- 14.17.1 Draft: Classical description

- 14.17.2 Draft: Quantum description
- 14.17.2.1 Draft: Magnetic dipole moment
- 14.17.2.2 Draft: Electric quadrupole moment
- 14.17.2.3 Draft: Shell model values
- 14.17.2.4 Draft: Values for deformed nuclei

- 14.17.3 Draft: Magnetic moment data
- 14.17.4 Draft: Quadrupole moment data

14.17 Draft: 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 Draft: Classical description

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

14.17.1.1 Draft: 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

This leads to the definition of the magnetic dipole moment as

In particular, a magnet wants to align itself with an external magnetic field

14.17.1.2 Draft: Electric quadrupole moment

Consider a nuclear charge distribution with charge density voltage

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

Since nuclei are so small compared to normal external fields, the
electric potential

where

The first integral in the expression above is just the net nuclear
charge electric dipole moment

in the

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

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

Note that the average of

For simplicity, the nasty fractions have been excluded from the definition of

That gives 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 confuse 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

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 Draft: 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 Draft: 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

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

(14.27) |

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

(14.28) |

The net magnetic dipole moment operator of the complete nucleus is

(14.29) |

Now assume that the nucleus is placed in an external magnetic field

The fact that that is possible is a consequence of small perturbation theory, as covered in addendum {A.38}.

However, it is not immediately clear what nuclear wave function

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

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

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.38}.

The most important consequence of those ideas is that

That is not very easy to see from the general expression for the magnetic moment, cluttered as it is withNuclei with spin zero do not have magnetic dipole moments.

14.17.2.2 Draft: 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

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

(14.32) |

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 spinNuclei with spin zero or with spin one-half do not have electric quadrupole moments.

It makes nuclei with spin

14.17.2.3 Draft: 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

while for an even-odd nucleus

The unit

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

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

Evaluation, {D.78}, gives

where

The expectation value

Next consider the case that there are not one but

In particular consider the case that there are hole

, a state of
angular momentum

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

14.17.2.4 Draft: 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

If a rotational band has a minimum spin

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

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 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

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

14.17.3 Draft: Magnetic moment data

Figure 14.42 shows ground state magnetic moments in units
of the nuclear magneton

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

The main excuses that are offered are:

- 1.
- The
- factorsand 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 - 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
, 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

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 14.42 that that just does not happen.

Even-even nuclei do not have magnetic moments in their ground state.
Figure 14.43 shows the magnetic moments of the first exited

14.17.4 Draft: Quadrupole moment data

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

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

where

Well, you might be able to find a smaller square somewhere. For
example, the square for lithium-6, straight above doubly-magic

Back to reality. Note that many nuclei in the

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 14.45 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

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

Note in figure 14.45 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 14.44 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.