### 12.3 Lad­ders

This sec­tion starts the quest to fig­ure out every­thing that the fun­da­men­tal com­mu­ta­tion re­la­tions mean for an­gu­lar mo­men­tum. It will first be ver­i­fied that any an­gu­lar mo­men­tum can al­ways be de­scribed us­ing eigen­states with def­i­nite val­ues of square an­gu­lar mo­men­tum and an­gu­lar mo­men­tum . Then it will be found that these an­gu­lar mo­men­tum states oc­cur in groups called lad­ders.

To start with the first one, the math­e­mat­i­cal con­di­tion for a com­plete set of eigen­states to ex­ist is that the an­gu­lar mo­men­tum op­er­a­tors and com­mute. They do; us­ing the com­mu­ta­tor ma­nip­u­la­tions of chap­ter 4.5.4), it is eas­ily found that:

So math­e­mat­ics says that eigen­states of and ex­ist sat­is­fy­ing
 (12.2) (12.3)

and that are com­plete in the sense that any state can be de­scribed in terms of these .

Un­for­tu­nately the eigen­states , ex­cept for states, do not sat­isfy re­la­tions like (12.2) for or . The prob­lem is that and do not com­mute with . But and do com­mute with , and you might won­der if that is still worth some­thing. To find out, mul­ti­ply, say, the zero com­mu­ta­tor by :

Now take the sec­ond term to the right hand side of the equa­tion, not­ing that with just a num­ber that can be moved up-front, to get:

Look­ing a bit closer at this equa­tion, it shows that the com­bi­na­tion sat­is­fies the same eigen­value prob­lem for as it­self. In other words, the mul­ti­pli­ca­tion by does not af­fect the square an­gu­lar mo­men­tum at all.

To be picky, that is not quite true if would be zero, be­cause zero is not an eigen­state of any­thing. How­ever, such a thing only hap­pens if there is no an­gu­lar mo­men­tum; (it would make an eigen­state of with eigen­value zero in ad­di­tion to an eigen­state of {D.63}). Ex­cept for that triv­ial case, does not af­fect square an­gu­lar mo­men­tum. And nei­ther does or any com­bi­na­tion of the two.

An­gu­lar mo­men­tum in the -​di­rec­tion is af­fected by and by , since they do not com­mute with like they do with . Nor is it pos­si­ble to find any lin­ear com­bi­na­tion of and that does com­mute with . What is the next best thing? Well, it is pos­si­ble to find two com­bi­na­tions, to wit

 (12.4)

that sat­isfy the “com­mu­ta­tor eigen­value prob­lems”:

These two turn out to be quite re­mark­able op­er­a­tors.

Like and , their com­bi­na­tions and leave alone. To ex­am­ine what the op­er­a­tor does with the lin­ear mo­men­tum in the -​di­rec­tion, mul­ti­ply its com­mu­ta­tor re­la­tion above by an eigen­state :

Or, tak­ing the sec­ond term to the right hand side of the equa­tion and not­ing that by de­f­i­n­i­tion ,

That is a stun­ning re­sult, as it shows that is an eigen­state with an­gu­lar mo­men­tum in­stead of . In other words, adds ex­actly one unit to the an­gu­lar mo­men­tum, turn­ing an state into a one!

If you ap­ply an­other time, you get a state of still higher an­gu­lar mo­men­tum , and so on, like the rungs on a lad­der. This is graph­i­cally il­lus­trated for some ex­am­ples in fig­ures 12.1 and 12.2. The process even­tu­ally comes to an halt at some top rung where 0. It has to, be­cause the an­gu­lar mo­men­tum in the -​di­rec­tion can­not just keep grow­ing for­ever: the square an­gu­lar mo­men­tum in the -​di­rec­tion only must stay less than the to­tal square an­gu­lar mo­men­tum in all three di­rec­tions {N.27}.

The sec­ond lad­der op­er­a­tor works in much the same way, but it goes down the lad­der; its deducts one unit from the an­gu­lar mo­men­tum in the -​di­rec­tion at each ap­pli­ca­tion. pro­vides the sec­ond stile to the lad­ders, and must ter­mi­nate at some bot­tom rung .