4.2.2.3 So­lu­tion an­gub-c

Ques­tion:

Ac­tu­ally, based on the de­rived eigen­func­tion, $C(r,\theta)e^{{\rm i}{m}\phi}$, would any macro­scopic par­ti­cle ever be at a sin­gle mag­netic quan­tum num­ber in the first place? In par­tic­u­lar, what can you say about where the par­ti­cle can be found in an eigen­state?

An­swer:

The square mag­ni­tude of the wave func­tion gives the prob­a­bil­ity of find­ing the par­ti­cle. The square mag­ni­tude,

$$\left\vert C(r,\theta)e^{{\rm i}{m}\phi}\right\vert^2 = \vert C(r,\theta)\vert^2,$$

is in­de­pen­dent of $\phi$. So to be in a state of def­i­nite an­gu­lar mo­men­tum, the par­ti­cle must be at all sides of the axis with equal prob­a­bil­ity. A macro­scopic par­ti­cle will at any given time be at a sin­gle an­gle com­pared to the axis, not at all an­gles at once. So, a macro­scopic par­ti­cle will have in­de­ter­mi­nacy in an­gu­lar mo­men­tum, just like it has in­de­ter­mi­nacy in po­si­tion, lin­ear mo­men­tum, en­ergy, etcetera.

Since the prob­a­bil­ity dis­tri­b­u­tion of an eigen­state is in­de­pen­dent of $\phi$, it is called ax­isym­met­ric around the $z$-​axis. Note that the wave func­tion it­self is only ax­isym­met­ric if $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, in other words, if the an­gu­lar mo­men­tum in the $z$-​di­rec­tion is zero. Eigen­states with dif­fer­ent an­gu­lar mo­men­tum look the same if you just look at the prob­a­bil­ity dis­tri­b­u­tion.