Sub­sec­tions


14.12 Draft: Shell model

The liq­uid drop model gives a very use­ful de­scrip­tion of many nu­clear prop­er­ties. It helps un­der­stand al­pha de­cay quite well. Still, it has def­i­nite lim­i­ta­tions. Quan­tum prop­er­ties such as the sta­bil­ity of in­di­vid­ual nu­clei, spin, mag­netic mo­ment, and gamma de­cay can sim­ply not be ex­plained us­ing a clas­si­cal liq­uid model with a cou­ple of sim­ple fixes ap­plied.

His­tor­i­cally, a ma­jor clue about a suit­able quan­tum model came from the magic num­bers. Nu­clei tend to be un­usu­ally sta­ble if the num­ber of pro­tons and/or neu­trons is one of the

\begin{displaymath}
\mbox{magic numbers:}\quad 2, 8, 20, 28, 50, 82, 126, \ldots
\end{displaymath} (14.14)

The higher magic num­ber val­ues are quite clearly seen in pro­ton pair and neu­tron pair re­moval graphs like fig­ures 14.7 and 14.8 in sec­tion 14.8.

If an ad­di­tional pro­ton is added to a nu­cleus with a magic num­ber of pro­tons, or an ad­di­tional neu­tron to a nu­cleus with a magic num­ber of neu­trons, then that ad­di­tional nu­cleon is much more weakly bound.

The dou­bly magic ${}\fourIdx{4}{2}{}{}{\rm {He}}$ he­lium-4 nu­cleus, with 2 pro­tons and 2 neu­trons, is a good ex­am­ple. It has more than three times the bind­ing en­ergy of ${}\fourIdx{3}{2}{}{}{\rm {He}}$ he­lium-3, which merely has a magic num­ber of pro­tons. Still, if you try to add an­other pro­ton or neu­tron to he­lium-4, it will not be bound at all, it will be ejected in less than 10$\POW9,{-21}$ sec­onds.

That is very rem­i­nis­cent of the elec­tron struc­ture of the he­lium atom. The two elec­trons in the he­lium atom are very tightly bound, mak­ing he­lium into an in­ert no­ble gas. In fact, it takes 25 eV of en­ergy to re­move an elec­tron from a he­lium atom. How­ever, for lithium, with one more elec­tron, the third elec­tron is very loosely bound, and read­ily given up in chem­i­cal re­ac­tions. It takes only 5.4 eV to re­move the third elec­tron from lithium. Sim­i­lar ef­fects ap­pear for the other no­ble gasses, neon with 10 elec­trons, ar­gon with 18, kryp­ton with 36, etcetera. The num­bers 2, 10, 18, 36, ..., are magic for elec­trons in atoms.

For atoms, the un­usual sta­bil­ity could be ex­plained in chap­ter 5.9 by ig­nor­ing the di­rect in­ter­ac­tions be­tween elec­trons. It was as­sumed that for each elec­tron, the com­pli­cated ef­fects of all the other elec­trons could be mod­eled by some av­er­age po­ten­tial that the elec­tron moves in. That ap­prox­i­ma­tion pro­duced sin­gle-elec­tron en­ergy eigen­func­tions for the elec­trons. They then had to oc­cupy these sin­gle-elec­tron states one by one on ac­count of Pauli’s ex­clu­sion prin­ci­ple. No­ble gasses com­pletely fill up an en­ergy level, re­quir­ing any ad­di­tional elec­trons to go into the next avail­able, sig­nif­i­cantly higher en­ergy level. That greatly de­creases the bind­ing en­ergy of these ad­di­tional elec­trons com­pared to those al­ready there.

The sim­i­lar­ity sug­gests that the pro­tons and neu­trons in nu­clei might be de­scribed sim­i­larly. There are now two types of par­ti­cles but in the ap­prox­i­ma­tion that each par­ti­cle is not di­rectly af­fected by the oth­ers it does not make much of a dif­fer­ence. Also, an­ti­sym­metriza­tion re­quire­ments only ap­ply when the par­ti­cles are iden­ti­cal, ei­ther both pro­tons or both neu­trons. There­fore, pro­tons and neu­trons can be treated com­pletely sep­a­rately. Their in­ter­ac­tions oc­cur only in­di­rectly through what­ever is used for the av­er­age po­ten­tial that they move in. The next sub­sec­tions work out a model along these lines.


14.12.1 Draft: Av­er­age po­ten­tial

The first step will be to iden­tify a suit­able av­er­age po­ten­tial for the nu­cle­ons. One ob­vi­ous dif­fer­ence dis­tin­guish­ing nu­clei from atoms is that the Coulomb po­ten­tial is not go­ing to hack it. In the elec­tron struc­ture of an atom the elec­trons re­pel each other, and the only rea­son the atom stays to­gether is that there is a nu­cleus to at­tract the elec­trons. But in­side a nu­cleus, the nu­cle­ons all at­tract each other and there is no ad­di­tional at­trac­tive core. In­deed, a Coulomb po­ten­tial like the one used for the elec­trons in atoms would get only the first magic num­ber, 2, right, pre­dict­ing 10, in­stead of 8, to­tal par­ti­cles for a filled sec­ond en­ergy level.

A bet­ter po­ten­tial is needed. Now in the cen­ter of a nu­cleus, the at­trac­tive forces come from all di­rec­tions and the net force will be zero by sym­me­try. Away from the cen­ter, the net force will be di­rected in­wards to­wards the cen­ter to keep the nu­cle­ons to­gether in­side the nu­cleus. The sim­plest po­ten­tial that de­scribes this is the har­monic os­cil­la­tor one. For that po­ten­tial, the in­ward force is sim­ply pro­por­tional to the dis­tance from the cen­ter. That makes the po­ten­tial en­ergy $V$ pro­por­tional to the square dis­tance from the cen­ter, as sketched in fig­ure 14.13a.

Fig­ure 14.13: Ex­am­ple av­er­age nu­clear po­ten­tials: (a) har­monic os­cil­la­tor, (b) im­pen­e­tra­ble sur­face, (c) Woods-Saxon, (d) Woods-Saxon for pro­tons.
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,70...
...]{\em (c)}}
\put(150,70){\makebox(0,0)[t]{\em (d)}}
\end{picture}
\end{figure}

The en­ergy eigen­val­ues of the har­monic os­cil­la­tor are

\begin{displaymath}
E_n = \left(n+{\textstyle\frac{1}{2}}\right)\hbar\omega
\qquad n = 1, 2, 3, \ldots %
\end{displaymath} (14.15)

Also, in spher­i­cal co­or­di­nates the en­ergy eigen­func­tions of the har­monic os­cil­la­tor can be taken to be of the form, {D.76},
\begin{displaymath}
\psi_{nlmm_s}^{\rm ho} = R_{nl}^{\rm ho}(r) Y_l^m(\theta,\p...
...s, l{-}1, l \\
m_s=\pm{\textstyle\frac{1}{2}}
\end{array} %
\end{displaymath} (14.16)

Here $l$ is the az­imuthal quan­tum num­ber that gives the square or­bital an­gu­lar mo­men­tum of the state as $l(l+1)\hbar^2$; $m$ is the mag­netic quan­tum num­ber that gives the or­bital an­gu­lar mo­men­tum in the di­rec­tion of the ar­bi­trar­ily cho­sen $z$-​axis as $m\hbar$, and $m_s$ is the spin quan­tum num­ber that gives the spin an­gu­lar mo­men­tum of the nu­cleon in the $z$-​di­rec­tion as $m_s\hbar$. The spin-up state with $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}$ is com­monly in­di­cated by a post­fix ${\uparrow}$, and sim­i­larly the spin-down one $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-{\textstyle\frac{1}{2}}$ by ${\downarrow}$. The de­tails of the func­tions $R_{nl}^{\rm {ho}}$ and $Y_l^m$ are of no par­tic­u­lar in­ter­est.

(It may be noted that the above spher­i­cal eigen­func­tions are dif­fer­ent from the Carte­sian ones de­rived in chap­ter 4.1, ex­cept for the ground state. How­ever, the spher­i­cal eigen­func­tions at a given en­ergy level can be writ­ten as com­bi­na­tions of the Carte­sian ones at that level, and vice-versa. So there is no fun­da­men­tal dif­fer­ence be­tween the two. It just works out that the spher­i­cal ver­sions are much more con­ve­nient in the rest of the story.)

Com­pared to the Coulomb po­ten­tial of the hy­dro­gen elec­tron as solved in chap­ter 4.3, the ma­jor dif­fer­ence is in the num­ber of en­ergy states at a given en­ergy level $n$. While for the Coulomb po­ten­tial the az­imuthal quan­tum num­ber $l$ can have any value from 0 to $n-1$, for the har­monic os­cil­la­tor $l$ must be odd or even de­pend­ing on whether $n-1$ is odd or even.

It does not make a dif­fer­ence for the low­est en­ergy level $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1; in that case only $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is al­lowed for ei­ther po­ten­tial. And since the num­ber of val­ues of the mag­netic quan­tum num­ber $m$ at a given value of $l$ is $2l+1$, there is only one pos­si­ble value for $m$. That means that there are only two dif­fer­ent en­ergy states at the low­est en­ergy level, cor­re­spond­ing to $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$ re­spec­tively $-\frac12$. Those two states ex­plain the first magic num­ber, 2. Two nu­cle­ons of a given type can oc­cupy the low­est en­ergy level; any fur­ther ones of that type must go into a higher level.

In par­tic­u­lar, he­lium-4 has the low­est en­ergy level for pro­tons com­pletely filled with its two pro­tons, and the low­est level for neu­trons com­pletely filled with its two neu­trons. That makes he­lium-4 the first dou­bly-magic nu­cleus. It is just like the two elec­trons in the he­lium atom com­pletely fill the low­est en­ergy level for elec­trons, mak­ing he­lium the first no­ble gas.

At the sec­ond en­ergy level $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, where the Coulomb po­ten­tial al­lows both $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, only $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is al­lowed for the har­monic os­cil­la­tor. So the num­ber of states avail­able at en­ergy level $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 is less than that of the Coulomb po­ten­tial. In par­tic­u­lar, the az­imuthal quan­tum num­ber $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 al­lows $2l+1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 val­ues of the mag­netic quan­tum num­ber $m$, times 2 val­ues for the spin quan­tum num­ber $m_s$. There­fore, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 at $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 cor­re­sponds to 3 times 2, or 6 en­ergy states. Com­bined with the two $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 states at en­ergy level $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, that gives a to­tal of 8. The sec­ond magic num­ber 8 has been ex­plained! It re­quires 8 nu­cle­ons of a given type to fill the low­est two en­ergy lev­els.

It makes oxy­gen-16 with 8 pro­tons and 8 neu­trons the sec­ond dou­bly-magic nu­cleus. Note that for the elec­trons in atoms, the sec­ond en­ergy level would also in­clude two $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 states. That is why the sec­ond no­ble gas is neon with 10 elec­trons, and not oxy­gen with 8.

Be­fore check­ing the other magic num­bers, first a prob­lem with the above pro­ce­dure of count­ing states must be ad­dressed. It is too easy. Every­body can eval­u­ate $2l+1$ and mul­ti­ply by 2 for the spin states! To make it more chal­leng­ing, physi­cists adopt the so-called spec­tro­scopic no­ta­tion in which they do not tell you the value of $l$. In­stead, they tell you a let­ter like maybe p, and you are then sup­posed to fig­ure out your­self that $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. The scheme is:

\begin{displaymath}
\mbox{s, p, d, f, g, h, i, [j], k, \ldots}
\qquad\Longrightarrow\qquad
l=0,1,2,3,4,5,6,7,8,\ldots
\end{displaymath}

The lat­ter part is mostly al­pha­betic, but by con­ven­tion j is not in­cluded. How­ever, my ref­er­ences on nu­clear physics do in­clude j; that is great be­cause it pro­vides ad­di­tional chal­lenge. Us­ing spec­tro­scopic no­ta­tions, the sec­ond en­ergy level states are reno­tated as

\begin{displaymath}
\psi_{21mm_s}
\qquad\Longrightarrow\qquad
2p
\end{displaymath}

where the 2 in­di­cates the value of $n$ giv­ing the en­ergy level. The ad­di­tional de­pen­dence on the mag­netic quan­tum num­bers $m$ and $m_s$ is kept hid­den from the unini­ti­ated. (To be fair, as long as there is no pre­ferred di­rec­tion to space, these quan­tum num­bers are phys­i­cally not of im­por­tance. If an ex­ter­nal mag­netic field is ap­plied, it pro­vides di­rec­tion­al­ity, and mag­netic quan­tum num­bers do be­come rel­e­vant.)

How­ever, physi­cists fig­ured that this would not pro­vide chal­lenge enough, since most stu­dents al­ready prac­ticed it for atoms. The above no­ta­tion fol­lows the one that physi­cists use for atoms. In this no­ta­tion, the lead­ing num­ber is $n$, the en­ergy level of the sim­plest the­o­ret­i­cal model. To pro­vide more chal­lenge, for nu­clei physi­cist re­place the lead­ing num­ber with a count of states at that an­gu­lar mo­men­tum. For ex­am­ple, physi­cists de­note 2p above by 1p, be­cause it is the low­est en­ergy p states. Damn what the­o­ret­i­cal en­ergy level it is. For still more chal­lenge, while most physi­cists start count­ing from one, some start from zero, mak­ing it 0p. How­ever, since it gives the au­thor of this book a headache to count an­gu­lar mo­men­tum states up­wards be­tween shells, this book will mostly fol­low the atomic con­ven­tion, and the lead­ing digit will in­di­cate $n$, the har­monic os­cil­la­tor en­ergy level. The of­fi­cial eigen­func­tion des­ig­na­tions will be listed in the fi­nal re­sults where ap­pro­pri­ate. Most but not all ref­er­ences will fol­low the of­fi­cial des­ig­na­tions.

In these terms, the en­ergy lev­els and num­bers of states for the har­monic os­cil­la­tor po­ten­tial are as shown in fig­ure 14.14. The third en­ergy level has 2 3s states and 10 3d states. Added to the 8 from the first two en­ergy lev­els, that brings the to­tal count to 20, the third magic num­ber.

Fig­ure 14.14: Nu­clear en­ergy lev­els for var­i­ous as­sump­tions about the av­er­age nu­clear po­ten­tial. The signs in­di­cate the par­ity of the states.
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,27...
...\tt 40?}}
\put(243,199){\makebox(0,0)[ct]{\tt 70?}}
\end{picture}
\end{figure}

Un­for­tu­nately, this is where it stops. The fourth en­ergy level should have only 8 states to reach the next magic num­ber 28, but in re­al­ity the fourth har­monic os­cil­la­tor level has 6 4p states and 14 4f ones. Still, get­ting 3 magic num­bers right seems like a good start.

The log­i­cal next step is to try to im­prove upon the har­monic os­cil­la­tor po­ten­tial. In an av­er­age nu­cleus, it can be ex­pected that the net force on a nu­cleon pretty much av­er­ages out to zero every­where ex­cept in a very thin layer at the outer sur­face. The rea­son is that the nu­clear forces are very short range; there­fore the forces seem to come equally from all di­rec­tions un­less the nu­cleon is very close to the sur­face. Only right at the sur­face do the par­ti­cles ex­pe­ri­ence a net in­ward at­trac­tion be­cause of the deficit of par­ti­cles be­yond the sur­face to pro­vide the full com­pen­sat­ing out­ward force. This sug­gests a pic­ture in which the nu­cle­ons do not ex­pe­ri­ence a net force within the con­fines of the nu­cleus. How­ever, at the sur­face, the po­ten­tial ramps up very steeply. As an ide­al­iza­tion the po­ten­tial be­yond the sur­face can be taken in­fi­nite.

That rea­son­ing re­sults in the im­pen­e­tra­ble-shell po­ten­tial shown in fig­ure 14.13. It too is an­a­lyt­i­cally solv­able, {D.77}. The en­ergy lev­els are shown in fig­ure 14.14. Un­for­tu­nately, it does not help any ex­plain­ing the fourth magic num­ber 28.

It does help un­der­stand why the shell model works at all, [[15]]. That is not at all ob­vi­ous; for a long time physi­cists re­ally be­lieved it would not work. For the elec­trons in an atom, the nu­cleus at least pro­duces some po­ten­tial that is in­de­pen­dent of the rel­a­tive po­si­tions of the elec­trons. In a nu­cleus, there is noth­ing: the po­ten­tial ex­pe­ri­enced by the nu­cle­ons is com­pletely de­pen­dent on rel­a­tive nu­cleon po­si­tions and spins. So what rea­son­able jus­ti­fi­ca­tion could there pos­si­bly be to as­sume that the nu­cle­ons act as if they move in an av­er­age po­ten­tial that is in­de­pen­dent of the other nu­cle­ons? How­ever, first as­sume that the only po­ten­tial en­ergy is the one that keeps the nu­cle­ons within the ex­per­i­men­tal nu­clear ra­dius. That is the im­pen­e­tra­ble shell model. In that case, the en­ergy eigen­func­tions are purely ki­netic en­ergy ones, and these have a shell struc­ture. Now re­store the ac­tual com­plex in­ter­ac­tions be­tween nu­cle­ons. You would at first guess that these should greatly change the en­ergy eigen­states. But if they re­ally do that, it would bring in large amounts of un­oc­cu­pied ki­netic en­ergy states. That would pro­duce a sig­nif­i­cant in­crease in ki­netic en­ergy, and that is not pos­si­ble be­cause the bind­ing en­ergy is fairly small com­pared to the ki­netic en­ergy. In par­tic­u­lar, there­fore, re­mov­ing the last nu­cleon should not re­quire an en­ergy very dif­fer­ent from a shell model value re­gard­less of how­ever com­plex the true po­ten­tial en­ergy re­ally is.

Of course, the im­pen­e­tra­ble-shell po­ten­tial too is open to crit­i­cism. A nu­cleus has maybe ten nu­cle­ons along a di­am­e­ter. Surely the thick­ness of the sur­face layer can­not rea­son­ably be much less than the spac­ing be­tween nu­cle­ons. Or much less than the range of the nu­clear forces, for that mat­ter. Also, the po­ten­tial should not be in­fi­nite out­side the nu­cleus; nu­cle­ons do es­cape from, or en­ter nu­clei with­out in­fi­nite en­ergy. The truth is clearly some­where in be­tween the har­monic os­cil­la­tor and im­pen­e­tra­ble shell po­ten­tials. A more re­al­is­tic po­ten­tial along such lines is the “Woods-Saxon” po­ten­tial

\begin{displaymath}
V = - \frac{V_0}{1+e^{(r-a)/d}} + \mbox{constant}
\end{displaymath}

which is sketched in fig­ure 14.13c. For pro­tons, there is an ad­di­tional re­pul­sive Coulomb po­ten­tial that will be max­i­mum at the cen­ter of the sphere and de­creases to zero pro­por­tional to 1$\raisebox{.5pt}{$/$}$$r$ out­side the nu­cleus. That gives a com­bined po­ten­tial as sketched in fig­ure 14.13d. Note that the Coulomb po­ten­tial is not short-range like the nu­cleon-nu­cleon at­trac­tions; its non­triv­ial vari­a­tion is not just re­stricted to a thin layer at the nu­clear sur­face.

Typ­i­cal en­ergy lev­els are sketched in fig­ure 14.14. As ex­pected, they are some­where in be­tween the ex­treme cases of the har­monic os­cil­la­tor and the im­pen­e­tra­ble shell.

The signs be­hind the re­al­is­tic en­ergy lev­els in 14.14 de­note the pre­dicted “par­ity” of the states. Par­ity is a very help­ful math­e­mat­i­cal quan­tity for study­ing nu­clei. The par­ity of a wave func­tion is one,” or “pos­i­tive, or even, if the wave func­tion stays the same when the pos­i­tive di­rec­tion of the three Carte­sian axes is in­verted. That re­places every ${\skew0\vec r}$ in the wave func­tion by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. The par­ity is mi­nus one,”, or “neg­a­tive, or odd, if the wave func­tion merely changes sign un­der an exes in­ver­sion. Par­ity is un­cer­tain when the wave func­tion changes in any other way; how­ever, nu­clei have def­i­nite par­ity as long as the weak force of beta de­cay does not play a role. It turns out that s, d, g, ...states have pos­i­tive par­ity while p, f, h, ...states have neg­a­tive par­ity, {D.14} or {D.76}. There­fore, the har­monic os­cil­la­tor shells have al­ter­nat­ingly pos­i­tive and neg­a­tive par­ity.

For the wave func­tions of com­plete nu­clei, the net par­ity is the prod­uct of the par­i­ties, (tak­ing them to be one or mi­nus one), of the in­di­vid­ual nu­cle­ons. Now physi­cist can ex­per­i­men­tally de­duce the par­ity of nu­clei in var­i­ous ways. It turns out that the par­i­ties of the nu­clei up to the third magic num­ber agree per­fectly with the val­ues pre­dicted by the en­ergy lev­els of fig­ure 14.14. (Only three un­sta­ble, ar­ti­fi­cially cre­ated, nu­clei dis­agree.) It re­ally ap­pears that the model is onto some­thing.

Un­for­tu­nately, the fourth magic num­ber re­mains un­ex­plained. In fact, any rea­son­able spher­i­cally sym­met­ric spa­tial po­ten­tial will not get the fourth magic num­ber right. There are 6 4p states and 14 4f ones; how could the ad­di­tional 8 states needed for the next magic num­ber 28 ever be ex­tracted from that? Twid­dling with the shape of a purely spa­tial po­ten­tial is not enough.


14.12.2 Draft: Spin-or­bit in­ter­ac­tion

Even­tu­ally, Mayer in the U.S., and in­de­pen­dently Jensen and his co-work­ers in Ger­many, con­cluded that spin had to be in­volved in ex­plain­ing the magic num­bers above 20. To un­der­stand why, con­sider the six 4p and four­teen 4f en­ergy states at the fourth en­ergy level of the har­monic os­cil­la­tor model. Clearly, the six 4p states can­not pro­duce the eight states of the en­ergy shell needed to ex­plain the next magic num­ber 28. And nei­ther can the four­teen 4f states, un­less for some rea­son they split into two dif­fer­ent groups whose en­ergy is no longer equal.

Why would they split? In non­quan­tum terms, all four­teen states have or­bital and spin an­gu­lar mo­men­tum vec­tors of ex­actly the same lengths. What is dif­fer­ent be­tween states is only the di­rec­tion of these vec­tors. And the ab­solute di­rec­tions can­not be rel­e­vant since the physics can­not de­pend on the ori­en­ta­tion of the axis sys­tem in which it is viewed. What it can de­pend on is the rel­a­tive align­ment be­tween the or­bital and spin an­gu­lar mo­men­tum vec­tors. This rel­a­tive align­ment is char­ac­ter­ized by the dot prod­uct be­tween the two vec­tors.

There­fore, the log­i­cal way to get an en­ergy split­ting be­tween states with dif­fer­ently aligned or­bital and spin an­gu­lar mo­men­tum is to pos­tu­late an ad­di­tional con­tri­bu­tion to the Hamil­ton­ian of the form

\begin{displaymath}
\Delta H \propto - {\skew 4\widehat{\vec L}}\cdot {\skew 6\widehat{\vec S}}
\end{displaymath}

Here ${\skew 4\widehat{\vec L}}$ is the or­bital an­gu­lar mo­men­tum vec­tor and ${\skew 6\widehat{\vec S}}$ the spin one. A con­tri­bu­tion to the Hamil­ton­ian of this type is called an spin-or­bit in­ter­ac­tion, be­cause it cou­ples spin with or­bital an­gu­lar mo­men­tum. Spin-or­bit in­ter­ac­tion was al­ready known from im­proved de­scrip­tions of the en­ergy lev­els of the hy­dro­gen atom, ad­den­dum {A.39}. How­ever, that elec­tro­mag­netic ef­fect is far too small to ex­plain the ob­served spin-or­bit in­ter­ac­tion in nu­clei. Also, it would get the sign of the cor­rec­tion wrong for neu­trons.

While nu­clear forces re­main in­com­pletely un­der­stood, there is no doubt that it is these much stronger forces, and not elec­tro­mag­netic ones, that pro­vide the mech­a­nism. Still, in anal­ogy to the elec­tronic case, the con­stant of pro­por­tion­al­ity is usu­ally taken to in­clude the net force $\partial{V}$$\raisebox{.5pt}{$/$}$$\partial{r}$ on the nu­cleon and an ad­di­tional fac­tor 1$\raisebox{.5pt}{$/$}$$r$ to turn or­bital mo­men­tum into ve­loc­ity. None of that makes a dif­fer­ence for the har­monic os­cil­la­tor po­ten­tial, for which the net ef­fect is still just a con­stant. Ei­ther way, next the strength of the re­sult­ing in­ter­ac­tion is ad­justed to match the ex­per­i­men­tal en­ergy lev­els.

To cor­rectly un­der­stand the ef­fect of spin-or­bit in­ter­ac­tion on the en­ergy lev­els of nu­cle­ons is not quite triv­ial. Con­sider the four­teen $\psi_{43mm_s}$ 4f states. They have or­bital an­gu­lar mo­men­tum in the cho­sen $z$-​di­rec­tion $m\hbar$, with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ -3,-2,-1,0,1,2,3, and spin an­gu­lar mo­men­tum $m_s\hbar$ with $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm\frac12$. Naively, you might as­sume that the spin-or­bit in­ter­ac­tion low­ers the en­ergy of the six states for which $m$ and $m_s$ have the same sign, raises it for the six where they have the op­po­site sign, and leaves the en­ergy of the two states with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 the same. That is not true. The prob­lem is that the spin-or­bit in­ter­ac­tion ${\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}$ in­volves $\L _x$ and $\L _y$, and these do not com­mute with $\L _z$ re­gard­less of how you ori­ent the axis sys­tem. And the same for ${\widehat S}_x$ and ${\widehat S}_y$.

With spin-or­bit in­ter­ac­tion, en­ergy eigen­func­tions of nonzero or­bital an­gu­lar mo­men­tum no longer have def­i­nite or­bital mo­men­tum $L_z$ in a cho­sen $z$-​di­rec­tion. And nei­ther do they have def­i­nite spin $S_z$ in such a di­rec­tion.
There­fore the en­ergy eigen­func­tions can no longer be taken to be of the form $R_{nl}(r)Y_l^m(\theta,\phi){\updownarrow}$. They have un­cer­tainty in both $m$ and $m_s$, so they will be com­bi­na­tions of states $R_{nl}(r)Y_l^m(\theta,\phi){\updownarrow}$ with vary­ing val­ues of $m$ and $m_s$.

How­ever, con­sider the net an­gu­lar mo­men­tum op­er­a­tor

\begin{displaymath}
{\skew 6\widehat{\vec J}}\equiv {\skew 4\widehat{\vec L}}+ {\skew 6\widehat{\vec S}}
\end{displaymath}

If you ex­pand its square mag­ni­tude,

\begin{displaymath}
{\widehat J}^2 = ({\skew 4\widehat{\vec L}}+ {\skew 6\wideh...
...widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}+ {\widehat S}^2
\end{displaymath}

you see that the spin-or­bit term can be writ­ten in terms of the square mag­ni­tudes of or­bital, spin, and net an­gu­lar mo­men­tum op­er­a­tors:

\begin{displaymath}
- {\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}= ...
...rac{1}{2}}\left[{\widehat J}^2 - \L ^2 - {\widehat S}^2\right]
\end{displaymath}

There­fore com­bi­na­tion states that have def­i­nite square net an­gu­lar mo­men­tum $J^2$ re­main good en­ergy eigen­func­tions even in the pres­ence of spin-or­bit in­ter­ac­tion.

Now a quick re­view is needed of the weird way in which an­gu­lar mo­menta com­bine into net an­gu­lar mo­men­tum in quan­tum me­chan­ics, chap­ter 12.7. In clas­si­cal me­chan­ics, the sum of an an­gu­lar mo­men­tum vec­tor with length $L$ and one with length $S$ could have any com­bined length $J$ in the range $\vert L-S\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $J$ $\raisebox{-.3pt}{$\leqslant$}$ $L+S$, de­pend­ing on the an­gle be­tween the vec­tors. How­ever, in quan­tum me­chan­ics, the length of the fi­nal vec­tor must be quan­tized as $\sqrt{j(j+1)}\hbar$ where the quan­tum num­ber $j$ must sat­isfy $\vert l-s\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $j$ $\raisebox{-.3pt}{$\leqslant$}$ $l+s$ and must change in in­te­ger amounts. In par­tic­u­lar, since the spin is given as $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, the net an­gu­lar mo­men­tum quan­tum num­ber $j$ can ei­ther be $l-\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ or $l+\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. (If $l$ is zero, the first pos­si­bil­ity is also ruled out, since square an­gu­lar mo­men­tum can­not be neg­a­tive.)

For the 4f en­ergy level $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, so the square net an­gu­lar mo­men­tum quan­tum num­ber $j$ can only be $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ or $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. And for a given value of $j$, there are $2j+1$ val­ues for the quan­tum num­ber $m_j$ giv­ing the net an­gu­lar mo­men­tum in the cho­sen $z$-​di­rec­tion. That means that there are six states with $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ and eight states with $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. The to­tal is four­teen, still the same num­ber of in­de­pen­dent states at the 4f level. In fact, the four­teen states of def­i­nite net an­gu­lar mo­men­tum $j$ can be writ­ten as lin­ear com­bi­na­tions of the four­teen $R_{nl}Y_l^m{\updownarrow}$ states. (Fig­ure 12.5 shows such com­bi­na­tions up to $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2; item 2 in chap­ter 12.8 gives a gen­eral for­mula.) Pic­to­ri­ally,

\begin{displaymath}
\mbox{7 4f${\uparrow}$\ and 7 4f${\downarrow}$\ states}
\q...
...ightarrow\qquad
\mbox{6 4f$_{5/2}$\ and 8 4f$_{7/2}$\ states}
\end{displaymath}

where the spec­tro­scopic con­ven­tion is to show the net an­gu­lar mo­men­tum $j$ as a sub­script for states in which its value is un­am­bigu­ous.

The spin-or­bit in­ter­ac­tion raises the en­ergy of the six 4f$_{5/2}$ states, but low­ers it for the eight 4f$_{7/2}$ states. In fact, from above, for any state of def­i­nite square or­bital and square net an­gu­lar mo­men­tum,

\begin{displaymath}
- {\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}= ...
...ex\hbox{\the\scriptfont0 2}\kern.05em \\
\end{array} \right.
\end{displaymath}

The eight 4f$_{7/2}$ states of low­ered en­ergy form the shell that is filled at the fourth magic num­ber 28.

Fig­ure 14.15: Schematic ef­fect of spin-or­bit in­ter­ac­tion on the en­ergy lev­els. The or­der­ing within bands is re­al­is­tic for neu­trons. The des­ig­na­tion be­hind the equals sign is the of­fi­cial one. (As­sum­ing count­ing starts at 1).
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,27...
... 70}}
\put(240,212.4){\makebox(0,0)[ct]{\tt\bf 82}}
\end{picture}
\end{figure}

Fig­ure 14.15 shows how the spin-or­bit split­ting of the en­ergy lev­els gives rise to the re­main­ing magic num­bers. In the fig­ure, the co­ef­fi­cient of the spin or­bit term was sim­ply taken to vary lin­early with the en­ergy level $n$. The de­tails de­pend on whether it is neu­trons or pro­tons, and may vary from nu­cleus to nu­cleus. Es­pe­cially for the higher en­ergy bands the Coulomb re­pul­sion has an in­creas­ingly large ef­fect on the en­er­gies of pro­tons.

The ma­jor shells, ter­mi­nated by magic num­bers, are shown as grey bands. In the num­ber­ing sys­tem fol­lowed here, a sub­shell with a dif­fer­ent num­ber as the oth­ers in the same ma­jor shell comes from a dif­fer­ent har­monic os­cil­la­tor en­ergy level. Fig­ure 14.15 also shows the of­fi­cial enu­mer­a­tion of the states. You be the judge which num­ber­ing sys­tem makes the most sense to you.

As sketched in fig­ure 14.15, spin-or­bit in­ter­ac­tion pushes the 5g$_{9/2}$ states down into the band that ends at magic num­ber 50. How­ever, the en­ergy gap be­tween be­tween the 5g$_{9/2}$ states and the 4...states in the band is rel­a­tively large. That is why you might think of 40 as a semi-magic num­ber if you want. For ex­am­ple, one good rea­son to con­sider this is fig­ure 14.19 dis­cussed later.

The de­tailed or­der­ing of the sub­shells above 50 varies with au­thor and even for a sin­gle au­thor. There is no unique an­swer, be­cause the shell model is only a sim­ple ap­prox­i­ma­tion to a sys­tem that does not fol­low sim­ple rules when ex­am­ined closely enough. Still, a spe­cific or­der­ing must be adopted if the shell model is to be com­pared to the data. This book will use the or­der­ings:

pro­tons:
1s$_{1/2}$
2p$_{3/2}$ 2p$_{1/2}$
3d$_{5/2}$ 3s$_{1/2}$ 3d$_{3/2}$
4f$_{7/2}$
4p$_{3/2}$ 4f$_{5/2}$ 4p$_{1/2}$ 5g$_{9/2}$
5g$_{7/2}$ 5d$_{5/2}$ 6h$_{11/2}$ 5d$_{3/2}$ 5s$_{1/2}$
6h$_{9/2}$ 6f$_{7/2}$ 6f$_{5/2}$ 6p$_{3/2}$ 6p$_{1/2}$ 7i$_{13/2}$
neu­trons:
1s$_{1/2}$
2p$_{3/2}$ 2p$_{1/2}$
3d$_{5/2}$ 3s$_{1/2}$ 3d$_{3/2}$
4f$_{7/2}$
4p$_{3/2}$ 4f$_{5/2}$ 4p$_{1/2}$ 5g$_{9/2}$
5d$_{5/2}$ 5g$_{7/2}$ 5s$_{1/2}$ 5d$_{3/2}$ 6h$_{11/2}$
6f$_{7/2}$ 6h$_{9/2}$ 6p$_{3/2}$ 6f$_{5/2}$ 7i$_{13/2}$ 6p$_{1/2}$
7g$_{9/2}$ 7d$_{5/2}$ 7i$_{11/2}$ 7g$_{7/2}$ 7s$_{1/2}$ 7d$_{3/2}$8j$_{15/2}$

The or­der­ing for pro­tons fol­lows [35, ta­ble 7-1], but not [35, p. 223], to Z=92, and then [30], whose ta­ble seems to come from Mayer and Jensen. The or­der­ing for neu­trons fol­lows [35], with the sub­shells be­yond 136 taken from [[10]]. How­ever, the 7i$_{13/2}$ and 6p$_{1/2}$ states were swapped since the shell fill­ing [35, ta­ble 7-1] makes a lot more sense if you do. The same swap is also found in [39, p. 255], fol­low­ing Klinken­berg, while [30, p. 155] puts the 7i$_{13/2}$ sub­shell even far­ther down be­low the 6p$_{3/2}$ state.


14.12.3 Draft: Ex­am­ple oc­cu­pa­tion lev­els

The pur­pose of this sec­tion is to ex­plore how the shell model works out for sam­ple nu­clei.

Fig­ure 14.16: En­ergy lev­els for dou­bly-magic oxy­gen-16 and neigh­bors. [pdf]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
% vertical spacing 137...
...\frac{29}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Fig­ure 14.16 shows ex­per­i­men­tal en­ergy spec­tra of var­i­ous nu­clei at the left. The en­ergy val­ues are in MeV. The ground state is de­fined to be the zero level of en­ergy. The length and color of the en­ergy lines in­di­cates the spin of the nu­cleus, and the par­ity is in­di­cated by a plus or mi­nus sign. Some im­por­tant spin val­ues are also listed ex­plic­itly. Yel­low lines in­di­cate states for which no unique spin and/or par­ity are de­ter­mined or are es­tab­lished with reser­va­tions. At the right in the fig­ure, a sketch of the oc­cu­pa­tion lev­els ac­cord­ing to the shell model is dis­played for easy ref­er­ence.

The top of the fig­ure shows data for oxy­gen-16, the nor­mal oxy­gen that makes up 99.8% of the oxy­gen in the at­mos­phere. Oxy­gen-16 is a dou­bly-magic nu­cleus with 8 pro­tons and 8 neu­trons. As the right-hand di­a­gram in­di­cates, these com­pletely fill up the low­est two ma­jor shells.

As the left-hand spec­trum shows, the oxy­gen-16 nu­cleus has zero net spin in the ground state. That is ex­actly what the shell model pre­dicts. In fact, it is a con­se­quence of quan­tum me­chan­ics that:

Com­pletely filled sub­shells have zero net an­gu­lar mo­men­tum.
Since the shell model says all shells are filled, the zero spin fol­lows. The shell model got the first one right. In­deed, it passes this test with fly­ing col­ors for all dou­bly-magic nu­clei.

Next,

Sub­shells with an even num­ber of nu­cle­ons have even par­ity.
That is just a con­se­quence of the fact that even if the sub­shell is a neg­a­tive par­ity one, neg­a­tive par­i­ties mul­ti­ply out pair­wise to pos­i­tive ones. Since all sub­shells of oxy­gen-16 con­tain an even num­ber of nu­cle­ons, the com­bined par­ity of the com­plete oxy­gen-16 nu­cleus should be pos­i­tive. It is. And it is for the other dou­bly-magic nu­clei.

The shell model im­plies that a dou­bly-magic nu­cleus like oxy­gen-16 should be be par­tic­u­larly sta­ble. So it should re­quire a great deal of en­ergy to ex­cite it. In­deed it does: fig­ure 14.16 shows that ex­cit­ing oxy­gen-16 takes over 6 MeV of en­ergy.

Fol­low­ing the shell model pic­ture, one ob­vi­ous way to ex­cite the nu­cleus would be to kick a sin­gle pro­ton or neu­tron out of the 2p$_{1/2}$ sub­shell into the next higher en­ergy 3d$_{5/2}$ sub­shell. The net re­sult is a nu­cleon with spin 5/2 in the 3d$_{5/2}$ sub­shell and one re­main­ing nu­cleon with spin 1/2 in the 2p$_{1/2}$ sub­shell. Quan­tum me­chan­ics al­lows these two nu­cle­ons to com­bine their spins into a net spin of ei­ther $\frac52+\frac12$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 or $\frac52-\frac12$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2. In ad­di­tion, since the nu­cleon kicked into the 3f$_{5/2}$ changes par­ity, so should the com­plete nu­cleus. And in­deed, there is an ex­cited level a bit above 6 MeV with a spin 3 and odd par­ity, a 3$\POW9,{-}$ level. It ap­pears the shell model may be onto some­thing.

Still, the ex­ited 0$\POW9,{+}$ state sug­gests there may be a bit more to the story. In a shell model ex­pla­na­tion, the par­ity of this state would re­quire a pair of nu­cle­ons to be kicked up. In the ba­sic shell model, it would seem that this should re­quire twice the en­ergy of kick­ing up one nu­cleon. Not all nu­clear ex­ci­ta­tions can be ex­plained by the ex­ci­ta­tion of just one or two nu­cle­ons, es­pe­cially if the mass num­ber gets over 50 or the ex­ci­ta­tion en­ergy high enough. This will be ex­plored in sec­tion 14.13. How­ever, be­fore sum­mar­ily dis­miss­ing a shell model ex­pla­na­tion for this state, first con­sider the fol­low­ing sec­tions on pair­ing and con­fig­u­ra­tion mix­ing.

Next con­sider oxy­gen-17 and flu­o­rine-17 in fig­ure 14.16. These two are ex­am­ples of so-called mir­ror nu­clei; they have the num­bers of pro­tons and neu­trons re­versed. Oxy­gen-17 has 8 pro­tons and 9 neu­trons while its twin flu­o­rine-17 has 9 pro­tons and 8 neu­trons. The sim­i­lar­ity in en­ergy lev­els be­tween the two il­lus­trates the idea of charge sym­me­try: nu­clear forces are the same if the pro­tons are turned into neu­trons and vice versa. (Of course, this swap does mess up the Coulomb forces, but Coulomb forces are not very im­por­tant for light nu­clei.)

Each of these two nu­clei has one more nu­cleon in ad­di­tion to an oxy­gen-16 core. Since the filled sub­shells of the oxy­gen-16 core have zero spin, the net nu­clear spin should be that of the odd nu­cleon in the 3d$_{5/2}$ sub­shell. And the par­ity should be even, since the odd nu­cleon is in an even par­ity shell. And in­deed each ground state has the pre­dicted spin of 5/2 and even par­ity. Chalk up an­other two for the shell model.

This is a big test for the shell model, be­cause if a dou­bly-magic-plus-one nu­cleus did not have the pre­dicted spin and par­ity of the fi­nal odd nu­cleon, there would be no rea­son­able way to ex­plain it. For­tu­nately, all nu­clei of this type pass the test.

For both oxy­gen-17 and flu­o­rine-17, there is also a low-en­ergy $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ ex­cited state, likely cor­re­spond­ing to kick­ing the odd nu­cleon up to the next mi­nor shell, the 3s$_{1/2}$ one. And so there is an ex­cited $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state, for kick­ing up the nu­cleon to the 3d$_{3/2}$ state in­stead.

How­ever, from the shell model, in par­tic­u­lar fig­ure 14.15, you would ex­pect the spac­ing be­tween the 3d$_{5/2}$ and 3s$_{1/2}$ sub­shells to be more than that be­tween the 3s$_{1/2}$ and 3d$_{3/2}$ ones. Clearly it is not. One con­sid­er­a­tion not in a shell model with a straight­for­ward av­er­age po­ten­tial is that a nu­cleon in an un­usu­ally far-out s or­bit could be closer to the other nu­cle­ons in lower or­bits than one in a far-out p or­bit; the s or­bit has larger val­ues near the cen­ter of the nu­cleus, {N.8}. While the shell model gets a con­sid­er­able num­ber of things right, it is cer­tainly not a very ac­cu­rate model.

Then there are the odd par­ity states. These are not so easy to un­der­stand: they re­quire a nu­cleon to be kicked up past a ma­jor shell bound­ary. That should re­quire a lot of en­ergy ac­cord­ing to the ideas of the shell model. It seems to make them hard to rec­on­cile with the much higher en­ergy of the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state. Some thoughts on these states will be given in the next sub­sec­tion.

The fourth nu­cleus in fig­ure 14.16 is ni­tro­gen-14. This is an odd-odd nu­cleus, with both an odd num­ber of pro­tons and of neu­trons. The odd pro­ton and odd neu­tron are in the 2p$_{1/2}$ shell, so each has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. Quan­tum me­chan­ics al­lows the two to com­bine their spins into a triplet state of net spin one, like they do in deu­terium, or in a sin­glet state of spin zero. In­deed the ground state is a 1$\POW9,{+}$ one like deu­terium. The low­est ex­cited state is a 0$\POW9,{+}$ one.

The most ob­vi­ous way to fur­ther ex­cite the nu­cleus with min­i­mal en­ergy would be to kick up a nu­cleon from the 2p$_{3/2}$ sub­shell to the 2p$_{1/2}$ one. That fills the 2p$_{1/2}$ shell, mak­ing its net spin zero. How­ever, there is now a “hole,” a miss­ing par­ti­cle, in the 2p$_{3/2}$ shell.

Holes in an oth­er­wise filled sub­shell have the same pos­si­ble an­gu­lar mo­men­tum val­ues as par­ti­cles in an oth­er­wise empty shell.
There­fore the hole must have the spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ of a sin­gle par­ti­cle. This can com­bine with the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ of the odd nu­cleon of the op­po­site type to ei­ther spin 1 or spin 2. A rel­a­tively low en­ergy 1$\POW9,{+}$ state can be ob­served in the ex­per­i­men­tal spec­trum.

The next higher 0$\POW9,{-}$ state would re­quire a par­ti­cle to cross a ma­jor shell bound­ary. Then again, the en­ergy of this ex­cited state is quite sub­stan­tial at 5 MeV. It seems sim­pler to as­sume that a 1s$_{1/2}$ nu­cleon is kicked to the 2p$_{1/2}$ shell than that a 2p$_{3/2}$ nu­cleon is kicked to the 3d$_{5/2}$ one. In the lat­ter case, it seems harder to ex­plain why the four odd nu­cle­ons would want to par­tic­u­larly com­bine their spins to zero. And you could give an ar­gu­ment based on the ideas of the next sub­sec­tion that 4 odd nu­cle­ons is a lot.


14.12.4 Draft: Shell model with pair­ing

This sec­tion ex­am­ines some nu­clei with more than a sin­gle nu­cleon in an un­filled shell.

Fig­ure 14.17: Nu­cleon pair­ing ef­fect. [pdf]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
% vertical spacing 137...
...\frac{29}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Con­sider first oxy­gen-18 in fig­ure 14.17, with both an even num­ber of pro­tons and an even num­ber of neu­trons. As al­ways, the filled sub­shells have no an­gu­lar mo­men­tum. That leaves the two 3d$_{5/2}$ neu­trons. These could have com­bined in­te­ger spin from 0 to 5 if they were dis­tin­guish­able par­ti­cles. How­ever, the two neu­trons are iden­ti­cal fermi­ons, and the wave func­tion must be an­ti­sym­met­ric with re­spect to their ex­change. It can be seen from chap­ter 12.8 item 3, or more sim­ply from ta­ble 12.1, that only the 0, 2, and 4 com­bined spins are al­lowed. Still, that leaves three pos­si­bil­i­ties for the net spin of the en­tire nu­cleus.

Now the ba­sic shell model is an “in­de­pen­dent par­ti­cle model:” there are no di­rect in­ter­ac­tions be­tween the par­ti­cles. Each par­ti­cle moves in a given av­er­age po­ten­tial, re­gard­less of what the oth­ers are do­ing. There­fore, if the shell model as cov­ered so far would be strictly true, all three spin states 0, 2, and 4 of oxy­gen-18 should have equal en­ergy. Then the ground state should be any com­bi­na­tion of these spins. But that is un­true. The ground-state has zero spin:

All even-even nu­clei have zero spin and even par­ity in the ground state.
There are zero known ex­cep­tions to this rule among ei­ther the sta­ble or un­sta­ble nu­clei.

So physi­cists have con­cluded that be­sides the av­er­age po­ten­tial in­cluded in the shell model, there must be an ad­di­tional pair­ing en­ergy that makes nu­cle­ons of the same type want to com­bine pair­wise into states of zero spin. In or­der to treat this ef­fect math­e­mat­i­cally with­out los­ing the ba­sic shell model, the pair­ing en­ergy must be treated as a rel­a­tively small per­tur­ba­tion to the shell model en­ergy. The­o­ries that do so are be­yond the scope of this book, al­though the gen­eral ideas of per­tur­ba­tion the­o­ries can be found in ad­den­dum {A.38}. Here it must be suf­fice to note that the pair­ing ef­fect ex­ists and is due to in­ter­ac­tions be­tween nu­cle­ons not in­cluded in the ba­sic shell model po­ten­tial.

There­fore the ba­sic shell model will from here on be re­ferred to as the “un­per­turbed” shell model. The “per­turbed shell model” will re­fer to the shell model in which ad­di­tional en­ergy cor­rec­tions are as­sumed to ex­ist that ac­count for non­triv­ial in­ter­ac­tions be­tween in­di­vid­ual nu­cle­ons. These cor­rec­tions will not be ex­plic­itly dis­cussed, but some of their ef­fects will be demon­strated by means of ex­per­i­men­tal en­ergy spec­tra.

If the pair­ing en­ergy is a rel­a­tively small per­tur­ba­tion to the shell model, then for oxy­gen-18 you would ex­pect that be­sides the zero spin ground state, the other pos­si­bil­i­ties of spin 2 and 4 would show up as low-ly­ing ex­cited states. In­deed the ex­per­i­men­tal spec­trum in fig­ure 14.17 shows 2$\POW9,{+}$ and 4$\POW9,{+}$ states of the right spin and par­ity, though their en­ergy is ob­vi­ously not so very low. To put it in con­text, the von Weizsäcker for­mula puts the pair­ing en­ergy at 22$\raisebox{.5pt}{$/$}$$\sqrt{A}$ MeV, which would be of the rough or­der of 5 MeV for oxy­gen-18.

If one neu­tron of the pair is kicked up to the 3s$_{1/2}$ state, a 2$\POW9,{+}$ or 3$\POW9,{+}$ state should re­sult. This will re­quire the pair to be bro­ken up and a sub­shell bound­ary to be crossed. A po­ten­tial 2$\POW9,{+}$ can­di­date is present in the spec­trum.

Like for oxy­gen-16, there is again an ex­cited 0$\POW9,{+}$ state of rel­a­tively low en­ergy. In this case how­ever, its en­ergy seems rather high in view that the two 3d$_{5/2}$ neu­trons could sim­ply be kicked up across the mi­nor shell bound­ary to the very nearby 3s$_{1/2}$ shell. An ex­pla­na­tion can be found in the fact that physi­cists have con­cluded that:

The pair­ing en­ergy in­creases with the an­gu­lar mo­men­tum of the sub­shell.
When the neu­tron pair is kicked from the 3d$_{5/2}$ shell to the 3s$_{1/2}$, its pair­ing en­ergy de­creases. There­fore this ex­ci­ta­tion re­quires ad­di­tional en­ergy be­sides the cross­ing of the mi­nor shell bound­ary.

It seems there­fore that the per­turbed shell model can give a plau­si­ble ex­pla­na­tion for the var­i­ous fea­tures of the en­ergy spec­trum. How­ever, care must be taken not to at­tach too much fi­nal­ity to such ex­pla­na­tions. Sec­tion 14.13 will give a very dif­fer­ent take on the ex­cited states of oxy­gen-18. Pre­sum­ably, nei­ther ex­pla­na­tion will be very ac­cu­rate. Only ad­di­tional con­sid­er­a­tions be­yond mere en­ergy lev­els can de­cide which ex­pla­na­tion gives the bet­ter de­scrip­tion of the ex­cited states.

The pur­pose in this sec­tion is to ex­am­ine what fea­tures seem to have a rea­son­able ex­pla­na­tion within a shell model con­text, not how ab­solutely ac­cu­rate that ex­pla­na­tion re­ally is.

Con­sider again the 0$\POW9,{+}$ ex­cited state of oxy­gen-16 in fig­ure 14.16 as dis­cussed in the pre­vi­ous sub­sec­tion. Some of the en­ergy needed for a pair of 2p$_{1/2}$ nu­cle­ons to cross the ma­jor shell bound­ary to the 3d$_{5/2}$ sub­shell will be com­pen­sated for by the higher pair­ing en­ergy in the new sub­shell. It still seems cu­ri­ous that the state would end up be­low the 3$\POW9,{-}$ one, though.

Sim­i­larly, the rel­a­tively low en­ergy $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state in oxy­gen-17 and flu­o­rine-17 can now be made a bit more plau­si­ble. To ex­plain the neg­a­tive par­ity, a nu­cleon must be kicked across the ma­jor shell bound­ary from the 2p$_{1/2}$ sub­shell to the 3d$_{5/2}$ one. That should re­quire quite a bit of en­ergy, but this will in part be com­pen­sated for by the fact that pair­ing now oc­curs at higher an­gu­lar mo­men­tum.

So what to make of the next $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state? One pos­si­bil­ity is that a 2p$_{1/2}$ nu­cleon is kicked to the 3s$_{1/2}$ sub­shell. The three spins could then com­bine into 5/2, [30, p. 131]. If true how­ever, this would be a quite sig­nif­i­cant vi­o­la­tion of the ba­sic ideas of the per­turbed shell model. Just con­sider: it re­quires break­ing up the 2p$_{1/2}$ pair and kick­ing one of the two neu­trons across both a ma­jor shell bound­ary and a sub­shell one. That would re­quire less en­ergy than the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ ex­ci­ta­tion in which the odd nu­cleon is merely kicked over two sub­shell bound­aries and no pair is bro­ken up? An al­ter­na­tive that is more con­sis­tent with the per­turbed shell model ideas would be that the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ex­ci­ta­tion is like the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ one, but with an ad­di­tional par­tial break up of the re­sult­ing pair. The en­ergy seems still low.

How about nu­clei with an odd num­ber of neu­trons and/or pro­tons in a sub­shell that is greater than one? For these:

The odd-par­ti­cle shell model pre­dicts that even if the num­ber of nu­cle­ons in a sub­shell is odd, in the ground state all nu­cle­ons ex­cept the fi­nal odd one still com­bine into spher­i­cally sym­met­ric states of zero spin.
That leaves only the fi­nal odd nu­cleon to pro­vide any nonzero spin and cor­re­spond­ing non­triv­ial elec­tro­mag­netic prop­er­ties.

Fig­ure 14.17 shows the ex­am­ple of oxy­gen-19, with three neu­trons in the un­filled 3d$_{5/2}$ sub­shell. The odd-par­ti­cle shell model pre­dicts that the first two neu­trons still com­bine into a state of zero spin like in oxy­gen-18. That leaves only the spin of the third neu­tron. And in­deed, the to­tal nu­clear spin of oxy­gen-18 is ob­served to be 5/2 in the ground state, the spin of this odd neu­tron. The odd-par­ti­cle shell model got it right.

It is im­por­tant to rec­og­nize that the odd-par­ti­cle shell model only ap­plies to the ground state. This is not al­ways suf­fi­ciently stressed. The­o­ret­i­cally, three 3d$_{5/2}$ neu­trons can com­bine their spins not just to spin 5/2, but also to 3/2 or 9/2 while still sat­is­fy­ing the an­ti­sym­metriza­tion re­quire­ment, ta­ble 12.1. And in­deed, the oxy­gen-19 en­ergy spec­trum in fig­ure 14.17 shows rel­a­tively low en­ergy $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ states. To ex­plain the en­er­gies of these states would re­quire com­pu­ta­tion us­ing an ac­tual per­turbed shell model, rather than just the odd-par­ti­cle as­sump­tion that such a model will lead to per­fect pair­ing of even num­bers of nu­cle­ons.

It is also im­por­tant to rec­og­nize that the odd-par­ti­cle shell model is only a pre­dic­tion. It does fail for a fair num­ber of nu­clei. That is true even ex­clud­ing the very heavy nu­clei for which the shell model does not ap­ply pe­riod. For ex­am­ple, note in fig­ure 14.17 how close to­gether are the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ en­ergy lev­els. You might guess that the or­der of those two states could eas­ily be re­versed for an­other nu­cleus. And so it can; there are a num­ber of nu­clei in which the spins com­bine into a net spin one unit less than that of the last odd nu­cleon. While the un­per­turbed shell model does not fun­da­men­tally fail for such nu­clei, (be­cause it does not pre­dict the spin at all), the ad­di­tional odd-par­ti­cle as­sump­tion does.

It should be noted that dif­fer­ent terms are used in lit­er­a­ture for the odd-par­ti­cle shell model. The term “shell model with pair­ing” is ac­cu­rate and un­der­stand­able, so that is not used. Some au­thors use the term “ex­treme in­de­pen­dent par­ti­cle model.” You read that right. While the un­per­turbed shell model is an in­de­pen­dent par­ti­cle model, the shell model with pair­ing has be­come a de­pen­dent par­ti­cle model: there are now pos­tu­lated di­rect in­ter­ac­tions be­tween the nu­cle­ons caus­ing them to pair. So what bet­ter way to con­fuse stu­dents than to call a de­pen­dent par­ti­cle model an ex­treme inde­pen­dent par­ti­cle model? How­ever, this term is too bla­tantly wrong even for some physi­cists. So, some other books use in­stead “ex­treme sin­gle-par­ti­cle model,” and still oth­ers use “one-par­ti­cle shell model.” Un­for­tu­nately, it is fun­da­men­tally a mul­ti­ple-par­ti­cle model. You can­not have par­ti­cle in­ter­ac­tions with a sin­gle par­ti­cle. Only physi­cists would come up with three dif­fer­ent names for the same model and get it wrong in each sin­gle case. This book uses the term odd-par­ti­cle shell model, (with odd in dic­tio­nary rather than math­e­mat­i­cal sense), since it is not wrong and sounds much like the other names be­ing bandied around. (The of­fi­cial names could be fixed up by adding the word al­most, like in ex­treme al­most in­de­pen­dent par­ti­cle model. This book will not go there, but you could sub­sti­tute as­ymp­tot­i­cally” for “al­most to sound more sci­en­tific.)

While the odd-par­ti­cle model ap­plies only to the ground state, some ex­cited states can still be de­scribed as purely odd-par­ti­cle ef­fects. In par­tic­u­lar, for the oxy­gen-19 ex­am­ple, the odd 3d$_{5/2}$ neu­tron could be kicked up to the 3s$_{1/2}$ sub­shell with no fur­ther changes. That would leave the two re­main­ing 3d$_{5/2}$ neu­trons with zero spin, and the nu­cleus with the new spin 1/2 of the odd neu­tron. In­deed a low-ly­ing $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state is ob­served. (Be­cause of the an­ti­sym­metriza­tion re­quire­ment, this state can­not re­sult from three neu­trons in the 3d$_{5/2}$ sub­shell.)

It may fur­ther be noted that pair­ing is not re­ally the right quan­tum term. If two nu­cle­ons have paired into the com­bi­na­tion of zero net spin, the next two can­not just en­ter the same com­bi­na­tion with­out vi­o­lat­ing the an­ti­sym­metriza­tion re­quire­ments be­tween the pairs. What re­ally hap­pens is that all four as a group com­bine into a state of zero spin. How­ever, every­one uses the term pair­ing, and so will this book.

Fig­ure 14.18: En­ergy lev­els for neigh­bors of dou­bly-magic cal­cium-40. [pdf]
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...\frac{29}{2}}$}}
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Ex­am­ples that high­light the per­tur­ba­tion ef­fects of the shell model are shown in fig­ure 14.18. These nu­clei have un­filled 4d$_{7/2}$ shells. Since that is a ma­jor shell with no sub­shells, nu­cleon tran­si­tions to dif­fer­ent shells re­quire quite a bit of en­ergy.

First ob­serve that all three nu­clei have a fi­nal odd 4f$_{7/2}$ nu­cleon and a cor­re­spond­ing ground state spin of 7/2 just like the odd-par­ti­cle shell model says they should. And the net nu­clear par­ity is neg­a­tive like that of the odd nu­cleon. That is quite grat­i­fy­ing.

As far as cal­cium-41 is con­cerned, one ob­vi­ous min­i­mal-en­ergy ex­ci­ta­tion would be that the odd neu­tron is kicked up from the 4f$_{7/2}$ shell to the 4p$_{3/2}$ shell. This will pro­duce a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ex­cited state. Such a state does in­deed ex­ist and it has rel­a­tively high en­ergy, as you would ex­pect from the fact that a ma­jor shell bound­ary must be crossed.

An­other ob­vi­ous min­i­mal-en­ergy ex­ci­ta­tion would be that a nu­cleon is kicked up from the filled 3d$_{3/2}$ shell to pair up with the odd nu­cleon al­ready in the 4f$_{7/2}$ shell. This re­quires again that a ma­jor shell bound­ary is crossed, though some en­ergy can be re­cov­ered by the fact that the new nu­cleon pair­ing is now at higher spin. Since here a nu­cleon changes shells from the pos­i­tive par­ity 3d$_{3/2}$ sub­shell to the neg­a­tive 4f$_{7/2}$ one, the nu­clear par­ity re­verses and the ex­cited state will be a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ one. Such a state is in­deed ob­served.

The un­sta­ble mir­ror twin of cal­cium-41, scan­dium-41 has en­ergy lev­els that are very much the same.

Next con­sider cal­cium-43. The odd-par­ti­cle shell model cor­rectly pre­dicts that in the ground state, the first two 4f$_{7/2}$ neu­trons pair up into zero spin, leav­ing the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ spin of the third neu­tron as the net nu­clear spin. How­ever, even al­low­ing for the an­ti­sym­metriza­tion re­quire­ments, the three 4f$_{7/2}$ neu­trons could in­stead com­bine into spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, or $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 15}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, ta­ble 12.1. A low-en­ergy $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ex­cited state, one unit of spin less than the ground state, is in­deed ob­served. A $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state is just above it. On the other hand, the low­est known $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state has more en­ergy than the low­est $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ one. Then again, con­sider the spin val­ues that are not pos­si­ble for the three neu­trons if they stay in the 4f$_{7/2}$ shell. The first $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 13}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 17}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ states oc­cur at en­er­gies well be­yond the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 15}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ one, and the first $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state only ap­pears at 2.6 MeV.

The low­est $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state en­ergy is half that of the one for cal­cium-41. Ap­par­ently, the 3d$_{3/2}$ neu­tron would rather pair up with 3 other at­tract­ing neu­trons in the 4f$_{7/2}$ shell than with just one. That seems rea­son­able enough. The over­all pic­ture seems in en­cour­ag­ing agree­ment with the per­turbed shell model ideas.

Scan­dium-43 has one pro­ton and two neu­trons in the 4f$_{7/2}$ shells. The odd-par­ti­cle model pre­dicts that in the ground state, the two neu­trons com­bine into zero spin. How­ever, the an­ti­sym­metriza­tion re­quire­ment al­lows ex­cited spins of 2, 4, and 6 with­out any nu­cle­ons chang­ing shells. The low­est ex­cited spin value 2 can com­bine with the 7/2 spin of the odd pro­ton into ex­cited nu­clear states from $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ up to $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$. Rel­a­tively low-ly­ing $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ states, but not a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ one, are ob­served. (The low­est-ly­ing po­ten­tial $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state is at 1.9 MeV. The low­est ly­ing po­ten­tial $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state is at 3.3 MeV, though there are 4 states of un­known spin be­fore that.)

Note how low the low­est $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state has sunk. That was maybe not quite un­pre­dictable. Two pro­tons plus two neu­trons in the 4f$_{7/2}$ shells have to obey less an­ti­sym­metriza­tion re­quire­ments than four pro­tons do, while the at­trac­tive nu­clear forces be­tween the four are about the same ac­cord­ing to charge in­de­pen­dence.

The dif­fer­ence be­tween the en­ergy lev­els of scan­dium-41 ver­sus scan­dium-43 is dra­matic. Af­ter all, the un­per­turbed shell model would al­most com­pletely ig­nore the two ad­di­tional neu­trons that scan­dium-43 has. Pro­tons and neu­trons are solved for in­de­pen­dently in the model. It brings up a point that is of­ten not suf­fi­ciently em­pha­sized in other ex­po­si­tions of nu­clear physics. The odd-par­ti­cle shell model is not an only the last odd par­ti­cle is im­por­tant model. It is a “the last odd par­ti­cle pro­vides the ground-state spin and elec­tro­mag­netic prop­er­ties, be­cause the other par­ti­cles are paired up in spher­i­cally sym­met­ric states” model. The the­o­ret­i­cal jus­ti­fi­ca­tion for the model, which is weak enough as it is al­ready, only ap­plies to the sec­ond state­ment.


14.12.5 Draft: Con­fig­u­ra­tion mix­ing

To bet­ter un­der­stand the shell model and its lim­i­ta­tions, com­bi­na­tions of states must be con­sid­ered.

Take once again the ex­cited 0$\POW9,{+}$ state of oxy­gen-16 shown in fig­ure 14.16. To cre­ate this state within the shell model pic­ture, a pair of 2p$_{1/2}$ nu­cle­ons must be kicked up to the 3d$_{5/2}$ sub­shell. Since that re­quires a ma­jor shell bound­ary cross­ing by two nu­cle­ons, it should take a con­sid­er­able amount of en­ergy. Some of it will be re­cov­ered by the fact that the nu­cleon pair­ing now oc­curs at higher an­gu­lar mo­men­tum. But there is an­other ef­fect.

First of all, there are two ways to do it: ei­ther the 2p$_{1/2}$ pro­tons or the two 2p$_{1/2}$ neu­trons can be kicked up. One pro­duces an ex­cited wave func­tion that will be in­di­cated by $\psi_{2p}$ and the other by $\psi_{2n}$. Be­cause of charge sym­me­try, and be­cause the Coulomb force is mi­nor for light nu­clei, these two states should have very nearly the same en­ergy.

Quan­tum me­chan­ics al­lows for lin­ear com­bi­na­tions of the two wave func­tions:

\begin{displaymath}
\Psi = c_1 \psi_{2p} + c_2 \psi_{2n}
\end{displaymath}

Within the strict con­text of the un­per­turbed shell model, it does not make a dif­fer­ence. That model as­sumes that the nu­cle­ons do not in­ter­act di­rectly with each other, only with an av­er­age po­ten­tial. There­fore the com­bi­na­tion should still have the same en­ergy as the in­di­vid­ual states.

But now con­sider the pos­si­bil­ity that both the pro­tons and the neu­trons would be in the 3d$_{5/2}$ sub­shell. In that case, surely you would agree that these four, mu­tu­ally at­tract­ing, nu­cle­ons in the same spa­cial or­bits would sig­nif­i­cantly in­ter­act and lower their en­ergy. Even if the un­per­turbed shell model ig­nores that.

Of course, the four nu­cle­ons are not all in the 3d$_{5/2}$ state; that would re­quire four ma­jor shell cross­ing and make things worse. Each com­po­nent state has only two nu­cle­ons in the 3d$_{5/2}$ sub­shell. How­ever, quan­tum me­chan­i­cal un­cer­tainty makes the two states in­ter­act through twi­light terms, chap­ter 5.3. These act in some sense as if all four nu­cle­ons are in­deed in the 3d$_{5/2}$ sub­shell at the same time. It has the weird ef­fect that the right com­bi­na­tion of the states $\psi_{2p}$ and $\psi_{2n}$ can have sig­nif­i­cantly less en­ergy than the low­est of the two in­di­vid­ual states. That is par­tic­u­larly true if the two orig­i­nal states have about the same en­ergy, as they have here.

The amount of en­ergy low­er­ing is hard to pre­dict. It de­pends on the amount of nu­cleon po­si­tions that have a rea­son­able prob­a­bil­ity for both states and the amount of in­ter­ac­tion of the nu­cle­ons. In­tu­ition still sug­gests it should be quite con­sid­er­able. And there is a more solid ar­gu­ment. If the strictly un­per­turbed shell model ap­plies, there should be two 0$\POW9,{+}$ en­ergy states with al­most the same en­ergy; one for pro­tons and one for neu­trons. How­ever, if there is sig­nif­i­cant twi­light in­ter­ac­tion be­tween the two, the en­ergy of one of the pair will be pushed way down and the other way up. There is no known sec­ond ex­cited 0$\POW9,{+}$ state with al­most the same en­ergy as the first one for oxy­gen-16.

Fig­ure 14.19: 2$\POW9,{+}$ ex­ci­ta­tion en­ergy of even-even nu­clei. [pdf][con]
\begin{figure}\centering
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\begin{picture}(405,56...
...put(12,0){\makebox(0,0)[bl]{1.5}}
}
\end{picture}}
\end{picture}
\end{figure}

Of course, a weird ex­cited state at 6 MeV in a nu­cleus is not such a big deal. But there is more. Con­sider fig­ure 14.19. It gives the ex­ci­ta­tion en­ergy of the low­est 2$\POW9,{+}$ state for all even-even nu­clei.

For all nu­clei ex­cept the crossed-out ones, the 2$\POW9,{+}$ state is the low­est ex­cited state of all. That seems cu­ri­ous al­ready. Why would the low­est ex­cited state not be a 0$\POW9,{+}$ one for a lot of even-even nu­clei? Based on the shell model you would as­sume there are two ways to ex­cite an even-even nu­cleus with min­i­mal en­ergy. The first way would be to kick a pair of nu­cle­ons up to the next sub­shell. That would cre­ate a 0$\POW9,{+}$ ex­cited state. It could re­quire very lit­tle en­ergy if the sub­shells are close to­gether.

The al­ter­na­tive way to ex­cite an even-even nu­cleus with min­i­mal en­ergy would break up a pair, but leave them in the same sub­shell. This would at the min­i­mum cre­ate a 2$\POW9,{+}$ state. (For par­tially filled shells of high enough an­gu­lar mo­men­tum, it may also be pos­si­ble to re­con­fig­ure the nu­cle­ons into a dif­fer­ent state that still has zero an­gu­lar mo­men­tum, but that does not af­fect the ar­gu­ment.) Break­ing the pair­ing should re­quire an ap­pre­cia­ble amount of en­ergy, on the MeV level. So why is the 2$\POW9,{+}$ state al­most in­vari­ably the low­est en­ergy one?

Then there is the mag­ni­tude of the 2$\POW9,{+}$ en­ergy lev­els. In fig­ure 14.19 the en­er­gies have been nor­mal­ized with the von Weizsäcker value for the pair­ing en­ergy,

\begin{displaymath}
\frac{2 C_p}{A^{C_e}}
\end{displaymath}

You would ex­pect all squares to have roughly the full size, show­ing that it takes about the von Weizsäcker en­ergy to break up the pair. Dou­bly magic nu­clei are quite happy to obey. Singly magic nu­clei seem a bit low, but hey, the break-up is usu­ally only par­tial, you know.

But for nu­clei that are not close to any magic num­ber for ei­ther pro­tons and neu­trons all hell breaks loose. Break-up en­er­gies one to two or­ders of mag­ni­tude less than the von Weizsäcker value are com­mon. How can the pair­ing en­ergy just sud­denly stop to ex­ist?

Fig­ure 14.20: Col­lec­tive mo­tion ef­fects. [pdf]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
% vertical spacing 137...
...\frac{29}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Con­sider a cou­ple of ex­am­ples in fig­ure 14.20. In case of ruthe­nium-104, it takes a measly 0.36 MeV to ex­cite the 2$\POW9,{+}$ state. But there are 10 dif­fer­ent ways to com­bine the four 5g$_{9.2}$ pro­tons into a 2$\POW9,{+}$ state, ta­ble 12.1. Kick up the pair of pro­tons from the 4p$_{1/2}$ shell, and there are an­other 10 2$\POW9,{+}$ states. The four 5g$_{7/2}$ neu­trons can pro­duce an­other 10 of them. Kick up a pair of neu­trons from the 5d$_{5/2}$ sub­shell, and there is 10 more. Pre­sum­ably, all these states will have sim­i­lar en­ergy. And there might be many other low-en­ergy ways to cre­ate 2$\POW9,{+}$ states, [30, pp. 135-136].

Con­sider now the fol­low­ing sim­plis­tic model. As­sume that the nu­cleus can be in any of $Q$ dif­fer­ent global states of the same en­ergy,

\begin{displaymath}
\psi_1, \psi_2, \psi_3, \ldots, \psi_Q
\end{displaymath}

Watch what hap­pens when such states are mixed to­gether. The en­ergy fol­lows from the Hamil­ton­ian co­ef­fi­cients

\begin{displaymath}
\begin{array}{ccccc}
E_{1} \equiv \langle\psi_1\vert H\psi...
...E_{Q} \equiv \langle\psi_Q\vert H\psi_Q\rangle \\
\end{array}\end{displaymath}

By as­sump­tion, the en­ergy lev­els $E_1,E_2,\ldots$ of the states are all about the same, and if the un­per­turbed shell model was ex­act, the per­tur­ba­tions $\varepsilon_{..}$ would all be zero. But since the shell model is only a rough ap­prox­i­ma­tion of what is go­ing on in­side nu­clei, the shell model states will not be true en­ergy eigen­func­tions. There­fore the co­ef­fi­cients $\varepsilon_{..}$ will surely not be zero, though what they will be is hard to say.

To get an idea of what can hap­pen, as­sume for now that the $\varepsilon_{..}$ are all equal and neg­a­tive. In that case, fol­low­ing sim­i­lar ideas as in chap­ter 5.3, a state of low­ered en­ergy ex­ists that is an equal com­bi­na­tion of each of the $Q$ in­di­vid­ual ex­cited states; its en­ergy will be lower than the orig­i­nal states by an amount $(Q-1)\varepsilon$. Even if $\varepsilon$ is rel­a­tively small, that will be a sig­nif­i­cant amount if the num­ber $Q$ of states with the same en­ergy is large.

Of course, the co­ef­fi­cients $\varepsilon_{..}$ will not all be equal and neg­a­tive. Pre­sum­ably they will vary in both sign and mag­ni­tude. In­ter­ac­tions be­tween states will also be lim­ited by sym­me­tries. (If states com­bine into an equiv­a­lent state that is merely ro­tated in space, there is no en­ergy low­er­ing.) Still, the low­est ex­ci­ta­tion en­ergy will be de­fined by the largest neg­a­tive ac­cu­mu­la­tion of shell model er­rors that is pos­si­ble.

The pic­ture that emerges then is that the 2$\POW9,{+}$ ex­ci­ta­tion for ruthe­nium-104, and most other nu­clei in the rough range 50 $\raisebox{.3pt}{$<$}$ $A$ $\raisebox{.3pt}{$<$}$ 150, is not just a mat­ter of just one or two nu­cle­ons chang­ing. It ap­par­ently in­volves the col­lab­o­ra­tive mo­tion of a large num­ber of nu­cle­ons. This would be quite a chal­lenge to de­scribe in the con­text of the shell model. There­fore physi­cists have de­vel­oped dif­fer­ent mod­els, ones that al­low for col­lec­tive mo­tion of the en­tire nu­cleus, like in sec­tion 14.13.

When the en­ergy of the ex­ci­ta­tion hits zero, the bot­tom quite lit­er­ally drops out of the shell model. In fact, even if the en­ergy merely be­comes low, the shell model must crash. If en­ergy states are al­most de­gen­er­ate, the slight­est thing will throw the nu­cleus from one to the other. In par­tic­u­lar, small per­tur­ba­tion the­ory shows that orig­i­nally small ef­fects blow up as the rec­i­p­ro­cal of the en­ergy dif­fer­ence, ad­den­dum {A.38}. Physi­cists have found that nu­clei in the rough ranges 150 $\raisebox{.3pt}{$<$}$ $A$ $\raisebox{.3pt}{$<$}$ 190 and $A$ $\raisebox{.3pt}{$>$}$ 220 ac­quire an in­trin­sic non­spher­i­cal shape, fun­da­men­tally in­val­i­dat­ing the shell model as cov­ered here. More phys­i­cally, as fig­ure 14.19 sug­gests, it hap­pens for pretty much all heavy nu­clei ex­cept ones close to the magic lines. The en­ergy spec­trum of a typ­i­cal nu­cleus in the non­spher­i­cal range, hafnium-176, is shown in fig­ure 14.20.


14.12.6 Draft: Shell model fail­ures

The pre­vi­ous sub­sec­tion al­ready in­di­cated two cases in which the shell model has ma­jor prob­lems with the ex­cited states. But in a num­ber of cases the shell model may also pre­dict an in­cor­rect ground state. Fig­ure 14.21 shows some typ­i­cal ex­am­ples.

Fig­ure 14.21: Fail­ures of the shell model. [pdf]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
% vertical spacing 137...
...\frac{29}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

In case of ti­ta­nium-47, the shell model pre­dicts that there will be five neu­trons in an un­filled 4f$_{7/2}$ sub­shell. It is be­lieved that this is in­deed cor­rect [35, p. 224]. The un­per­turbed shell model makes no pre­dic­tions about the nu­clear spin. How­ever, the odd-par­ti­cle shell model says that in the ground state the nu­clear spin should be that of the odd neu­tron, $\frac72$. But it is not, the spin is $\frac52$. The pair­ing of the even num­ber of neu­trons in the 4f$_{7/2}$ shell is not com­plete. While un­for­tu­nate, this is re­ally not that sur­pris­ing. The per­tur­ba­tion Hamil­ton­ian used to de­rive the pre­dic­tion of nu­cleon pair­ing is a very crude one. It is quite com­mon to see sub­shells with at least three par­ti­cles and three holes (three places for ad­di­tional par­ti­cles) end up with a unit less spin than the odd-par­ti­cle model pre­dicts. It al­most hap­pened for oxy­gen-19 in fig­ure 14.17.

In fact, 5 par­ti­cles in a shell in which the sin­gle-par­ti­cle spin is $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ can com­bine their spin into a va­ri­ety of net val­ues. Ta­ble 12.1 shows that $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 15}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ are all pos­si­ble. Com­pared to that, the odd-par­ti­cle pre­dic­tion does not seem that bad. Note that the pre­dicted state of spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ has only slightly more en­ergy than the ground state. On the other hand, other states that might be pro­duced through the com­bined spin of the five neu­trons have much more en­ergy.

Flu­o­rine-19 shows a more fun­da­men­tal fail­ure of the shell model. The shell model would pre­dict that the odd pro­ton is in the 3d$_{5/2}$ state, giv­ing the nu­cleus spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ and even par­ity. In fact, it should be just like flu­o­rine-17 in fig­ure 14.16. For the un­per­turbed shell model, the ad­di­tional two neu­trons should not make a sig­nif­i­cant dif­fer­ence. But the nu­clear spin is $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, and that means that the odd pro­ton must be in the 3s$_{1/2}$ state. A look at fig­ure 14.15 shows that the un­per­turbed shell model can­not qual­i­ta­tively ex­plain this swap­ping of the two states.

It is the the­o­reti­cian’s loss, but the ex­per­i­men­tal­ist’s gain. The fact that flu­o­rine has spin one-half makes it a pop­u­lar tar­get for nu­clear mag­netic res­o­nance stud­ies. Spin one-half nu­clei are easy to an­a­lyze and they do not have non­triv­ial elec­tric fields that mess up the nice sharp sig­nals in nu­clei with larger spin.

And maybe the the­o­reti­cian can take some com­fort in the fact that this com­plete fail­ure is rare among the light nu­clei. In fact, the main other ex­am­ple is flu­o­rine-19’s mir­ror twin neon-19. Also, there is an ex­cited state with the cor­rect spin and par­ity just above the ground state. But no funny busi­ness here; if you are go­ing to call flu­o­rine-19 al­most right, you have to call flu­o­rine-17 al­most wrong.

Note also how low the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ex­cited state has be­come. Maybe this can be some­what un­der­stood from the fact that the kicked-up 2p$_{1/2}$ pro­ton is now in a sim­i­lar spa­tial or­bit with three other nu­cle­ons, rather than just one like in the case of flu­o­rine-17. In any case, it would surely re­quire a rather so­phis­ti­cated per­turbed shell model to de­scribe it, one that in­cludes nu­cle­ons of both type in the per­tur­ba­tion.

And note that for­mu­lat­ing a per­turbed shell model from phys­i­cal prin­ci­ples is not easy any­way, be­cause the ba­sic shell model al­ready in­cludes the in­ter­ac­tions be­tween nu­cle­ons in an av­er­age sense. The per­tur­ba­tions must not just iden­tify the in­ter­ac­tions, but more im­por­tantly, what part of these in­ter­ac­tions is still miss­ing from the un­per­turbed shell model.

For the highly un­sta­ble beryl­lium-11 and ni­tro­gen-11 mir­ror nu­clei, the shell model gets the spin right, but the par­ity wrong! In shell model terms, a change of par­ity re­quires the cross­ing of a ma­jor shell bound­ary. Beryl­lium-11 is known to be a “halo nu­cleus,” a nu­cleus whose ra­dius is no­tice­ably larger than that pre­dicted by the liq­uid drop for­mula (14.9). This is as­so­ci­ated with a gross in­equal­ity be­tween the num­ber of pro­tons and neu­trons. Beryl­lium-11 has only 4 pro­tons, but 7 neu­trons; far too many for such a light nu­cleus. Beryl­lium-13 with 9 neu­trons pre­sum­ably starts to sim­ply throw the bums out. Beryl­lium-11 does not do that, but it keeps one neu­tron at arms length. The halo of beryl­lium-11 is a sin­gle neu­tron one. (That of its beta-de­cay par­ent lithium-11 is a two-neu­tron one. Such a nu­cleus is called “Bor­romean,” af­ter the three in­ter­lock­ing rings in the shield of the princes of Bor­romeo. Like the rings, the three-body sys­tem lithium-9 plus two neu­trons hangs to­gether but if any of the three is re­moved, the other two fall apart too. Both lithium-10 and the dineu­tron are not bound.) Halo nu­cle­ons tend to pre­fer states of low or­bital an­gu­lar mo­men­tum, be­cause in clas­si­cal terms it re­duces the ki­netic en­ergy they need for an­gu­lar mo­tion. The po­ten­tial en­ergy is less sig­nif­i­cant so far out. In shell model terms, the beryl­lium-11 neu­tron has the 3s$_{1/2}$ state avail­able to go to; that state does in­deed have the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ spin and pos­i­tive par­ity ob­served. Very lit­tle seems to be known about ni­tro­gen-11 at the time of writ­ing; no en­ergy lev­els, no elec­tric quadru­pole mo­ment (but nei­ther is there for beryl­lium-11). It is hard to do ex­per­i­ments at your leisure on a nu­cleus that lives for less than 10$\POW9,{-21}$ s.

For much heav­ier nu­clei, the sub­shells are of­ten very close to­gether. Also, un­like for the 3d$_{5/2}$ and 3s$_{1/2}$ states, the shell model of­ten does not pro­duce an un­am­bigu­ous or­der­ing for them. In that case, it is up to you whether you want to call it a fail­ure if a par­ti­cle does not fol­low what­ever am­bigu­ous or­der­ing you have adopted.

Se­le­nium-77 il­lus­trates a more fun­da­men­tal rea­son why the odd par­ti­cle may end up in the wrong state. The fi­nal odd neu­tron would nor­mally be the third one in the 5g$_{9/2}$ state. That would give the nu­cleus a net spin of $\frac92$ and pos­i­tive par­ity. There is in­deed a low-ly­ing ex­cited state like that. (It is just above a $\frac72$ one that might be an ef­fect of in­com­plete pair­ing.) How­ever, the nu­cleus finds that if it pro­motes a neu­tron from the 4p$_{1/2}$ shell to the 5g$_{9/2}$ one just above, that neu­tron can pair up at higher an­gu­lar mo­men­tum, low­er­ing the over­all nu­clear en­ergy. That leaves the odd neu­tron in the 4p$_{1/2}$ state, giv­ing the nu­cleus a net spin of $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ and neg­a­tive par­ity. Pro­mo­tion hap­pens quite of­ten if there are more than 32 nu­cle­ons of a given type and there is a state of lower spin im­me­di­ately be­low the one be­ing filled.

Tan­ta­lum-181 is an ex­am­ple nu­cleus that is not spher­i­cal. For it, the shell model sim­ply does not ap­ply as de­rived here. So there is no need to worry about it. Which is a good thing, be­cause it does not seem easy to jus­tify a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ ground state based on the shell model. As noted in the pre­vi­ous sub­sec­tion, non­spher­i­cal nu­clei ap­pear near the sta­ble line for mass num­bers of about 150 to 190 and above 220. There are also a few with mass num­bers be­tween 20 and 30.

Pre­ston & Bhaduri [35, p. 224ff] give an ex­ten­sive ta­ble of nu­cle­ons with odd mass num­ber, list­ing shell oc­cu­pa­tion num­bers and spin. No­table is iron-57, be­lieved to have three neu­trons in the 4p$_{3/2}$ shell as the shell model says, but with a net nu­clear spin of $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$. Since the three neu­trons can­not pro­duce that spin, in a shell model ex­pla­na­tion the 6 pro­tons in the 4f$_{7/2}$ shell will need to con­tribute. Sim­i­larly neodymium-149 with, maybe, 7 neu­trons in the 6f$_{7/2}$ shell has an un­ex­pected $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ground state. Pal­la­dium-101 with 5 neu­trons in the 5d$_{5/2}$ shell has an un­ex­pected spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ ac­cord­ing to the ta­ble; how­ever, the more re­cent data of [3] list the nu­cleus at the ex­pected $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ value. In gen­eral the ta­ble shows that the ground state spin val­ues of spher­i­cal nu­clei with odd mass num­bers are al­most all cor­rectly pre­dicted if you know the cor­rect oc­cu­pa­tion num­bers of the shells. How­ever, pre­dict­ing those num­bers for heavy nu­clei is of­ten non­triv­ial.