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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ 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.

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 ${\left\vert 1\:1\right\rangle}$ net mo­men­tum state is found in terms of prod­uct states as:

$${{\left\vert 1\:1\right\rangle}}_{ab} = 1 \times {{\left\ve... ...lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$$

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 ${\left\vert\:0\right\rangle}$ sin­glet state is found as:

$${{\left\vert\:0\right\rangle}}_{ab} = \sqrt{\leavevmode \k... ...lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$$

Sim­i­larly the sec­ond col­umn gives the mid­dle rung ${\left\vert 1\:0\right\rangle}$ 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 ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$ prod­uct state equals

$${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scri... ...iptfont0 2}\kern.05em}\; {{\left\vert 1\:0\right\rangle}}_{ab}$$

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 $j_b=\frac12$.

Fig­ure 12.6: Cleb­sch-Gor­dan co­ef­fi­cients when the sec­ond an­gu­lar mo­men­tum con­tri­bu­tion has az­imuthal quan­tum num­ber $j_b=1$.

How about the Cleb­sch-Gor­dan co­ef­fi­cients to com­bine other lad­ders than the spins of two spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ 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.65} 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.66}.