The paramagnetism of bulk materials
We would like now to describe the phenomenon of the paramagnetism of bulk materials. Suppose we have a substance whose atoms have permanent magnetic moments, for example a crystal like copper sulfate. In the crystal there are copper ions whose inner electron shells have a net angular momentum and a net magnetic moment. So the copper ion is an object which has a permanent magnetic moment. Let’s say just a word about which atoms have magnetic moments and which ones don’t. Any atom, like sodium for instance, which has an odd number of electrons, will have a magnetic moment. Sodium has one electron in its unfilled shell. This electron gives the atom a spin and a magnetic moment. Ordinarily, however, when compounds are formed the extra electrons in the outside shell are coupled together with other electrons whose spin directions are exactly opposite, so that all the angular momenta and magnetic moments of the valence electrons usually cancel out. That’s why, in general, molecules do not have a magnetic moment. Of course if you have a gas of sodium atoms, there is no such cancellation.1  Also, if you have what is called in chemistry a “free radical”—an object with an odd number of valence electrons—then the bonds are not completely satisfied, and there is a net angular momentum.  In most bulk materials there is a net magnetic moment only if there are atoms present whose inner electron shell is not filled. Then there can be a net angular momentum and a magnetic moment. Such atoms are found in the “transition element” part of the periodic table—for instance, chromium, manganese, iron, nickel, cobalt, palladium, and platinum are elements of this kind. Also, all of the rare earth elements have unfilled inner shells and permanent magnetic moments. There are a couple of other strange things that also happen to have magnetic moments, such as liquid oxygen, but we will leave it to the chemistry department to explain the reason.  Now suppose that we have a box full of atoms or molecules with permanent moments—say a gas, or a liquid, or a crystal. We would like to know what happens if we apply an external magnetic field. With no magnetic field, the atoms are kicked around by the thermal motions, and the moments wind up pointing in all directions. But when there is a magnetic field, it acts to line up the little magnets; then there are more moments lying toward the field than away from it. The material is “magnetized.”  We define the magnetization $\FLPM$ of a material as the net magnetic moment per unit volume, by which we mean the vector sum of all the atomic magnetic moments in a unit volume. If there are $N$ atoms per unit volume and their average moment is $\av{\FLPmu}$ then $\FLPM$ can be written as $N$ times the average atomic moment: \begin{equation} \label{Eq:II:35:8} \FLPM=N\av{\FLPmu}. \end{equation} The definition of $\FLPM$ corresponds to the definition of the electric polarization $\FLPP$ of Chapter 10.  The classical theory of paramagnetism is just like the theory of the dielectric constant we showed you in Chapter 11. One assumes that each of the atoms has a magnetic moment $\FLPmu$, which always has the same magnitude but which can point in any direction. In a field $\FLPB$, the magnetic energy is $-\FLPmu\cdot\FLPB=-\mu B\cos\theta$, where $\theta$ is the angle between the moment and the field. From statistical mechanics, the relative probability of having any angle is $e^{-\text{energy}/kT}$, so angles near zero are more likely than angles near $\pi$. Proceeding exactly as we did in Section 11-3, we find that for small magnetic fields $\FLPM$ is directed parallel to $\FLPB$ and has the magnitude \begin{equation} \label{Eq:II:35:9} M=\frac{N\mu^2B}{3kT}. \end{equation} [See Eq. (11.20).] This approximate formula is correct only for $\mu B/kT$ much less than one.  We find that the induced magnetization—the magnetic moment per unit volume—is proportional to the magnetic field. This is the phenomenon of paramagnetism. You will see that the effect is stronger at lower temperatures and weaker at higher temperatures. When we put a field on a substance, it develops, for small fields, a magnetic moment proportional to the field. The ratio of $M$ to $B$ (for small fields) is called the magnetic susceptibility.  Now we want to look at paramagnetism from the point of view of quantum mechanics. We take first the case of an atom with a spin of $1/2$. In the absence of a magnetic field the atoms have a certain energy, but in a magnetic field there are two possible energies, one for each value of $J_z$. For $J_z=+\hbar/2$, the energy is changed by the magnetic field by the amount \begin{equation} \label{Eq:II:35:10} \Delta U_1=+g\biggl(\frac{q_e\hbar}{2m}\biggr)\cdot\frac{1}{2}\cdot B. \end{equation} (The energy shift $\Delta U$ is positive for an atom because the electron charge is negative.) For $J_z=-\hbar/2$, the energy is changed by the amount \begin{equation} \label{Eq:II:35:11} \Delta U_2=-g\biggl(\frac{q_e\hbar}{2m}\biggr)\cdot\frac{1}{2}\cdot B. \end{equation} To save writing, let’s set \begin{equation} \label{Eq:II:35:12} \mu_0=g\biggl(\frac{q_e\hbar}{2m}\biggr)\cdot\frac{1}{2}; \end{equation} then \begin{equation} \label{Eq:II:35:13} \Delta U=\pm\mu_0B. \end{equation} The meaning of $\mu_0$ is clear: $-\mu_0$ is the $z$-component of the magnetic moment in the up-spin case, and $+\mu_0$ is the $z$-component of the magnetic moment in the down-spin case.  Now statistical mechanics tells us that the probability that an atom is in one state or another is proportional to \begin{equation*} e^{-(\text{Energy of state})/kT}. \end{equation*} With no magnetic field the two states have the same energy; so when there is equilibrium in a magnetic field, the probabilities are proportional to \begin{equation} \label{Eq:II:35:14} e^{-\Delta U/kT}. \end{equation} The number of atoms per unit volume with spin up is \begin{equation} \label{Eq:II:35:15} N_{\text{up}}=ae^{-\mu_0B/kT}, \end{equation} and the number with spin down is \begin{equation} \label{Eq:II:35:16} N_{\text{down}}=ae^{+\mu_0B/kT}. \end{equation} The constant $a$ is to be determined so that \begin{equation} \label{Eq:II:35:17} N_{\text{up}}+N_{\text{down}}=N, \end{equation} the total number of atoms per unit volume. So we get that \begin{equation} \label{Eq:II:35:18} a=\frac{N}{e^{+\mu_0B/kT}+e^{-\mu_0B/kT}}. \end{equation}  What we are interested in is the average magnetic moment along the $z$-axis. The atoms with spin up will contribute a moment of $-\mu_0$, and those with spin down will have a moment of $+\mu_0$; so the average moment is \begin{equation} \label{Eq:II:35:19} \av{\mu} = \frac{N_{\text{up}}(-\mu_0)+N_{\text{down}}(+\mu_0)}{N}. \end{equation}  The magnetic moment per unit volume $M$ is then $N\av{\mu}$. Using Eqs. (35.15), (35.16), and (35.17), we get that \begin{equation} \label{Eq:II:35:20} M=N\mu_0\, \frac{e^{+\mu_0B/kT}-e^{-\mu_0B/kT}}{e^{+\mu_0B/kT}+e^{-\mu_0B/kT}}. \end{equation} This is the quantum-mechanical formula for $M$ for atoms with $j=1/2$. Incidentally, this formula can also be written somewhat more concisely in terms of the hyperbolic tangent function: \begin{equation} \label{Eq:II:35:21} M=N\mu_0\,\tanh\frac{\mu_0B}{kT}. \end{equation}  A plot of $M$ as a function of $B$ is given in Fig. 35-7. When $B$ gets very large, the hyperbolic tangent approaches $1$, and $M$ approaches the limiting value $N\mu_0$. So at high fields, the magnetization saturates. We can see why that is; at high enough fields the moments are all lined up in the same direction. In other words, they are all in the spin-down state, and each atom contributes the moment $\mu_0$.  In most normal cases—say, for typical moments, room temperatures, and the fields one can normally get (like $10{,}000$ gauss)—the ratio $\mu_0B/kT$ is about $0.002$. One must go to very low temperatures to see the saturation. For normal temperatures, we can usually replace $\tanh x$ by $x$, and write \begin{equation} \label{Eq:II:35:22} M=\frac{N\mu_0^2B}{kT}. \end{equation}  Just as we saw in the classical theory, $M$ is proportional to $B$. In fact, the formula is almost exactly the same, except that there seems to be a factor of $1/3$ missing. But we still need to relate the $\mu_0$ in our quantum formula to the $\mu$ that appears in the classical result, Eq. (35.9).  In the classical formula, what appears is $\mu^2=\FLPmu\cdot\FLPmu$, the square of the vector magnetic moment, or \begin{equation} \label{Eq:II:35:23} \FLPmu\cdot\FLPmu=\biggl(g\,\frac{q_e}{2m}\biggr)^2\FLPJ\cdot\FLPJ. \end{equation} We pointed out in the last chapter that you can very likely get the right answer from a classical calculation by replacing $\FLPJ\cdot\FLPJ$ by $j(j+1)\hbar^2$. In our particular example, we have $j=1/2$, so \begin{equation*} j(j+1)\hbar^2=\tfrac{3}{4}\hbar^2. \end{equation*} Substituting this for $\FLPJ\cdot\FLPJ$ in Eq. (35.23), we get \begin{equation*} \FLPmu\cdot\FLPmu=\biggl(g\,\frac{q_e}{2m}\biggr)^2 \frac{3\hbar^2}{4}, \end{equation*} or in terms of $\mu_0$, defined in Eq. (35.12), we get \begin{equation*} \FLPmu\cdot\FLPmu=3\mu_0^2. \end{equation*} Substituting this for $\mu^2$ in the classical formula, Eq. (35.9), does indeed reproduce the correct quantum formula, Eq. (35.22).  The quantum theory of paramagnetism is easily extended to atoms of any spin $j$. The low-field magnetization is \begin{equation} \label{Eq:II:35:24} M=Ng^2\,\frac{j(j+1)}{3}\,\frac{\mu_B^2B}{kT}, \end{equation} where \begin{equation} \label{Eq:II:35:25} \mu_B=\frac{q_e\hbar}{2m} \end{equation} is a combination of constants with the dimensions of a magnetic moment. Most atoms have moments of roughly this size. It is called the Bohr magneton. The spin magnetic moment of the electron is almost exactly one Bohr magneton. 