Why position and momentum operators are both continuous spectrum while angular momentum is discrete?

Question

We know that position $\hat{r}$ and momentum $\hat{p}$ are both continuous spectrum operators, i.e. $$\hat{r}|r'\rangle=r'|r'\rangle, \quad \hat{p}|p'\rangle=p'|p'\rangle.$$ But the angular operator $\hat{L}=\hat{r}\times\hat{p}$ is not: $$\hat{L}^2|l\rangle=\hbar^2 l(l+1)|l\rangle.$$ Could anyone give some explanation?

Answer 1

One possible answer: boundary conditions.

If we think of the classical to quantum transition to be from point-like particles to waves which describe probability densities, then the position and momentum of a free particle does not have any explicit boundary conditions. That is, there is no restriction on the specific character of the probability density wave.

On the other hand, for a particle in orbit, with well-defined angular momentum, when replacing the point particle with a wave means the wave has to satisfy certain conditions for stability - namely, it must be continuous. That restricts the wavelengths to only be certain multiples of the radius - and viola, discreteness.

Of course, we aren't actually talking about analogous systems here - I said "free particle" for position and momentum, whereas the angular momentum quantum number only occurs for bound states, but the origin of both is boundary conditions. The free particle has none, and therefore there are no conditions set on the continuity of the waves. The bound particle must stay bound, which results in restrictions on what wavelengths of waves can actually be present.

Answer 2

In general for a bounded state, its energy spectrum is discrete. Free particles on the other hand has continuous energy spectrum. This is true for a rotating motion as it is also bounded, i.e. the starting point is exactly equal to the end point upon rotation by $2\pi$, therefore its eigenvalues are also discrete.