## Sunday, 20 January 2008

### Rotation

Angular momentum is something that took me a while to appreciate. But as one has to deal with it a lot (orbitals, rotating molecules, spectroscopy terms) it makes sense to take a closer look at it.

I think the problem is if one thinks too quantum mechanically. And then it's difficult to understand what's happening. Angular momentum can be represented by a vector that points in the axis of rotation, its length corresponds to the absolute value of the momentum. We can write:

$\vec{L} = \begin{pmatrix} L_x \\ L_y \\ L_z \end{pmatrix}$

Because of the uncertainty principle we cannot know all the components. In a spherical system, except of having all three bits of information, we have two: L2 and Lz. This can be derived from the fact that the corresponding operators commute and must have a set of identical eigenfunctions. We cannot exactly determine Lx and Ly because their operators do not commute with Lz. In other words: We know how fast our particle rotates and we have partial information about the direction.

Actually quantum mechanics imposes a second restriction (but still: keep thinking about a rotating particle). L2 and Lz are quantised, hence "quantum mechanics". We have

$\vec{L}^2 = l(l+1)\hbar^2
l=0,1,2,...$

and

$L_z = m\hbar
m=-l,-l+1, ..., l$

One thing you notice is that:

$|L_z|\leq l\hbar < \sqrt{l(l+1)}\hbar = |\vec{L}|$

In other words there will always remain something for Lx and Ly and the total momentum cannot be clearly determined. But if you move to classical mechanics (large l), the difference will vanish.

In QM strong motion means a large derivative which in turn means nodal planes (if you consider real functions [1]). This works here, too: l is the number of nodal planes the wave function has on a given sphere. m is the number of nodal planes you pass when you move around the z-axis.

[1] Complex functions can oscillate without nodal planes. eix always has the absolute value of 1 but a non-zero derivative. It's a rotation in the complex plane.