c/A wavefunction on the sphere
with|L|= 11~andLz= 8~, shown
using the color conventions
defined in figure u, p. 914.
knowing all three components of an angular momentum vector si-
multaneously. The most general statement about this is the follow-
ing theorem:The angular momentum vector in quantum physics
The most that can be known about a (nonzero) orbital angular
momentum vector is its magnitude and one of its three vector com-
ponents. Both are quantized in units of~.To see why this is true, consider the example wavefunction shown
in figure c. This is the like the quantum moat of figure a, p. 919,
but extended to one more dimension. If we slice the sphere in any
plane perpendicular to thezaxis, we get an 8-cycle circular rainbow
exactly like figure a. This is required becauseLz= 8~. But if we
take a slice perpendicular to some other axis, such as theyaxis, we
don’t get a circular rainbow as we would for a state with a definite
value ofLy. It is obviously not possible to get circular rainbows for
slices perpendicular to more than one axis.
For those with a taste for rigor, here is a complete argument:
Theorem: On the sphere, if a wavefunction has definite values of
bothLz andLx, then it is a wavefunction that is constant every-
where, soL= 0.
Lemma 1: If the component of`A along a certain axis A has a
definite value and is nonzero, then (a) Ψ = 0 at the poles, and (b)
Ψ is of the formAei`Aφon any circle in a plane perpendicular to the
axis. Part a holds becauseL= 0 ifr⊥= 0. For b, see p. 919.
Lemma 2: If the component ofLalong a certain axis has a definite
value and is zero, then Ψ is constant in any plane perpendicular to
that axis. This follows from lemma 1 in the case where`A= 0.
Case I:`zand`xare both nonzero. We have Ψ = 0 at the poles
along both thexaxis and thez axis. Thez-axis pole is a point
on the great circle perpendicular to thexaxis, and vice versa, so
applying 1b,A= 0 and Ψ vanishes on both of these great circles.
But now if we apply 1b along any slice perpendicular to either axis,
we get Ψ = 0 everywhere on that slice, so Ψ = 0 everywhere.
Case II:`z and `x are both zero. By lemma 2, Ψ is a constant
everywhere.
Case III: One component is zero and the other nonzero. Let`zbe
the one that is zero. By 1a, Ψ = 0 at thex-axis pole, so by 2, Ψ = 0
on the great circle perpendicular toz. But then 1b tells us that
Ψ = 0 everywhere.922 Chapter 13 Quantum Physics