Touching base(s) with Angular Momentum and Spin¶

The first problem we'll attempt to solve is that of energies of electrons in presence of magnetic fields. It's not necessarily the most general or interesting problem, but the idea of measuring (changes of) energy states of things is very popular in physics. To do that, we should talk a bit about angular momentum in quantum mechanics. We can start with classical mechanics first, where angular momentum is simply defined as

$$ \vec{L} = \vec{r} \times \vec{p}. $$

If we know the momentum and position of an object relative to some reference point, it's easy to know it's angular momentum. Just to get classical physics out of the way, I'll get a little bit ahead of myself and also point out that angular momentum can be shown to be related to magnetic moment $\vec{m}$ which describes the strength of a magnetic dipole. In classical physics the relationship between these two is

$$ \vec{m} = \gamma \vec{L}, $$

where $\gamma$ is the gyromagnetic ratio. If you want a hand-wavy explanation for why are these two related, consider how does a magnetic dipole get produced: From a circulating current. So a massive charge that flies around in circles will have angular momentum because...ehm...it has mass and flies around in circles, and it also produces a magnetic dipole because the circling around of charge forms a current loop.

However, we want to deal with quantum mechanics, so we need operators that will act on a wave function (or a state).

What follows is not that hard to derive (it's usually done in the first quantum mechanics class in undergrad, so you don't need more than first-year college math) but it's tedious, especially if you're allergic to spherical coordinates. To get started here, we need to know only a few algebraic properties of the operators which I will spell out in the next few paragraphs. If this is new to you, your two options are to either take my word for it (not recommended), or do a quick Google search for angular momentum in quantum mechanics (first good hit for me was this) and at least skim through what you find. I'll wait.

Done? Good. So, the angular momentum operator (in position representation) is

$$ \hat{\mathbf{L}} = -i \hbar (\vec{r} \times

abla), $$

where $

abla$ is the nabla operator. In this case (we represent the wavefunctions in real-space), the momentum operator is $-i\hbar

abla$ (position derivative of the wavefunction with some pre-factors) and the position operator is just multiplication by the position, so the operator $\hat{\mathbf{L}}$ follows relatively closely it's classical counterpart. But that's where the similarities end. In classical mechanics, nothing is stopping you from knowing all the components of the angular momentum $\vec{L} = \left( L_x, L_y, L_z \right)$ but in quantum mechanics, the uncertainty principle poops on our parade.

It can be shown that only commuting properties can be measured at the same time with arbitrarily small uncertainty. If we calculate the commutator $\left[ \hat{X}, \hat{Y} \right] = \hat{X}\hat{Y} - \hat{Y}\hat{X}$ of the angular momentum component operators, we'll see that individual components don't fall into this category:

$$ \left[ \hat{L}_l, \hat{L}_m \right] =i\hbar \sum_n \varepsilon_{lmn}\hat{L}_n, $$

where $l,m,n = \{x,y,z\}$ and $\varepsilon_{lmn}$ is the Levi-Civita symbol (proof is left as an excercise to the reader). This means that you cannot measure more than one component at a time (or rather, you can but it will be subject to quantum-mechanical uncertainty). Imagine it as the angular momentum vector rotating along an axis, drawing out a cone surface while doing so. You can determine the cone side length (the magnitude) and the cone angle/height (the projection along that axis). The uncertainty principle forbids you to know where on the cone surface does the angular momentum lie.

The other important ingredient to what follows is the magnitude of angular momentum $\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$, which fortunately does commute with the components:

$$\left[\hat{L}^2, \hat{L}_x \right] = \left[\hat{L}^2, \hat{L}_y \right] = \left[\hat{L}^2, \hat{L}_z \right] = 0.$$

Proof of this one is obvious if you do the commutation relations for individual components. As one would expect from quantum mechanics, the angular momentum is quantized and has corresponding quantum numbers for magnitude $l \in \{ 0, \frac{1}{2}, 1, \frac{3}{2}, 2,\ldots \}$ and projection along one axis (conventionally, the z-axis) $m \in \{ -l, -l+1,\ldots,l-1, l \}$.