- 14.13.1 Draft: Classical liquid drop
- 14.13.2 Draft: Nuclear vibrations
- 14.13.3 Draft: Nonspherical nuclei
- 14.13.4 Draft: Rotational bands

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

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 [40, 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

(14.17) |

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

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

(14.18) |

Vibration with

Vibration occurs only for

(14.20) |

(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, [36, p. 345, 576]. This however
affects

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

(14.22) |

In particular, for light nuclei, the predicted energy is about
3

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, [36, p. 342, 354] or small, [31, 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

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

And indeed, oxygen-18 in figure 14.17 shows a nicely
compact triplet of this kind at about twice the energy of the lowest

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.

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

Now the theoretically-ideal vibrating nucleus would have a 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

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

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 macroscopic

nuclei also show a nice

As subsection 14.13.1 showed, liquid drops can also vibrate
according to spherical harmonics

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

If that was true, essentially all nuclei would be spherical. A mass number of 150 corresponds to about

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 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:

where

Consider now first a rough ballpark of the energies involved. Since

For a typical nonspherical nucleus like hafnium-177 in figure 14.24, taking the intrinsic state to be the ground state with

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, [31, 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

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

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

For example, the predicted energy of the

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 6

It is found that the next higher single-particle state has magnetic
quantum number

The low-lying

The general approach as outlined above has been extraordinarily successful in explaining the excited states of deformed nuclei, [31, 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:

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,
[40, 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

Next, the kinetic energy of rotation is, since

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

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

where the

However, the sees

the complete nuclear momentum. If

The expression for the kinetic energy of nuclear rotation then becomes

As an example, consider the

Much larger values of

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

This can be thought of as a consequence of the fact that the

As an example, consider erbium-164. The ground state band in the top
right of figure 14.27 consists of the states

Other bands have been identified that are build upon vibrational
intrinsic states. (A

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,

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

For spherical nuclei, the vibrational model also predicts a

In the figure, nuclei marked with V

have R

have the sequence

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

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 [40, p. 416], but since they do not list the energies of the second anomalous band, that value was taken from [36, 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.