Subsections


14.13 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 14.18, and many other even-even nuclei with such energies may be found in figure 14.17. 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 14.17, they are the very small squares. Examples are hafnium-176 in figure 14.18 and tantalum-181 in figure 14.19.

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 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.42}. 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.42}:

\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 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}^...
...prox 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 14.17. 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 14.15 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 14.20: An excitation energy ratio for even-even nuclei. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...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 14.20. 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. Nuclei that have a 2$\POW9,{+}$ lowest excited state, immediately followed by a 0$\POW9,{+}$, 2$\POW9,{+}$, 4$\POW9,{+}$ triplet, in any order, are marked with a V for vibrational. Green squares marked with a V have in addition the rough energy ratio of 2 between the second and first excited energies predicted by the vibrating liquid drop model. These requirements are clearly nontrivial, so it is encouraging that so many nuclei except the very heavy ones satisfy them. Still, many even-even nuclei do not. Obviously liquid-drop vibrations are only a part of the complete story.

Figure 14.21: Textbook vibrating nucleus tellurium-120. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Much heavier vibrating nuclei than oxygen-18 are tellurium-120 in figure 14.21 as well as the earlier example of ruthenium-104 in figure 14.18. 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.

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 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 14.20. 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 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 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 ...
...+1)
\qquad (j_{\rm min} \ne {\textstyle\frac{1}{2}})
$} %
\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 14.22: Rotational bands of hafnium-177. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{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 14.22, taking the intrinsic state to be the ground state with $j_{\rm {min}}$ equal to $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, the state $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ 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 Basic rotational bands

Consider the spectrum of the deformed nucleus hafnium-177 in figure 14.22. 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 13}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, ... And indeed, a sequence of such excited states can be identified in the spectrum, as shown in the top right of figure 14.22. 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ rotating state relative to the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 13}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, ...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 14.22. The lower energies agree very well with the experimental values. Of course, the agreement of the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ levels is automatic, but that of the higher levels is not.

For example, the predicted energy of the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ 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-.2em\lower.4ex\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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ground state of definite nuclear spin $j$. Increasing angular momentum then gives rise to the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 13}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$, ... rotational band build on this ground state.

It is found that the next higher single-particle state has magnetic quantum number $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ and even parity. If the odd neutron is kicked into that state, it produces a low-energy $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ excited nuclear state. Adding rotational motion to this intrinsic state produces the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 13}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$, ...rotational band shown in the middle left of figure 14.22. (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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state is believed to be a result of promotion, where a neutron from a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ single-particle state is kicked up to the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ 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 14.22. 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 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 $\big\vert j_{\rm {min}}\big\rangle $ with magnetic quantum number $j_{\rm {min}}$ along the axis and a state $\big\vert-j_{\rm {min}}\big\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...
...\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} =...
...J_{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 $\big\vert\pm{j}_{\rm {min}}\big\rangle $ into some combination of states $\big\vert\pm{j}_{\rm {min}}\pm1\big\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 $\big\vert j_{\rm {min}}-1\big\rangle $ and $\big\vert-j_{\rm {min}}+1\big\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 ...
...\right\}
\quad j_{\rm min} = {\textstyle\frac{1}{2}}
$} %
\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 14.23: Ground state rotational band of tungsten-183. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

As an example, consider the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ ground state rotational band of tungsten-183, figure 14.23. 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ 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 14.23 show, the prediction is quite good. Note in particular that the energy interval between the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ states is less than that between the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ states. Without the alternating term, there would be no way to explain that.

Figure 14.24: Rotational bands of aluminum-25. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Much larger values of $a$ are observed for lighter nuclei. As an example, consider aluminum-25 in figure 14.24. 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ band in the bottom left of figure 14.24. For this band $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$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 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 $\big\vert j_{\rm {min}}\big\rangle $ and $\big\vert-j_{\rm {min}}\big\rangle $ states of the previous subsubsection become equal. Their odd or even combinations must be constrained to prevent them from canceling each other.

Figure 14.25: Rotational bands of erbium-164. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

As an example, consider erbium-164. The ground state band in the top right of figure 14.25 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 14.26: Ground state rotational band of magnesium-24. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Light even-even nuclei can also be deformed and show rotational bands. As an example, figure 14.26 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 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 14.20, 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 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 14.27: Rotational bands of osmium-190. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Figure 14.27 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.