned} P_n = \ket{n}\bra{n}. \end{aligned} \]
These give an infinite CSCO; we can label all of the energy eigenstates uniquely as a binary string \( \ket{00100...} \). This is obviously not efficient, but it does meet the mathematical definition of a CSCO. So there is no algorithm for determining a unique, best CSCO for a given problem; in practice, we look for the smallest CSCO we can find, and commuting with the Hamiltonian is a plus.
How do we know when we do not have a CSCO? The main requirement for such a set is that measuring all of the operators in it will result in a unique state, which means that there must be no degeneracy in the eigenvalues when we measure everything. Of course, an individual operator can still have degeneracy, as long as some other operator breaks the degeneracy, e.g. our set of unique eigenkets may contain the states \( \ket{a,b,c_1} \) and \( \ket{a,b,c_2} \), where \( c_1 \neq c_2 \). If we find two states with degenerate eigenvalues for every operator in our set, then we don't have a CSCO, and must find some new operator to add.
Now you should ask the question: how do we decide whether there is an unbroken degeneracy? Let's take the example of a spin-1/2 particle in a harmonic oscillator potential. We can describe the state of the particle by its energy and spin eigenvalues \( \ket{n,\pm} \). But suppose that we don't know that we're dealing with a spin-1/2 particle, and only know about the energy label \( \ket{n} \).
To describe this system, I need to introduce the direct product, denoted by the symbol \( \otimes \), which can combine two kets together: if \( \ket{\alpha} \) is defined in Hilbert space \( \mathcal{H}_1 \), and \( \ket{\beta} \) is defined in Hilbert space \( \mathcal{H}_2 \), then their direct product \( \ket{\alpha} \otimes \ket{\beta} \) is a new ket, \( \ket{\gamma} \), which is defined on the product Hilbert space \( \mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2 \). For an electron moving through space, we can consider its overall state to be given by the direct product of position and spin states,
\[ \begin{aligned} \ket{\vec{x}, \pm} = \ket{\vec{x}} \otimes \ket{\pm}. \end{aligned} \]
where if there's no risk of ambiguity, we often just write the product state as the combined ket on the left.
The product Hilbert space is spanned by all possible combined states coming from one space or the other. A two-dimensional example might help to clarify things somewhat. Suppose we have two spin-1/2 particles in two separate boxes \( A \) and \( B \), with no interactions. If the basis states for each particle are \( \ket{\pm}_A \) and \( \ket{\pm}_B \), then for the product state we can write the states \( \ket{\pm}_A \otimes \ket{\pm}_B \), four basis states in total; the dimension of a product Hilbert space is the product of the dimensions of the individual spaces.
Back to the electron in a harmonic oscillator, in which we're unaware of the electron spin. We prepare our system in an energy eigenstate, which will also have some unknown spin wavefunction:
\[ \begin{aligned} \ket{\psi} = (\frac{3}{5} \ket{0} + \frac{4}{5} \ket{1}) \otimes (\alpha \ket{+} + \beta \ket{-}) \\ = \frac{3}{5} \alpha \ket{0,+} + \frac{3}{5} \beta \ket{0,-} + \frac{4}{5} \alpha \ket{1,+} + \frac{4}{5} \beta \ket{1,-}. \end{aligned} \]
What is the probability of observing the ground state? Well, since we don't know about the spin in our experiment, we can only measure the overlap with the energy eigenstate, which means we are implicitly summing over the spin outcomes:
\[ \begin{aligned} |\sprod{0}{\psi}|^2 = |\sprod{0,+}{\psi}|^2 + |\sprod{0,-}{\psi}|^2 \\ = |\frac{3}{5} \alpha|^2 + |\frac{3}{5} \beta|^2 \\ = \frac{9}{25} (|\alpha|^2 + |\beta|^2) = \frac{9}{25} \end{aligned} \]
by normalization. This is the same as the probability of the simple harmonic oscillator without any spin! If the only observable we look at is the energy, then it's impossible to tell that we're missing anything. So the question of whether we have found a CSCO is in fact an experimental question. If we perform different experiments on a system, measure different observables, then we can try to see if we're missing any information, i.e. if a state we previously thought was a completely determined eigenstate splits into multiple possible outcomes.
To foreshadow a bit, a good example is given by the hydrogen atom, which is usually first studied as a CSCO with labels \( \ket{n,l,m} \), corresponding to the energy level and the angular momentum state of the orbital electron. Indeed, experimentally (and ignoring hyperfine transitions), the emission spectrum of hydrogen is well-described by states labelled by only those three quantum numbers. But if we put our atoms in a weak magnetic field, we observe a splitting of the \( \ket{n,l,m} \) spectral lines into several different energy levels! This tells us that our initial set of quantum numbers wasn't complete, and indeed we must add the spin of the orbital electron as an observable in order to recover a CSCO for hydrogen.
Now we're ready to tackle the Schrödinger equation in three dimensions. To make full use of rotational symmetry and angular momentum, we will restrict our attention to spherically symmetric potentials,
\[ \begin{aligned} V(\vec{r}) = V(r). \end{aligned} \]
Since \( \hat{L}^2 \) only depends on \( \theta \) and \( \phi \), this ensures that \( \hat{L}^2 \) commutes with the Hamiltonian. We also know from the commutation relations that any of the angular momentum components \( \hat{L}_i \) will commute with \( \hat{L}^2 \), and since they're all also functions of angles only, they commute with \( \hat{H} \) too. Thus, we can choose as a complete set of commuting observables
\[ \begin{aligned} \textrm{CSCO}: \{\hat{H}, \hat{L}{}^2, \hat{L}_z\}. \end{aligned} \]
Our eigenstates can thus be written \( \ket{E, l, m} \), with the usual eigenvalue equations applying for the angular momentum states,
\[ \begin{aligned} \hat{L}_z \ket{E,l,m} = \hbar m \ket{E,l,m} \\ \hat{L}{}^2 \ket{E,l,m} = \hbar^2 l(l+1) \ket{E,l,m}. \end{aligned} \]
Our wavefunctions can thus be written as \( \psi_{Elm}(r,\theta,\phi) \). Let's write out the time-independent Schrödinger equation satisfied by these states:
\[ \begin{aligned} -\frac{\hbar^2}{2m} \left[ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r} \right) - \frac{l(l+1)}{r^2} \right] \psi_{Elm}(r,\theta,\phi) = (E - V(r)) \psi_{Elm}(r,\theta,\phi). \end{aligned} \]
This is a separable differential equation; in fact, because we've used the fact that we're looking for eigenstates of \( \hat{L}^2 \), it's only a differential equation in \( r \). This means that the complete solution to the wavefunction is separable into radial and angular components:
\[ \begin{aligned} \psi_{Elm}(r,\theta,\phi) = R_{El} (r) Y_l^m (\theta, \phi), \end{aligned} \]
where the radial wavefunction can't depend on \( m \) since that eigenvalue doesn't appear in the equation above. The angular dependence is totally separate, and will be common to all solutions for a spherically symmetric potential.
Although they don't appear in the Schrödinger equation, the \( Y_l^m \) functions are still tightly constrained. Since they describe all of the angular dependence of the wavefunction, we can isolate them by writing
\[ \begin{aligned} \sprod{\theta, \phi}{l,m} = Y_l^m (\theta, \phi), \end{aligned} \]
where I'm ignoring the radial/energy label since it doesn't affect the angular functions at all. This lets us obtain constraints on the \( Y_l^m \) functions from the eigenvalue equations for angular momentum. First, we see that
\[ \begin{aligned} \bra{\theta,\phi} \hat{L}_z \ket{l,m} = m \hbar \sprod{\theta, \phi}{l,m}. \end{aligned} \]
As we found above, acting on the left with \( \hat{L}_z \) gives \( -i \hbar \partial / \partial \phi \), so we have
\[ \begin{aligned} -i\hbar \frac{\partial}{\partial \phi} Y_l^m (\theta,\phi) = m \hbar Y_l^m (\theta,\phi). \end{aligned} \]
Acting with \( \hat{L}^2 \) similarly gives a somewhat more complicated differential equation,
\[ \begin{aligned} \left[ \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial}{\partial \theta} \right) + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} + l(l+1) \right] Y_l^m(\theta, \phi) = 0. \end{aligned} \]
Finally, the fact that the \( Y_l^m \) are eigenfunctions gives them some nice properties, in particularly the orthonormality of the eigenstates leads to a completeness relation: since
\[ \begin{aligned} \sprod{l',m'}{l,m} = \delta_{l,l'} \delta_{m,m'}, \end{aligned} \]
we have the identity
\[ \begin{aligned} \int_0^{2\pi} d\phi \int_{-1}^1 d(\cos \theta)\ [Y_{l'}^{m'}]^\star (\theta,\phi) Y_l^m(\theta, \phi) = \delta_{l,l'} \delta_{m,m'}. \end{aligned} \]
We can also form a completeness sum over \( l \) and \( m \):
\[ \begin{aligned} \sum_{l=0}^\infty \sum_{m=-l}^l Y_l^m (\theta', \phi') [Y_l^m]^\star (\theta, \phi) = \delta(\phi - \phi') \delta(\cos \theta - \cos \theta'). \end{aligned} \]
Now let's solve for the harmonics themselves. Starting with the \( \hat{L}^2 \) equation alone, we notice that it is a separable differential equation once again: we may write
\[ \begin{aligned} Y_l^m (\theta,\phi) = T(\theta) P(\phi) \end{aligned} \]
and then obtain
\[ \begin{aligned} \frac{1}{T(\theta)}\left[ \sin \theta \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial}{\partial \theta} \right) + l(l+1) \sin^2 \theta \right] T(\theta) = - \frac{1}{P(\phi)} \frac{\partial^2}{\partial \phi^2} P(\phi). \end{aligned} \]
The left-hand side is a function only of \( \theta \), while the right-hand side depends just on \( \phi \), so we set them both equal to a constant, \( m^2 \). The \( \phi \) equation is easy:
\[ \begin{aligned} \left[ \frac{\partial^2}{\partial \phi^2} + m^2 \right] P(\phi) = 0 \\ \Rightarrow P(\phi) = e^{\pm i m \phi}. \end{aligned} \]
The sign ambiguity is fixed by the other equation we've been ignoring, namely the \( \hat{L}_z \) eigenvalue equation above, which requires \( -i\hbar \partial P(\phi) / \partial \phi = m \hbar P(\phi) \), and so
\[ \begin{aligned} P(\phi) = e^{im \phi}. \end{aligned} \]
As expected, only certain quantized values of \( m \) are allowed, because the wavefunction must be single-valued, i.e. \( P(\phi) = P(\phi + 2\pi) \). This requires \( m \) to be an integer.
Notice that we would have expected just from the commutation relations that \( m \) could have been an integer or a half-integer; we see now that the latter are not allowed for orbital angular momentum, since they would give us a wavefunction which is not single-valued. Only the intrinsic spin angular momentum carried by a particle may be half-integer valued. In particular, the spherical harmonics which we're about to derive are nonsensical for half-integer \( l \).
Now let's go back to the \( \theta \) equation. It will be helpful to let \( u = \cos \theta \), in which case
\[ \begin{aligned} \frac{\partial}{\partial u} = -\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}, \end{aligned} \]
leading to
\[ \begin{aligned} \left[ \frac{\partial}{\partial u} \left( (1-u^2) \frac{\partial}{\partial u} \right) + \left( l(l+1) - \frac{m^2}{1-u^2} \right) \right] T(u) = 0 \\ (1-u^2) \frac{\partial^2 T}{\partial u^2} - 2u \frac{\partial T}{\partial u} + \left[ l(l+1) - \frac{m^2}{1-u^2} \right] T(u) = 0. \end{aligned} \]
For the case \( m=0 \), this is the Legendre differential equation; its solutions are the Legendre polynomials \( P_n(u) \). A shorthand way to write the polynomials, which isn't too cumbersome for at least the first few, is Rodrigues's formula,
\[ \begin{aligned} P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \left[ (x^2 - 1)^n \right]. \end{aligned} \]
The first few Legendre polynomials are worth knowing:
\[ \begin{aligned} P_0(x) = 1 \\ P_1(x) = x \\ P_2(x) = \frac{1}{2} (3x^2 - 1) \\ P_3(x) = \frac{1}{2} (5x^3 - 3x) \end{aligned} \]
For \( m \neq 0 \), the equation above is known as the associated Legendre differential equation; its solutions are known as associated Legendre polynomials, but they're a bit less commonly encountered. As you should expect, for integer \( l \) solutions to this equation are only possible for \( |m| < l \), recovering our other limit from the commutation relations. If you're trying to look up solutions for non-zero \( m \) to do something practical, you'd be much better off trying to look for the complete spherical harmonics themselves.
We've so far ignored the overall normalization of the states, but that's fixed by the integral completeness relation between the \( Y_l^m \)'s. I'll skip those details, but the result is, for the \( m=0 \) states,
\[ \begin{aligned} Y_l^0(\theta, \phi) = \sqrt{\frac{2l+1}{4\pi}} P_l(\cos \theta). \end{aligned} \]
The formula for the general spherical harmonics can actually be written in a fairly compact way, but it's sufficiently impractical that I won't reproduce it here; look for Sakurai (3.6.37) if you're curious. These days, if you're doing something serious involving the spherical harmonics, you'll be better off with a table or a software package like Mathematica, instead of trying to derive them yourself from recursion relations. If you do use a table, you'll likely only find the spherical harmonics for \( m \geq 0 \) tabulated; this is because you can get the negative-\( m \) harmonics through the useful identity
\[ \begin{aligned} Y_l^{-m}(\theta, \phi) = (-1)^m [Y_l^m(\theta, \phi)]^\star. \end{aligned} \]
It is very helpful to know at least what the first few spherical harmonics look like explicitly, so let's write them down:
\[ \begin{aligned} Y_0^0(\theta, \phi) = \frac{1}{\sqrt{4\pi}} \\ Y_1^0(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos \theta \\ Y_1^1(\theta, \phi) = -\sqrt{\frac{3}{8\pi}} e^{i\phi} \sin \theta \\ Y_2^0(\theta, \phi) = \frac{1}{4}\sqrt{\frac{5}{\pi}} (3 \cos^2 \theta - 1) \\ Y_2^1(\theta, \phi) = -\frac{1}{2} \sqrt{\frac{15}{2\pi}} e^{i\phi} \sin \theta \cos \theta \\ Y_2^2(\theta, \phi) = \frac{1}{4} \sqrt{\frac{15}{2\pi}} e^{2i\phi} \sin^2 \theta \end{aligned} \]
You can explore some plots of the various spherical harmonics in space here; the \( m=0 \) plots you've probably seen before, maybe in a chemistry class as the electron orbitals in an atom.
There are "spectroscopic labels" corresponding to each of the \( l \), which are historical labels that you have likely heard before:
| l | spec. label |
|---|---|
| 0 | s |
| 1 | p |
| 2 | d |
| 3 | f |
| 4 | g |
| ... | ... |
Fun fact: the first four label names are short for "sharp", "principal", "diffuse", and "fundamental" (I've also seen "fine" for the last label, which is what I said in class, but "fundamental" seems best supported by historical literature.) These aren't particularly intuitive so they probably won't help you remember the notation, unfortunately. After "f" the labels just go in alphabetical order, and no longer stand for anything.
Why did I even bother introducing the Legendre polynomials? Often we will consider physical systems where \( l \) is a conserved quantum number but \( m \) is not, in which case it is very helpful to sum over product of spherical harmonics. These sums obey the addition formula,
\[ \begin{aligned} \sum_{m=-l}^{l} Y_l^m(\theta_1, \phi_1) [Y_l^m]^\star(\theta_2, \phi_2) = \frac{2l+1}{4\pi} P_l(\cos \omega) \end{aligned} \]
where
\[ \begin{aligned} \cos \omega = \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 \cos(\phi_1-\phi_2) \end{aligned} \]
If \( \phi_1 = \phi_2 \), then we can identify \( \omega = \theta_1 - \theta_2 \). We can think of the spherical harmonics themselves as a sort of generalization of trigonometric functions to two spherical angles instead of one polar angle; this is then a generalization of a familiar trig identity. It's also worth noting that the Legendre polynomials satisfy \( P_l(1) = 1 \) for any \( l \), which means that the sum of spherical harmonics with the same angles is particularly simple:
\[ \begin{aligned} \sum_{m=-l}^{l} |Y_l^m(\theta, \phi)|^2 = \frac{2l+1}{4\pi}. \end{aligned} \]
This is a useful identity for normalizing wavefunctions which contain spherical harmonics.
As it turns out, the spherical harmonics have distinctive properties under parity. What does a parity transformation look like in three dimensions? In terms of the rectilinear coordinates, everything just flips sign at once: \( x \rightarrow -x, y \rightarrow -y, z \rightarrow -z \). In spherical coordinates, we have instead
\[ \begin{aligned} \hat{P} \ket{r} = \ket{r} \\ \hat{P} \ket{\theta} = \ket{\pi - \theta} \\ \hat{P} \ket{\phi} = \ket{\phi + \pi}. \end{aligned} \]
The addition of \( \pi \) to \( \phi \) accounts for flipping the signs of \( x \) and \( y \), while the \( \theta \) transformation gives us the angle from the downwards-pointing \( z \)-axis. Notice that this last condition can also be written \( \cos \theta \rightarrow - \cos \theta \).
Now, remember that we can write the spherical harmonics as separate functions of \( \theta \) and \( \phi \):
\[ \begin{aligned} Y_l^m(\theta, \phi) = P_l^m (\cos \theta) e^{im \phi}, \end{aligned} \]
where \( P_l^m \) are associated Legendre polynomials. It turns out that these functions contain polynomials with either only even or only odd powers of \( \cos \theta \), times some factors of \( \sin \theta \) which are invariant. As a result, the parity transformation in \( \theta \) has a definite effect:
\[ \begin{aligned} \hat{P} P_l^m(\cos \theta) = P_l^m(-\cos \theta) = (-1)^{l-|m|} P_l^m(\cos \theta). \end{aligned} \]
As for the transformation of the azimuthal part, that's much easier to see:
\[ \begin{aligned} \hat{P} e^{im\phi} = e^{im(\phi + \pi)} = (-1)^m e^{im\phi}. \end{aligned} \]
Putting things together, we find
\[ \begin{aligned} \hat{P} Y_l^m (\theta, \phi) = (-1)^{l-|m|} (-1)^m Y_l^m(\theta, \phi) = (-1)^l Y_l^m(\theta, \phi) \end{aligned} \]
where the factor of \( (-1)^{m-|m|} \) either has an exponent of 0 or \( -2m \), both of which give \( 1 \) since \( m \) must be an integer here. Thus, spherical harmonics with \( l=0,2,4... \) are even under parity, while \( l=1,3,5... \) are odd, independent of \( m \). This will be a very useful property when we come to consider transition matrix elements; if we are studying a system with a Hamiltonian that respects parity, then many possible transitions can be immediately ruled out.
Once again, there is a close analogy to trigonometric functions here; the trigonometric functions \( \sin \theta \) and \( \cos \theta \) are also eigenfunctions of parity, with
\[ \begin{aligned} \hat{P} \sin \theta = -\sin \theta \\ \hat{P} \cos \theta = \cos \theta. \end{aligned} \]