12.5 A warn­ing about an­gu­lar mo­men­tum

Nor­mally, eigen­states are in­de­ter­mi­nate by a com­plex num­ber of mag­ni­tude one. If you so de­sire, you can mul­ti­ply any nor­mal­ized eigen­state by a num­ber of unit mag­ni­tude of your own choos­ing, and it is still a nor­mal­ized eigen­state. It is im­por­tant to re­mem­ber that in an­a­lyt­i­cal ex­pres­sions in­volv­ing an­gu­lar mo­men­tum, you are not al­lowed to do this.

As an ex­am­ple, con­sider a pair of spin 1/2 par­ti­cles, call them $a$ and $b$, in the sin­glet state, in which their spins can­cel and there is no net an­gu­lar mo­men­tum. It was noted in chap­ter 5.5.6 that this state takes the form

$${{\left\vert\:0\right\rangle}}_{ab} = \frac{{{\left\vert\l... ...\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b}{\sqrt 2}$$

(This sec­tion will use kets rather than ar­rows for spin states.) But if you were al­lowed to ar­bi­trar­ily change the de­f­i­n­i­tion of say the spin state ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_a$ by a mi­nus sign, then the mi­nus sign in the sin­glet state above would turn in a plus sign. The given ex­pres­sion for the sin­glet state, with its mi­nus sign, is only cor­rect if you use the right nor­mal­iza­tion fac­tors for the in­di­vid­ual states.

It all has to do with the lad­der op­er­a­tors ${\widehat J}^+$ and ${\widehat J}^-$. They are very con­ve­nient for analy­sis, but to make that eas­i­est, you would like to know ex­actly what they do to the an­gu­lar mo­men­tum states ${\left\vert j\:m\right\rangle}$. What you have seen so far is that ${\widehat J}^+{\left\vert j\:m\right\rangle}$ pro­duces a state with the same square an­gu­lar mo­men­tum, and with an­gu­lar mo­men­tum in the $z$-​di­rec­tion equal to $(m+1)\hbar$. In other words, ${\widehat J}^+{\left\vert j\:m\right\rangle}$ is some mul­ti­ple of a suit­ably nor­mal­ized eigen­state ${\left\vert j\:m{+}1\right\rangle}$;

$${\widehat J}^+ {\left\vert j\:m\right\rangle} = C {\left\vert j\:m{+}1\right\rangle}$$

where the num­ber $C$ is the mul­ti­ple. What is that mul­ti­ple? Well, from the mag­ni­tude of ${\widehat J}^+{\left\vert j\:m\right\rangle}$, de­rived ear­lier in (12.6) you know that its square mag­ni­tude is

$$\vert C\vert^2 = j(j+1)\hbar^2 - m^2 \hbar^2 - m\hbar^2.$$

But that still leaves $C$ in­de­ter­mi­nate by a fac­tor of unit mag­ni­tude. Which would be very in­con­ve­nient in the analy­sis of an­gu­lar mo­men­tum.

To re­solve this co­nun­drum, re­stric­tions are put on the nor­mal­iza­tion fac­tors of the an­gu­lar mo­men­tum states ${\left\vert j\:m\right\rangle}$ in lad­ders. It is re­quired that the nor­mal­iza­tion fac­tors are cho­sen such that the lad­der op­er­a­tor con­stants are pos­i­tive real num­bers. That re­ally leaves only one nor­mal­iza­tion fac­tor in an en­tire lad­der freely se­lec­table, say the one of the top rung.

Most of the time, this is not a big deal. Only when you start try­ing to get too clever with an­gu­lar mo­men­tum nor­mal­iza­tion fac­tors, then you want to re­mem­ber that you can­not re­ally choose them to your own lik­ing.

The good news is that in this con­ven­tion, you know pre­cisely what the lad­der op­er­a­tors do {D.64}:

$${\widehat J}^+ {\left\vert j\:m\right\rangle} = \hbar \sqr... ...ig) - m\big(1+m\big)} \; {\left\vert j\:m{+}1\right\rangle} %$$ (12.9)

$${\widehat J}^- {\left\vert j\:m\right\rangle} = \hbar \sqr... ...ig) + m\big(1-m\big)} \; {\left\vert j\:m{-}1\right\rangle} %$$ (12.10)