Orientations of Quantised Angular Momentum

This method is a little more hand-wavy than some and doesn't really lead to a robust answer to the question for spin angular momentum specifically - but it is reasonable enough to get a feel for the ideas.

We want to understand why there are $2L+1$ different orientations (projections) of the quantised angular momentum $L$.

Consider a particle moving on the surface of a sphere. The quantum mechanics of this system are well documented in any textbook (e.g Atkins), and the solutions are the spherical harmonics. The energy of the particle is given by:

$$ E = \frac{l(l+1)\hbar^2}{2I} $$

Where $l$ is the angular momentum quantum number and $I$ is the moment of inertia of the particle. Classically, if we consider the same system, the energy of the particle is given by:

$$ E = \frac{L^2}{2I} $$

Where $L$ is the angular momentum of the particle. Comparison of the two equations leads to the identification of the angular momentum, $L$, as:

$$ L = \sqrt{l(l+1)}\hbar ~( = \vert \vec{L} \vert) $$

Physically, we can interpret this as the length of the angular momentum vector, $\vert\vec{L}\vert$, of the particle. This angular momentum vector could represent any angular momentum, though we have explicitly compared to the classical angular momentum $L$ (so we won't see the half-integer angular momentum fall beautifully out of the equations).

We can represent this angular momentum in Cartesian coordinates, with components along each coordinate axis:

$$ \vec{L} = L_x \,\vec{i} + L_y\,\vec{j} + L_z\,\vec{k} $$

If you consider the observation of the angular momentum of a particle, you find that you can simultaneously observe only the square of the total angular momentum $L^2$ and one of the components. By convention, this component is the z-component, $L_z$. By analogy with the simpler system, the particle on a ring, you can identify $L_z$ as:

$$ L_z = m \hbar, ~\text{where}~ m = 0, \pm 1, \pm 2, ... $$

This looks like we are getting somewhere familiar: we know that we expect $m = -1, 0, 1$ for $l=1$ (e.g. the three p-orbitals). But we have not seen a reason to limit the value of $m$, yet.

Consider now a vector $\vec{L}$ and it's z-component, $L_z$. Clearly, the z-component cannot be longer than the original vector. So we have:

$$ L_z \leq \vert \vec{L} \vert $$

We know from earlier that we have expressions for both $\vert \vec{L} \vert$ and $L_z$, so:

$$ m\hbar \leq \sqrt{l(l+1)}\hbar $$

Dividing by $\hbar$ and squaring:

$$ m^2 \leq l(l+1) $$

We know that $m = 0, \pm 1, \pm 2, ...$ and that $l = 0, 1, 2, ...$

The inequality above is what brings the restriction on $m$. For example, in the case that $l=1$:

$$ \begin{align} l(l+1) &= 2 \\ m^2 &\leq 2 \\ m &\leq \pm \sqrt{2} ~(\pm 1.414) \end{align} $$

Thus we can see that the only values of $m$ that satisfy $m^2 \leq l(l+1)$ for $l=1$ are $m = -1, 0$ and 1, i.e three values of $m$.

This is the basis of the rule that there are $2l+1$ different values of $m$ for a given integer $l$.

We can also go a bit further and show that for a given integer value of $l$, the largest integer $m$ that satisfies $m^2 \leq l(l+1)$ is always $\vert m \vert = l$. One way to do this is a satisfying proof by contradiction. Suppose that the largest integer $m$ that satisifes $m^2 \leq l(l+1)$ is actually $\vert m \vert = l+1$ (i.e. the next integer along to what we expect). Then we have:

$$ \begin{align} (l+1)^2 &\leq l(l+1) \\ l^2 + 2l + 1 &\leq l^2 + l \\ l + 1 &\leq 0 \end{align} $$

The last line obviously isn't true for any $l = 0, 1, 2 ...$. So $\vert m \vert = l+1$ was too big.

Now we try taking the next integer down, i.e. $\vert m \vert = l$, and running the same argument:

$$ \begin{aligned} l^2 &\leq l(l+1) \\ l^2 &\leq l^2 + l \\ 0 &\leq l \end{aligned} $$

This is now obviously true for any value of $l = 0, 1, 2, ...$.

So the largest integer $m$ that satisfies $m^2 \leq l(l+1)$ is $\vert m \vert = l$.

Thus we conclude that the values of $m$ (the z-component of our quantised angular momentum) that are possible for a given value of $l$ are:

$$ m = 0, \pm 1, \pm 2, ... \pm l $$

Which leads directly to the rule that a given angular momentum $L$ has $2L+1$ projections. You can think of the it as:

Each set of positive and negative integers gives the $2L$.
The zero gives the additional projection, which is the $+1$
Overall it is $2L+1$.

A more general way to show this, which also suggests/predicts the existence of half-integer angular momentum, is a bit more involved and uses a lot of angular momentum algebra and ladder operators...