Subsections


14.17 Electromagnetic Moments

The most important electromagnetic property of nuclei is their net charge. It is what keeps the electrons in atoms and molecules together. However, nuclei are not really electric point charges. They have a small size. In a spatially varying electric field most respond somewhat different than a point charge. It is said that they have an electric quadrupole moment. Also, most nuclei act like little electromagnets. It is said that they have a “magnetic dipole moment.” These properties are important for applications like NMR and MRI, and for experimentally examining nuclei.


14.17.1 Classical description

This subsection explains the magnetic dipole and electric quadrupole moments from a classical point of view.


14.17.1.1 Magnetic dipole moment

The most basic description of an electromagnet is charges going around in circles. It can be seen from either classical or quantum electromagnetics that the strength of an electromagnet is proportional to the angular momentum $\vec{L}$ of the charges times the ratio of their charge $q$ to their mass $m$, chapter 13.2 or 13.4.

This leads to the definition of the magnetic dipole moment as

\begin{displaymath}
\vec \mu \equiv \frac{q}{2m} \vec L
\end{displaymath}

In particular, a magnet wants to align itself with an external magnetic field $\skew2\vec{\cal B}_{\rm {ext}}$. The energy involved in this alignment is

\begin{displaymath}
- \vec\mu \cdot \skew2\vec{\cal B}_{\rm ext}
\end{displaymath}


14.17.1.2 Electric quadrupole moment

Consider a nuclear charge distribution with charge density $\rho_c$ placed in an external electrical potential, or voltage $\varphi$. The potential energy due to the external field is

\begin{displaymath}
V = \int \varphi \rho_c { \rm d}^3{\skew0\vec r}
\end{displaymath}

It may be noted that since nuclear energies are of the order of MeV, an external field is not going to change the nuclear charge distribution $\rho$. It would need to have a million volt drop over a couple of femtometers to make a dent in it. Unless you shoot very high energy charged particles at the nucleus, that is not going to happen. Also, the current discussion assumes that the external field is steady or at least quasi-steady. That should be reasonable in many cases, as nuclear internal time scales are very fast.

Since nuclei are so small compared to normal external fields, the electric potential $\varphi$ can be well represented by a Taylor series. That gives the potential energy as

\begin{displaymath}
V = \varphi_0 \int \rho_c { \rm d}^3{\skew0\vec r}
+ \s...
...l r_j}\right)_0
\int r_i r_j \rho { \rm d}^3{\skew0\vec r}
\end{displaymath}

where $(r_1,r_2,r_3)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(x,y,z)$ are the three components of position and 0 indicates that the derivative is evaluated at the origin, the center of the nucleus.

The first integral in the expression above is just the net nuclear charge $q$. This makes the first term exactly the same as the potential energy of a point charge. The second integral defines the electric dipole moment in the $i$-​direction. It is nonzero if on average the charge is shifted somewhat towards one side of the nucleus. But nuclei do not have nonzero electric dipole moments. The reason is that nuclei have definite parity; the wave function is either the same or the same save for a minus sign when you look at the opposite side of the nucleus. Since the probability of a proton to be found at a given position is proportional to the square magnitude of the wave function, it is just as likely to be found at one side as the other one. (That should really be put more precisely for the picky. The dipole contribution of any set of positions of the protons is canceled by an opposite contribution from the set of opposite nucleon positions.)

The last integral in the expression for the potential energy defines the quadrupole matrix or tensor. You may note a mathematical similarity with the moment of inertia matrix of a solid body in classical mechanics. Just like there, the quadrupole matrix can be simplified by rotating the coordinate system to principal axes. That rotation gets rid of the integrals $\int{r}_ir_j\rho{\rm d}^3{\skew0\vec r}$ for $i$ $\raisebox{.2pt}{$\ne$}$ $j$, so what is left is

\begin{displaymath}
V = V_{\rm pc}
+ \frac12 \left(\frac{\partial^2\varphi}{...
...rtial z^2}\right)_0
\int z^2 \rho { \rm d}^3{\skew0\vec r}
\end{displaymath}

where the first term is the potential of the point charge.

Note that the average of $x^2$, $y^2$, and $z^2$ is $\frac13r^2$. It is convenient to subtract that average in each integral. The subtraction does not change the value of the potential energy. The reason is that the sum of the three second order derivatives of the external field $\varphi$ is zero due to Maxwell”s first equation, chapter 13.2. All that then leads to a definition of an electric quadrupole moment for a single axis, taken to be the $z$-​axis, as

\begin{displaymath}
Q \equiv \frac{1}{e} \int (3 z^2 - r^2) \rho { \rm d}^3{\skew0\vec r}
\end{displaymath}

For simplicity, the nasty fractions have been excluded from the definition of $Q$. Also, it has been scaled with the charge $e$ of a single proton.

That gives $Q$ units of square length, which is easy to put in context. Recall that nuclear sizes are of the order of a few femtometer. So the SI unit square femtometer, fm$\POW9,{2}$ or 10$\POW9,{-30}$ m$\POW9,{2}$, works quite nicely for the quadrupole moment $Q$ as defined. It is therefore needless to say that most sources do not use it. They use the “barn,” a non-SI unit equal to 10$\POW9,{-28}$ m$\POW9,{2}$. The reason is historical; during the second world war some physicists figured that the word barn would hide the fact that work was being done on nuclear bombs from the Germans. Of course, that did not work since so few memos and reports are one-word ones. However, physicists discovered that it did help confusing students, so the term has become very widely used in the half century since then. Also, unlike a square femtometer, the barn is much too large compared to a typical nuclear cross section, producing all these sophisticated looking tiny decimal fractions.

To better understand the likely values of the quadrupole moment, consider the effect of the charge distribution of a single proton. If the charge distribution is spherically symmetric, the averages of $x^2$, $y^2$ and $z^2$ are equal, making $Q$ zero. However, consider the possibility that the charge distribution is not spherical, but an ellipsoid of revolution, a “spheroid.”. If the axis of symmetry is the $z$-​axis, and the charge distribution hugs closely to that axis, the spheroid will look like a cigar or zeppelin. Such a spheroid is called“prolate.” The value of $Q$ is then about ${\textstyle\frac{2}{5}}$ of the square nuclear radius $R$. If the charge distribution stays close to the $xy$-​plane, the spheroid will look like a flattened sphere. Such a spheroid is called “oblate.” In that case the value of $Q$ is about $-{\textstyle\frac{2}{5}}$ of the square nuclear radius. Either way, the values of $Q$ are noticeably less than the square nuclear radius.

It may be noted that the quadrupole integrals also pop up in the description of the electric field of the nucleus itself. Far from the nucleus, the deviations in its electric field from that of a point charge are proportional to the same integrals, compare chapter 13.3.3.


14.17.2 Quantum description

Quantum mechanics makes for some changes to the classical description of the electromagnetic moments. Angular momentum is quantized, and spin must be included.


14.17.2.1 Magnetic dipole moment

As the classical description showed, the strength of an electromagnet is essentially the angular momentum of the charges going around, times the ratio of their charge to their mass. In quantum mechanics angular momentum comes in units of $\hbar$. Also, for nuclei the charged particles are protons with charge $e$ and mass $m_{\rm p}$. Therefore, a good unit to describe magnetic strengths in terms of is the so-called “nuclear magneton”

\begin{displaymath}
\fbox{$\displaystyle
\mu_{\rm N}\equiv \frac{e\hbar}{2m_{\rm p}}
$} %
\end{displaymath} (14.26)

In those terms, the magnetic magnetic dipole moment operator of a single proton is

\begin{displaymath}
\frac{1}{\hbar} {\skew 4\widehat{\vec L}}_{\rm {p}} \mu_{\rm N}
\end{displaymath}

But quantum mechanics brings in a complication, chapter 13.4. Protons have intrinsic angular momentum, called spin. That also acts as an electromagnet. In addition the magnetic strength per unit of angular momentum is different for spin than for orbital angular momentum. The factor that it is different is called the proton $g$-​factor $g_p$. That then makes the total magnetic dipole moment operator of a single proton equal to

\begin{displaymath}
{\skew 4\widehat{\skew{-.5}\vec\mu}}_{\rm p} = \frac{1}{\h...
...t{\vec S}}\right) \mu_{\rm N}
\qquad g_{\rm p} \approx 5.59
\end{displaymath} (14.27)

The above value of the proton $g$-​factor is experimental.

Neutrons do not have charge and therefore their orbital motion creates no magnetic moment. However, neutrons do create a magnetic moment through their spin:

\begin{displaymath}
{\skew 4\widehat{\skew{-.5}\vec\mu}}_{\rm n} = \frac{1}{\h...
...widehat{\vec S}}\mu_{\rm N}
\qquad g_{\rm n} \approx - 3.83
\end{displaymath} (14.28)

The reason is that the neutron consists of three charged quarks; they produce a net magnetic moment even if they do not produce a net charge.

The net magnetic dipole moment operator of the complete nucleus is

\begin{displaymath}
{\skew 4\widehat{\skew{-.5}\vec\mu}}= \frac{1}{\hbar}
\l...
...A g_{\rm n} {\skew 6\widehat{\vec S}}_i
\right] \mu_{\rm N}
\end{displaymath} (14.29)

where $i$ is the nucleon number, the first $Z$ being protons and the rest neutrons.

Now assume that the nucleus is placed in an external magnetic field ${\cal B}$ and take the $z$-​axis in the direction of the field. Because nuclear energies are so large, external electromagnetic fields are far too weak to change the quantum structure of the nucleus; its wave function remains unchanged to a very good approximation. However, the field does produce a tiny change in the energy levels of the quantum states. These may be found using expectation values:

\begin{displaymath}
\Delta E = \langle \Psi\vert{-}{\widehat\mu}_z {\cal B}\vert \Psi\rangle
\end{displaymath}

The fact that that is possible is a consequence of small perturbation theory, as covered in addendum {A.37}.

However, it is not immediately clear what nuclear wave function $\Psi$ to use in the expectation value above. Because of the large values of nuclear energies, a nucleus is affected very little by its surroundings. It behaves essentially as if it is isolated in empty space. That means that while the nuclear energy may depend on the magnitude of the nuclear spin ${\skew 6\widehat{\vec J}}$, (i.e. the net nuclear angular momentum), it does not depend on its direction. In quantum terms, the energy does not depend on the component ${\widehat J}_z$ in the chosen $z$-​direction. So, what should be used in the above expectation value to find the change in the energy of a nucleus in a state of spin $j$? States with definite values of $J_z$? Linear combinations of such states? You get a difference answer depending on what you choose.

Now a nucleus is a composite structure, consisting of protons or neutrons, each contributing to the net magnetic moment. However, the protons and neutrons themselves are composite structures too, each consisting of three quarks. Yet at normal energy levels protons and neutrons act as elementary particles, whose magnetic dipole moment is a scalar multiple $g\mu_{\rm N}$ of their spin. Their energies in a magnetic field split into two values, one for the state with $J_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\hbar$ and the other with $J_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12\hbar$. One state corresponds to magnetic quantum number $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, the other to $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12$.

The same turns out to be true for nuclei; they too behave as elementary particles as long as their wave functions stay intact. In a magnetic field, the original energy level of a nucleus with spin $j$ splits into equally spaced levels corresponding to nuclear magnetic quantum numbers $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j,j{-}1,\ldots,-j$. The numerical value of the magnetic dipole moment $\mu$ is therefore defined to be the expectation value of ${\widehat\mu}_z$ in the nuclear state in which $m_j$ has its largest value $j$, call it the $\vert jj\rangle$ state:

\begin{displaymath}
\fbox{$\displaystyle
\mu \equiv \langle jj \vert{\widehat\mu}_z \vert jj \rangle
$} %
\end{displaymath} (14.30)

The fact that nuclei would behave so simple is related to the fact that nuclei are essentially in empty space. That implies that the complete wave function of a nucleus in the ground state, or another energy eigenstate, will vary in a very simple way with angular direction. Furthermore, that variation is directly given by the angular momentum of the nucleus. A brief discussion can be found in chapter 7.3 and its note. See also the discussion of the Zeeman effect, and in particular the weak Zeeman effect, in addendum {A.37}.

The most important consequence of those ideas is that

Nuclei with spin zero do not have magnetic dipole moments.
That is not very easy to see from the general expression for the magnetic moment, cluttered as it is with $g$-​factors. However, zero spin means on a very fundamental level that the complete wave function of a nucleus is independent of direction, chapter 4.2.3. A magnetic dipole strength requires directionality, there must be a north pole and a south pole. That cannot occur for nuclei of spin zero.


14.17.2.2 Electric quadrupole moment

The definition of the electric quadrupole moment follows the same ideas as that of the magnetic dipole moment. The numerical value of the quadrupole moment is defined as the expectation value of $3z^2-r^2$, summed over all protons, in the state in which the net nuclear magnetic quantum number $m_j$ equals the nuclear spin $j$:

\begin{displaymath}
\fbox{$\displaystyle
Q \equiv \langle jj \vert\sum_{i=1}^Z 3 z_i^2 - r_i^2\vert jj \rangle
$} %
\end{displaymath} (14.31)

Note that there is a close relation with the spherical harmonics;

\begin{displaymath}
3 z^2 - r^2 = \sqrt{\frac{16\pi}{5}} r^2 Y_2^0
\end{displaymath} (14.32)

That is important because it implies that
Nuclei with spin zero or with spin one-half do not have electric quadrupole moments.
To see why, note that the expectation value involves the absolute square of the wave function. Now if you multiply two wave functions together that have an angular dependence corresponding to a spin $j$, mathematically speaking you get pretty much the angular dependence of two particles of spin $j$. That cannot become more than an angular dependence of spin $2j$, in other words an angular dependence with terms proportional to $Y_{2j}^m$. Since the spherical harmonics are mutually orthonormal, $Y_{2j}^m$ integrates away against $Y_2^0$ for $j$ $\raisebox{-.3pt}{$\leqslant$}$ $\frac12$.

It makes nuclei with spin $\leavevmode\kern.03em
\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ popular for nuclear magnetic resonance studies. Without the perturbing effects due to quadrupole interaction with the electric field, they give nice sharp signals. Also of course, analysis is easier with only two spin states and no quadrupole moment.


14.17.2.3 Shell model values

According to the odd-particle shell model, all even-even nuclei have spin zero and therefore no magnetic or electric moments. That is perfectly correct.

For nuclei with an odd mass number, the model says that all nucleons except for the last odd one are paired up in spherically symmetric states of zero spin that produce no magnetic moment. Therefore, the magnetic moment comes from the last proton or neutron. To get it, according to the second last subsubsection, what is needed is the expectation value of the magnetic moment operator ${\widehat\mu}_z$ as given there. Assume the shell that the odd nucleon is in has single-particle net momentum $j$. According to the definition of magnetic moment, the magnetic quantum number must have its maximum value $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$. Call the corresponding state the $\psi_{nljj}$ one because the spectroscopic notation is useless as always. In particular for an odd-even nucleus,

\begin{displaymath}
\mu = \frac{1}{\hbar}
\langle\psi_{nljj}\vert L_z + g_p {\widehat S}_z\vert\psi_{nljj}\rangle \mu_{\rm N}
\end{displaymath}

while for an even-odd nucleus

\begin{displaymath}
\mu
= \frac{1}{\hbar} \langle\psi_{nljj}\vert g_n {\widehat S}_z\vert\psi_{nljj}\rangle \mu_{\rm N}
\end{displaymath}

The unit $\mu_{\rm N}$ is the nuclear magneton. The expectation values can be evaluated by writing the state $\psi_{nljj}$ in terms of the component states $\psi_{nlmm_s}$ of definite angular momentum $\L _z$ and spin ${\widehat S}_z$ following chapter 12.8, 2.

It is then found that for an odd proton, the magnetic moment is

\begin{displaymath}
\fbox{$\displaystyle
\begin{array}{ll}
j = l - {\texts...
...}} \left(2j - 1 + g_p\right) \mu_{\rm N}
\end{array}
$} %
\end{displaymath} (14.33)

while for an odd neutron
\begin{displaymath}
\fbox{$\displaystyle
\begin{array}{ll}
j = l - {\texts...
... {\textstyle\frac{1}{2}} g_n \mu_{\rm N}
\end{array}
$} %
\end{displaymath} (14.34)

These are called the “Schmidt values.”

Odd-odd nuclei are too messy to be covered here, even if the Nordheim rules would be reliable.

For the quadrupole moments of nuclei of odd mass number, filled shells do not produce a quadrupole moment, because they are spherically symmetric. Consider now first the case that there is a single proton in an otherwise empty shell with single-particle momentum $j$. Then the magnetic moment of the nucleus can be found as the one of that proton:

\begin{displaymath}
Q = Q_{\rm p} = \langle\psi_{nljj}\vert 3 z^2 - r^2\vert\psi_{nljj}\rangle
\end{displaymath}

Evaluation, {D.79}, gives
\begin{displaymath}
\fbox{$\displaystyle
Q_{\rm p} = - \frac{2j-1}{2j+2} \langle r^2\rangle
$} %
\end{displaymath} (14.35)

where $\langle{r}^2\rangle$ is the expectation value of $r^2$ for the proton. Note that this is zero as it should if the spin $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$. Since the spin $j$ must be half-integer, zero spin is not a consideration. For all other values of $j$, the one-proton quadrupole moment is negative.

The expectation value $\langle{r}^2\rangle$ can hardly be much more than the square nuclear radius, excepting maybe halo nuclei. A reasonable guess would be to assume that the proton is homogeneously distributed within the nuclear radius $R$. That gives a ballpark value

\begin{displaymath}
\langle r^2\rangle \approx {\textstyle\frac{3}{5}} R^2
\end{displaymath}

Next consider the case that there are not one but $I$ $\raisebox{-.5pt}{$\geqslant$}$ 3 protons in the unfilled shell. The picture of the odd-particle shell model as usually painted is: the first $I-1$ protons are pairwise combined in spherically symmetric states and the last odd proton is in a single particle state, blissfully unaware of the other protons in the shell. In that case, the quadrupole moment would self evidently be the same as for one proton in the shell. But as already pointed out in section 14.12.4, the painted picture is not really correct. For one, it does not satisfy the antisymmetrization requirement for all combinations on protons. There really are $I$ protons in the shell sharing one wave function that produces a net spin equal to $j$.

In particular consider the case that there are $2j$ protons in the shell. Then the wave function takes the form of a filled shell, having no quadrupole moment, plus a hole, a state of angular momentum $j$ for the missing proton. Since a proton hole has minus the charge of a proton, the quadrupole moment for a single hole is opposite to that of one proton:

\begin{displaymath}
\fbox{$\displaystyle
Q_{2j\rm p} = - Q_{\rm p}
$} %
\end{displaymath} (14.36)

In other words, the quadrupole moment for a single hole is predicted to be positive. For $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, a single proton also means a single hole, so the quadrupole moment must, once more, be zero. It has been found that the quadrupole moment changes linearly with the odd number of protons, [30, p, 129]. Therefore for shells with more than one proton and more than one hole, the quadrupole moment is in between the one-proton and one-hole values. It follows that the one-proton value provides an upper bound to the magnitude of the quadrupole moment for any number of protons in the shell.

Since neutrons have no charge, even-odd nuclei would in the simplest approximation have no quadrupole moment at all. However, consider the odd neutron and the spherical remainder of the nucleus as a two-body system going around their common center of gravity. In that picture, the charged remainder of the nucleus will create a quadrupole moment. The position vector of the remainder of the nucleus is about 1$\raisebox{.5pt}{$/$}$$A$ times shorter than that of the odd neutron, so quadratic lengths are a factor 1$\raisebox{.5pt}{$/$}$$A^2$ shorter. However, the nucleus has $Z$ times as much charge as a single proton. Therefore you expect nuclei with an odd neutron to have about $Z$$\raisebox{.5pt}{$/$}$$A^2$ times the quadrupole moment of the corresponding nucleus with an odd proton instead of an odd neutron. For heavy nuclei, that would still be very much smaller than the magnetic moment of a similar odd-even nucleus.


14.17.2.4 Values for deformed nuclei

For deformed nuclei, part of the angular momentum is due to rotation of the nucleus as a whole. In particular, for the ground state rotational band of deformed even-even nuclei, all angular momentum is in rotation of the nucleus as a whole. This is orbital angular momentum. Protons with orbital angular momentum produce a magnetic dipole moment equal to their angular momentum, provided the dipole moment is expressed in terms of the nuclear magneton $\mu_{\rm N}$. Uncharged neutrons do not produce a dipole moment from orbital angular momentum. Therefore. the magnetic dipole moment of the nucleus is about

\begin{displaymath}
\fbox{$\displaystyle
\mbox{even-even, ground state band:...
...{R}} j \mu_{\rm N}\quad g_{\rm{R}} \approx \frac{Z}{A}
$} %
\end{displaymath} (14.37)

where the $g$-​factor reflects the relative amount of the nuclear angular momentum that belongs to the protons. This also works for vibrational nuclei, since their angular momentum too is in global motion of the nucleus.

If a rotational band has a minimum spin $j_{\rm {min}}$ that is not zero, the dipole moment is, [39, p. 392],

\begin{displaymath}
\fbox{$\displaystyle
\mu =
\left[
g_{\rm{R}} j + \fr...
...rac{Z}{A} \qquad j_{\rm min}\ne{\textstyle\frac{1}{2}}
$} %
\end{displaymath} (14.38)

where $g_{\rm {int}}j_{\rm {min}}\mu_{\rm N}$ reflects an internal magnetic dipole strength. If $j_{\rm {min}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, the top of the first ratio has an additional term that has a magnitude proportional to $2j+1$ and alternates in sign.

The quadrupole moment of deformed nuclei is typically many times larger than that of a shell model one. According to the shell model, all protons except at most one are in spherical orbits producing no quadrupole moment. But if the nucleus is deformed, typically into about the shape of some spheroid instead of a sphere, then all protons contribute. Such a nucleus has a very large “intrinsic” quadrupole moment $Q_{\rm {int}}$.

However, that intrinsic quadrupole moment is not the one measured. For example, many heavy even-even nuclei have very distorted intrinsic shapes but all even-even nuclei have a measured quadrupole moment that is zero in their ground state. That is a pure quantum effect. Consider the state in which the axis of the nucleus is aligned with the $z$-​direction. In that state a big quadrupole moment would be observed due to the directional charge distribution. But there are also states in which the nucleus is aligned with the $x$-​direction, the $y$-​direction, and any other direction for that matter. No big deal classically: you just grab hold of the nucleus and measure its quadrupole moment. But quantum mechanics makes the complete wave function a linear combination of all these different possible orientations; in fact an equal combination of them by symmetry. If all directions are equal, there is no directionality left; the measured quadrupole moment is zero. Also, directionality means angular momentum in quantum mechanics; if all directions are equal the spin is zero. Grabbing hold of the nucleus means adding directionality, adding angular momentum. That creates an excited state.

A simple known system that shows such effects is the hydrogen atom. Classically the atom is just an electron and a proton at opposite sides of their center of gravity. If they are both on the $z$-​axis, say, that system would have a nonzero quadrupole moment. But such a state is not an exact energy eigenstate, far from it. It interacts with states in which the direction of the connecting line is different. By symmetry, the ground state is the one in which all directions have the same probability. The atom has become spherically symmetric. Still, the atom has not become intrinsically spherically symmetric; the wave function is not of a form like $\psi_1(r_{\rm {e}})\psi_2(r_{\rm {p}})$. The positions of electron and proton are still correlated, {A.5}.

A model of a spheroidal nucleus produces the following relationship between the intrinsic quadrupole moment and the one that is measured:

\begin{displaymath}
\fbox{$\displaystyle
Q = \frac{3j_{\rm min}^2 - j(j+1)}{(j+1)(2j+3)} Q_{\rm int}
$} %
\end{displaymath} (14.39)

where $j_{\rm {min}}$ is the angular momentum of the nucleus when it is not rotating. Derivations may be found in [39] or [35]. It can be seen that when the nucleus is not rotating, the measured quadrupole moment is much smaller than the intrinsic one unless the angular momentum is really large. When the nucleus gets additional rotational angular momentum, the measured quadrupole moment decreases even more and eventually ends up with the opposite sign.


14.17.3 Magnetic moment data

Figure 14.40 shows ground state magnetic moments in units of the nuclear magneton $\mu_{\rm N}$. Even-even nuclei do not have magnetic moments in their ground state, so they are not shown. The red and blue horizontal lines are the Schmidt values predicted by the shell model. They differ in whether spin subtracts from or adds to the net angular momentum $j$ to produce the orbital momentum $l$. Red dots should be on the red lines, blue dots on the blue lines. For black dots, no confident prediction of the orbital angular momentum could be made. The values have an error of no more than about 0.1 $\mu_{\rm N}$, based on a subjective evaluation of both reported errors as well as differences between results obtained by different studies for the same number. These differences are often much larger than the reported errors for the individual numbers.

Figure 14.40: Magnetic dipole moments of the ground-state nuclei. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...6.6,165){\picfrbox(34,8){\scriptsize$j=7$}}
}
\end{picture}
\end{figure}

One good thing to say about it all is that the general magnitude is well predicted. Few nuclei end up outside the Schmidt lines. (Rhodium-103, a stable odd-even $\leavevmode\kern.03em
\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{-}$ nucleus, is a notable exception.) Also, some nuclei are actually on their line. And the others tend to at least be on the right side of the cloud. The bad news is, of course, that the agreement is only qualitatively.

The main excuses that are offered are:

1.
The $g$-​factors $g_{\rm {p}}$ and $g_{\rm {n}}$ describe the effectiveness of proton and neutron spins in generating magnetic moments in free space. They may be reduced when these nucleons are inside a nucleus. Indeed, it seems reasonable enough to assume that the motion of the quarks that make up the protons and neutrons could be affected if there are other quarks nearby. Reduction of the $g$-​factors drives the Schmidt lines towards each other, and that can clearly reduce the average errors. Unfortunately, different nuclei would need different reductions to obtain quantitative agreement.

2.
Collective motion. If some of the angular momentum is into collective motion, it tends to drift the magnetic moment towards about $\frac12{j}\mu_{\rm N}$, compare (14.38). To compute the effect requires the internal magnetic moment of the nucleus to be known. For some nuclei, fairly good magnetic moments can be obtained by using the Schmidt values for the internal magnetic moment, [39, p. 393].

For odd-odd nuclei, the data average out to about $0.5j$ nuclear magnetons, with a standard deviation of about one magneton. These average values are shown as yellow lines in figure 14.40. Interestingly enough, the average is like a collective rotation, (14.37).

According to the shell model, two odd particles contribute to the spin and magnetic moment of odd-odd nuclei. So they could have significantly larger spins and magnetic moments than odd mass nuclei. Note from the data in figure 14.40 that that just does not happen.

Figure 14.41: 2$\POW9,{+}$ magnetic moment of even-even nuclei. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...-12){\makebox(0,0)[bl]{2}}
}
\end{picture}}
\end{picture}
\end{figure}

Even-even nuclei do not have magnetic moments in their ground state. Figure 14.41 shows the magnetic moments of the first exited 2$\POW9,{+}$ state of these nuclei. The values are in fairly good agreement with the prediction (14.37) of collective motion that the magnetic moment equals $Zj$$\raisebox{.5pt}{$/$}$$A$ nuclear magnetons. Bright green squares are correct. Big deviations occur only near magic numbers. The maximum error in the shown data is about a quarter of a nuclear magneton, subjectively evaluated.


14.17.4 Quadrupole moment data

If you love the shell model, you may want to skip this subsection. It is going to get a beating.

Figure 14.42: Electric quadrupole moment. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...){\makebox(0,0)[bl]{$<0$}}
}
\end{picture}}
\end{picture}
\end{figure}

The prediction of the shell model is relatively straightforward. The electric quadrupole moment of a single proton in an unfilled shell of high angular momentum can quite well be ballparked as

\begin{displaymath}
Q_{\rm p ballpark} \sim {\textstyle\frac{3}{5}} R^2
\end{displaymath}

where $R$ is the nuclear radius computed from (14.9). This value corresponds to the area of the square marked “a proton’s” in the legend of figure 14.42. As discussed in subsection 14.17.2.3, if there are more protons in the shell, the magnitude is less, though the sign will eventually reverse. If the angular momentum is not very high, the magnitude is less. If there is no odd proton, the magnitude will be almost zero. So, essentially all squares in figure 14.42 must be smaller, most a lot smaller, and those on lines of even $Z$ very much smaller, than the single proton square in the legend...

Well, you might be able to find a smaller square somewhere. For example, the square for lithium-6, straight above doubly-magic $\fourIdx{4}{2}{}{}{\rm {He}}$, has about the right size and the right color, blue. The data shown have a subjectively estimated error of up to 40%, [sic], and the area of the squares gives the scaled quadrupole moment. Nitrogen-14, straight below doubly-magic $\fourIdx{16}{8}{}{}{\rm {O}}$, has a suitably small square of the right color, red. So does potassium-39 with one proton less than doubly-magic $\fourIdx{40}{20}{}{}{\rm {Ca}}$. Bismuth-209, with one more proton than $\fourIdx{208}{82}{}{}{\rm {Pb}}$ has a relatively small square of the right color. Some nuclei on magic proton number lines have quite small scaled quadrupole moments, though hardly almost zero as they should. Nuclei one proton above magic proton numbers tend to be of the right color, blue, as long as their squares are small. Nuclei one proton below the magic proton numbers should be red; however, promotion can mess that up.

Back to reality. Note that many nuclei in the $Z$ $\raisebox{.3pt}{$<$}$ 82, $N$ $\raisebox{.3pt}{$>$}$ 82 wedge, and above $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 82, as well as various other nuclei, especially away from the stable line, have quadrupole moments that are very many times larger than the ballpark for a single proton. That is simply not possible unless many or all protons contribute to the quadrupole moment. The odd-particle shell model picture of a spherically symmetric nuclear core plus an odd proton, and maybe a neutron, in nonspherical orbits hanging on is completely wrong for these nuclei. These nuclei have a global shape that simply is not spherical. And because the shell model was derived based on a spherical potential, its results are invalid for these nuclei. They are the deformed nuclei that also showed up in figures 14.17 and 14.20. It is the quadrupole moment that shows that it was not just an empty excuse to exclude these nuclei in shell model comparisons. The measured quadrupole moments show without a shadow of a doubt that the shell model cannot be valid.

You might however wonder about the apparently large amount in random scatter in the quadrupole moments of these nuclei. Does the amount of deformation vary that randomly? Before that can be answered, a correction to the data must be applied. Measured quadrupole moments of a deformed nucleus are often much too small for the actual nuclear deformation. The reason is uncertainty in the angular orientation of these nuclei. In particular, nuclei with spin zero have complete uncertainty in orientation. Such nuclei have zero measured quadrupole moment regardless how big the deformation of the nucleus is. Nuclei with spin one-half still have enough uncertainty in orientation to measure as zero.

Figure 14.43: Electric quadrupole moment corrected for spin. [pdf]
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Figure 14.43 shows what happens if you try to estimate the intrinsic quadrupole moment of the nuclei in absence of uncertainty in angular orientation. For nuclei whose spin is at least one, the estimate was made based on the measured value using (14.39), with both $j_{\rm {min}}$ and $j$ equal to the spin. This assumes that the intrinsic shape is roughly spheroidal. For shell-model nuclei, this also roughly corrects for the spin effect, though it overcorrects to some extent for nuclei of low spin.

To estimate the intrinsic quadrupole moment of nuclei with zero ground state spin, including all even-even nuclei, the quadrupole moment of the lowest excited 2$\POW9,{+}$ state was used, if it had been measured. For spin one-half the lowest $\leavevmode\kern.03em
\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ state was used. In either case, $j_{\rm {min}}$ was taken to be the spin of the ground state and $j$ that of the excited state. Regrettably, these estimates do not make much sense if the nucleus is not a rotating one.

Note in figure 14.43 how much more uniform the squares in the regions of deformed nuclei have become. And that the squares of nuclei of spin zero and one-half have similar sizes. These nuclei were not really more spherical; it was just hidden from experiments.

The observed intrinsic quadrupole moments in the regions of deformed nuclei correspond to roughly 20% radial deviation from the spherical value. Clearly, that means quite a large change in shape.

It may be noted that figure 14.42 leaves out magnesium-23, whose reported quadrupole moment of 1.25 barn is far larger that that of similar nuclei. If this value is correct, clearly magnesium-23 must be a halo nucleus with two protons outside a neon-21 core.