12.8 Some im­por­tant re­sults

This sec­tion gives some re­sults that are used fre­quently in quan­tum analy­sis, but usu­ally not ex­plic­itly stated.

First a note on no­ta­tions. It is fairly com­mon to use the let­ter $l$ for or­bital an­gu­lar mo­men­tum, $s$ for spin, and $j$ for com­bi­na­tions of or­bital and an­gu­lar mo­men­tum. This sub­sec­tion will fol­low these con­ven­tions where ap­pro­pri­ate.

If all pos­si­ble an­gu­lar mo­men­tum states are filled with a fermion, the re­sult­ing an­gu­lar mo­men­tum is zero and the wave func­tion is spher­i­cally sym­met­ric. For ex­am­ple, con­sider the sim­pli­fied case that there is one spin­less fermion in each spher­i­cal har­monic at a given az­imuthal quan­tum num­ber $l$. Then it is easy to see from the form of the spher­i­cal har­mon­ics that the com­bined wave func­tion is in­de­pen­dent of the an­gu­lar po­si­tion around the $z$-​axis. And all spher­i­cal har­mon­ics at that $l$ are filled what­ever you take to be the $z$-​axis. This makes no­ble gasses into the equiv­a­lent of bil­liard balls. More gen­er­ally, if there is one fermion for every pos­si­ble di­rec­tion of the an­gu­lar mo­men­tum, by sym­me­try the net an­gu­lar mo­men­tum can only be zero.

If a spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ fermion has or­bital an­gu­lar mo­men­tum quan­tum num­ber $l$, net (or­bital plus spin) an­gu­lar mo­men­tum quan­tum num­ber $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l+{\textstyle\frac{1}{2}}$, and net mo­men­tum in the $z$-​di­rec­tion quan­tum num­ber $m_j$, its net state is given in terms of the in­di­vid­ual or­bital and spin states as:

{\left\vert j\:m_j\right...
+ \sqrt{\frac{j-m_j}{2j}}Y_l^{m_j+\frac12}{\downarrow}

If the net spin is $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l-{\textstyle\frac{1}{2}}$, as­sum­ing that $l$ $\raisebox{.3pt}{$>$}$ 0, that be­comes

j=l-{\textstyle\frac{1}{2}}: \quad\!\!
{\left\vert j\:m_j\...
+ \sqrt{\frac{j+1+m_j}{2j+2}}Y_l^{m_j+\frac12}{\downarrow}

Note that if the net an­gu­lar mo­men­tum is un­am­bigu­ous, the or­bital and spin mag­netic quan­tum num­bers $m$ and $m_s$ are in gen­eral un­cer­tain.

For iden­ti­cal par­ti­cles, an im­por­tant ques­tion is how the Cleb­sch-Gor­dan co­ef­fi­cients change un­der par­ti­cle ex­change:

\langle j_{ab} m_{ab} \vert{\left\vert j_a\:m_a\right\rangl...
...ert j_b\:m_b\right\rangle} {\left\vert j_a\:m_a\right\rangle}

For $j_a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j_b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, this ver­i­fies that the triplet states $j_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 are sym­met­ric, and the sin­glet state $j_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is an­ti­sym­met­ric. More gen­er­ally, states with the max­i­mum net an­gu­lar mo­men­tum $j_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j_a+j_b$ and whole mul­ti­ples of 2 less are sym­met­ric un­der par­ti­cle ex­change. States that are odd amounts less than the max­i­mum are an­ti­sym­met­ric un­der par­ti­cle ex­change.

When the net an­gu­lar mo­men­tum state is swapped with one of the com­po­nent states, the re­la­tion is

\lefteqn{\langle j_{ab} m_{ab} \vert{\left\vert j_a\:m_a\righ...
...ngle} {\left\vert j_b\:\rule[2.5pt]{5pt}{.5pt}m_b\right\rangle}

This is of in­ter­est in fig­ur­ing out what states pro­duce zero net an­gu­lar mo­men­tum, $j_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. In that case, the right hand side is zero un­less $j_b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j_a$ and $m_b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-m_a$; and then $\langle{j}_am_a\vert{\left\vert\:0\right\rangle}{\left\vert j_a\:m_a\right\rangle}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. You can only cre­ate zero an­gu­lar mo­men­tum from a pair of par­ti­cles that have the same square an­gu­lar mo­men­tum; also, only prod­uct states with zero net an­gu­lar mo­men­tum in the $z$-​di­rec­tion are in­volved.