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 $\big\vert j\:m\big\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 $\big\vert j\:m\big\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, {\wideha...
...J}^2 = {\widehat J}_x^2 + {\widehat J}_y^2 + {\widehat J}_z^2
\end{displaymath}

So mathematics says that eigenstates $\big\vert j\:m\big\rangle $ of ${\widehat J}_z$ and ${\widehat J}^2$ exist satisfying
 $\displaystyle {\widehat J}_z \big\vert j\:m\big\rangle$ $\textstyle =$ $\displaystyle J_z \big\vert j\:m\big\rangle \qquad
\mbox{where \emph{by definition} $J_z = m\hbar$}$  (12.2)
 $\displaystyle {\widehat J}^2 \big\vert j\:m\big\rangle$ $\textstyle =$ $\displaystyle J^2 \big\vert j\:m\big\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 $\big\vert j\:m\big\rangle $.

Unfortunately the eigenstates $\big\vert j\:m\big\rangle $, except for $\big\vert\:0\big\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 $\big\vert j\:m\big\rangle $:

\begin{displaymath}[{\widehat J}^2,{\widehat J}_x]\big\vert j\:m\big\rangle = ({...
... {\widehat J}_x {\widehat J}^2) \big\vert j\:m\big\rangle = 0
\end{displaymath}

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

\begin{displaymath}
{\widehat J}^2\left({\widehat J}_x \big\vert j\:m\big\rang...
... = J^2 \left({\widehat J}_x \big\vert j\:m\big\rangle \right)
\end{displaymath}

Looking a bit closer at this equation, it shows that the combination ${\widehat J}_x\big\vert j\:m\big\rangle $ satisfies the same eigenvalue problem for ${\widehat J}^2$ as $\big\vert j\:m\big\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\big\vert j\:m\big\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 $\big\vert j\:m\big\rangle $ an eigenstate of ${\widehat J}_x$ with eigenvalue zero in addition to an eigenstate of ${\widehat J}_z$ {D.64}). 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

\begin{displaymath}
{\widehat J}^+ \equiv {\widehat J}_x + {\rm i}{\widehat J}...
...idehat 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...
...ad
[{\widehat J}_z, {\widehat J}^-] = -\hbar{\widehat J}^-.
\end{displaymath}

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 $\big\vert j\:m\big\rangle $:

\begin{displaymath}
({\widehat J}_z {\widehat J}^+ - {\widehat J}^+ {\widehat ...
...m\big\rangle = \hbar{\widehat J}^+ \big\vert j\:m\big\rangle
\end{displaymath}

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

\begin{displaymath}
{\widehat J}_z \left({\widehat J}^+ \big\vert j\:m\big\ran...
...)\hbar \left({\widehat J}^+ \big\vert j\:m\big\rangle \right)
\end{displaymath}

That is a stunning result, as it shows that ${\widehat J}^+\big\vert j\:m\big\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 $\big\vert j\:m\big\rangle $ state into a $\big\vert j\:m\!+\!1\big\rangle $ one!

Figure 12.1: Example bosonic ladders.
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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 $\big\vert j\:m{+}2\big\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}^+\big\vert j\:m_{\rm {max}}\big\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 $m_{\rm {min}}$.