Angular Momentum
We review the basics of angular momentum and explain how quantum mechanics restricts its allowed, measurable values in order to retain its finite resolution of information it contains.
Quantum Mechanics so Far
Photons
For a while now we've been discussing how Planck's constant $h$ sets a finite resolution on the information carrying capacity of space and time. In some sense, h represents an irreducible or atomic unit of information.
The first major example of this was photons: a quantization of the electromagnetic field into particles whose energy came in discrete units. Depending on the wavelength of light $\lambda$, this quantum of energy was proportional to $h$:
$$\mathfrak{e}(\lambda) =\frac{hc}{\lambda}.$$
Light waves only exist in integer multiples of photons. In particular, a light wave with $n$ such photons had total energy:
$$E(n,\lambda) = n\frac{hc}{\lambda}.$$
Atomic Electrons
The next major example we saw was provided by the electrons bound to an atom. For the case of hydrogen, we saw that their energy was also dependent upon a positive, whole number. The energy of individual states in a hydrogen atom is given by
\begin{equation}\label{En}E(n) = \frac{13.6\,\mathrm{eV}}{n^{2}}.\end{equation}
The base energy level of 13.6 eV depends, of course, on $h$, albeit in a complicated manner.
Today we give a third example of how Quantum Mechanics imposes integral constraints on physical systems: angular momentum.
Angular Momentum
While $hc$ gave us an easy to remember relationship between energy and length,
$$hc = 1240\,\mathrm{eV}\cdot\mathrm{nm},$$
we now have enough exposure to $h$ to warrant looking at it's value by itself. That value is:
$$h = 6.62607015 \times 10^{-34} \, \mathrm{J}\cdot\mathrm{s}.$$
$h$, in other words, has units of energy times time. $h$, as it turns out, has precisely the same units of angular momentum:
$$\overrightarrow{L} = \overrightarrow{r}\times\overrightarrow{p}.$$
Like ordinary momentum, angular momentum is conserved physical observable that can, amongst other things, quantify how fast something is spinning or revolving around something else. The magnitude of $\overrightarrow{L}$ is given by
$$|\overrightarrow{L}| = rp\sin\phi,$$
where $\phi$ represents the angle between the direction of a given particle's position and its momentum. In other words, how quickly it's changing direction as a function of time.
One can study classical angular momentum for months on end - there is an enormous amount of things to say - but for our purposes here, we can think of it linearly: more rotation means higher angular momentum.
There are two kinds of rotation that we'll concern ourselves with at the atomic scale. Particles can spin - that is rotate about themselves, or they can orbit some other particle. Today we'll discuss the latter.
Electrons in large atoms can pick up some rotational angular momentum, and as it turns out, nature constrains the amount of angular momentum they're allowed to have.
Orbital Angular Momentum is Quantized
The amount of angular momentum a particle like an electron is allowed to have is quantized in units of $h$. More precisely, the allowed values of orbital angular momentum are:
\begin{equation}\label{ell}\ell(n) = \frac{nh}{2\pi}.\end{equation}
This may be somewhat intuitive. $h$ represents the smallest unit of angular momentum - a finite resolution of rotation - and nature allows $h$ units of angular momentum per rotation through $2\pi$ radians. $n$ of course is any whole number, which by convention we can choose to be positive.
Thus an electron orbiting a nucleus in an atom is afforded only a discrete number of possible rotational speeds:
$$\ell:\quad \frac{h}{2\pi},\;\frac{2h}{2\pi},\;\frac{3h}{2\pi},\;\frac{4h}{2\pi},\;\cdots$$
Chemists have a notation for these values of angular momentum, they call them orbitals, and these orbits have labels
$$\mathrm{orbital:}\quad s,\; p, \; d,\; f,\; \cdots$$
respectively.
Indeed, the periodic table can be organized into a grid, where the rows correspond to different energy levels as in \eqref{En}, and the columns correspond to different values of angular momentum as in \eqref{ell}.
Why angular momentum is quantized by \eqref{ell} is a question we will eventually get to, but there's an important consequence of this quantization that is worth pointing out.
Observing Angular Momentum
Angular momentum is a vector quantity, which means it has a magnitude and a direction. So far we've only discussed its total magnitude$^{1}$, which is restricted to the values in \eqref{ell}.
Actually observing $\overrightarrow{L}$ means measuring both is magnitude and direction. Typically this is one by measuring the projection of $\overrightarrow{L}$ along some direction using a detector:
$$\overrightarrow{L}\cdot\widehat{n} = |\overrightarrow{L}|\cos\theta.$$
This is very similar to measuring photon polarization, although the model to keep in mind is an electron orbiting a nucleus.
The trouble here is that arbitrary $\theta$ represents an arbitrarily small fraction of the total angular momentum. This is not allowed by Quantum Mechanics$^{2}$. Even when projected along a given direction, \eqref{ell} are the only possible observations of angular momentum that can be made.
Quantum Mechanics limits angular resolution
This has a very curious interpretation in terms of the angle $\theta$. By restricting the allowed, observable values of angular momentum, Quantum Mechanics has imposed a finite resolution on the possible observations of $\theta$!
For an electron with total angular momentum $\ell(n_{\sf max})$, Quantum Mechanics restricts the allowed values of $\theta$, so that
$$\frac{nh}{2\pi} = \frac{n_{\sf max}h}{2\pi}\cos\theta.$$
This simplifies to
\begin{equation}\label{angles}\cos\theta = \frac{n}{n_{\sf max}},\end{equation}
or
$$\theta = \arccos \left(\frac{n}{n_{\sf max}}\right),$$
where $n$ must be an integer such that
$$|n|\leq n_{\sf max}.$$
For very large values of $n_{\sf max}$, we have very large, total angular momentum. This also means that $\theta$ can be observed with very high precision. Examples of large angular momenta include the orbit of the Moon about the Earth, a carousel or even the swing of a baseball bat.
For smaller values of $n_{\sf max}$ - such as those you'd find in a typical atom - the resolution on $\theta$ is fairly restricted.
Observable Angles
Let's do some examples familiar from chemistry.
For the $s$-orbitals$^{3}$, $\ell = 0$, so there's nothing to say.
For the $p$-orbitals, $n_{\sf max} = 1$, so \eqref{angles} tells us that $n = 1,0,-1$, or
$$\theta_{p} = 0^{\circ}, \; 90^{\circ},\; 180^{\circ}.$$
For the $d$-orbitals, $n_{\sf max} = 2$, so \eqref{angles} tells us that $n = 2,1,0,-1,-2$, or
$$\theta_{d} = 0^{\circ}, \;60^{\circ}\; 90^{\circ},\; 120^{\circ},\;180^{\circ}.$$
Note that there are two electrons permitted per energy level in a given atom, so each orbital has double the allowed configurations. Hence Neon can have the electron configuration:
$$[Ne] = 1s^{2}2s^{2}2p^{6}.$$
Summary
Not only does Quantum Mechanics constrain the angular momentum of objects like electrons to values
$$\ell(n) = \frac{nh}{2\pi},$$
it also constrains any and all observations projected along any direction to these values as well.
We've interpreted this as a finite resolving power we have one measuring the angle at which a particle is spinning. What if the ``actual'' direction of that particle, the ``actual'' angle, is not represented as one that satisfies \eqref{angles}?
Then just like the ``odd photon'' left out in the polarization example, nature merely answers these questions using probability and statistics. As with photon polarization, the ``actual'' angle will be represented in the average measurement, although the individual measurements are confined to those of the form $\ell(n)$.
Exercises
Exercise 1 : In a Hydrogen atom, an electron can be excited to orbit around the proton in a $p$-orbital, giving it one unit of $\frac{h}{2\pi}$. Therefore we can observe the electron having angular momentum values:
$$\frac{h}{2\pi}\quad\mathrm{or}\quad 0\quad \mathrm{or}\quad -\frac{h}{2\pi}.$$
In particular we can observe it to not be rotating. Despite this fact, each of these three configurations has precisely the same energy. Explain how this could be.
Exercise 2 : Carbon Dioxide, CO$_{2}$, is a linear molecule in the sense that both oxygens are double bonded to the carbon, yielding in a effect a molecule whose physical arrangement is akin to that of a straight rod.
To a first approximation, the energy of a rotating CO$_{2}$ molecule depends on how fast it's rotating.
$$E(\ell) = E_{0} \ell(\ell+1),$$
Where $E_{0}$ is
$$E_{0} \approx 5 \times 10^{-5}\; \mathrm{eV}.$$
As with the electron energy levels in hydrogen, rotational modes of molecules can be excited or relaxed by the absorption or emission of photons, respectively. Based on these energy levels, what wavelengths of photons would you expect to see coming from a container full of pure CO$_{2}$?
What are the wavelengths associated to the transitions you computed above? Where are they on the spectrum of electromagnetic radiation?
Exercise 3 : While there are some complications, electrons in atoms follow the same basic structure as the hydrogen atom we saw last time. Rows of the periodic table correspond to the maximum energy level of the experienced by an electron:
$$E(n) = \frac{E_{0}}{n^{2}}.$$
Columns in the periodic table correspond to the maximum angular momentum of those electrons:
$$\ell(m)= \frac{mh}{2\pi}.$$
As chemistry permits 2 electrons per orbital, go through the periodic table and assign element a value for $n$ and $m$. Note where $s$, $p$, $d$ and $f$ orbitals live on the periodic table. Comment on the relationship between $n$ and $m$.
$^{1}$ : When we study the origin of this quantization, we'll learn that the observable quantity for the magnitude of $\overrightarrow{L}$ is actually the square of this vector, $L^{2} = \overrightarrow{L}\cdot\overrightarrow{L}$.
$^{2}$ : It's like measuring an arbitrarily small fraction of a photon with a polarization filter, it's just not well-defined!
$^{3}$ : That that electrons can be in orbit around a proton with zero angular momentum is not physically sensible, classically. This is something that only Quantum Mechanics can provide.
