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 $j$ can ei­ther be a non­neg­a­tive whole num­ber or half of one, but noth­ing else. And it will show that the mag­netic quan­tum num­ber $m$ must range from $-j$ to $+j$ 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 $m_{\rm {max}}$, ${\widehat J}^+{\left\vert l\:m\right\rangle}$ has to be zero for $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {max}}$. More specif­i­cally, its mag­ni­tude $\left\vert{\widehat J}^+{\left\vert j\:m\right\rangle}\right\vert$ must be zero. The square mag­ni­tude is given by the in­ner prod­uct with it­self:

$$\left\vert{\widehat J}^+{\left\vert j\:m\right\rangle}\righ... ...{\widehat J}^+{\left\vert j\:m\right\rangle}\bigg\rangle = 0.$$

Now be­cause of the com­plex con­ju­gate that is used in the left hand side of an in­ner prod­uct, (see chap­ter 2.3), ${\widehat J}^+$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\widehat J}_x+{\rm i}{\widehat J}_y$ goes to the other side of the prod­uct as ${\widehat J}^-$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\widehat J}_x-{\rm i}{\widehat J}_y$, and you must have

$$\left\vert{\widehat J}^+{\left\vert j\:m\right\rangle}\righ... ...t J}^-{\widehat J}^+{\left\vert j\:m\right\rangle}\bigg\rangle$$

That op­er­a­tor prod­uct can be mul­ti­plied out:

$${\widehat J}^-{\widehat J}^+\equiv({\widehat J}_x-{\rm i}{\... ...({\widehat J}_x{\widehat J}_y - {\widehat J}_y{\widehat J}_x),$$

but ${\widehat J}_x^2+{\widehat J}_y^2$ is the square an­gu­lar mo­men­tum ${\widehat J}^2$ ex­cept for ${\widehat J}_z^2$, and the term within the paren­the­ses is the com­mu­ta­tor $[{\widehat J}_x,{\widehat J}_y]$ which is ac­cord­ing to the fun­da­men­tal com­mu­ta­tion re­la­tions equal to ${\rm i}\hbar{\widehat J}_z$, so

$${\widehat J}^-{\widehat J}^+ = {\widehat J}^2 - {\widehat J}_z^2 - \hbar{\widehat J}_z %$$ (12.5)

The ef­fect of each of the op­er­a­tors in the left hand side on a state ${\left\vert j\:m\right\rangle}$ is known and the in­ner prod­uct can be fig­ured out:

$$\left\vert{\widehat J}^+{\left\vert j\:m\right\rangle}\right\vert^2 = j(j+1)\hbar^2 - m^2 \hbar^2 - m\hbar^2 %$$ (12.6)

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

$$j(j+1)\hbar^2 - m_{\rm max}^2 \hbar^2 - m_{\rm max}\hbar^2 = 0$$

There are two pos­si­ble so­lu­tions to this qua­dratic equa­tion for $m_{\rm {max}}$, to wit $m_{\rm {max}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$ or $-m_{\rm {max}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j+1$. The sec­ond so­lu­tion is im­pos­si­ble since it al­ready would have the square $z$ an­gu­lar mo­men­tum ex­ceed the to­tal square an­gu­lar mo­men­tum. So un­avoid­ably,

$$m_{\rm max} = j$$

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

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

$${\widehat J}^+{\widehat J}^- = {\widehat J}^2 - {\widehat J}_z^2 + \hbar{\widehat J}_z %$$ (12.7)

and then

$$\left\vert{\widehat J}^-{\left\vert j\:m\right\rangle}\right\vert^2 = j(j+1)\hbar^2 - m^2 \hbar^2 + m\hbar^2 %$$ (12.8)

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

$$m_{\rm min} = -j$$

It is nice and sym­met­ric; lad­ders run from $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$j$ up to $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$, 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 $j$ and $m$ can be. For, since each step on a lad­der in­creases the mag­netic quan­tum num­ber $m$ by one unit, you have for the to­tal num­ber of steps up from bot­tom to top:

$$\mbox{total number of steps } = m_{\rm max} - m_{\rm min} = 2j$$

But the num­ber of steps is a whole num­ber, and so the az­imuthal quan­tum $j$ must ei­ther be a non­neg­a­tive in­te­ger, such as 0, 1, 2, ..., or half of one, such as $\frac12$, $\frac32$, ...

In­te­ger $j$ 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 $\Delta$ par­ti­cles.

Note that if $j$ is a half-in­te­ger, then so are the cor­re­spond­ing val­ues of $m$, since $m$ starts from $\vphantom{0}\raisebox{1.5pt}{$-$}$$j$ and in­creases in unit steps. See again fig­ures 12.1 and 12.2 for some ex­am­ples. Also note that lad­ders ter­mi­nate just be­fore $z$-​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 $Y^m_3$ one in fig­ure 12.1 with ${\widehat J}^+$ and then come down the sec­ond one us­ing ${\widehat J}^-$. The rea­son is that the states ${\left\vert j\:m\right\rangle}$ are eigen­states of the op­er­a­tors ${\widehat J}^-{\widehat J}^+$, (12.5), and ${\widehat J}^+{\widehat J}^-$, (12.7), so go­ing up with ${\widehat J}^+$ and then down again with ${\widehat J}^-$, 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.