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 ${\left\vert j\:m\right\rangle}$ eigen­states with def­i­nite val­ues of square an­gu­lar mo­men­tum $J^2$ and $z$ an­gu­lar mo­men­tum $J_z$. 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 ${\left\vert j\:m\right\rangle}$ to ex­ist is that the an­gu­lar mo­men­tum op­er­a­tors ${\widehat J}^2$ and ${\widehat J}_z$ com­mute. They do; us­ing the com­mu­ta­tor ma­nip­u­la­tions of chap­ter 4.5.4), it is eas­ily found that:

\begin{displaymath}[{\widehat J}^2, {\widehat J}_x]=
[{\widehat J}^2, {\widehat...
... J}^2 = {\widehat J}_x^2 + {\widehat J}_y^2 + {\widehat J}_z^2

So math­e­mat­ics says that eigen­states ${\left\vert j\:m\right\rangle}$ of ${\widehat J}_z$ and ${\widehat J}^2$ ex­ist sat­is­fy­ing
 $\displaystyle {\widehat J}_z {\left\vert j\:m\right\rangle}$ $\textstyle =$ $\displaystyle J_z {\left\vert j\:m\right\rangle} \qquad
\mbox{where \emph{by definition} $J_z = m\hbar$}$  (12.2)
 $\displaystyle {\widehat J}^2 {\left\vert j\:m\right\rangle}$ $\textstyle =$ $\displaystyle J^2 {\left\vert j\:m\right\rangle} \qquad
\mbox{where \emph{by definition}
$J^2 = j(j+1)\hbar^2$\ and $j\mathrel{\raisebox{-1pt}{$\geqslant$}}0$}%
$  (12.3)

and that are com­plete in the sense that any state can be de­scribed in terms of these ${\left\vert j\:m\right\rangle}$.

Un­for­tu­nately the eigen­states ${\left\vert j\:m\right\rangle}$, ex­cept for ${\left\vert\:0\right\rangle}$ states, do not sat­isfy re­la­tions like (12.2) for ${\widehat J}_x$ or ${\widehat J}_y$. The prob­lem is that ${\widehat J}_x$ and ${\widehat J}_y$ do not com­mute with ${\widehat J}_z$. But ${\widehat J}_x$ and ${\widehat J}_y$ do com­mute with ${\widehat J}^2$, and you might won­der if that is still worth some­thing. To find out, mul­ti­ply, say, the zero com­mu­ta­tor $[{\widehat J}^2,{\widehat J}_x]$ by ${\left\vert j\:m\right\rangle}$:

\begin{displaymath}[{\widehat J}^2,{\widehat J}_x]{\left\vert j\:m\right\rangle}...
...idehat J}_x {\widehat J}^2) {\left\vert j\:m\right\rangle} = 0

Now take the sec­ond term to the right hand side of the equa­tion, not­ing that ${\widehat J}^2{\left\vert j\:m\right\rangle}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $J^2{\left\vert j\:m\right\rangle}$ with $J^2$ just a num­ber that can be moved up-front, to get:

{\widehat J}^2\left({\widehat J}_x {\left\vert j\:m\right\r...
...J^2 \left({\widehat J}_x {\left\vert j\:m\right\rangle}\right)

Look­ing a bit closer at this equa­tion, it shows that the com­bi­na­tion ${\widehat J}_x{\left\vert j\:m\right\rangle}$ sat­is­fies the same eigen­value prob­lem for ${\widehat J}^2$ as ${\left\vert j\:m\right\rangle}$ it­self. In other words, the mul­ti­pli­ca­tion by ${\widehat J}_x$ does not af­fect the square an­gu­lar mo­men­tum $J^2$ at all.

To be picky, that is not quite true if ${\widehat J}_x{\left\vert j\:m\right\rangle}$ 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 ${\left\vert j\:m\right\rangle}$ an eigen­state of ${\widehat J}_x$ with eigen­value zero in ad­di­tion to an eigen­state of ${\widehat J}_z$ {D.63}). Ex­cept for that triv­ial case, ${\widehat J}_x$ does not af­fect square an­gu­lar mo­men­tum. And nei­ther does ${\widehat J}_y$ or any com­bi­na­tion of the two.

An­gu­lar mo­men­tum in the $z$-​di­rec­tion is af­fected by ${\widehat J}_x$ and by ${\widehat J}_y$, since they do not com­mute with ${\widehat J}_z$ like they do with ${\widehat J}^2$. Nor is it pos­si­ble to find any lin­ear com­bi­na­tion of ${\widehat J}_x$ and ${\widehat J}_y$ that does com­mute with ${\widehat J}_z$. What is the next best thing? Well, it is pos­si­ble to find two com­bi­na­tions, to wit

{\widehat J}^+ \equiv {\widehat J}_x + {\rm i}{\widehat J}_...
...widehat J}^- \equiv {\widehat J}_x - {\rm i}{\widehat J}_y , %
\end{displaymath} (12.4)

that sat­isfy the “com­mu­ta­tor eigen­value prob­lems”:

\begin{displaymath}[{\widehat J}_z, {\widehat J}^+]= \hbar{\widehat J}^+
\quad ...
[{\widehat J}_z, {\widehat J}^-] = -\hbar{\widehat J}^-.

These two turn out to be quite re­mark­able op­er­a­tors.

Like ${\widehat J}_x$ and ${\widehat J}_y$, their com­bi­na­tions ${\widehat J}^+$ and ${\widehat J}^-$ leave $J^2$ alone. To ex­am­ine what the op­er­a­tor ${\widehat J}^+$ does with the lin­ear mo­men­tum in the $z$-​di­rec­tion, mul­ti­ply its com­mu­ta­tor re­la­tion above by an eigen­state ${\left\vert j\:m\right\rangle}$:

({\widehat J}_z {\widehat J}^+ - {\widehat J}^+ {\widehat J...
...t\rangle} = \hbar{\widehat J}^+ {\left\vert j\:m\right\rangle}

Or, tak­ing the sec­ond term to the right hand side of the equa­tion and not­ing that by de­f­i­n­i­tion ${\widehat J}_z{\left\vert j\:m\right\rangle}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m\hbar{\left\vert j\:m\right\rangle}$,

{\widehat J}_z \left({\widehat J}^+ {\left\vert j\:m\right\...
...bar \left({\widehat J}^+ {\left\vert j\:m\right\rangle}\right)

That is a stun­ning re­sult, as it shows that ${\widehat J}^+{\left\vert j\:m\right\rangle}$ is an eigen­state with $z$ an­gu­lar mo­men­tum $J_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(m+1)\hbar$ in­stead of $m\hbar$. In other words, ${\widehat J}^+$ adds ex­actly one unit $\hbar$ to the $z$ an­gu­lar mo­men­tum, turn­ing an ${\left\vert j\:m\right\rangle}$ state into a ${\left\vert j\:m\!+\!1\right\rangle}$ one!

Fig­ure 12.1: Ex­am­ple bosonic lad­ders.
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\strut a graviton?}}}

Fig­ure 12.2: Ex­am­ple fermi­onic lad­ders.
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If you ap­ply ${\widehat J}^+$ an­other time, you get a state of still higher $z$ an­gu­lar mo­men­tum ${\left\vert j\:m{+}2\right\rangle}$, 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 $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {max}}$ where ${\widehat J}^+{\left\vert j\:m_{\rm {max}}\right\rangle}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. It has to, be­cause the an­gu­lar mo­men­tum in the $z$-​di­rec­tion can­not just keep grow­ing for­ever: the square an­gu­lar mo­men­tum in the $z$-​di­rec­tion only must stay less than the to­tal square an­gu­lar mo­men­tum in all three di­rec­tions {N.27}.

The sec­ond lad­der op­er­a­tor ${\widehat J}^-$ works in much the same way, but it goes down the lad­der; its deducts one unit $\hbar$ from the an­gu­lar mo­men­tum in the $z$-​di­rec­tion at each ap­pli­ca­tion. ${\widehat J}^-$ pro­vides the sec­ond stile to the lad­ders, and must ter­mi­nate at some bot­tom rung $m_{\rm {min}}$.