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:

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

So mathematics says that eigen­states $\big\vert j\:m\big\rangle$ of ${\widehat J}_z$ and ${\widehat J}^2$ exist satisfying

$${\widehat J}_z \big\vert j\:m\big\rangle = J_z \big\vert j\:m\big\rangle \qquad \mbox{where {\em by definition} $J_z = m\hbar$}$$ (12.2)

$${\widehat J}^2 \big\vert j\:m\big\rangle = J^2 \big\vert j\:m\big\rangle \qquad \mbox{where {\em by definition} $J^2 = j(j+1)\hbar^2$\ and $j\mathrel{\raisebox{-1pt}{$\geqslant$}}0$}%$$ (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 $\big\vert j\:m\big\rangle$:

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

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:

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

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

$${\widehat J}^+ \equiv {\widehat J}_x + {\rm i}{\widehat J}... ...dehat J}^- \equiv {\widehat J}_x - {\rm i}{\widehat J}_y , %$$ (12.4)

that satisfy the “commutator eigenvalue problems”:

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

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

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

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

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

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.

Figure 12.2: Example fermionic ladders.

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 $m_{\rm {min}}$.