Subsections


14.12 Shell model

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

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

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

The higher magic number values are quite clearly seen in proton pair and neutron pair removal graphs like figures 14.5 and 14.6 in section 14.9.

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

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

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

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

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


14.12.1 Average potential

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

A better potential is needed. Now in the center of a nucleus, the attractive forces come from all directions and the net force will be zero by symmetry. Away from the center, the net force will be directed inwards towards the center to keep the nucleons together inside the nucleus. The simplest potential that describes this is the harmonic oscillator one. For that potential, the inward force is simply proportional to the distance from the center. That makes the potential energy $V$ proportional to the square distance from the center, as sketched in figure 14.11a.

Figure 14.11: Example average nuclear potentials: (a) harmonic oscillator, (b) impenetrable surface, (c) Woods-Saxon, (d) Woods-Saxon for protons.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...(c)}}
\put(150,70){\makebox(0,0)[t]{\em (d)}}
\end{picture}
\end{figure}

The energy eigenvalues of the harmonic oscillator are

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

Also, in spherical coordinates the energy eigenfunctions of the harmonic oscillator can be taken to be of the form, {D.77},
\begin{displaymath}
\psi_{nlmm_s}^{\rm ho} = R_{nl}^{\rm ho}(r) Y_l^m(\theta,\...
...l{-}1, l \\
m_s=\pm{\textstyle\frac{1}{2}}
\end{array} %
\end{displaymath} (14.16)

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

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

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

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

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

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

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

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

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

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

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

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

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

In these terms, the energy levels and numbers of states for the harmonic oscillator potential are as shown in figure 14.12. The third energy level has 2 3s states and 10 3d states. Added to the 8 from the first two energy levels, that brings the total count to 20, the third magic number.

Figure 14.12: Nuclear energy levels for various assumptions about the average nuclear potential. The signs indicate the parity of the states.
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\centering
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\begin{picture}(...
...?}}
\put(243,199){\makebox(0,0)[ct]{\tt 70?}}
\end{picture}
\end{figure}

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

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

That reasoning results in the impenetrable-shell potential shown in figure 14.11. It too is analytically solvable, {D.78}. The energy levels are shown in figure 14.12. Unfortunately, it does not help any explaining the fourth magic number 28.

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

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

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

which is sketched in figure 14.11c. For protons, there is an additional repulsive Coulomb potential that will be maximum at the center of the sphere and decreases to zero proportional to 1$\raisebox{.5pt}{$/$}$$r$ outside the nucleus. That gives a combined potential as sketched in figure 14.11d. Note that the Coulomb potential is not short-range like the nucleon-nucleon attractions; its nontrivial variation is not just restricted to a thin layer at the nuclear surface.

Typical energy levels are sketched in figure 14.12. As expected, they are somewhere in between the extreme cases of the harmonic oscillator and the impenetrable shell.

The signs behind the realistic energy levels in 14.12 denote the predicted “parity” of the states. Parity is a very helpful mathematical quantity for studying nuclei. The parity of a wave function is one,” or “positive, or even, if the wave function stays the same when the positive direction of the three Cartesian axes is inverted. That replaces every ${\skew0\vec r}$ in the wave function by $\vphantom0\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. The parity is minus one,”, or “negative, or odd, if the wave function merely changes sign under an exes inversion. Parity is uncertain when the wave function changes in any other way; however, nuclei have definite parity as long as the weak force of beta decay does not play a role. It turns out that s, d, g, ...states have positive parity while p, f, h, ...states have negative parity, {D.14} or {D.77}. Therefore, the harmonic oscillator shells have alternatingly positive and negative parity.

For the wave functions of complete nuclei, the net parity is the product of the parities, (taking them to be one or minus one), of the individual nucleons. Now physicist can experimentally deduce the parity of nuclei in various ways. It turns out that the parities of the nuclei up to the third magic number agree perfectly with the values predicted by the energy levels of figure 14.12. (Only three unstable, artificially created, nuclei disagree.) It really appears that the model is onto something.

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


14.12.2 Spin-orbit interaction

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

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

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

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

Here ${\skew 4\widehat{\vec L}}$ is the orbital angular momentum vector and ${\skew 6\widehat{\vec S}}$ the spin one. A contribution to the Hamiltonian of this type is called an spin-orbit interaction, because it couples spin with orbital angular momentum. Spin-orbit interaction was already known from improved descriptions of the energy levels of the hydrogen atom, addendum {A.38}. However, that electromagnetic effect is far too small to explain the observed spin-orbit interaction in nuclei. Also, it would get the sign of the correction wrong for neutrons.

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

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

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

However, consider the net angular momentum operator

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

If you expand its square magnitude,

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

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

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

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

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

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

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

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

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

\begin{displaymath}
- {\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}=...
...for } j=l+{\textstyle\frac{1}{2}} \\
\end{array}
\right.
\end{displaymath}

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

Figure 14.13: Schematic effect of spin-orbit interaction on the energy levels. The ordering within bands is realistic for neutrons. The designation behind the equals sign is the official one. (Assuming counting starts at 1).
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Figure 14.13 shows how the spin-orbit splitting of the energy levels gives rise to the remaining magic numbers. In the figure, the coefficient of the spin orbit term was simply taken to vary linearly with the energy level $n$. The details depend on whether it is neutrons or protons, and may vary from nucleus to nucleus. Especially for the higher energy bands the Coulomb repulsion has an increasingly large effect on the energies of protons.

The major shells, terminated by magic numbers, are shown as grey bands. In the numbering system followed here, a subshell with a different number as the others in the same major shell comes from a different harmonic oscillator energy level. Figure 14.13 also shows the official enumeration of the states. You be the judge which numbering system makes the most sense to you.

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

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

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


14.12.3 Example occupation levels

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

Figure 14.14: Energy levels for doubly-magic oxygen-16 and neighbors. [pdf]
\begin{figure}
\centering
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% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Figure 14.14 shows experimental energy spectra of various nuclei at the left. The energy values are in MeV. The ground state is defined to be the zero level of energy. The length and color of the energy lines indicates the spin of the nucleus, and the parity is indicated by a plus or minus sign. Some important spin values are also listed explicitly. Yellow lines indicate states for which no unique spin and/or parity are determined or are established with reservations. At the right in the figure, a sketch of the occupation levels according to the shell model is displayed for easy reference.

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

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

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

Next,

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

The shell model implies that a doubly-magic nucleus like oxygen-16 should be be particularly stable. So it should require a great deal of energy to excite it. Indeed it does: figure 14.14 shows that exciting oxygen-16 takes over 6 MeV of energy.

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

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

Next consider oxygen-17 and fluorine-17 in figure 14.14. These two are examples of so-called mirror nuclei; they have the numbers of protons and neutrons reversed. Oxygen-17 has 8 protons and 9 neutrons while its twin fluorine-17 has 9 protons and 8 neutrons. The similarity in energy levels between the two illustrates the idea of charge symmetry: nuclear forces are the same if the protons are turned into neutrons and vice versa. (Of course, this swap does mess up the Coulomb forces, but Coulomb forces are not very important for light nuclei.)

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

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

For both oxygen-17 and fluorine-17, there is also a low-energy $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ excited state, likely corresponding to kicking the odd nucleon up to the next minor shell, the 3s$_{1/2}$ one. And so there is an excited $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state, for kicking up the nucleon to the 3d$_{3/2}$ state instead.

However, from the shell model, in particular figure 14.13, you would expect the spacing between the 3d$_{5/2}$ and 3s$_{1/2}$ subshells to be more than that between the 3s$_{1/2}$ and 3d$_{3/2}$ ones. Clearly it is not. One consideration not in a shell model with a straightforward average potential is that a nucleon in an unusually far-out s orbit could be closer to the other nucleons in lower orbits than one in a far-out p orbit; the s orbit has larger values near the center of the nucleus, {N.8}. While the shell model gets a considerable number of things right, it is certainly not a very accurate model.

Then there are the odd parity states. These are not so easy to understand: they require a nucleon to be kicked up past a major shell boundary. That should require a lot of energy according to the ideas of the shell model. It seems to make them hard to reconcile with the much higher energy of the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state. Some thoughts on these states will be given in the next subsection.

The fourth nucleus in figure 14.14 is nitrogen-14. This is an odd-odd nucleus, with both an odd number of protons and of neutrons. The odd proton and odd neutron are in the 2p$_{1/2}$ shell, so each has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. Quantum mechanics allows the two to combine their spins into a triplet state of net spin one, like they do in deuterium, or in a singlet state of spin zero. Indeed the ground state is a 1$\POW9,{+}$ one like deuterium. The lowest excited state is a 0$\POW9,{+}$ one.

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

Holes in an otherwise filled subshell have the same possible angular momentum values as particles in an otherwise empty shell.
Therefore the hole must have the spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ of a single particle. This can combine with the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ of the odd nucleon of the opposite type to either spin 1 or spin 2. A relatively low energy 1$\POW9,{+}$ state can be observed in the experimental spectrum.

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


14.12.4 Shell model with pairing

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

Figure 14.15: Nucleon pairing effect. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Consider first oxygen-18 in figure 14.15, with both an even number of protons and an even number of neutrons. As always, the filled subshells have no angular momentum. That leaves the two 3d$_{5/2}$ neutrons. These could have combined integer spin from 0 to 5 if they were distinguishable particles. However, the two neutrons are identical fermions, and the wave function must be antisymmetric with respect to their exchange. It can be seen from chapter 12.8 item 3, or more simply from table 12.1, that only the 0, 2, and 4 combined spins are allowed. Still, that leaves three possibilities for the net spin of the entire nucleus.

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

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

So physicists have concluded that besides the average potential included in the shell model, there must be an additional pairing energy that makes nucleons of the same type want to combine pairwise into states of zero spin. In order to treat this effect mathematically without losing the basic shell model, the pairing energy must be treated as a relatively small perturbation to the shell model energy. Theories that do so are beyond the scope of this book, although the general ideas of perturbation theories can be found in addendum {A.37}. Here it must be suffice to note that the pairing effect exists and is due to interactions between nucleons not included in the basic shell model potential.

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

If the pairing energy is a relatively small perturbation to the shell model, then for oxygen-18 you would expect that besides the zero spin ground state, the other possibilities of spin 2 and 4 would show up as low-lying excited states. Indeed the experimental spectrum in figure 14.15 shows 2$\POW9,{+}$ and 4$\POW9,{+}$ states of the right spin and parity, though their energy is obviously not so very low. To put it in context, the von Weizsäcker formula puts the pairing energy at 22$\raisebox{.5pt}{$/$}$$\sqrt{A}$ MeV, which would be of the rough order of 5 MeV for oxygen-18.

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

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

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

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

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

Consider again the 0$\POW9,{+}$ excited state of oxygen-16 in figure 14.14 as discussed in the previous subsection. Some of the energy needed for a pair of 2p$_{1/2}$ nucleons to cross the major shell boundary to the 3d$_{5/2}$ subshell will be compensated for by the higher pairing energy in the new subshell. It still seems curious that the state would end up below the 3$\POW9,{-}$ one, though.

Similarly, the relatively low energy $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state in oxygen-17 and fluorine-17 can now be made a bit more plausible. To explain the negative parity, a nucleon must be kicked across the major shell boundary from the 2p$_{1/2}$ subshell to the 3d$_{5/2}$ one. That should require quite a bit of energy, but this will in part be compensated for by the fact that pairing now occurs at higher angular momentum.

So what to make of the next $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ state? One possibility is that a 2p$_{1/2}$ nucleon is kicked to the 3s$_{1/2}$ subshell. The three spins could then combine into 5/2, [30, p. 131]. If true however, this would be a quite significant violation of the basic ideas of the perturbed shell model. Just consider: it requires breaking up the 2p$_{1/2}$ pair and kicking one of the two neutrons across both a major shell boundary and a subshell one. That would require less energy than the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ excitation in which the odd nucleon is merely kicked over two subshell boundaries and no pair is broken up? An alternative that is more consistent with the perturbed shell model ideas would be that the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ excitation is like the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ one, but with an additional partial break up of the resulting pair. The energy seems still low.

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

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

Figure 14.15 shows the example of oxygen-19, with three neutrons in the unfilled 3d$_{5/2}$ subshell. The odd-particle shell model predicts that the first two neutrons still combine into a state of zero spin like in oxygen-18. That leaves only the spin of the third neutron. And indeed, the total nuclear spin of oxygen-18 is observed to be 5/2 in the ground state, the spin of this odd neutron. The odd-particle shell model got it right.

It is important to recognize that the odd-particle shell model only applies to the ground state. This is not always sufficiently stressed. Theoretically, three 3d$_{5/2}$ neutrons can combine their spins not just to spin 5/2, but also to 3/2 or 9/2 while still satisfying the antisymmetrization requirement, table 12.1. And indeed, the oxygen-19 energy spectrum in figure 14.15 shows relatively low energy $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ states. To explain the energies of these states would require computation using an actual perturbed shell model, rather than just the odd-particle assumption that such a model will lead to perfect pairing of even numbers of nucleons.

It is also important to recognize that the odd-particle shell model is only a prediction. It does fail for a fair number of nuclei. That is true even excluding the very heavy nuclei for which the shell model does not apply period. For example, note in figure 14.15 how close together are the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ energy levels. You might guess that the order of those two states could easily be reversed for another nucleus. And so it can; there are a number of nuclei in which the spins combine into a net spin one unit less than that of the last odd nucleon. While the unperturbed shell model does not fundamentally fail for such nuclei, (because it does not predict the spin at all), the additional odd-particle assumption does.

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

While the odd-particle model applies only to the ground state, some excited states can still be described as purely odd-particle effects. In particular, for the oxygen-19 example, the odd 3d$_{5/2}$ neutron could be kicked up to the 3s$_{1/2}$ subshell with no further changes. That would leave the two remaining 3d$_{5/2}$ neutrons with zero spin, and the nucleus with the new spin 1/2 of the odd neutron. Indeed a low-lying $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state is observed. (Because of the antisymmetrization requirement, this state cannot result from three neutrons in the 3d$_{5/2}$ subshell.)

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

Figure 14.16: Energy levels for neighbors of doubly-magic calcium-40. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
...9}{2}}$}}
\put(400,4){\makebox(0,0)[bl]{?}}
\end{picture}
\end{figure}

Examples that highlight the perturbation effects of the shell model are shown in figure 14.16. These nuclei have unfilled 4d$_{7/2}$ shells. Since that is a major shell with no subshells, nucleon transitions to different shells require quite a bit of energy.

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

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

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

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

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

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

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

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

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


14.12.5 Configuration mixing

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

Take once again the excited 0$\POW9,{+}$ state of oxygen-16 shown in figure 14.14. To create this state within the shell model picture, a pair of 2p$_{1/2}$ nucleons must be kicked up to the 3d$_{5/2}$ subshell. Since that requires a major shell boundary crossing by two nucleons, it should take a considerable amount of energy. Some of it will be recovered by the fact that the nucleon pairing now occurs at higher angular momentum. But there is another effect.

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

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

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

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

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

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

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

Figure 14.17: 2$\POW9,{+}$ excitation energy of even-even nuclei. [pdf]
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Of course, a weird excited state at 6 MeV in a nucleus is not such a big deal. But there is more. Consider figure 14.17. It gives the excitation energy of the lowest 2$\POW9,{+}$ state for all even-even nuclei.

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

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

Then there is the magnitude of the 2$\POW9,{+}$ energy levels. In figure 14.17 the energies have been normalized with the von Weizsäcker value for the pairing energy,

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

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

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

Figure 14.18: Collective motion effects. [pdf]
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Consider a couple of examples in figure 14.18. In case of ruthenium-104, it takes a measly 0.36 MeV to excite the 2$\POW9,{+}$ state. But there are 10 different ways to combine the four 5g$_{9.2}$ protons into a 2$\POW9,{+}$ state, table 12.1. Kick up the pair of protons from the 4p$_{1/2}$ shell, and there are another 10 2$\POW9,{+}$ states. The four 5g$_{7/2}$ neutrons can produce another 10 of them. Kick up a pair of neutrons from the 5d$_{5/2}$ subshell, and there is 10 more. Presumably, all these states will have similar energy. And there might be many other low-energy ways to create 2$\POW9,{+}$ states, [30, pp. 135-136].

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

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

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

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

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

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

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

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

When the energy of the excitation hits zero, the bottom quite literally drops out of the shell model. In fact, even if the energy merely becomes low, the shell model must crash. If energy states are almost degenerate, the slightest thing will throw the nucleus from one to the other. In particular, small perturbation theory shows that originally small effects blow up as the reciprocal of the energy difference, addendum {A.37}. Physicists have found that nuclei in the rough ranges 150 $\raisebox{.3pt}{$<$}$ $A$ $\raisebox{.3pt}{$<$}$ 190 and $A$ $\raisebox{.3pt}{$>$}$ 220 acquire an intrinsic nonspherical shape, fundamentally invalidating the shell model as covered here. More physically, as figure 14.17 suggests, it happens for pretty much all heavy nuclei except ones close to the magic lines. The energy spectrum of a typical nucleus in the nonspherical range, hafnium-176, is shown in figure 14.18.


14.12.6 Shell model failures

The previous subsection already indicated two cases in which the shell model has major problems with the excited states. But in a number of cases the shell model may also predict an incorrect ground state. Figure 14.19 shows some typical examples.

Figure 14.19: Failures of the shell model. [pdf]
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In case of titanium-47, the shell model predicts that there will be five neutrons in an unfilled 4f$_{7/2}$ subshell. It is believed that this is indeed correct [35, p. 224]. The unperturbed shell model makes no predictions about the nuclear spin. However, the odd-particle shell model says that in the ground state the nuclear spin should be that of the odd neutron, $\frac72$. But it is not, the spin is $\frac52$. The pairing of the even number of neutrons in the 4f$_{7/2}$ shell is not complete. While unfortunate, this is really not that surprising. The perturbation Hamiltonian used to derive the prediction of nucleon pairing is a very crude one. It is quite common to see subshells with at least three particles and three holes (three places for additional particles) end up with a unit less spin than the odd-particle model predicts. It almost happened for oxygen-19 in figure 14.15.

In fact, 5 particles in a shell in which the single-particle spin is $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ can combine their spin into a variety of net values. Table 12.1 shows that $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 11}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, and $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 15}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ are all possible. Compared to that, the odd-particle prediction does not seem that bad. Note that the predicted state of spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ has only slightly more energy than the ground state. On the other hand, other states that might be produced through the combined spin of the five neutrons have much more energy.

Fluorine-19 shows a more fundamental failure of the shell model. The shell model would predict that the odd proton is in the 3d$_{5/2}$ state, giving the nucleus spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ and even parity. In fact, it should be just like fluorine-17 in figure 14.14. For the unperturbed shell model, the additional two neutrons should not make a significant difference. But the nuclear spin is $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, and that means that the odd proton must be in the 3s$_{1/2}$ state. A look at figure 14.13 shows that the unperturbed shell model cannot qualitatively explain this swapping of the two states.

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

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

Note also how low the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ excited state has become. Maybe this can be somewhat understood from the fact that the kicked-up 2p$_{1/2}$ proton is now in a similar spatial orbit with three other nucleons, rather than just one like in the case of fluorine-17. In any case, it would surely require a rather sophisticated perturbed shell model to describe it, one that includes nucleons of both type in the perturbation.

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

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

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

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

Tantalum-181 is an example nucleus that is not spherical. For it, the shell model simply does not apply as derived here. So there is no need to worry about it. Which is a good thing, because it does not seem easy to justify a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ ground state based on the shell model. As noted in the previous subsection, nonspherical nuclei appear near the stable line for mass numbers of about 150 to 190 and above 220. There are also a few with mass numbers between 20 and 30.

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