Sub­sec­tions


14.13 Draft: Col­lec­tive Struc­ture

Some nu­clear prop­er­ties are dif­fi­cult to ex­plain us­ing the shell model ap­proach as cov­ered here. There­fore physi­cists have de­vel­oped dif­fer­ent mod­els.

For ex­am­ple, nu­clei may have ex­cited states with un­ex­pect­edly low en­ergy. One ex­am­ple is ruthe­nium-104 in fig­ure 14.20, and many other even-even nu­clei with such en­er­gies may be found in fig­ure 14.19. If you try to ex­plain the ex­ci­ta­tion en­ergy within a shell model con­text, you are led to the idea that many shell model ex­ci­ta­tions com­bine forces, as in sec­tion 14.12.5.

Then there are nu­clei for which the nor­mal shell model does not work at all. They are called the non­spher­i­cal or de­formed nu­clei. Among the line of most sta­ble nu­clei, they are roughly the “rare earth” lan­thanides and the ex­tremely heavy ac­tinides that are de­formed. In terms of the mass num­ber, the ranges are about 150 $\raisebox{.3pt}{$<$}$ $A$ $\raisebox{.3pt}{$<$}$ 190 and 220 $\raisebox{.3pt}{$<$}$ $A$. (How­ever, var­i­ous very un­sta­ble lighter nu­clei are quite non­spher­i­cal too. None of this is writ­ten in stone.) In terms of fig­ure 14.19, they are the very small squares. Ex­am­ples are hafnium-176 in fig­ure 14.20 and tan­ta­lum-181 in fig­ure 14.21.

It seems clear that many or all nu­clei par­tic­i­pate in these ef­fects. Try­ing to ex­plain such or­ga­nized mas­sive nu­cleon par­tic­i­pa­tion based on a per­turbed ba­sic shell model alone would be very dif­fi­cult, and math­e­mat­i­cally un­sound in the case of de­formed nu­clei. A com­pletely dif­fer­ent ap­proach is de­sir­able.

Nu­clei with many nu­cle­ons and densely spaced en­ergy lev­els bear some sim­i­lar­ity to macro­scopic sys­tems. Based on that idea, physi­cists had an­other look at the clas­si­cal liq­uid drop model for nu­clei. That model was quite suc­cess­ful in ex­plain­ing the size and ground state en­ergy lev­els of nu­clei in sec­tion 14.10.

But liq­uid drops are not nec­es­sar­ily sta­tic; they can vi­brate. Vi­brat­ing states pro­vide a model for low-en­ergy ex­cited states in which the nu­cle­ons as a group par­tic­i­pate non­triv­ially. Fur­ther­more, the vi­bra­tions can be­come un­sta­ble, pro­vid­ing a model for per­ma­nent nu­clear de­for­ma­tion or nu­clear fis­sion. De­formed nu­clei can dis­play ef­fects of ro­ta­tion of the nu­clei. This sec­tion will give a ba­sic de­scrip­tion of these ef­fects.


14.13.1 Draft: Clas­si­cal liq­uid drop

This sec­tion re­views the me­chan­ics of a clas­si­cal liq­uid drop, like say a droplet of wa­ter. How­ever, there will be one ad­di­tional ef­fect in­cluded that you would be un­likely to see in a drop of wa­ter: it will be as­sumed that the liq­uid con­tains dis­trib­uted pos­i­tively charged ions. This is needed to al­low for the very im­por­tant desta­bi­liz­ing ef­fect of the Coulomb forces in a nu­cleus.

It will be as­sumed that the nu­clear liq­uid is ho­mo­ge­neous through­out. That is a some­what doubt­ful as­sump­tion for a model of a nu­cleus; there is no a pri­ori rea­son to as­sume that the pro­ton and neu­tron mo­tions are the same. But a two-liq­uid model, such as found in [39, p. 183ff], is be­yond the cur­rent cov­er­age.

It will fur­ther be as­sumed that the nu­clear liq­uid pre­serves its vol­ume. This as­sump­tion is con­sis­tent with the for­mula (14.9) for the nu­clear ra­dius, and it greatly sim­pli­fies the clas­si­cal analy­sis.

The von Weizsäcker for­mula showed that the nu­clear po­ten­tial en­ergy in­creases with the sur­face area. The rea­son is that nu­cle­ons near the sur­face of the nu­cleus are not sur­rounded by a full set of at­tract­ing neigh­bor­ing nu­cle­ons. Macro­scop­i­cally, this ef­fect is ex­plained as “sur­face ten­sion.” Sur­face ten­sion is de­fined as in­creased po­ten­tial en­ergy per unit sur­face area. (The work in ex­pand­ing the length of a rec­tan­gu­lar sur­face area must equal the in­creased po­ten­tial en­ergy of the sur­face mol­e­cules. From that it is seen that the sur­face ten­sion is also the ten­sion force at the perime­ter of the sur­face per unit length.)

Us­ing the sur­face term in the von Weizsäcker for­mula (14.10) and (14.9), the nu­clear equiv­a­lent of the sur­face ten­sion is

\begin{displaymath}
\sigma = \frac{C_s}{4\pi R_A^2}
\end{displaymath} (14.17)

The $C_d$ term in the von Weizsäcker for­mula might also be af­fected by the nu­clear sur­face area be­cause of its un­avoid­able ef­fect on the nu­clear shape, but to sim­plify things this will be ig­nored.

The sur­face ten­sion wants to make the sur­face of the drop as small as pos­si­ble. It can do so by mak­ing the drop spher­i­cal. How­ever, this also crowds the pro­tons to­gether the clos­est, and the Coulomb re­pul­sions re­sist that. So the Coulomb term fights the trend to­wards a spher­i­cal shape. This can cause heavy nu­clei, for which the Coulomb term is big, to fis­sion into pieces. It also makes lighter nu­clei less re­sis­tant to de­for­ma­tion, pro­mot­ing nu­clear vi­bra­tions or even per­ma­nent de­for­ma­tions. To in­clude the Coulomb term in the analy­sis of a clas­si­cal drop of liq­uid, it can be as­sumed that the liq­uid is charged, with to­tal charge $Ze$.

In­fin­i­tes­i­mal vi­bra­tions of such a liq­uid drop can be an­a­lyzed, {A.43}. It is then seen that the drop can vi­brate around the spher­i­cal shape with dif­fer­ent nat­ural fre­quen­cies. For a sin­gle mode of vi­bra­tion, the ra­dial dis­place­ment of the sur­face of the drop away from the spher­i­cal value takes the form

\begin{displaymath}
\delta = \varepsilon l \sin(\omega t - \varphi) \bar Y_l^m(\theta,\phi)
\end{displaymath} (14.18)

Here $\varepsilon{l}$ is the in­fin­i­tes­i­mal am­pli­tude of vi­bra­tion, $\omega$ the fre­quency, and $\varphi$ a phase an­gle. Also $\theta$ and $\phi$ are the co­or­di­nate an­gles of a spher­i­cal co­or­di­nate sys­tem with its ori­gin at the cen­ter of the drop, N.3. The $\bar{Y}_l^m$ are es­sen­tially the spher­i­cal har­mon­ics of or­bital an­gu­lar mo­men­tum fame, chap­ter 4.2.3. How­ever, in the clas­si­cal analy­sis it is more con­ve­nient to use the real ver­sion of the $Y_l^m$. For $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, there is no change, and for $m$ $\raisebox{.2pt}{$\ne$}$ 0 they can be ob­tained from the com­plex ver­sion by tak­ing $Y_l^m\pm{Y}_l^{-m}$ and di­vid­ing by $\sqrt2$ or $\sqrt2{\rm i}$ as ap­pro­pri­ate.

Vi­bra­tion with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is not pos­si­ble, be­cause it would mean that the ra­dius in­creased or de­creased every­where, ($Y_0^0$ is a con­stant), which would change the vol­ume of the liq­uid. Mo­tion with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is pos­si­ble, but it can be seen from the spher­i­cal har­mon­ics that this cor­re­sponds to trans­la­tion of the drop at con­stant ve­loc­ity, not to vi­bra­tion.

Vi­bra­tion oc­curs only for $l$ $\raisebox{-.5pt}{$\geqslant$}$ 2, and the fre­quency of vi­bra­tion is then, {A.43}:

\begin{displaymath}
\omega = \sqrt{\frac{E_{s,l}^2}{\hbar^2} \frac{1}{A} -
\frac{E_{c,l}^2}{\hbar^2} \frac{Z^2}{A^2}} %
\end{displaymath} (14.19)

The con­stants $E_{s,l}$ and $E_{c,l}$ ex­press the rel­a­tive strengths of the sur­face ten­sion and Coulomb re­pul­sions, re­spec­tively. The val­ues of these con­stants are, ex­pressed in en­ergy units,
\begin{displaymath}
E_{s,l} =
\frac{\hbar c}{R_A} \sqrt{\frac{(l-1)l(l+2)}{3}\...
...\frac{2(l-1)l}{2l+1}\frac{e^2}{4\pi\epsilon_0R_Am_{\rm p}c^2}}
\end{displaymath} (14.20)

The most im­por­tant mode of vi­bra­tion is the one at the low­est fre­quency, which means $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2. In that case the nu­mer­i­cal val­ues of the con­stants are
\begin{displaymath}
E_{s,2} \approx 35 \mbox{ MeV}
\qquad
E_{c,2} \approx 5.1 \mbox{ MeV}
\end{displaymath} (14.21)

Of course a nu­cleus with a lim­ited num­ber of nu­cle­ons and en­ergy lev­els is not a clas­si­cal sys­tem with count­less mol­e­cules and en­ergy lev­els. The best you may hope for that there will be some rea­son­able qual­i­ta­tive agree­ment be­tween the two.

It turns out that the liq­uid drop model sig­nif­i­cantly over­es­ti­mates the sta­bil­ity of nu­clei with re­spect to rel­a­tively small de­vi­a­tions from spher­i­cal. How­ever, it does much a bet­ter job of es­ti­mat­ing the sta­bil­ity against the large scale de­for­ma­tions as­so­ci­ated with nu­clear fis­sion.

Also, the in­er­tia of a nu­cleus can be quite dif­fer­ent from that of a liq­uid drop, [35, p. 345, 576]. This how­ever af­fects $E_{s,l}$ and $E_{c,l}$ equally, and so it does not fun­da­men­tally change the bal­ance be­tween sur­face ten­sion and Coulomb re­pul­sions.


14.13.2 Draft: Nu­clear vi­bra­tions

In the pre­vi­ous sub­sec­tion, the vi­bra­tional fre­quen­cies of nu­clei were de­rived us­ing a clas­si­cal liq­uid drop model. They ap­ply to vi­bra­tions of in­fi­nitely small am­pli­tude, hence in­fin­i­tes­i­mal en­ergy.

How­ever, for a quan­tum sys­tem like a nu­cleus, en­ergy should be quan­tized. In par­tic­u­lar, just like the vi­bra­tions of the elec­tro­mag­netic field come in pho­tons of en­ergy $\hbar\omega$, you ex­pect vi­bra­tions of mat­ter to come in “phonons” of en­ergy $\hbar\omega$. Plug­ging in the clas­si­cal ex­pres­sion for the low­est fre­quency gives

\begin{displaymath}
E_{\rm vibration} = \sqrt{E_{s,2}^2 \frac{1}{A} - E_{c,2}^2...
...\approx 35 \mbox{ MeV}
\quad
E_{c,2} \approx 5.1 \mbox{ MeV}
\end{displaymath} (14.22)

That is in the ball­park of ex­ci­ta­tion en­er­gies for nu­clei, sug­gest­ing that col­lec­tive mo­tion of the nu­cle­ons is some­thing that must be con­sid­ered.

In par­tic­u­lar, for light nu­clei, the pre­dicted en­ergy is about 35$\raisebox{.5pt}{$/$}$$\sqrt{A}$ MeV, com­pa­ra­ble to the von Weizsäcker ap­prox­i­ma­tion for the pair­ing en­ergy, 22$\raisebox{.5pt}{$/$}$$\sqrt{A}$ MeV. There­fore, it is in the ball­park to ex­plain the en­ergy of the 2$\POW9,{+}$ ex­ci­ta­tion of light even-even nu­clei in fig­ure 14.19. The pre­dicted en­er­gies are how­ever def­i­nitely too high. That re­flects the fact men­tioned in the pre­vi­ous sub­sec­tion that the clas­si­cal liq­uid drop over­es­ti­mates the sta­bil­ity of nu­clei with re­spect to small de­for­ma­tions. Note also the big dis­crete ef­fects of the magic num­bers in the fig­ure. Such quan­tum ef­fects are com­pletely missed in the clas­si­cal liq­uid drop model.

It should also be pointed out that now that the en­ergy is quan­tized, the ba­sic as­sump­tion that the am­pli­tude of the vi­bra­tions is in­fin­i­tes­i­mal is vi­o­lated. A quick ball­park shows the peak quan­tized sur­face de­flec­tions to be in the fm range, which is not re­ally small com­pared to the nu­clear ra­dius. If the am­pli­tude was in­deed in­fin­i­tes­i­mal, the nu­cleus would elec­tri­cally ap­pear as a spher­i­cally sym­met­ric charge dis­tri­b­u­tion. Whether you want to call the de­vi­a­tions from that pre­dic­tion ap­pre­cia­ble, [35, p. 342, 354] or small, [30, p. 152], nonzero val­ues should cer­tainly be ex­pected.

As far as the spin is con­cerned, the clas­si­cal per­tur­ba­tion of the sur­face of the drop is given in terms of the spher­i­cal har­mon­ics $Y_2^m$. The over­all mass dis­tri­b­u­tion has the same de­pen­dence on an­gu­lar po­si­tion. In quan­tum terms you would as­so­ciate such an an­gu­lar de­pen­dence with an az­imuthal quan­tum num­ber 2 and even par­ity, hence with a 2$\POW9,{+}$ state. It all seems to fit to­gether rather nicely.

There is more. You would ex­pect the pos­si­bil­ity of a two-phonon ex­ci­ta­tion at twice the en­ergy. Phonons, like pho­tons, are bosons; if you com­bine two of them in a set of sin­gle-par­ti­cle states of square an­gu­lar mo­men­tum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, then the net square an­gu­lar mo­men­tum can be ei­ther 0, 2, or 4, ta­ble 12.2. So you would ex­pect a triplet of ex­cited 0$\POW9,{+}$, 2$\POW9,{+}$, and 4$\POW9,{+}$ states at roughly twice the en­ergy of the low­est 2$\POW9,{+}$ ex­cited state.

And in­deed, oxy­gen-18 in fig­ure 14.17 shows a nicely com­pact triplet of this kind at about twice the en­ergy of the low­est 2$\POW9,{+}$ state. Ear­lier, these states were seem­ingly sat­is­fac­to­rily ex­plained us­ing sin­gle-nu­cleon ex­ci­ta­tions, or com­bi­na­tions of a few of them. Now how­ever, the liq­uid drop the­ory ex­plains them, also seem­ingly sat­is­fac­tory, as mo­tion of the en­tire nu­cleus!

That does not nec­es­sar­ily mean that one the­ory must be wrong and one right. There is no doubt that nei­ther the­ory has any real ac­cu­racy for this nu­cleus. The com­plex ac­tual dy­nam­ics is quite likely to in­clude non­triv­ial as­pects of each the­ory. The ques­tion is whether the the­o­ries can rea­son­ably pre­dict cor­rect prop­er­ties of the nu­cleus, re­gard­less of the ap­prox­i­mate way that they ar­rive at those pre­dic­tions. More­over, a nu­clear physi­cist would al­ways want to look at the de­cay of the ex­cited states, as well as their elec­tro­mag­netic prop­er­ties where avail­able, be­fore clas­si­fy­ing their na­ture. That how­ever is be­yond the scope of this book.

Fig­ure 14.22: An ex­ci­ta­tion en­ergy ra­tio for even-even nu­clei. [pdf][con]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
\begin{picture}(405,56...
...2){R: $0^+$, $2^+$, $4^+$, $6^+$}
}
\end{picture}}
\end{picture}
\end{figure}

Many, but by no means all, even-even nu­clei show sim­i­lar vi­bra­tional char­ac­ter­is­tics. That is il­lus­trated in fig­ure 14.22. This fig­ure shows the ra­tio of the sec­ond ex­cited en­ergy level $E_2$, re­gard­less of spin, di­vided by the en­ergy $E_{2^+}$ of the low­est 2$\POW9,{+}$ ex­cited state. Nor­mally $E_{2^+}$ is the low­est ex­cited state of all, so $E_2$ will be larger, mak­ing the ra­tio greater than one. That are the nu­clei we are in­ter­ested in here. He­lium 4, the lit­tle black square at the $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 in­ter­sec­tion, is of no in­ter­est.

Now the the­o­ret­i­cally-ideal vi­brat­ing nu­cleus would have a $2^+$ low­est ex­cited state, cor­re­spond­ing to a sin­gle phonon of vi­bra­tion. It would also have a triplet of states with two phonons, where ide­ally speak­ing all three of these states should each have an en­ergy $E_2$ that is ex­actly twice the en­ergy $E_{2^+}$ of the state with one phonon. That makes the ideal en­ergy ra­tio $E_2/E_{2^+}$ equal to 2. The nu­clei that do have that en­ergy ra­tio are shown as bright green squares in fig­ure 14.22. These bright green squares are likely can­di­dates for nu­clei with vi­brat­ing ex­cited states of low en­ergy. But in ad­di­tion, for the ideal vi­brat­ing nu­cleus, the triplet of ex­ited two-phonon states must con­sist of ex­actly one $0^+$ state, ex­actly one $2^+$ state, and ex­actly one $4^+$ state, be­cause that are the only three states that the two phonons can pro­duce. The nu­clei four which the low­est ex­cited state is $2^+$ one, and the next three states con­sist of one $0^+$ state, one $2^+$ state, and one $4^+$ state, are marked with an V in fig­ure 14.22. So bright green squares in fig­ure 14.22 with an V in them are surely nu­clei will­ing to vi­brate. Any­thing else would be too much of a co­in­ci­dence to be­lieve. You can see that nu­clei with vi­brat­ing ex­cited states are quite com­mon be­low $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 82 neu­trons, or near magic num­bers.

Still, many even-even nu­clei do not seem to have vi­bra­tional ex­cited states. But many of those who do not still have that un­ex­pected low­est ex­ited state that has spin $2^+$ and very lit­tle en­ergy. Ob­vi­ously, then, liq­uid-drop vi­bra­tions must be only a part of the story.

Fig­ure 14.23: Text­book vi­brat­ing nu­cleus tel­lurium-120. [pdf]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
% vertical spacing 137...
...\frac{29}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Much heav­ier vi­brat­ing nu­clei than oxy­gen-18 are tel­lurium-120 in fig­ure 14.23 as well as the ear­lier ex­am­ple of ruthe­nium-104 in fig­ure 14.20. Both nu­clei have again a fairly com­pact 0$\POW9,{+}$, 2$\POW9,{+}$, 4$\POW9,{+}$ triplet at roughly twice the en­ergy of the low­est 2$\POW9,{+}$ ex­cited state. But these more macro­scopic nu­clei also show a nice 0$\POW9,{+}$, 2$\POW9,{+}$, 3$\POW9,{+}$, 4$\POW9,{+}$, 6$\POW9,{+}$ quin­tu­plet at very roughly three times the en­ergy of the low­est 2$\POW9,{+}$ state. Yes, three iden­ti­cal phonons of spin 2 can have a com­bined spin of 0, 2, 3, 4, and 6, but not 1 or 5 (c.f. 12.2).

As sub­sec­tion 14.13.1 showed, liq­uid drops can also vi­brate ac­cord­ing to spher­i­cal har­mon­ics $Y_l^m$ for $l$ $\raisebox{.3pt}{$>$}$ 2. The low­est such pos­si­bil­ity $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 has spin 3 and neg­a­tive par­ity. Vi­bra­tion of this type is called “oc­tu­pole vi­bra­tion,” while $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 is re­ferred to as “quadru­pole vi­bra­tion.” For very light nu­clei, the en­ergy of oc­tu­pole vi­bra­tion is about twice that of the quadru­pole type. That would put the 3$\POW9,{-}$ oc­tu­pole vi­bra­tion right in the mid­dle of the two-phonon quadru­pole triplet. How­ever, for heav­ier nu­clei the 3$\POW9,{-}$ state will be rel­a­tively higher, since the en­ergy re­duc­tion due to the Coulomb term is rel­a­tively smaller in the case $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3. In­deed, the first 3$\POW9,{-}$ states for tel­lurium-120 and ruthe­nium-104 are found well above the quadru­pole quin­tu­plet. The low­est 3$\POW9,{-}$ state for much lighter oxy­gen-18 is rel­a­tively lower.


14.13.3 Draft: Non­spher­i­cal nu­clei

The clas­si­cal liq­uid drop model pre­dicts that the nu­cleus can­not main­tain a spher­i­cal ground state if the desta­bi­liz­ing Coulomb en­ergy ex­ceeds the sta­bi­liz­ing nu­clear sur­face ten­sion. In­deed, from elec­tro­mag­netic mea­sure­ments, it is seen that many very heavy nu­clei do ac­quire a per­ma­nent non­spher­i­cal shape. These are called “de­formed nu­clei”.

They are roughly the red squares and yel­low squares marked with R in fig­ure 14.22. Near the sta­ble line, their mass num­ber ranges are from about 150 to 190 and above 220. But many un­sta­ble much lighter nu­clei are de­formed too.

The liq­uid drop model, in par­tic­u­lar (14.19), pre­dicts that the nu­clear shape be­comes un­sta­ble at

\begin{displaymath}
\frac{Z^2}{A} = \frac{E_{s,2}^2}{E_{c,2}^2} \approx 48
\end{displaymath}

If that was true, es­sen­tially all nu­clei would be spher­i­cal. A mass num­ber of 150 cor­re­sponds to about $Z^2$$\raisebox{.5pt}{$/$}$$A$ equal to 26. How­ever, as pointed out in sub­sec­tion 14.13.1, the liq­uid drop model over­es­ti­mates the sta­bil­ity with re­spect to rel­a­tively small de­for­ma­tions. How­ever, it does a fairly good job of ex­plain­ing the sta­bil­ity with re­spect to large ones. That ex­plains why the de­for­ma­tion of the de­formed nu­clei does not progress un­til they have fis­sioned into pieces.

Physi­cists have found that most de­formed nu­clei can be mod­eled well as spher­oids, i.e. el­lip­soids of rev­o­lu­tion. The nu­cleus is no longer as­sumed to be spher­i­cally sym­met­ric, but still ax­i­ally sym­met­ric. Com­pared to spher­i­cal nu­clei, there is now an ad­di­tional nondi­men­sion­al num­ber that will af­fect the var­i­ous prop­er­ties: the ra­tio of the lengths of the two prin­ci­pal axes of the spher­oid. That com­pli­cates analy­sis. A sin­gle the­o­ret­i­cal num­ber now be­comes an en­tire set of num­bers, de­pend­ing on the value of the nondi­men­sion­al pa­ra­me­ter. For some nu­clei fur­ther­more, ax­ial sym­me­try is in­suf­fi­cient and a model of an el­lip­soid with three un­equal axes is needed. In that case there are two nondi­men­sion­al pa­ra­me­ters. Things get much messier still then.


14.13.4 Draft: Ro­ta­tional bands

Vi­bra­tion is not the only semi-clas­si­cal col­lec­tive mo­tion that nu­clei can per­form. De­formed nu­clei can also ro­tate as a whole. This sec­tion gives a sim­pli­fied semi-clas­si­cal de­scrip­tion of it.


14.13.4.1 Draft: Ba­sic no­tions in nu­clear ro­ta­tion

Clas­si­cally speak­ing, the ki­netic en­ergy of a solid body due to ro­ta­tion around an axis is $T_{\rm {R}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12{\cal I}_{\rm {R}}\omega^2$, where ${\cal I}_{\rm {R}}$ is the mo­ment of in­er­tia around the axis and $\omega$ the an­gu­lar ve­loc­ity. Quan­tum me­chan­ics does not use an­gu­lar ve­loc­ity but an­gu­lar mo­men­tum $J$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\cal I}_{\rm {R}}\omega$, and in these terms the ki­netic en­ergy is $T_{\rm {R}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $J^2$$\raisebox{.5pt}{$/$}$$2{\cal I}_{\rm {R}}$. Also, the square an­gu­lar mo­men­tum $J^2$ of a nu­cleus is quan­tized to be $\hbar^2j(j+1)$ where $j$ is the net spin of the nu­cleus, i.e. the az­imuthal quan­tum num­ber of its net an­gu­lar mo­men­tum.

There­fore, the ki­netic en­ergy of a nu­cleus due to its over­all ro­ta­tion be­comes:

\begin{displaymath}
\fbox{$\displaystyle
T_{\rm{R}} = \frac{\hbar^2}{2{\cal I}...
...ern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em)
$} %
\end{displaymath} (14.23)

Here $j_{\rm {min}}$ is the az­imuthal quan­tum num­ber of the “in­trin­sic state” in which the nu­cleus is not ro­tat­ing as a whole. The an­gu­lar mo­men­tum of this state is in the in­di­vid­ual nu­cle­ons and not avail­able for nu­clear ro­ta­tion, so it must be sub­tracted. The to­tal en­ergy of a state with spin $j$ is then

\begin{displaymath}
E_j = E_{\rm min} + T_{\rm {R}}
\end{displaymath}

where $E_{\rm {min}}$ is the en­ergy of the in­trin­sic state.

Fig­ure 14.24: Ro­ta­tional bands of hafnium-177. [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 now first a rough ball­park of the en­er­gies in­volved. Since $j$ is in­te­ger or half in­te­ger, the ro­ta­tional en­ergy comes in dis­crete amounts of $\hbar^2$$\raisebox{.5pt}{$/$}$$2{\cal I}_{\rm {R}}$. The clas­si­cal value for the mo­ment of in­er­tia ${\cal I}_{\rm {R}}$ of a rigid sphere of mass $m$ and ra­dius $R$ is $\frac25mR^2$. For a nu­cleus the mass $m$ is about $A$ times the pro­ton mass and the nu­clear ra­dius is given by (14.9). Plug­ging in the num­bers, the ball­park for ro­ta­tional en­er­gies be­comes

\begin{displaymath}
\frac{35}{A^{5/3}} \left[ j(j+1)-j_{\rm min}(j_{\rm min}+1) \right]
\mbox{ MeV}
\end{displaymath}

For a typ­i­cal non­spher­i­cal nu­cleus like hafnium-177 in fig­ure 14.24, tak­ing the in­trin­sic state to be the ground state with $j_{\rm {min}}$ equal to $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, the state $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ with an ad­di­tional unit of spin due to nu­clear ro­ta­tion would have a ki­netic en­ergy of about 0.06 MeV. The low­est ex­cited state is in­deed a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ one, but its en­ergy above the ground state is about twice 0.06 MeV. A nu­cleus is not at all like a rigid body in clas­si­cal me­chan­ics. It has al­ready been pointed out in the sub­sec­tion on nu­clear vi­bra­tions that in many ways a nu­cleus is much like a clas­si­cal fluid. Still, it re­mains true that ro­ta­tional en­er­gies are small com­pared to typ­i­cal sin­gle-nu­cleon ex­ci­ta­tion en­er­gies. There­fore ro­ta­tional ef­fects must be in­cluded if the low-en­ergy ex­cited states are to be un­der­stood.

To bet­ter un­der­stand the dis­crep­ancy in ki­netic en­ergy, drop the du­bi­ous as­sump­tion that the nu­clear ma­te­r­ial is a rigid solid. Pic­ture the nu­cleus in­stead as a spher­oid shape ro­tat­ing around an axis nor­mal to the axis of sym­me­try. As far as the in­di­vid­ual nu­cle­ons are con­cerned, this shape is stand­ing still be­cause the nu­cle­ons are go­ing so much faster than the nu­clear shape. A typ­i­cal nu­cleon has a ki­netic en­ergy in the or­der of 20 MeV, not a tenth of a MeV, and it is so much lighter than the en­tire nu­cleus to boot. Still, on the larger time scale of the nu­clear ro­ta­tions, the nu­cle­ons do fol­low the over­all mo­tion of the nu­clear shape, com­pare chap­ter 7.1.5. To de­scribe this, con­sider the nu­clear sub­stance to be an ideal liq­uid, one with­out in­ter­nal vis­cos­ity. With­out vis­cos­ity, the nu­clear liq­uid will not pick up the over­all ro­ta­tion of the nu­clear shape, so if the nu­clear shape is spher­i­cal, the nu­clear liq­uid will not be af­fected at all. This re­flect the fact that

Nu­clear ro­ta­tions can only be ob­served in nu­clei with a non­spher­i­cal equi­lib­rium state, [30, p. 142].

But if the ro­tat­ing nu­clear shape is not spher­i­cal, the nu­clear liq­uid can­not be at rest. Then it will still have to move ra­di­ally in­wards or out­wards to fol­low the chang­ing nu­clear sur­face ra­dius at a given an­gu­lar po­si­tion. This will in­volve some an­gu­lar mo­tion too, but it will re­main lim­ited. (Tech­ni­cally speak­ing, the mo­tion will re­main ir­ro­ta­tional, which means that the curl of the ve­loc­ity field will re­main zero.) In the liq­uid pic­ture, the mo­ment of in­er­tia has no phys­i­cal mean­ing and is sim­ply de­fined by the re­la­tion $T_{\rm {R}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12{\cal I}_{\rm {R}}\omega^2$, with $T_{\rm {R}}$ the ki­netic en­ergy of the liq­uid. If typ­i­cal num­bers are plugged into this pic­ture, [30, p. 145], you find that the pre­dicted ro­ta­tional en­er­gies are now too high. There­fore the con­clu­sion must be that the nu­clear sub­stance be­haves like some­thing in be­tween a solid and an ideal liq­uid, at least as far as nu­clear ro­ta­tions are con­cerned. Fairly good val­ues for the mo­ment of in­er­tia can be com­puted by mod­el­ing the nu­cleon pair­ing ef­fect us­ing a su­per­fluid model, [39, pp. 493ff]


14.13.4.2 Draft: Ba­sic ro­ta­tional bands

Con­sider the spec­trum of the de­formed nu­cleus hafnium-177 in fig­ure 14.24. At first the spec­trum seems a mess. How­ever, take the ground state to be an in­trin­sic state with a spin $j_{\rm {min}}$ equal to $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. Then you would ex­pect that there would also be en­ergy lev­els with the nu­cleus still in the same state but ad­di­tion­ally ro­tat­ing as a whole. Since quan­tum me­chan­ics re­quires that $j$ in­creases in in­te­ger steps, the ro­tat­ing ver­sions of the ground state should have spin $j$ equal to any one of $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 13}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, ... And in­deed, a se­quence of such ex­cited states can be iden­ti­fied in the spec­trum, as shown in the top right of fig­ure 14.24. Such a se­quence of en­ergy states is called a “ro­ta­tional band.” Note that all states in the band have the same par­ity. That is ex­actly what you would ex­pect based on the clas­si­cal pic­ture of a ro­tat­ing nu­cleus: the par­ity op­er­a­tor is a purely spa­tial one, so mere ro­ta­tion of the nu­cleus in time should not change it.

How about quan­ti­ta­tive agree­ment with the pre­dicted ki­netic en­er­gies of ro­ta­tion (14.23)? Well, as seen in the pre­vi­ous sub­sub­sec­tion, the ef­fec­tive mo­ment of in­er­tia is hard to find the­o­ret­i­cally. How­ever, it can be com­puted from the mea­sured en­ergy of the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ro­tat­ing state rel­a­tive to the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ground state us­ing (14.23). That pro­duces a mo­ment of in­er­tia equal to 49% of the cor­re­spond­ing solid sphere value. Then that value can be used to com­pute the en­er­gies of the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 13}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, ...states, us­ing again (14.23). The en­er­gies ob­tained in this way are in­di­cated by the spin-col­ored tick marks on the axis in the top-right graph of fig­ure 14.24. The lower en­er­gies agree very well with the ex­per­i­men­tal val­ues. Of course, the agree­ment of the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ lev­els is au­to­matic, but that of the higher lev­els is not.

For ex­am­ple, the pre­dicted en­ergy of the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state is 0.251 MeV, and the ex­per­i­men­tal value is 0.250 MeV. That is just a frac­tion of a per­cent er­ror, which is very much non­triv­ial. For higher ro­ta­tional en­er­gies, the ex­per­i­men­tal en­er­gies do grad­u­ally be­come some­what lower than pre­dicted, but noth­ing ma­jor. There are many ef­fects that could ex­plain the low­er­ing, but an im­por­tant one is “cen­trifu­gal stretch­ing.” As noted, a nu­cleus is not re­ally a rigid body, and un­der the cen­trifu­gal forces of rapid ro­ta­tion, it can stretch a bit. This in­creases the ef­fec­tive mo­ment of in­er­tia and hence low­ers the ki­netic en­ergy, (14.23).

How about all these other ex­cited en­ergy lev­els of hafnium-177? Well, first con­sider the na­ture of the ground state. Since hafnium-177 does not have a spher­i­cal shape, the nor­mal shell model does not ap­ply to it. In par­tic­u­lar, the nor­mal shell model would have the hafnium-177’s odd neu­tron alone in the 6f$_{5/2}$ sub­shell; there­fore it of­fers no log­i­cal way to ex­plain the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ ground state spin. How­ever, the Schrö­din­ger equa­tion can solved us­ing a non­spher­i­cal, but still ax­i­ally sym­met­ric, po­ten­tial to find suit­able sin­gle par­ti­cle states. Us­ing such states, it turns out that the fi­nal odd neu­tron goes into a state with mag­netic quan­tum num­ber $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$2 around the nu­clear axis and odd par­ity. (For a non­spher­i­cal po­ten­tial, the square an­gu­lar mo­men­tum no longer com­mutes with the Hamil­ton­ian and both $l$ and $j$ be­come un­cer­tain.) With ro­ta­tion, or bet­ter, with un­cer­tainty in ax­ial ori­en­ta­tion, this state gives rise to the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ground state of def­i­nite nu­clear spin $j$. In­creas­ing an­gu­lar mo­men­tum then gives rise to the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 13}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, ... ro­ta­tional band build on this ground state.

It is found that the next higher sin­gle-par­ti­cle state has mag­netic quan­tum num­ber $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ and even par­ity. If the odd neu­tron is kicked into that state, it pro­duces a low-en­ergy $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ ex­cited nu­clear state. Adding ro­ta­tional mo­tion to this in­trin­sic state pro­duces the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 13}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$, ...ro­ta­tional band shown in the mid­dle left of fig­ure 14.24. (Note that for this band, the ex­per­i­men­tal en­er­gies are larger than pre­dicted. Cen­trifu­gal stretch­ing is cer­tainly not the only ef­fect caus­ing de­vi­a­tions from the sim­ple the­ory.) In this case, the es­ti­mated mo­ment of in­er­tia is about 64% of the solid sphere value. There is no rea­son to as­sume that the mo­ment of in­er­tia re­mains the same if the in­trin­sic state of the nu­cleus changes. How­ever, clearly the value must re­main sen­si­ble.

The low-ly­ing $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state is be­lieved to be a re­sult of pro­mo­tion, where a neu­tron from a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ sin­gle-par­ti­cle state is kicked up to the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state where it can pair up with the 105th neu­tron al­ready there. Its ro­tat­ing ver­sions give rise to the ro­ta­tional band in the mid­dle right of fig­ure 14.24. The mo­ment of in­er­tia is about 45% of the solid sphere value. The last two bands have mo­ments of in­er­tia of 54% and 46%, in the ex­pected ball­park.

The gen­eral ap­proach as out­lined above has been ex­tra­or­di­nar­ily suc­cess­ful in ex­plain­ing the ex­cited states of de­formed nu­clei, [30, p. 156].


14.13.4.3 Draft: Bands with in­trin­sic spin one-half

The semi-clas­si­cal ex­pla­na­tion of ro­ta­tional bands was very sim­plis­tic. While it works fine if the in­trin­sic spin of the ro­tat­ing nu­clear state is at least one, it de­vel­ops prob­lems if it be­comes one-half or zero. The most tricky case is spin one-half.

De­spite less than stel­lar re­sults in the past, the dis­cus­sion of the prob­lem will stick with a semi-clas­si­cal ap­proach. Re­call first that an­gu­lar mo­men­tum is a vec­tor. In vec­tor terms, the to­tal an­gu­lar mo­men­tum of the nu­cleus con­sists of ro­ta­tional an­gu­lar mo­men­tum and in­trin­sic an­gu­lar mo­men­tum of the non­ro­tat­ing nu­cleus:

\begin{displaymath}
\vec J = \vec J_{\rm rot} + \vec J_{\rm min}
\end{displaymath}

Now in the ex­pres­sion for ro­ta­tional en­ergy, (14.23) it was im­plic­itly as­sumed that the square an­gu­lar mo­men­tum of the nu­cleus is the sum of the square an­gu­lar mo­men­tum of ro­ta­tion plus the square an­gu­lar mo­men­tum of the in­trin­sic state. But clas­si­cally the Pythagorean the­o­rem shows that this is only true if the two an­gu­lar mo­men­tum vec­tors are or­thog­o­nal.

In­deed, a more care­ful quan­tum treat­ment, [39, pp. 356-389], gives rise to a semi-clas­si­cal pic­ture in which the axis of the ro­ta­tion is nor­mal to the axis of sym­me­try of the nu­cleus. In terms of the in­vis­cid liq­uid model of sub­sub­sec­tion 14.13.4.1, ro­ta­tion about an axis of sym­me­try does not do any­thing. That leaves only the in­trin­sic an­gu­lar mo­men­tum for the com­po­nent of an­gu­lar mo­men­tum along the axis of sym­me­try. The mag­netic quan­tum num­ber of this com­po­nent is $j_{\rm {min}}$, equal to the spin of the in­trin­sic state. Cor­rect that: the di­rec­tion of the axis of sym­me­try should not make a dif­fer­ence. There­fore, the com­plete wave func­tion should be an equal com­bi­na­tion of a state ${\left\vert j_{\rm {min}}\right\rangle}$ with mag­netic quan­tum num­ber $j_{\rm {min}}$ along the axis and a state ${\left\vert-j_{\rm {min}}\right\rangle}$ with mag­netic quan­tum $-j_{\rm {min}}$.

Next, the ki­netic en­ergy of ro­ta­tion is, since $\vec{J}_{\rm {rot}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vec{J}-\vec{J}_{\rm {min}}$,

\begin{displaymath}
\frac{1}{2{\cal I}_{\rm {R}}} \vec J_{\rm rot}^2
= \frac{1...
..._{\rm min}
+ \frac{1}{2{\cal I}_{\rm {R}}} \vec J_{\rm min}^2
\end{displaymath}

As long as the mid­dle term in the right hand side av­er­ages away, the nor­mal for­mula (14.23) for the en­ergy of the ro­ta­tional states is fine. This hap­pens if there is no cor­re­la­tion be­tween the an­gu­lar mo­men­tum vec­tors $\vec{J}$ and $\vec{J}_{\rm {min}}$, be­cause then op­po­site and par­al­lel align­ments will can­cel each other.

But not too quick. There is an ob­vi­ous cor­re­la­tion since the ax­ial com­po­nents are equal. The term can be writ­ten out in com­po­nents to give

\begin{displaymath}
\frac{1}{{\cal I}_{\rm {R}}} \vec J\cdot\vec J_{\rm min} =
...
..._xJ_{x,\rm min} + J_yJ_{y,\rm min} + J_zJ_{z,\rm min}
\right]
\end{displaymath}

where the $z$-​axis is taken as the axis of sym­me­try of the nu­cleus. Now think of these com­po­nents as quan­tum op­er­a­tors. The $z$ com­po­nents are no prob­lem: since the mag­netic quan­tum num­ber is con­stant along the axis, this term will just shift the all en­ergy lev­els in the band by the same amount, leav­ing the spac­ings be­tween en­ergy lev­els the same.

How­ever, the $x$ and $y$ com­po­nents have the ef­fect of turn­ing a state ${\left\vert\pm{j}_{\rm {min}}\right\rangle}$ into some com­bi­na­tion of states ${\left\vert\pm{j}_{\rm {min}}\pm1\right\rangle}$, chap­ter 12.11. Since there is no ro­ta­tional mo­men­tum in the ax­ial di­rec­tion, $\vec{J}$ and $\vec{J}_{\rm {min}}$ have quite sim­i­lar ef­fects on the wave func­tion, but it is not the same, for one be­cause $\vec{J}$ sees the com­plete nu­clear mo­men­tum. If $j_{\rm {min}}$ is 1 or more, the ef­fects re­main in­con­se­quen­tial: then the pro­duced states are part of a dif­fer­ent vi­bra­tional band, with dif­fer­ent en­er­gies. A bit of in­ter­ac­tion be­tween states of dif­fer­ent en­ergy is usu­ally not a big deal, chap­ter 5.3. But if $j_{\rm {min}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$ then ${\left\vert j_{\rm {min}}-1\right\rangle}$ and ${\left\vert-j_{\rm {min}}+1\right\rangle}$ are part of the state it­self. In that case, the $x$ and $y$ com­po­nents of the $\vec{J}\cdot\vec{J}_{\rm {min}}$ term pro­duces a con­tri­bu­tion to the en­ergy that does not av­er­age out, and the larger $\vec{J}$ is, the larger the con­tri­bu­tion.

The ex­pres­sion for the ki­netic en­ergy of nu­clear ro­ta­tion then be­comes

\begin{displaymath}
\fbox{$\displaystyle
T_{\rm{R}} = \frac{\hbar^2}{2{\cal I}...
...kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em
$} %
\end{displaymath} (14.24)

where $a$ is a con­stant. Note that the ad­di­tional term is al­ter­nat­ingly pos­i­tive and neg­a­tive. Av­er­aged over the en­tire ro­ta­tional band, the ad­di­tional term does still pretty much van­ish.

Fig­ure 14.25: Ground state ro­ta­tional band of tung­sten-183. [pdf]
\begin{figure}\centering
\setlength{\unitlength}{1pt}
% vertical spacing 137...
...\frac{29}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

As an ex­am­ple, con­sider the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ground state ro­ta­tional band of tung­sten-183, fig­ure 14.25. To com­pute the ro­ta­tional ki­netic en­er­gies in the band us­ing (14.24) re­quires the val­ues of both ${\cal I}_{\rm {R}}$ and $a$. The mea­sured en­er­gies of the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ states above the ground state can be used to do so. That pro­duces a mo­ment of in­er­tia equal to 45% of the solid sphere value and a nondi­men­sion­al con­stant $a$ equal to 0.19. Next then the for­mula can be used to pre­dict the en­er­gies of the re­main­ing states in the band. As the ax­ial tick marks in the right graph of fig­ure 14.25 show, the pre­dic­tion is quite good. Note in par­tic­u­lar that the en­ergy in­ter­val be­tween the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ 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 is less than that be­tween the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ states. With­out the al­ter­nat­ing term, there would be no way to ex­plain that.

Fig­ure 14.26: Ro­ta­tional bands of alu­minum-25. [pdf]
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Much larger val­ues of $a$ are ob­served for lighter nu­clei. As an ex­am­ple, con­sider alu­minum-25 in fig­ure 14.26. This nu­cleus has been stud­ied in de­tail, and a num­ber of bands with an in­trin­sic spin 1/2 have been iden­ti­fied. Par­tic­u­larly in­ter­est­ing is the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ band in the bot­tom left of fig­ure 14.26. For this band $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$3.2, and that is big enough to change the or­der of the states in the band! For this nu­cleus, the mo­ments of in­er­tia are 70%, 96%, 107%, 141% and 207% re­spec­tively of the solid sphere value.


14.13.4.4 Draft: Bands with in­trin­sic spin zero

The case that the in­trin­sic state has spin zero is par­tic­u­larly im­por­tant, be­cause all even-even nu­clei have a 0$\POW9,{+}$ ground state. For bands build on a zero-spin in­trin­sic state, the thing to re­mem­ber is that the only val­ues in the band are even spin and even par­ity: 0$\POW9,{+}$, 2$\POW9,{+}$, 4$\POW9,{+}$, ...

This can be thought of as a con­se­quence of the fact that the ${\left\vert j_{\rm {min}}\right\rangle}$ and ${\left\vert-j_{\rm {min}}\right\rangle}$ states of the pre­vi­ous sub­sub­sec­tion be­come equal. Their odd or even com­bi­na­tions must be con­strained to pre­vent them from can­cel­ing each other.

Fig­ure 14.27: Ro­ta­tional bands of er­bium-164. [pdf]
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As an ex­am­ple, con­sider er­bium-164. The ground state band in the top right of fig­ure 14.27 con­sists of the states 0$\POW9,{+}$, 2$\POW9,{+}$, 4$\POW9,{+}$, ...as ex­pected. The en­er­gies ini­tially agree well with the the­o­ret­i­cal pre­dic­tion (14.23) shown as tick marks. For ex­am­ple, the pre­dic­tion for the 4$\POW9,{+}$ level has less than 2% er­ror. De­vi­a­tions from the­ory show up at higher an­gu­lar mo­menta, which may have to do with cen­trifu­gal stretch­ing.

Other bands have been iden­ti­fied that are build upon vi­bra­tional in­trin­sic states. (A $\beta$ or beta vi­bra­tion main­tains the as­sumed in­trin­sic ax­ial sym­me­try of the nu­cleus, a $\gamma$ or gamma one does not.) Their en­ergy spac­ings fol­low (14.23) again well. The mo­ments of in­er­tia are 46%, 49% and 61% of the solid sphere value.

Fig­ure 14.28: Ground state ro­ta­tional band of mag­ne­sium-24. [pdf]
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Light even-even nu­clei can also be de­formed and show ro­ta­tional bands. As an ex­am­ple, fig­ure 14.28 shows the ground state band of mag­ne­sium-24. The mo­ment of in­er­tia is 75% of the solid sphere value.

It may also be men­tioned that nu­clei with in­trin­sic spin zero com­bined with an oc­tu­pole vi­bra­tion can give rise to bands that are purely odd spin/odd par­ity ones, 1$\POW9,{-}$, 3$\POW9,{-}$, 5$\POW9,{-}$, ..., [39, p. 368]. The low­est odd par­ity states for er­bium-164 are 1$\POW9,{-}$ and 3$\POW9,{-}$ ones, with no 2$\POW9,{-}$ state in be­tween, for a re­ally high es­ti­mated mo­ment of in­er­tia of 148%, and a po­ten­tial 5$\POW9,{-}$ state at roughly, but not ac­cu­rately, the pre­dicted po­si­tion. Anom­alous bands that have the par­ity in­verted may also be pos­si­ble; hafnium-176 is be­lieved to have a cou­ple of ex­cited states like that, 0$\POW9,{-}$ at 1.819 MeV and 2$\POW9,{-}$ at 1.857 MeV, with a mo­ment of in­er­tia of 98%.

Cen­trifu­gal ef­fects can be se­vere enough to change the in­ter­nal struc­ture of the nu­cleus non­triv­ially. Typ­i­cally, zero-spin pair­ings be­tween nu­cle­ons may be bro­ken up, al­low­ing the nu­cle­ons to start ro­tat­ing along with the nu­cleus. That cre­ates a new band build on the changed in­trin­sic state. Physi­cists then de­fine the state of low­est en­ergy at a given an­gu­lar mo­men­tum as the “yrast state.” The term is not an acronym, but Swedish for that what ro­tates more. For a dis­cus­sion, a book for spe­cial­ists will need to be con­sulted.


14.13.4.5 Draft: Even-even nu­clei

All nu­clei with even num­bers of both pro­tons and neu­trons have a 0$\POW9,{+}$ ground state. For non­spher­i­cal ones, the ro­ta­tional model pre­dicts a ground state band of low-ly­ing 2$\POW9,{+}$, 4$\POW9,{+}$, 6$\POW9,{+}$, ...states. The ra­tio of the en­ergy lev­els of the 4$\POW9,{+}$ and 2$\POW9,{+}$ states is given by (14.23)

\begin{displaymath}
\left.\frac{\hbar^2}{2{\cal I}_{\rm {R}}} 4(4+1)\right/
\frac{\hbar^2}{2{\cal I}_{\rm {R}}} 2(2+1) = \frac{10}3
\end{displaymath}

For spher­i­cal nu­clei, the vi­bra­tional model also pre­dicts a 2$\POW9,{+}$ low­est ex­cited state, but the 4$\POW9,{+}$ ex­cited state is now part of a triplet, and the triplet has only twice the en­ergy of the 2$\POW9,{+}$ state. There­fore, if the ra­tio of the en­ergy of the sec­ond ex­cited state to the low­est 2$\POW9,{+}$ state is plot­ted, as done in fig­ure 14.22, then vi­brat­ing nu­clei should be in­di­cated by a value 2 (green) and ro­tat­ing nu­clei by a value 3.33 (red). If the fig­ure is ex­am­ined, it may be open to some doubt whether green squares are nec­es­sar­ily vi­bra­tional, but the red squares quite nicely lo­cate the ro­ta­tional ones.

In the fig­ure, nu­clei marked with V have 0$\POW9,{+}$, 2$\POW9,{+}$, and a 0$\POW9,{+}$, 2$\POW9,{+}$, 4$\POW9,{+}$ triplet as the low­est 5 en­ergy states, the triplet al­lowed to be in any or­der. Nu­clei marked with an R have the se­quence 0$\POW9,{+}$, 2$\POW9,{+}$, 4$\POW9,{+}$, and 6$\POW9,{+}$ as the low­est four en­ergy states. Note that this cri­te­rion misses the light ro­ta­tional nu­clei like mag­ne­sium-24; for light nu­clei the ro­ta­tional en­er­gies are not small enough to be well sep­a­rated from shell ef­fects. Near the sta­ble line, ro­tat­ing nu­clei are found in the ap­prox­i­mate mass num­ber ranges 20 $\raisebox{.3pt}{$<$}$ $A$ $\raisebox{.3pt}{$<$}$ 30, 150 $\raisebox{.3pt}{$<$}$ $A$ $\raisebox{.3pt}{$<$}$ 190, and 220 $\raisebox{.3pt}{$<$}$ $A$. How­ever, away from the sta­ble line ro­tat­ing nu­clei are also found at other mass num­bers.


14.13.4.6 Draft: Non­ax­ial nu­clei

While most nu­clei are well mod­eled as ax­i­ally sym­met­ric, some nu­clei are not. For such nu­clei, an el­lip­soidal model can be used with the three ma­jor axes all off dif­fer­ent length. There are now two nondi­men­sion­al axis ra­tios that char­ac­ter­ize the nu­cleus, rather than just one.

This ad­di­tional nondi­men­sion­al pa­ra­me­ter makes the spec­trum much more com­plex. In par­tic­u­lar, in ad­di­tion to the nor­mal ro­ta­tional bands, as­so­ci­ated anom­alous sec­ondary bands ap­pear. The first is a 2$\POW9,{+}$, 3$\POW9,{+}$, 4$\POW9,{+}$, ...one, the sec­ond a 4$\POW9,{+}$, 5$\POW9,{+}$, ...one, etcetera. The en­er­gies in these bands are not in­de­pen­dent, but re­lated to those in the pri­mary band.

Fig­ure 14.29: Ro­ta­tional bands of os­mium-190. [pdf]
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Fig­ure 14.29 shows an ex­am­ple. The pri­mary ground state band in the top right quickly de­vel­ops big de­vi­a­tions from the ax­i­ally sym­met­ric the­ory (14.23) val­ues (thin tick marks.) Com­pu­ta­tion us­ing the el­lip­soidal model for a suit­able value of the de­vi­a­tion from ax­ial sym­me­try is much bet­ter (thick tick marks.) The pre­dicted en­ergy lev­els of the first anom­alous band also agree well with the pre­dicted val­ues. The iden­ti­fi­ca­tion of the bands was taken from [39, p. 416], but since they do not list the en­er­gies of the sec­ond anom­alous band, that value was taken from [35, p. 388].

In the limit that the nu­clear shape be­comes ax­i­ally sym­met­ric, the anom­alous bands dis­ap­pear to­wards in­fi­nite en­ergy. In the limit that the nu­clear shape be­comes spher­i­cal, all states in the pri­mary bands ex­cept the low­est one also dis­ap­pear to in­fin­ity, as­sum­ing that the mo­ment of in­er­tia be­comes zero as the ideal liq­uid model says.