12.9 Mo­men­tum of par­tially filled shells

One very im­por­tant case of com­bin­ing an­gu­lar mo­menta oc­curs for both elec­trons in atoms and nu­cle­ons in nu­clei. In these prob­lems there are a num­ber of iden­ti­cal fermi­ons in sin­gle-par­ti­cle states that dif­fer only in the net (or­bital plus spin) mo­men­tum in the cho­sen $z$-​di­rec­tion. Loosely speak­ing, the sin­gle-par­ti­cle states are the same, just at dif­fer­ent an­gu­lar ori­en­ta­tions. Such a set of states is of­ten called a shell. The ques­tion is then: what com­bi­na­tions of the states are an­ti­sym­met­ric with re­spect to ex­change of the fermi­ons, and there­fore al­lowed? More specif­i­cally, what is their com­bined net an­gu­lar mo­men­tum?


Ta­ble 12.1: Pos­si­ble com­bined an­gu­lar mo­men­tum of iden­ti­cal fermi­ons in shells of sin­gle-par­ti­cle states that dif­fer in mag­netic quan­tum num­ber. The top shows odd num­bers of par­ti­cles, the bot­tom even num­bers.
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The an­swer is given in ta­ble 12.1, {D.67}. In it, $I$ is the num­ber of fermi­ons in the shell. Fur­ther $j^{\rm p}$ is the net an­gu­lar mo­men­tum of the sin­gle-par­ti­cle states that make up the shell. (Or the az­imuthal quan­tum num­ber of that an­gu­lar mo­men­tum re­ally.) Sim­i­larly the val­ues of $j$ in­di­cate the pos­si­ble net an­gu­lar mo­men­tum quan­tum num­bers of all $i$ fermi­ons com­bined. The main body of the ta­ble lists the mul­ti­plic­ity of sets with the given an­gu­lar mo­men­tum. Note that the ta­ble is split into odd and even num­bers of par­ti­cles. That sim­pli­fies the pre­sen­ta­tion, be­cause odd num­bers of par­ti­cles pro­duce only half-in­te­ger net an­gu­lar mo­men­tum, and even num­bers only in­te­ger net an­gu­lar mo­men­tum.

For ex­am­ple, con­sider a sin­gle par­ti­cle, $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, in a set of sin­gle-par­ti­cle states with an­gu­lar mo­men­tum $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. For a sin­gle par­ti­cle, the com­bined mo­men­tum $j$ is sim­ply the sin­gle par­ti­cle mo­men­tum $j^{\rm p}$, ex­plain­ing the sin­gle 1 in the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ col­umn. But note that the 1 stands for a set of states; the mag­netic net quan­tum num­ber $m^{\rm p}$ of the sin­gle par­ti­cle could still be any one of $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, ..., $-\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. All the ten states in this set have net an­gu­lar mo­men­tum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$.

Next as­sume that there are two par­ti­cles in the same $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ sin­gle-par­ti­cle states. Then if both par­ti­cles would be in the $m^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ sin­gle-par­ti­cle state, their com­bined an­gu­lar mo­men­tum in the $z$-​di­rec­tion $m$ would be 2 $\times$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9. Fol­low­ing the Cleb­sch-Gor­dan de­riva­tion shows that this state would have com­bined an­gu­lar mo­men­tum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9. But the two iden­ti­cal fermi­ons can­not be both in the $m^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ state; that vi­o­lates the Pauli ex­clu­sion prin­ci­ple. That is why there is no en­try in the $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9 col­umn. If the first par­ti­cle is in the $m^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ state, the sec­ond one can at most be in the $m^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ state, for a to­tal of $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8. More pre­cisely, the par­ti­cles would have to be in the an­ti­sym­met­ric com­bi­na­tion, or Slater de­ter­mi­nant, of these two states. That an­ti­sym­met­ric com­bi­na­tion can be seen to have com­bined an­gu­lar mo­men­tum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8. There are other com­bi­na­tions of states that also have $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8, but val­ues of $m$ equal to 7, 6, ..., $\vphantom0\raisebox{1.5pt}{$-$}$8, for a to­tal of 17 states. That set of 17 states is in­di­cated by the 1 in the $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8 col­umn.

It is also pos­si­ble for the two $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cles to com­bine their an­gu­lar mo­men­tum into smaller even val­ues of the to­tal an­gu­lar mo­men­tum $j$. In fact, it is pos­si­ble for the par­ti­cles to com­bine their an­gu­lar mo­menta so that they ex­actly can­cel one an­other; then the net an­gu­lar mo­men­tum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That is in­di­cated by the 1 in the $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 col­umn. Clas­si­cally you would say that the mo­men­tum vec­tors of the two par­ti­cles are ex­actly op­po­site, pro­duc­ing a zero re­sul­tant. In quan­tum me­chan­ics true an­gu­lar mo­men­tum vec­tors do not ex­ist due to un­cer­tainty of the com­po­nents, but com­plete can­ce­la­tion is still pos­si­ble.

The $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 set con­sists of just one state, be­cause $m$ can only be zero for a state with zero an­gu­lar mo­men­tum. The en­tire ta­ble row for two $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cles could in prin­ci­ple be de­rived by writ­ing out the ap­pro­pri­ate Cleb­sch-Gor­dan co­ef­fi­cients. But that would be one very big ta­ble.

If there are five $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cles. they can com­bine their an­gu­lar mo­menta into quite a wide va­ri­ety of net an­gu­lar mo­men­tum val­ues. For ex­am­ple, the 2 in the $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ col­umn in­di­cates that there are two sets of states with com­bined an­gu­lar mo­men­tum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. Each set has 6 mem­bers, be­cause for each set $m$ can be any one of $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, ..., $-\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. So there are a to­tal of 12 in­de­pen­dent com­bi­na­tion states that have net an­gu­lar mo­men­tum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$.

Note that a shell has $2j^{\rm p}+1$ dif­fer­ent sin­gle-par­ti­cle states, be­cause the mag­netic quan­tum num­ber $m^{\rm p}$ can have the val­ues $j^{\rm p}$, $j^{\rm p}{-}1$, ..., $\vphantom0\raisebox{1.5pt}{$-$}$$j^{\rm p}$. There­fore a shell can ac­com­mo­date up to $2j^{\rm p}+1$ fermi­ons ac­cord­ing to the ex­clu­sion prin­ci­ple. How­ever, the ta­ble only lists com­bined an­gu­lar mo­men­tum val­ues for up to $j^{\rm p}+\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cles. The rea­son is that any more is un­nec­es­sary. A given num­ber of holes in an oth­er­wise filled shell pro­duces the same com­bined an­gu­lar mo­men­tum val­ues as the same num­ber of par­ti­cles in an oth­er­wise empty shell. For ex­am­ple, two fermi­ons in a $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ shell, (zero holes), have the same com­bined an­gu­lar mo­men­tum as zero par­ti­cles: zero. In­deed, those two fermi­ons must be in the an­ti­sym­met­ric sin­glet state with spin zero. In gen­eral, a com­pletely filled shell has zero an­gu­lar mo­men­tum and is spher­i­cally sym­met­ric.


Ta­ble 12.2: Pos­si­ble com­bined an­gu­lar mo­men­tum of iden­ti­cal bosons.
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The same sit­u­a­tion for iden­ti­cal bosons is shown in ta­ble 12.2. For iden­ti­cal bosons there is no limit to the num­ber of par­ti­cles that can go into a shell. The ta­ble was ar­bi­trar­ily cut off at 9 par­ti­cles and a max­i­mum spin of 18.