### D.67 Mo­men­tum of shells

Ta­ble 12.1 was orig­i­nally taken from [36], who in turn took it from the book of Mayer and Jensen. How­ever, the fi­nal ta­ble con­tains three ty­pos, as can be seen from the fact that in three cases the num­bers of states do not add up to the cor­rect to­tal. (The er­rors are: for 3 par­ti­cles with spin 9/2, the 13/2 com­bined state is omit­ted, for 4 par­ti­cles with spin 9/2, the spin 8 state should be dou­ble, and for 4 par­ti­cles with spin 11/2, a spin 7 (dou­ble) state is miss­ing. Sim­i­larly, [5, p. 140] has the same miss­ing 13/2 com­bined state, and in ad­di­tion for 3 par­ti­cles with spin 7/2, there is a 1/2 state that should not be there.)

So ta­ble 12.1 was in­stead com­puter-gen­er­ated, and should there­fore be free of ty­pos. Since the pro­gram had to be writ­ten any­way, some more val­ues were gen­er­ated and are in ta­ble D.1.

De­duc­ing the ta­ble us­ing Cleb­sch-Gor­dan co­ef­fi­cients would be a messy ex­er­cise in­deed. A sim­pler pro­ce­dure, [31], will here be il­lus­trated for the ex­am­ple that the num­ber of fermi­ons is 3 and the an­gu­lar mo­men­tum of the sin­gle-par­ti­cle states is . Then the pos­si­bil­i­ties for the sin­gle-par­ti­cle an­gu­lar mo­men­tum in the -​di­rec­tion are , , , , , and . So there are 6 dif­fer­ent one par­ti­cle states, and these will give rise to 6!​3!(6-3)! = 20 dif­fer­ent an­ti­sym­met­ric states for 3 par­ti­cles, chap­ter 5.7.

The com­bi­na­tion states can be cho­sen to have def­i­nite val­ues of the com­bined an­gu­lar mo­men­tum and mo­men­tum in the -​di­rec­tion . In the ab­sence of any an­ti­sym­metriza­tion re­quire­ments, that can be seen from the way that states com­bine us­ing Cleb­sch-Gor­dan co­ef­fi­cients. And these states of def­i­nite com­bined an­gu­lar mo­men­tum must ei­ther be an­ti­sym­met­ric and al­low­able, or sym­met­ric and not al­lowed. The rea­son is that ex­chang­ing fermi­ons does not do any­thing phys­i­cally, since the fermi­ons are iden­ti­cal. So the an­gu­lar mo­men­tum and par­ti­cle ex­change op­er­a­tors com­mute. There­fore, the eigen­states of the an­gu­lar mo­men­tum op­er­a­tors can also be taken to be eigen­states of the par­ti­cle ex­change op­er­a­tors, which means ei­ther sym­met­ric (eigen­value 1) or an­ti­sym­met­ric (eigen­value 1).

Let be the to­tal mag­netic quan­tum num­ber of the 3 fermi­ons in any com­bi­na­tion of sin­gle-par­ti­cle states. First note that is the sum of the three val­ues of the in­di­vid­ual par­ti­cles. Next, the high­est that can be is , but the fermi­ons can­not all three be in the same state, only one can. Three fermi­ons need three dif­fer­ent states, so the high­est the com­bined can be is . This triplet of val­ues of gives ex­actly one an­ti­sym­met­ric com­bi­na­tion of states with . (There is only one Slater de­ter­mi­nant for three dif­fer­ent given states, chap­ter 5.7). Since the com­bined an­gu­lar mo­men­tum of this state in any ar­bi­trary di­rec­tion can never be ob­served to be more than , be­cause that would vi­o­late the above ar­gu­ment in a ro­tated co­or­di­nate sys­tem, it must be a state. The first con­clu­sion is there­fore that the an­gu­lar mo­menta can­not com­bine into a to­tal greater than . And since can­not be less than , there must be states with .

But note that if is a valid com­bi­na­tion of sin­gle-par­ti­cle states, then so should be the states with for the other val­ues of ; these can be thought of as fully equiv­a­lent states sim­ply ori­ented un­der a dif­fer­ent an­gle. That means that there are a to­tal of 10 com­bi­na­tion states with , in which is any one of , , ..., .

Next con­sider what com­bi­na­tions have . The only com­bi­na­tion of three dif­fer­ent val­ues that adds up to is . So there is only one com­bined state with . Since it was al­ready in­ferred above that there must be one such state with , that must be the only one. So ap­par­ently there is no state with : such a state would show up as a sec­ond state un­der the right ori­en­ta­tion.

There are two in­de­pen­dent pos­si­bil­i­ties to cre­ate a triplet of dif­fer­ent states with : or . One com­bi­na­tion of such a type is al­ready iden­ti­fied as be­ing a state, so the sec­ond must cor­re­spond to a state. Since the ori­en­ta­tion should again not make a dif­fer­ence, there must be a to­tal of 6 such states, one for each of the dif­fer­ent val­ues of in the range from to .

There are three ways to cre­ate a triplet of states with : , , and . Two of these are al­ready iden­ti­fied as be­ing and , so there must be one set of 4 states with .

That makes a to­tal of 20 states, so there must not be any states with . In­deed, there are only three ways to pro­duce : , , and , and each of these three states is al­ready as­signed to a value of .

It is tricky, but it works. And it is eas­ily put on a com­puter.

For bosons, the idea is the same, ex­cept that states with equal val­ues of can no longer be ex­cluded.