12.7 Clebsch-Gordan coefficients

In classical physics, combining angular momentum from different sources is easy; the net components in the $x$, $y$, and $z$ directions are simply the sum of the individual components. In quantum mechanics, things are trickier, because if the component in the $z$-​direction exists, those in the $x$ and $y$ directions do not. But the previous subsection showed how to the spin angular momenta of two spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particles could be combined. In similar ways, the angular momentum states of any two ladders, whatever their origin, can be combined into net angular momentum ladders. And then those ladders can in turn be combined with still other ladders, allowing net angular momentum states to be found for systems of arbitrary complexity.

The key is to be able to combine the angular momentum ladders from two different sources into net angular momentum ladders. To do so, the net angular momentum can in principle be described in terms of product states in which each source is on a single rung of its ladder. But as the example of the last section illustrated, such product states give incomplete information about the net angular momentum; they do not tell you what square net angular momentum is. You need to know what combinations of product states produce rungs on the ladders of the net angular momentum, like the ones illustrated in figure 12.3. In particular, you need to know the coefficients that multiply the product states in those combinations.

Figure 12.4: Clebsch-Gordan coefficients of two spin one half particles.
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...\kern.05em}}\big\rangle _b$}}
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These coefficients are called Clebsch-Gordan coefficients. The ones corresponding to figure 12.3 are tabulated in Figure 12.4. Note that there are really three tables of numbers; one for each rung level. The top, single number, table says that the $\big\vert 1\:1\big\rangle $ net momentum state is found in terms of product states as:

\begin{displaymath}
{\big\vert 1\:1\big\rangle }_{ab} = 1 \times {\big\vert\le...
...m\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b
\end{displaymath}

The second table gives the states with zero net angular momentum in the $z$-​direction. For example, the first column of the table says that the $\big\vert\:0\big\rangle $ singlet state is found as:

\begin{displaymath}
{\big\vert\:0\big\rangle }_{ab} =
\sqrt{\leavevmode \ker...
...m\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b
\end{displaymath}

Similarly the second column gives the middle rung $\big\vert 1\:0\big\rangle $ on the triplet ladder. The bottom table gives the bottom rung of the triplet ladder.

You can also read the tables horizontally {N.29}. For example, the first row of the middle table says that the ${\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/...
...rn-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b$ product state equals

\begin{displaymath}
{\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scrip...
...scriptfont0 2}\kern.05em}\; {\big\vert 1\:0\big\rangle }_{ab}
\end{displaymath}

That in turn implies that if the net square angular momentum of this product state is measured, there is a 50/50 chance of it turning out to be either zero, or the $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 (i.e. $2\hbar^2$) value. The $z$-​momentum will always be zero.

Figure 12.5: Clebsch-Gordan coefficients when the second angular momentum contribution has azimuthal quantum number $j_b=\frac12$.
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Figure 12.6: Clebsch-Gordan coefficients when the second angular momentum contribution has azimuthal quantum number $j_b=1$.
\begin{figure}
\centering
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...2)
{$\displaystyle {j_a=2 \atop j_b=1}$}}}
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\end{figure}

How about the Clebsch-Gordan coefficients to combine other ladders than the spins of two spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particles? Well, the same procedures used in the previous section work just as well to combine the angular momenta of any two angular momentum ladders, whatever their size. Just the thing for a long winter night. Or, if you live in Florida, you just might want to write a little computer program that does it for you {D.66} and outputs the tables in human-readable form {N.30}, like figures 12.5 and 12.6.

From the figures you may note that when two states with total angular momentum quantum numbers $j_a$ and $j_b$ are combined, the combinations have total angular quantum numbers ranging from $j_a+j_b$ to $\vert j_a-j_b\vert$. This is similar to the fact that when in classical mechanics two angular momentum vectors are combined, the combined total angular momentum $J_{ab}$ is at most $J_a+J_b$ and at least $\vert J_a-J_b\vert$. (The so-called “triangle inequality” for combining vectors.) But of course, $j$ is not quite a proportional measure of $J$ unless $J$ is large; in fact, $J$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{j(j+1)}\hbar$ {D.67}.