### 12.4 Pos­si­ble val­ues of an­gu­lar mo­men­tum

The fact that the an­gu­lar mo­men­tum lad­ders of the pre­vi­ous sec­tion must have a top and a bot­tom rung re­stricts the pos­si­ble val­ues that an­gu­lar mo­men­tum can take. This sec­tion will show that the az­imuthal quan­tum num­ber can ei­ther be a non­neg­a­tive whole num­ber or half of one, but noth­ing else. And it will show that the mag­netic quan­tum num­ber must range from to in unit in­cre­ments. In other words, the bosonic and fermi­onic ex­am­ple lad­ders in fig­ures 12.1 and 12.2 are rep­re­sen­ta­tive of all that is pos­si­ble.

To start, in or­der for a lad­der to end at a top rung , has to be zero for . More specif­i­cally, its mag­ni­tude must be zero. The square mag­ni­tude is given by the in­ner prod­uct with it­self:

Now be­cause of the com­plex con­ju­gate that is used in the left hand side of an in­ner prod­uct, (see chap­ter 2.3), goes to the other side of the prod­uct as , and you must have

That op­er­a­tor prod­uct can be mul­ti­plied out:

but is the square an­gu­lar mo­men­tum ex­cept for , and the term within the paren­the­ses is the com­mu­ta­tor which is ac­cord­ing to the fun­da­men­tal com­mu­ta­tion re­la­tions equal to , so
 (12.5)

The ef­fect of each of the op­er­a­tors in the left hand side on a state is known and the in­ner prod­uct can be fig­ured out:
 (12.6)

The ques­tion where an­gu­lar mo­men­tum lad­ders end can now be an­swered:

There are two pos­si­ble so­lu­tions to this qua­dratic equa­tion for , to wit or . The sec­ond so­lu­tion is im­pos­si­ble since it al­ready would have the square an­gu­lar mo­men­tum ex­ceed the to­tal square an­gu­lar mo­men­tum. So un­avoid­ably,

That is one of the things this sec­tion was sup­posed to show.

The low­est rung on the lad­der goes the same way; you get

 (12.7)

and then
 (12.8)

and the only ac­cept­able so­lu­tion for the low­est rung on the lad­ders is

It is nice and sym­met­ric; lad­ders run from up to , as the ex­am­ples in fig­ures 12.1 and 12.2 al­ready showed.

And in fact, it is more than that; it also lim­its what the quan­tum num­bers and can be. For, since each step on a lad­der in­creases the mag­netic quan­tum num­ber by one unit, you have for the to­tal num­ber of steps up from bot­tom to top:

But the num­ber of steps is a whole num­ber, and so the az­imuthal quan­tum must ei­ther be a non­neg­a­tive in­te­ger, such as 0, 1, 2, ..., or half of one, such as , , ...

In­te­ger val­ues oc­cur, for ex­am­ple, for the spher­i­cal har­mon­ics of or­bital an­gu­lar mo­men­tum and for the spin of bosons like pho­tons. Half-in­te­ger val­ues oc­cur, for ex­am­ple, for the spin of fermi­ons such as elec­trons, pro­tons, neu­trons, and par­ti­cles.

Note that if is a half-in­te­ger, then so are the cor­re­spond­ing val­ues of , since starts from and in­creases in unit steps. See again fig­ures 12.1 and 12.2 for some ex­am­ples. Also note that lad­ders ter­mi­nate just be­fore -​mo­men­tum would ex­ceed to­tal mo­men­tum.

It may also be noted that lad­ders are dis­tinct. It is not pos­si­ble to go up one lad­der, like the first one in fig­ure 12.1 with and then come down the sec­ond one us­ing . The rea­son is that the states are eigen­states of the op­er­a­tors , (12.5), and , (12.7), so go­ing up with and then down again with , or vice-versa, re­turns to the same state. For sim­i­lar rea­sons, if the tops of two lad­ders are or­tho­nor­mal, then so is the rest of their rungs.