## 34The Magnetism of Matter

Review: | Section 15-1, Vol. II, “The forces on a current loop; energy of a dipole.” |

### 34–1Diamagnetism and paramagnetism

In this chapter we are going to talk about the magnetic
properties of materials. The material which has the most striking
magnetic properties is, of course, iron. Similar magnetic properties are
shared also by the elements nickel, cobalt, and—at sufficiently low
temperatures (below $16^\circ$C)—by gadolinium, as well as by a number
of peculiar alloys. That kind of magnetism, called
*ferromagnetism*, is sufficiently striking and complicated that we
will discuss it in a special chapter. However, all ordinary substances
do show some magnetic effects, although very small ones—a thousand to
a million times less than the effects in ferromagnetic materials. Here
we are going to describe ordinary magnetism, that is to say, the
magnetism of substances other than the ferromagnetic ones.

This small magnetism is of two kinds. Some materials are
*attracted* toward magnetic fields; others are *repelled*.
Unlike the electrical effect in matter, which always causes dielectrics
to be attracted, there are two signs to the magnetic effect. These two
signs can be easily shown with the help of a strong electromagnet which
has one sharply pointed pole piece and one flat pole piece, as drawn in
Fig. 34-1. The magnetic field is much stronger near the
pointed pole than near the flat pole. If a small piece of material is
fastened to a long string and suspended between the poles, there will,
in general, be a small force on it. This small force can be seen by the
slight displacement of the hanging material when the magnet is turned
on. The few ferromagnetic materials are attracted very strongly toward
the pointed pole; all other materials feel only a very weak force. Some
are weakly attracted to the pointed pole; and some are weakly repelled.

The effect is most easily seen with a small cylinder of bismuth, which
is *repelled* from the high-field region. Substances which are
repelled in this way are called *diamagnetic*. Bismuth is one of
the strongest diamagnetic materials, but even with it, the effect is
still quite weak. Diamagnetism is always very weak. If a small piece
of aluminum is suspended between the poles, there is also a weak
force, but *toward* the pointed pole. Substances like aluminum
are called *paramagnetic*. (In such an experiment, eddy-current
forces arise when the magnet is turned on and off, and these can give
off strong impulses. You must be careful to look for the net
displacement after the hanging object settles down.)

We want now to describe briefly the mechanisms of these two effects.
First, in many substances the atoms have no permanent magnetic
moments, or rather, all the magnets within each atom balance out so
that the *net* moment of the atom is zero. The electron spins and
orbital motions all exactly balance out, so that any particular atom
has no average magnetic moment. In these circumstances, when you turn on
a magnetic field little extra currents are generated inside the atom by
induction. According to Lenz’s law, these currents are in such a direction as to oppose the
increasing field. So the induced magnetic moments of the atoms are
directed *opposite* to the magnetic field. This is the mechanism of
diamagnetism.

Then there are some substances for which the atoms do have a permanent
magnetic moment—in which the electron spins and orbits have a net
circulating current that is not zero. So besides the diamagnetic
effect (which is always present), there is also the possibility of
lining up the individual atomic magnetic moments. In this case, the
moments try to line up *with* the magnetic field (in the way the
permanent dipoles of a dielectric are lined up by the electric field),
and the induced magnetism tends to enhance the magnetic field. These
are the paramagnetic substances. Paramagnetism is generally fairly
weak because the lining-up forces are relatively small compared with
the forces from the thermal motions which try to derange the order. It
also follows that paramagnetism is usually sensitive to the
temperature. (The paramagnetism arising from the spins of the
electrons responsible for conduction in a metal constitutes an
exception. We will not be discussing this phenomenon here.) For
ordinary paramagnetism, the lower the temperature, the stronger the
effect. There is more lining-up at low temperatures when the deranging
effects of the collisions are less. Diamagnetism, on the other hand,
is more or less independent of the temperature. In any substance with
built-in magnetic moments there is a diamagnetic as well as a
paramagnetic effect, but the paramagnetic effect usually dominates.

In Chapter 11 we described a *ferroelectric* material,
in which all the electric dipoles get lined up by their own mutual
electric fields. It is also possible to imagine the magnetic analog of
ferroelectricity, in which all the atomic moments would line up and lock
together. If you make calculations of how this should happen, you will
find that because the magnetic forces are so much smaller than the
electric forces, thermal motions should knock out this alignment even at
temperatures as low as a few tenths of a degree Kelvin. So it would be
impossible at room temperature to have any permanent lining up of the
magnets.

On the other hand, this is exactly what does happen in iron—it does
get lined up. There is an effective force between the magnetic moments
of the different atoms of iron which is much, much greater than the
*direct magnetic* interaction. It is an indirect effect which can
be explained only by quantum mechanics. It is about ten thousand times
stronger than the direct magnetic interaction, and is what lines up
the moments in ferromagnetic materials. We discuss this special
interaction in a later chapter.

Now that we have tried to give you a qualitative explanation of
diamagnetism and paramagnetism, we must correct ourselves and say that
*it is not possible* to understand the magnetic effects of
materials in any honest way from the point of view of classical
physics. Such magnetic effects are a *completely
quantum-mechanical phenomenon*. It is, however, possible to make some
phoney classical arguments and to get some idea of what is going
on. We might put it this way. You can make some classical arguments
and get guesses as to the behavior of the material, but these
arguments are not “legal” in any sense because it is absolutely
essential that quantum mechanics be involved in every one of these
magnetic phenomena. On the other hand, there are situations, such as
in a plasma or a region of space with many free electrons, where the
electrons do obey the laws of classical mechanics. And in those
circumstances, some of the theorems from classical magnetism are
worthwhile. Also, the classical arguments are of some value for historical
reasons. The first few times that people were able to guess at the
meaning and behavior of magnetic materials, they used classical
arguments. Finally, as we have already illustrated, classical
mechanics can give us some useful guesses as to what might
happen—even though the really honest way to study this subject would
be to learn quantum mechanics first and then to understand the
magnetism in terms of quantum mechanics.

On the other hand, we don’t want to wait until we learn quantum mechanics inside out to understand a simple thing like diamagnetism. We will have to lean on the classical mechanics as kind of half showing what happens, realizing, however, that the arguments are really not correct. We therefore make a series of theorems about classical magnetism that will confuse you because they will prove different things. Except for the last theorem, every one of them will be wrong. Furthermore, they will all be wrong as a description of the physical world, because quantum mechanics is left out.

### 34–2Magnetic moments and angular momentum

The first theorem we want to prove from classical mechanics is the following: If an electron is moving in a circular orbit (for example, revolving around a nucleus under the influence of a central force), there is a definite ratio between the magnetic moment and the angular momentum. Let’s call $\FLPJ$ the angular momentum and $\FLPmu$ the magnetic moment of the electron in the orbit. The magnitude of the angular momentum is the mass of the electron times the velocity times the radius. (See Fig. 34-2.) It is directed perpendicular to the plane of the orbit. \begin{equation} \label{Eq:II:34:1} J=mvr. \end{equation} (This is, of course, a nonrelativistic formula, but it is a good approximation for atoms, because for the electrons involved $v/c$ is generally of the order of $e^2/\hbar c\approx1/137$, or about $1$ percent.)

The magnetic moment of the same orbit is the current times the area. (See Section 14-5.) The current is the charge per unit time which passes any point on the orbit, namely, the charge $q$ times the frequency of rotation. The frequency is the velocity divided by the circumference of the orbit; so \begin{equation*} I=q\,\frac{v}{2\pi r}. \end{equation*} The area is $\pi r^2$, so the magnetic moment is \begin{equation} \label{Eq:II:34:2} \mu=\frac{qvr}{2}. \end{equation} It is also directed perpendicular to the plane of the orbit. So $\FLPJ$ and $\FLPmu$ are in the same direction: \begin{equation} \label{Eq:II:34:3} \FLPmu=\frac{q}{2m}\,\FLPJ\:(\text{orbit}). \end{equation} Their ratio depends neither on the velocity nor on the radius. For any particle moving in a circular orbit the magnetic moment is equal to $q/2m$ times the angular momentum. For an electron, the charge is negative—we can call it $-q_e$; so for an electron \begin{equation} \label{Eq:II:34:4} \FLPmu=-\frac{q_e}{2m}\,\FLPJ\:(\text{electron orbit}). \end{equation}

That’s what we would expect classically and, miraculously enough, it
is also true quantum-mechanically. It’s one of those things. However,
if you keep going with the classical physics, you find other places
where it gives the wrong answers, and it is a great game to try to
remember which things are right and which things are wrong. We might
as well give you immediately what is true *in general* in quantum
mechanics. First, Eq. (34.4) is true for *orbital
motion*, but that’s
not the only magnetism that exists. The electron also has a spin
rotation about its own axis (something like the earth rotating on its
axis), and as a result of that spin it has both an angular momentum and
a magnetic moment. But for reasons that are purely
quantum-mechanical—there is no classical explanation—the ratio of
$\FLPmu$ to $\FLPJ$ for the electron spin is twice as large as it is for
orbital motion of the spinning electron:
\begin{equation}
\label{Eq:II:34:5}
\FLPmu=-\frac{q_e}{m}\,\FLPJ\:(\text{electron spin}).
\end{equation}

In any atom there are, generally speaking, several electrons and some
combination of spin and orbit rotations which builds up a total
angular momentum and a total magnetic moment. Although there is no
classical reason why it should be so, it is *always true* in
quantum mechanics that (for an isolated atom) the direction of the
magnetic moment is exactly opposite to the direction of the angular
momentum. The ratio of the two is not necessarily either $-q_e/m$
or $-q_e/2m$, but somewhere in between, because there is a mixture of the
contributions from the orbits and the spins. We can write
\begin{equation}
\label{Eq:II:34:6}
\FLPmu=-g\biggl(\frac{q_e}{2m}\biggr)\FLPJ,
\end{equation}
where $g$ is a factor which is characteristic of the state of the atom.
It would be $1$ for a pure orbital moment, or $2$ for a pure spin
moment, or some other number in between for a complicated system like an
atom. This formula does not, of course, tell us very much. It says that
the magnetic moment is *parallel to* the angular momentum, but can
have any magnitude. The form of Eq. (34.6) is convenient,
however, because $g$—called the “Landé $g$-factor”—is a
dimensionless constant whose magnitude is of the order of one. It is one
of the jobs of quantum mechanics to predict the $g$-factor for any
particular atomic state.

You might also be interested in what happens in nuclei. In nuclei
there are protons and neutrons which may move around in some kind of
orbit and at the same time, like an electron, have an intrinsic
spin. Again the magnetic moment is parallel to the angular
momentum. Only now the order of magnitude of the ratio of the two is
what you would expect for a *proton* going around in a circle,
with $m$ in Eq. (34.3) equal to the *proton* mass. Therefore it is usual to write for nuclei
\begin{equation}
\label{Eq:II:34:7}
\FLPmu=g\biggl(\frac{q_e}{2m_p}\biggr)\FLPJ,
\end{equation}
where $m_p$ is the mass of the proton, and $g$—called the
*nuclear* $g$-factor—is a number near
one, to be determined for each nucleus.

Another important difference for a nucleus is that the *spin*
magnetic moment of the proton does *not* have a $g$-factor of
$2$, as the electron does. For a proton, $g=2\cdot(2.79)$. Surprisingly
enough, the *neutron* also has a spin magnetic moment, and its
magnetic moment relative to its angular momentum is $2\cdot(-1.91)$. The
neutron, in other words, is not exactly “neutral” in the magnetic
sense. It is like a little magnet, and it has the kind of magnetic
moment that a rotating *negative* charge would have.

### 34–3The precession of atomic magnets

One of the consequences of having the magnetic moment proportional to
the angular momentum is that an atomic magnet placed in a magnetic field
will *precess*. First we will argue classically. Suppose that we
have the magnetic moment $\FLPmu$ suspended freely in a uniform magnetic
field. It will feel a torque $\FLPtau$, equal to $\FLPmu\times\FLPB$,
which tries to bring it in line with the field direction. But the atomic
magnet is a gyroscope—it has the angular momentum $\FLPJ$. Therefore
the torque due to the magnetic field will not cause the magnet to line
up. Instead, the magnet will *precess*, as we saw when we analyzed
a gyroscope in Chapter 20 of Volume I. The angular
momentum—and with it the magnetic moment—precesses about an axis
parallel to the magnetic field. We can find the rate of precession by
the same method we used in Chapter 20 of the first volume.

Suppose that in a small time $\Delta t$ the angular momentum changes
from $\FLPJ$ to $\FLPJ'$, as drawn in Fig. 34-3, staying always at
the same angle $\theta$ with respect to the direction of the magnetic
field $\FLPB$. Let’s call $\omega_p$ the angular velocity of the
precession, so that in the time $\Delta t$ the angle *of
precession*
is $\omega_p\,\Delta t$. From the geometry of the figure, we see that
the change of angular momentum in the time $\Delta t$ is
\begin{equation*}
\Delta J=(J\sin\theta)(\omega_p\,\Delta t).
\end{equation*}
So the rate of change of the angular momentum is
\begin{equation}
\label{Eq:II:34:8}
\ddt{J}{t}=\omega_pJ\sin\theta,
\end{equation}
which must be equal to the torque:
\begin{equation}
\label{Eq:II:34:9}
\tau=\mu B\sin\theta.
\end{equation}
The angular velocity of precession is then
\begin{equation}
\label{Eq:II:34:10}
\omega_p=\frac{\mu}{J}\,B.
\end{equation}

Substituting $\mu/J$ from Eq. (34.6), we see that for an atomic system \begin{equation} \label{Eq:II:34:11} \omega_p=g\,\frac{q_eB}{2m}; \end{equation} the precession frequency is proportional to $B$. It is handy to remember that for an atom (or electron) \begin{equation} \label{Eq:II:34:12} f_p=\frac{\omega_p}{2\pi}=(\text{$1.4$ megacycles/gauss})gB, \end{equation} and that for a nucleus \begin{equation} \label{Eq:II:34:13} f_p=\frac{\omega_p}{2\pi}=(\text{$0.76$ kilocycles/gauss})gB. \end{equation} (The formulas for atoms and nuclei are different only because of the different conventions for $g$ for the two cases.)

According to the *classical* theory, then, the electron
orbits—and spins—in an atom should precess in a magnetic field. Is
it also true quantum-mechanically? It is essentially true, but the
meaning of the “precession” is different. In quantum mechanics one
cannot talk about the *direction* of the angular momentum in the
same sense as one does classically; nevertheless, there is a very
close analogy—so close that we continue to call it “precession.”
We will discuss it later when we talk about the quantum-mechanical
point of view.

### 34–4Diamagnetism

Next we want to look at *dia*magnetism from the classical point
of view. It can be worked out in several ways, but one of the nice
ways is the following. Suppose that we slowly turn on a magnetic field
in the vicinity of an atom. As the magnetic field changes an
*electric* field is generated by magnetic induction. From
Faraday’s law, the line integral of $\FLPE$ around any closed path is
the rate of change of the magnetic flux through the path. Suppose we
pick a path $\Gamma$ which is a circle of radius $r$ concentric with
the center of the atom, as shown in Fig. 34-4. The average
tangential electric field $E$ around this path is given by
\begin{equation*}
E2\pi r=-\ddt{}{t}\,(B\pi r^2),
\end{equation*}
and there is a circulating electric field whose strength is
\begin{equation*}
E=-\frac{r}{2}\,\ddt{B}{t}.
\end{equation*}

The induced electric field acting on an electron in the atom produces a torque equal to $-q_eEr$, which must equal the rate of change of the angular momentum $dJ/dt$: \begin{equation} \label{Eq:II:34:14} \ddt{J}{t}=\frac{q_er^2}{2}\,\ddt{B}{t}. \end{equation} Integrating with respect to time from zero field, we find that the change in angular momentum due to turning on the field is \begin{equation} \label{Eq:II:34:15} \Delta J=\frac{q_er^2}{2}\,B. \end{equation} This is the extra angular momentum from the twist given to the electrons as the field is turned on.

This added angular momentum makes an extra magnetic moment which,
because it is an *orbital* motion, is just $-q_e/2m$ times the
angular momentum. The induced diamagnetic moment is
\begin{equation}
\label{Eq:II:34:16}
\Delta\mu=-\frac{q_e}{2m}\,\Delta J=-\frac{q_e^2r^2}{4m}\,B.
\end{equation}
The minus sign (as you can see is right by using Lenz’s law) means
that the added moment is opposite to the magnetic field.

We would like to write Eq. (34.16) a little differently.
The $r^2$ which appears is the radius from an axis through the atom parallel
to $\FLPB$, so if $\FLPB$ is along the $z$-direction, it is $x^2+y^2$.
If we consider spherically symmetric atoms (or average over atoms with
their natural axes in all directions) the average of $x^2+y^2$ is $2/3$
of the average of the square of the true radial distance from the center
*point* of the atom. It is therefore usually more convenient to
write Eq. (34.16) as
\begin{equation}
\label{Eq:II:34:17}
\Delta\mu=-\frac{q_e^2}{6m}\av{r^2}B.
\end{equation}

In any case, we have found an induced atomic moment proportional to the magnetic field $B$ and opposing it. This is diamagnetism of matter. It is this magnetic effect that is responsible for the small force on a piece of bismuth in a nonuniform magnetic field. (You could compute the force by working out the energy of the induced moments in the field and seeing how the energy changes as the material is moved into or out of the high-field region.)

We are still left with the problem: What is the mean square radius, $\av{r^2}$? Classical mechanics cannot supply an answer. We must go back and start over with quantum mechanics. In an atom we cannot really say where an electron is, but only know the probability that it will be at some place. If we interpret $\av{r^2}$ to mean the average of the square of the distance from the center for the probability distribution, the diamagnetic moment given by quantum mechanics is just the same as formula (34.17). This equation, of course, is the moment for one electron. The total moment is given by the sum over all the electrons in the atom. The surprising thing is that the classical argument and quantum mechanics give the same answer, although, as we shall see, the classical argument that gives Eq. (34.17) is not really valid in classical mechanics.

The same diamagnetic effect occurs even when an atom already has a permanent moment. Then the system will precess in the magnetic field. As the whole atom precesses, it takes up an additional small angular velocity, and that slow turning gives a small current which represents a correction to the magnetic moment. This is just the diamagnetic effect represented in another way. But we don’t really have to worry about that when we talk about paramagnetism. If the diamagnetic effect is first computed, as we have done here, we don’t have to worry about the fact that there is an extra little current from the precession. That has already been included in the diamagnetic term.

### 34–5Larmor’s theorem

We can already conclude something from our results so far. First of
all, in the classical theory the moment $\FLPmu$ was always
proportional to $\FLPJ$, with a given constant of proportionality for
a particular atom. There wasn’t any spin of the electrons, and the
constant of proportionality was always $-q_e/2m$; that is to say, in
Eq. (34.6) we should set $g=1$. The ratio of $\FLPmu$
to $\FLPJ$ was independent of the internal motion of the electrons. Thus,
according to the classical theory, all systems of electrons would
precess with *the same* angular velocity. (This is *not*
true in quantum mechanics.) This result is related to a theorem in
classical mechanics that we would now like to prove. Suppose we have a
group of electrons which are all held together by attraction toward a
central point—as the electrons are attracted by a nucleus. The
electrons will also be interacting with each other, and can, in
general, have complicated motions. Suppose you have solved for the
motions with *no* magnetic field and then want to know what the
motions would be *with* a weak magnetic field. The theorem says
that the motion with a weak magnetic field is always one of the
no-field solutions with an added rotation, about the axis of the
field, with the angular velocity $\omega_L=q_eB/2m$. (This is the same
as $\omega_p$, if $g=1$.) There are, of course, many possible
motions. The point is that for every motion without the magnetic field
there is a corresponding motion in the field, which is the original
motion plus a uniform rotation. This is called Larmor’s theorem,
and $\omega_L$ is called the *Larmor
frequency*.

We would like to show how the theorem can be proved, but we will let
you work out the details. Take, first, one electron in a central force
field. The force on it is just $\FLPF(r)$, directed toward the
center. If we now turn on a uniform magnetic field, there is an
additional force, $q\FLPv\times\FLPB$; so the total force is
\begin{equation}
\label{Eq:II:34:18}
\FLPF(r)+q\FLPv\times\FLPB.
\end{equation}
Now let’s look at the same system from a coordinate system rotating with
angular velocity $\omega$ about an axis through the center of force and
parallel to $\FLPB$. This is no longer an inertial system, so we have to
put in the proper pseudo forces—the
centrifugal and
Coriolis
forces we
talked about in Chapter 19 of Volume I. We found there
that in a frame rotating with angular velocity $\omega$, there is an
apparent *tangential* force proportional to $v_r$, the radial
component of velocity:
\begin{equation}
\label{Eq:II:34:19}
F_t=-2m\omega v_r.
\end{equation}
And there is an apparent radial force which is given by
\begin{equation}
\label{Eq:II:34:20}
F_r=m\omega^2r+2m\omega v_t,
\end{equation}
where $v_t$ is the tangential component of the velocity, measured
*in* the rotating frame. (The radial component $v_r$ for rotating
and inertial frames is the same.)

Now for small enough angular velocities (that is, if $\omega r\ll v_t$), we can
neglect the first term (centrifugal) in Eq. (34.20) in comparison
with the second (Coriolis). Then Eqs. (34.19)
and (34.20) can be written together as
\begin{equation}
\label{Eq:II:34:21}
\FLPF=-2m\FLPomega\times\FLPv.
\end{equation}
If we now *combine* a rotation and a magnetic field, we must add
the force in Eq. (34.21) to that in Eq. (34.18).
The total force is
\begin{equation}
\label{Eq:II:34:22}
\FLPF(r)+q\FLPv\times\FLPB+2m\FLPv\times\FLPomega
\end{equation}
[we reverse the cross product and the sign of Eq. (34.21) to
get the last term]. Looking at our result, we see that if
\begin{equation*}
2m\FLPomega=-q\FLPB
\end{equation*}
the two terms on the right cancel, and in the moving frame the only
force is $\FLPF(r)$. The motion of the electron is just the same as
with no magnetic field—and, of course, no rotation. We have proved
Larmor’s theorem for one electron. Since the proof assumes a
small $\omega$, it also means that the theorem is true only for weak
magnetic fields. The only thing we could ask you to improve on is to
take the case of many electrons mutually interacting with each other,
but all in the same central field, and prove the same theorem. So no
matter how complex an atom is, if it has a central field the theorem
is true. But that’s the end of the classical mechanics, because it
isn’t true in fact that the motions precess in that way. The
precession frequency $\omega_p$ of Eq. (34.11) is only
equal to $\omega_L$ if $g$ happens to be equal to $1$.

### 34–6Classical physics gives neither diamagnetism nor paramagnetism

Now we would like to demonstrate that according to classical mechanics
there can be no diamagnetism and no paramagnetism at all. It sounds
crazy—first, we have proved that there are paramagnetism,
diamagnetism, precessing orbits, and so on, and now we are going to
prove that it is all wrong. Yes!—We are going to prove that
*if* you follow the *classical* mechanics far enough, there
are no such magnetic effects—*they all cancel out*. If you
start a classical argument in a certain place and don’t go far enough,
you can get any answer you want. But the only legitimate and correct
proof shows that there is no magnetic effect whatever.

It is a consequence of classical mechanics that if you have any kind
of system—a gas with electrons, protons, and whatever—kept in a
box so that the whole thing can’t turn, there will be no magnetic
effect. It is possible to have a magnetic effect if you have an
isolated system, like a star held together by itself, which can start
rotating when you put on the magnetic field. But if you have a piece
of material that is held in place so that it can’t start spinning,
then there will be no magnetic effects. What we mean by holding down
the spin is summarized this way: At a given temperature we suppose
that there is *only one state* of thermal equilibrium. The
theorem then says that if you turn on a magnetic field and wait for
the system to get into thermal equilibrium, there will be no
paramagnetism or diamagnetism—there will be no induced magnetic
moment. Proof: According to statistical mechanics, the probability
that a system will have any given state of motion is proportional
to $e^{-U/kT}$, where $U$ is the energy of that motion. Now what is the
energy of motion? For a particle moving in a constant magnetic field,
the energy is the ordinary potential energy plus $mv^2/2$, with
nothing additional for the magnetic field. [You know that the forces
from electromagnetic fields are $q(\FLPE+\FLPv\times\FLPB)$, and that
the rate of work $\FLPF\cdot\FLPv$ is just $q\FLPE\cdot\FLPv$, which
is not affected by the magnetic field.] So the energy of a system,
whether it is in a magnetic field or not, is always given by the
kinetic energy plus the potential energy. Since the probability of any
motion depends only on the energy—that is, on the velocity and
position—it is the same whether or not there is a magnetic
field. For *thermal* equilibrium, therefore, the magnetic field
has no effect. If we have one system in a box, and then have another
system in a second box, this time with a magnetic field, the
probability of any particular velocity at any point in the first box
is the same as in the second. If the first box has no average
circulating current (which it will not have if it is in equilibrium
with the stationary walls), there is no average magnetic moment. Since
in the second box all the motions are the same, there is no average
magnetic moment there either. Hence, if the temperature is kept
constant and thermal equilibrium is re-established after the field is
turned on, there can be no magnetic moment induced by the
field—according to classical mechanics. We can only get a
satisfactory understanding of magnetic phenomena from quantum
mechanics.

Unfortunately, we cannot assume that you have a thorough understanding
of quantum mechanics, so this is hardly the place to discuss the
matter. On the other hand, we don’t always have to learn something
first by learning the exact rules and then by learning how they are
applied in different cases. Almost every subject that we have taken up
in this course has been treated in a different way. In the case of
electricity, we wrote the Maxwell equations on “Page One” and then
deduced all the consequences. That’s one way. But we will *not* now
try to begin a new “Page One,” writing the equations of quantum
mechanics and deducing everything from them. We will just have to tell
you some of the consequences of quantum mechanics, before you learn
where they come from. So here we go.

### 34–7Angular momentum in quantum mechanics

We have already given you a relation between the magnetic moment and
the angular momentum. That’s pleasant. But what do the magnetic moment
and the angular momentum *mean* in quantum mechanics? In quantum
mechanics it turns out to be best to define things like magnetic
moments in terms of the other concepts such as energy, in order to
make sure that one knows what it means. Now, it is easy to define a
magnetic moment in terms of energy, because the energy of a moment in
a magnetic field is, in the classical theory,
$\FLPmu\cdot\FLPB$. Therefore, the following definition has been taken
in quantum mechanics: If we calculate the energy of a system in a
magnetic field and we find that it is proportional to the field
strength (for small field), the coefficient is called the component of
magnetic moment in the direction of the field. (We don’t have to get
so elegant for our work now; we can still think of the magnetic moment
in the ordinary, to some extent classical, sense.)

Now we would like to discuss the idea of angular momentum in quantum
mechanics—or rather, the characteristics of what, in quantum
mechanics, is called angular momentum. You see, when you go to new
kinds of laws, you can’t just assume that each word is going to mean
exactly the same thing. You may think, say, “Oh, I know what angular
momentum is. It’s that thing that is changed by a torque.” But what’s
a torque? In quantum mechanics we have to have new definitions of old
quantities. It would, therefore, be legally best to call it by some
other name such as “quantangular momentum,” or something like that,
because it is the angular momentum as defined in quantum
mechanics. But if we can find a quantity in quantum mechanics which is
identical to our old idea of angular momentum when the system becomes
large enough, there is no use in inventing an extra word. We might as
well just call it angular momentum. With that understanding, this odd
thing that we are about to describe *is* angular momentum. It is
the thing which in a large system we recognize as angular momentum in
classical mechanics.

First, we take a system in which angular momentum is conserved, such as an atom all by itself in empty space. Now such a thing (like the earth spinning on its axis) could, in the ordinary sense, be spinning around any axis one wished to choose. And for a given spin, there could be many different “states,” all of the same energy, each “state” corresponding to a particular direction of the axis of the angular momentum. So in the classical theory, with a given angular momentum, there is an infinite number of possible states, all of the same energy.

It turns out in quantum mechanics, however, that several strange
things happen. First, the number of states in which such a system
*can exist* is limited—there is only a finite number. If the
system is small, the finite number is very small, and if the system is
large, the finite number gets very, very large. Second, we
*cannot* describe a “state” by giving the *direction* of
its angular momentum, but only by giving the *component* of the
angular momentum along some direction—say in the
$z$-direction. Classically, an object with a given total angular
momentum $J$ could have, for its $z$-component, any value from $+J$
to $-J$. But quantum-mechanically, the $z$-component of angular momentum
can have only certain discrete values. Any given system—a particular
atom, or a nucleus, or anything—with a given energy, has a
characteristic number $j$, and its $z$-component of angular momentum can
only be one of the following set of values:
\begin{equation}
\begin{aligned}
j\hbar&\\
(j-1)\hbar&\\
(j-2)\hbar&\\
\vdots\phantom{)\hbar}&\\
-(j-2)\hbar&\\
-(j-1)\hbar&\\
-j\hbar&\\
\end{aligned}
\label{Eq:II:34:23}
\end{equation}
The largest $z$-component is $j$ times $\hbar$; the next smaller is
one unit of $\hbar$ less, and so on down to $-j\hbar$. The number $j$
is called “the spin of the system.” (Some people call it the “total
angular momentum quantum number”; but we’ll call it the “spin.”)

You may be worried that what we are saying can only be true for some
“special” $z$-axis. But that is not so. For a system whose spin
is $j$, the component of angular momentum along *any* axis can have
only one of the values in (34.23). Although it is quite
mysterious, we ask you just to accept it for the moment. We will come
back and discuss the point later. You may at least be pleased to hear
that the $z$-component goes from some number to minus the *same*
number, so that we at least don’t have to decide which is the plus
direction of the $z$-axis. (Certainly, if we said that it went
from $+j$ to minus a different amount, that would be infinitely mysterious,
because we wouldn’t have been able to define the $z$-axis, pointing
the other way.)

Now if the $z$-component of angular momentum must go down by integers
from $+j$ to $-j$, then $j$ must be an integer. No! Not quite;
twice $j$ must be an integer. It is only the *difference* between $+j$
and $-j$ that must be an integer. So, in general, the spin $j$ is
either an integer or a half-integer, depending on whether $2j$ is even
or odd. Take, for instance, a nucleus like lithium, which has a spin
of three-halves, $j=3/2$. Then the angular momentum around the
$z$-axis, in units of $\hbar$, is one of the following:
\begin{equation*}
\begin{matrix}
+3/2\phantom{.}\\
+1/2\phantom{.}\\
-1/2\phantom{.}\\
-3/2.
\end{matrix}
\end{equation*}
There are four possible states, each of the same energy, if the
nucleus is in empty space with no external fields. If we have a system
whose spin is two, then the $z$-component of angular momentum has only
the values, in units of $\hbar$,
\begin{equation*}
\begin{matrix}
\phantom{-}2\phantom{.}\\
\phantom{-}1\phantom{.}\\
\phantom{-}0\phantom{.}\\
-1\phantom{.}\\
-2.
\end{matrix}
\end{equation*}
If you count how many states there are for a given $j$, there are
$(2j+1)$ possibilities. In other words, if you tell me the energy and
also the spin $j$, it turns out that there are exactly $(2j+1)$ states
with that energy, each state corresponding to one of the different
possible values of the $z$-component of the angular momentum.

We would like to add one other fact. If you pick out any atom of
known $j$ at random and measure the $z$-component of the angular momentum,
then you may get any one of the possible values, and each of the
values is *equally* likely. All of the states are in fact single
states, and each is just as good as any other. Each one has the same
“weight” in the world. (We are assuming that nothing has been done
to sort out a special sample.) This fact has, incidentally, a simple
classical analog. If you ask the same question classically: What is
the likelihood of a particular $z$-component of angular momentum if
you take a random sample of systems, all with the same total angular
momentum?—the answer is that all values from the maximum to the
minimum are equally likely. (You can easily work that out.) The
classical result corresponds to the equal probability of the
$(2j+1)$ possibilities in quantum mechanics.

From what we have so far, we can get another interesting and somewhat
surprising conclusion. In certain classical calculations the quantity
that appears in the final result is the *square* of the magnitude
of the angular momentum $\FLPJ$—in other words,
$\FLPJ\cdot\FLPJ$. It turns out that it is often possible to
*guess* at the correct quantum-mechanical formula by using the
classical calculation and the following simple rule: Replace
$J^2=\FLPJ\cdot\FLPJ$ by $j(j+1)\hbar^2$. This rule is commonly used,
and usually gives the correct result, but *not* always. We can
give the following argument to show why you might expect this rule to
work.

The scalar product $\FLPJ\cdot\FLPJ$ can be written as
\begin{equation*}
\FLPJ\cdot\FLPJ=J_x^2+J_y^2+J_z^2.
\end{equation*}
Since it is a scalar, it should be the same for any orientation of the
spin. Suppose we pick samples of any given atomic system at random and
make measurements of $J_x^2$, or $J_y^2$, or $J_z^2$, the
*average value* should be the same for each. (There is no special
distinction for any one of the directions.) Therefore, the average
of $\FLPJ\cdot\FLPJ$ is just equal to three times the average of any
component squared, say of $J_z^2$;
\begin{equation*}
\av{\FLPJ\cdot\FLPJ} = 3\av{J_z^2}.
\end{equation*}
But since $\FLPJ\cdot\FLPJ$ is the same for all orientations, its
average is, of course, just its constant value; we have
\begin{equation}
\label{Eq:II:34:24}
\FLPJ\cdot\FLPJ = 3\av{J_z^2}.
\end{equation}

If we now say that we will use the same equation for quantum mechanics, we can easily find $\av{J_z^2}$. We just have to take the sum of the $(2j+1)$ possible values of $J_z^2$, and divide by the total number; \begin{equation} \label{Eq:II:34:25} \av{J_z^2} = \frac {j^2+(j-1)^2+\dotsb+(-j+1)^2+(-j)^2} {2j+1}\,\hbar^2. \end{equation} \begin{gather} \label{Eq:II:34:25} \av{J_z^2} =\\[1ex] \frac{j^2+(j-1)^2+\dotsb+(-j+1)^2+(-j)^2} {2j+1}\,\hbar^2.\notag \end{gather} For a system with a spin of $3/2$, it goes like this: \begin{equation*} \av{J_z^2} = \frac {(3/2)^2+(1/2)^2+(-1/2)^2+(-3/2)^2} {4}\,\hbar^2=\frac{5}{4}\,\hbar^2. \end{equation*} \begin{gather*} \av{J_z^2} =\\[1.25ex] \frac{(3/2)^2+(1/2)^2+(-1/2)^2+(-3/2)^2} {4}\,\hbar^2\\[1.25ex] \kern{-1.75ex}=\frac{5}{4}\,\hbar^2. \end{gather*} We conclude that \begin{equation*} \FLPJ\cdot\FLPJ = 3\av{J_z^2} = 3\tfrac{5}{4}\hbar^2=\tfrac{3}{2}(\tfrac{3}{2}+1)\hbar^2. \end{equation*} We will leave it for you to show that Eq. (34.25), together with Eq. (34.24), gives the general result \begin{equation} \label{Eq:II:34:26} \FLPJ\cdot\FLPJ=j(j+1)\hbar^2. \end{equation} Although we would think classically that the largest possible value of the $z$-component of $\FLPJ$ is just the magnitude of $\FLPJ$—namely, $\sqrt{\FLPJ\cdot\FLPJ}$—quantum mechanically the maximum of $J_z$ is always a little less than that, because $j\hbar$ is always less than $\sqrt{j(j+1)}\hbar$. The angular momentum is never “completely along the $z$-direction.”

### 34–8The magnetic energy of atoms

Now we want to talk again about the magnetic moment. We have said that in quantum mechanics the magnetic moment of a particular atomic system can be written in terms of the angular momentum by Eq. (34.6); \begin{equation} \label{Eq:II:34:27} \FLPmu=-g\biggl(\frac{q_e}{2m}\biggr)\FLPJ, \end{equation} where $-q_e$ and $m$ are the charge and mass of the electron.

An atomic magnet placed in an external magnetic field will have an extra magnetic energy which depends on the component of its magnetic moment along the field direction. We know that \begin{equation} \label{Eq:II:34:28} U_{\text{mag}}=-\FLPmu\cdot\FLPB. \end{equation} Choosing our $z$-axis along the direction of $\FLPB$, \begin{equation} \label{Eq:II:34:29} U_{\text{mag}}=-\mu_zB. \end{equation} Using Eq. (34.27), we have that \begin{equation*} U_{\text{mag}}=g\biggl(\frac{q_e}{2m}\biggr)J_zB. \end{equation*} Quantum mechanics says that $J_z$ can have only certain values: $j\hbar$, $(j-1)\hbar$, …, $-j\hbar$. Therefore, the magnetic energy of an atomic system is not arbitrary; it can have only certain values. Its maximum value, for instance, is \begin{equation*} g\biggl(\frac{q_e}{2m}\biggr)\hbar jB. \end{equation*} The quantity $q_e\hbar/2m$ is usually given the name “the Bohr magneton” and written $\mu_B$: \begin{equation*} \mu_B=\frac{q_e\hbar}{2m}. \end{equation*} The possible values of the magnetic energy are \begin{equation*} U_{\text{mag}}=g\mu_BB\,\frac{J_z}{\hbar}, \end{equation*} where $J_z/\hbar$ takes on the possible values $j$, $(j-1)$, $(j-2)$, …, $(-j+1)$, $-j$.

In other words, the energy of an atomic system is changed when it is put in a magnetic field by an amount that is proportional to the field, and proportional to $J_z$. We say that the energy of an atomic system is “split into $2j+1$ levels” by a magnetic field. For instance, an atom whose energy is $U_0$ outside a magnetic field and whose $j$ is $3/2$, will have four possible energies when placed in a field. We can show these energies by an energy-level diagram like that drawn in Fig. 34-5. Any particular atom can have only one of the four possible energies in any given field $B$. That is what quantum mechanics says about the behavior of an atomic system in a magnetic field.

The simplest “atomic” system is a single electron. The spin of an electron is $1/2$, so there are two possible states: $J_z=\hbar/2$ and $J_z=-\hbar/2$. For an electron at rest (no orbital motion), the spin magnetic moment has a $g$-value of $2$, so the magnetic energy can be either $\pm\mu_BB$. The possible energies in a magnetic field are shown in Fig. 34-6. Speaking loosely we say that the electron either has its spin “up” (along the field) or “down” (opposite the field).

For systems with higher spins, there are more states. We can think that the spin is “up” or “down” or cocked at some “angle” in between, depending on the value of $J_z$.

We will use these quantum mechanical results to discuss the magnetic properties of materials in the next chapter.