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by Pravegaa Education

Chapter 8

Magnetism in Solids

Magnetic Moments, Fields and Susceptibility

Where an atomic moment comes from

Three things inside an atom can produce a magnetic moment.

  1. The orbital motion of the electrons. A circulating charge is a current loop, and a current loop is a magnetic dipole.
  2. The spin of the electrons. An intrinsic moment with no classical analogue, and in most solids the dominant contribution.
  3. The moment induced by the field itself. Applying a field changes the orbital motion, and the change always produces a moment opposing the field. This is diamagnetism, and unlike the first two it is present in every atom.

The first two exist whether or not a field is applied and give moments that align along the field: they are paramagnetic in character. The third exists only because a field is applied and gives a moment against it. Every solid has the third; only some have the first two.

The orbital moment.

An electron of charge \(-e\) circulating with speed \(v\) in an orbit of radius \(r\) completes one circuit in \(T = 2\pi r/v\), so the current is

\begin{equation} I = \frac{e}{T} = \frac{ev}{2\pi r}, \end{equation}
and the moment of the equivalent loop is \(\mu = IA\):
\begin{equation} \mu = \frac{ev}{2\pi r}\,\pi r^{2} = \frac{evr}{2}. \end{equation}
The orbital angular momentum is \(L = m_{e}vr\), so
\begin{equation} \boxed{\;\vb{\mu}_{L} = -\frac{e}{2m_{e}}\,\vb{L}\;} \qquad \gamma_{L} \equiv \left|\frac{\mu_{L}}{L}\right| = \frac{e}{2m_{e}} . \end{equation}
The minus sign is not decoration: because the electron is negative, its moment points opposite to its angular momentum, and every sign in the subject depends on remembering it.
The spin moment.

Experiment (and Dirac theory) gives spin twice the gyromagnetic ratio of orbital motion,

\begin{equation} \vb{\mu}_{S} = -g_{s}\,\frac{e}{2m_{e}}\,\vb{S}, \qquad g_{s} \simeq 2, \qquad \gamma_{S} = \frac{e}{m_{e}} . \end{equation}
The Bohr magneton.

Since \(L\) comes in units of \(\hbar\), the natural unit of atomic magnetism is

\begin{equation} \mu_{B} = \frac{e\hbar}{2m_{e}} = 9.274\times10^{-24}\ \text{J\,T}^{-1} = 5.788\times10^{-5}\ \text{eV\,T}^{-1}. \end{equation}
A useful number to carry: \(\mu_{B}B/k_{B} \approx 0.67\) K per tesla. Even in a 10 T laboratory field the magnetic energy is \(\mu_{B}B \approx 5.8\times10^{-4}\) eV, some forty-five times smaller than \(k_{B}T \approx 0.026\) eV at room temperature. Magnetic energies are tiny compared with thermal energies. This single fact explains why paramagnetic alignment is so feeble, why \(\chi \sim 10^{-3}\) at best, and why the existence of iron magnets at 1000 K demands an explanation that has nothing to do with magnetic forces between dipoles.
Dipole–dipole coupling cannot cause ferromagnetism
Two moments of one \(\mu_{B}\) a lattice spacing apart interact with energy \(\sim \mu_{0}\mu_{B}^{2}/4\pi a^{3}\), which for \(a = 2\,\Ang\) is \(1.1\times10^{-24}\) J — an equivalent temperature of only 0.08 K. Iron orders at 1043 K. Magnetic dipole coupling is four orders of magnitude too weak, and the ordering must come from somewhere else — the exchange interaction of the section on ferromagnetism, which is electrostatic in origin and magnetic only in its consequences.

\(\vb{B}\), \(\vb{H}\) and \(\vb{M}\)

Inside matter three vector fields are in play. \(\vb{M}\), the magnetisation, is the moment per unit volume. \(\vb{H}\) is the field intensity, fixed by the free currents in the coils. \(\vb{B}\) is the flux density, and it is what a moving charge actually feels. In SI units

\begin{equation} \vb{B} = \mu_{0}\left(\vb{H} + \vb{M}\right). \end{equation}
For a linear medium \(\vb{M} = \chi\vb{H}\), and then
\begin{equation} \vb{B} = \mu_{0}(1+\chi)\vb{H} = \mu_{0}\mu_{r}\vb{H} = \mu\vb{H}, \qquad \mu_{r} = 1 + \chi . \end{equation}
The susceptibility is defined in general as a derivative,
\begin{equation} \chi = \frac{\partial M}{\partial H}, \end{equation}
which matters because for ferromagnets \(M\) is not proportional to \(H\) and the ratio \(M/H\) is not a material constant.
Reading the sign and size of \(\chi\)
ClassSign of \(\chi\)Typical \(|\chi|\)\(T\) dependence
Diamagnetnegative\(10^{-6}\)–\(10^{-5}\)none
Paramagnet (ions)positive\(10^{-5}\)–\(10^{-3}\)\(\chi = C/T\)
Pauli paramagnet (metal)positive\(10^{-6}\)none
Ferromagnetpositive, huge\(10^{2}\)–\(10^{5}\)\(C/(T-T_{C})\), \(T>T_{C}\)
Antiferromagnetpositive, small\(10^{-4}\)\(C/(T+T_{N})\), \(T>T_{N}\)
Superconductor (type I)\(\chi = -1\) exactly\(1\)none
Three questions identify any material: is \(\chi\) positive or negative, how big is it, and how does it vary with \(T\)?
\(\chi\) is dimensionless in SI, but not in CGS
In SI, \(\vb{M}\) and \(\vb{H}\) share units of A m\(^{-1}\) and \(\chi\) is a pure number. Older literature and many examination questions use CGS, where \(\vb{B} = \vb{H} + 4\pi\vb{M}\) and susceptibilities are quoted per gram or per mole; the two conventions differ by \(4\pi\), so \(\chi_{\text{SI}} = 4\pi\,\chi_{\text{CGS}}\). If a problem quotes the Curie constant in “CGS units” it is not being decorative — the numerical answer depends on it. Check which convention is in force before substituting.

The perfect diamagnet

A type-I superconductor expels magnetic flux entirely (the Meissner effect), so \(\vb{B} = 0\) inside. Setting \(\vb{B} = 0\) in the relation above gives \(\vb{M} = -\vb{H}\), hence

\begin{equation} \chi_{\text{SC}} = -1, \qquad \mu_{r} = 0 . \end{equation}
This is the strongest diamagnetism possible: no material can have \(\chi < -1\), since that would require \(\vb{B}\) and \(\vb{H}\) to point in opposite directions. On an \(M\)–\(H\) plot drawn to the same scale in both axes, a type-I superconductor is the unique line of slope exactly \(-1\), an ordinary diamagnet has a small negative slope, and a paramagnet a small positive slope. That diagram is a recurring examination question, and \(\chi = -1\) is the only entry on it that is exact.

The five classes

The five types of magnetic order. The arrows show the arrangement of moments on neighbouring sites in the absence of an applied field.

{1.3}

TypeMoments without fieldNet \(M\) at \(H=0\)Example
Diamagneticnone0Cu, Bi, H\(_{2}\)O, Ar
Paramagneticpresent, randomly oriented0Al, Pt, O\(_{2}\), Gd\(^{3+}\) salts
Ferromagneticpresent, parallel\(\neq 0\)Fe, Co, Ni, Gd
Antiferromagneticpresent, antiparallel, equal0MnO, Cr, NiO
Ferrimagneticantiparallel, unequal\(\neq 0\)Fe\(_{3}\)O\(_{4}\), ferrites

Arrangements of moments at \(H=0\). The paramagnet has moments but no correlation between them; the ferromagnet aligns them; the antiferromagnet alternates equal moments and cancels exactly; the ferrimagnet alternates unequal moments and so retains a net magnetisation. A diamagnet (not shown) has no moments at all until a field is applied.

Paired electrons contribute no net spin moment
A frequently garbled statement. In a filled shell or a filled orbital the electrons occur in pairs of opposite spin, and their spin moments cancel exactly; the orbital moments cancel too, since for every occupied \(m_{\ell}\) there is an occupied \(-m_{\ell}\). A closed-shell atom therefore has \(L = S = J = 0\) and no permanent moment at all, which is why the inert gases, and ions like Na\(^{+}\) and Cl\(^{-}\), are purely diamagnetic. Permanent moments require unpaired electrons, and that is why paramagnetism is found in transition-metal and rare-earth ions with partly filled \(d\) or \(f\) shells.

The molecular oxygen case is often quoted as an “exception”, which invites the wrong conclusion. O\(_{2}\) is paramagnetic because molecular orbital theory puts two electrons singly into degenerate \(\pi^{*}\) orbitals with parallel spins. It is an exception to the expectation that a stable molecule has all electrons paired — not an exception to the rule that paramagnetism needs unpaired spins. The rule has no exceptions.

Diamagnetism

Diamagnetism is the only magnetic effect that every material has. It arises because switching on a field changes the orbital motion of the electrons, and by Lenz's law the change opposes the flux that caused it. The induced moment is therefore antiparallel to \(\vb{B}\), \(\chi < 0\), and the material is weakly repelled from a field gradient.

In an atom with unpaired electrons the paramagnetic response is one to three orders of magnitude larger and completely masks the diamagnetic one. So although diamagnetism is universal, it is observed only in materials with no permanent moments: closed-shell atoms and ions, most covalent molecules, and superconductors.

Larmor precession

Apply a field \(\vb{B}\) to an electron in a closed shell. The extra force is \(-e\,\vb{v}\times\vb{B}\), and its effect on a bound orbit is not to distort the orbit but to make the whole orbit precess about \(\vb{B}\) at the Larmor frequency

\begin{equation} \omega_{L} = \frac{eB}{2m_{e}} . \end{equation}
This is Larmor's theorem: for a spherically symmetric charge distribution, and to first order in \(B\), the motion in a frame rotating at \(\omega_{L}\) is the same as the motion without the field.

The precession is itself a circulation of charge and so a current. For \(Z\) electrons circulating with angular frequency \(\omega_{L}\), the induced current is the charge \(-Ze\) passing a point once per period,

\begin{equation} I = -Ze\,\frac{\omega_{L}}{2\pi} = -\frac{Ze^{2}B}{4\pi m_{e}} . \end{equation}
\(Z\) is a count of electrons, not a mass
In the diamagnetic susceptibility, \(Z\) is the number of electrons per atom contributing to the response — the atomic number, or in practice the number of core electrons. It is not the atomic mass, and confusing the two changes the answer by a factor of roughly two for light elements and more for heavy ones. The distinction also has physical content: diamagnetism grows with the number of electrons and with the size of their orbits, which is why bismuth and graphite are the strongest ordinary diamagnets and helium among the weakest.

The Langevin diamagnetic susceptibility

The moment of a current loop of area \(\pi\rho^{2}\), where \(\rho\) is the distance from the field axis, is \(\mu = I\pi\langle\rho^{2}\rangle\):

\begin{equation} \mu = -\frac{Ze^{2}B}{4\pi m_{e}}\,\pi\langle\rho^{2}\rangle = -\frac{Ze^{2}B}{4m_{e}}\,\langle\rho^{2}\rangle . \end{equation}
Now \(\langle\rho^{2}\rangle = \langle x^{2}\rangle + \langle y^{2}\rangle\) is a projection onto the plane perpendicular to \(\vb{B}\), whereas the quantity a quantum calculation naturally supplies is the full \(\langle r^{2}\rangle = \langle x^{2}\rangle + \langle y^{2}\rangle + \langle z^{2}\rangle\). For a spherically symmetric charge cloud the three are equal, so
\begin{equation} \langle \rho^{2}\rangle = \tfrac{2}{3}\langle r^{2}\rangle . \end{equation}
Substituting and multiplying by the number of atoms per unit volume \(N\) gives \(M = N\mu\), and with \(B \simeq \mu_{0}H\) for a weak response,
\begin{equation} \boxed{\; \chi_{\text{dia}} = -\frac{\mu_{0}NZe^{2}}{6m_{e}}\,\langle r^{2}\rangle \;} \end{equation}
What the formula says
  • The sign is negative and cannot be otherwise. There is no material parameter in the relation above that can change sign; \(N\), \(Z\), \(e^{2}\) and \(\langle r^{2}\rangle\) are all positive. Diamagnetism always opposes.
  • There is no temperature in the formula. Nothing is being aligned against thermal disorder — the moment is induced, not oriented — so \(\chi_{\text{dia}}\) is essentially independent of \(T\). This is the sharpest experimental distinction from paramagnetism.
  • It scales as \(\langle r^{2}\rangle\). The outermost electrons dominate, since \(r^{2}\) weights them heavily. Large, polarisable, many-electron atoms are the strongest diamagnets.
Worked example 8.1.
Estimate the diamagnetic susceptibility of solid argon, with \(N = 2.66\times10^{28}\) m\(^{-3}\), \(Z = 18\) and \(\langle r^{2}\rangle^{1/2} = 0.62\,\Ang\).

Solution. Substituting into the relation above with \(\langle r^{2}\rangle = (0.62\times10^{-10})^{2} = 3.84\times10^{-21}\) m\(^{2}\):

\begin{align*} \chi &= -\frac{(4\pi\times10^{-7})(2.66\times10^{28})(18)(1.602\times10^{-19})^{2} (3.84\times10^{-21})}{6\,(9.11\times10^{-31})}\\ &= -\frac{5.93\times10^{-35}}{5.47\times10^{-30}} = -1.1\times10^{-5}. \end{align*}
A number of order \(10^{-5}\), negative, and dimensionless, exactly as the classification table of the section on magnetic Moments, Fields and Susceptibility promises. Measured values for the solid inert gases lie in the range \(-1\) to \(-2 \times 10^{-5}\).

The two signatures of a diamagnet: \(M\) is linear in \(H\) with a small negative slope, and \(\chi\) does not vary with temperature. Compare the paramagnetic \(\chi = C/T\) hyperbola of the figure above.

Landau diamagnetism

The Langevin result applies to electrons bound in atoms. Conduction electrons in a metal are not bound, and one might expect them to give no diamagnetic contribution at all — classically they give none, by the Bohr–van Leeuwen theorem. Quantum mechanically the field quantises their orbital motion into Landau levels, and the resulting susceptibility is

\begin{equation} \chi_{\text{Landau}} = -\tfrac{1}{3}\,\chi_{\text{Pauli}}, \end{equation}
with \(\chi_{\text{Pauli}}\) the spin contribution derived in the section on paramagnetism. The free-electron gas is therefore net paramagnetic, with two thirds of the bare Pauli value. In a real metal the factor is modified by the band mass, since the orbital term depends on \(m^{*}\) while the spin term does not, and in bismuth — where \(m^{*}\) is very small — the Landau term wins outright and the metal is strongly diamagnetic.
Diamagnetism in one line each
  • Universal: present in every atom, since every atom has orbiting electrons.
  • Weak: \(\chi \sim -10^{-5}\), so it is visible only when nothing paramagnetic is present.
  • Temperature-independent: nothing is competing with thermal disorder.
  • Strongest in large-\(Z\), large-\(r\) systems: Bi, graphite, and (as a special case with \(\chi = -1\)) type-I superconductors.

Paramagnetism

A paramagnet has permanent moments that are randomly oriented at \(H=0\) and partially aligned by a field. The alignment is opposed by thermal agitation, and the balance between the two produces the characteristic \(1/T\) susceptibility known as Curie's law. Because the mechanism is orientation rather than induction, temperature enters in an essential way — the single sharpest contrast with diamagnetism.

The moment of a free ion: Hund's rules

Before any statistical mechanics we need the size of the moment. For an ion with a partly filled shell the electrons arrange themselves to minimise energy subject to the Pauli principle, and the resulting ground state is fixed by three rules, applied in order.

Hund's rules
  1. Maximise \(S\). Fill orbitals with parallel spins as far as the exclusion principle allows. Parallel spins keep electrons apart and lower the Coulomb repulsion.
  2. Then maximise \(L\), consistent with rule 1. Electrons orbiting in the same sense meet less often.
  3. Then fix \(J\) by spin–orbit coupling:
    \begin{equation*} J = |L - S| \quad\text{if the shell is }\textit{less}\text{ than half full}, \qquad J = L + S \quad\text{if }\textit{more}\text{ than half full}. \end{equation*}
    For an exactly half-filled shell \(L = 0\) and \(J = S\), so the two prescriptions agree.

The magnetic moment of the ion is then

\begin{equation} \vb{\mu} = -g_{J}\,\mu_{B}\,\vb{J}, \qquad \mu_{\text{eff}} = g_{J}\sqrt{J(J+1)}\;\mu_{B} \equiv p\,\mu_{B}, \end{equation}
where \(p\) is the effective magneton number and \(g_{J}\) is the Landé \(g\)-factor,
\begin{equation} \boxed{\; g_{J} = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} \;} \end{equation}
The two limits are worth memorising: pure spin (\(L=0\), \(J=S\)) gives \(g_{J}=2\); pure orbital (\(S=0\), \(J=L\)) gives \(g_{J}=1\).
\(\sqrt{J(J+1)}\), not \(J\)
The moment that enters the susceptibility is the root-mean-square value \(g_{J}\mu_{B}\sqrt{J(J+1)}\), because it comes from \(\langle\mu^{2}\rangle\) in the high-temperature expansion. The saturation moment — the largest value \(\mu_{z}\) can take, and the one that matters at low temperature or high field — is \(g_{J}\mu_{B}J\). These are different numbers, and using the wrong one is the commonest error in this section. For Gd\(^{3+}\) (\(J = 7/2\), \(g=2\)) they are \(7.94\mu_{B}\) and \(7\mu_{B}\) respectively.
Worked example 8.2.
Find the ground state, \(g_{J}\) and \(p\) for a free ion with an \(f^{7}\) configuration (e.g. Gd\(^{3+}\), Eu\(^{2+}\)).

Solution. The \(f\) shell has \(\ell = 3\), so \(m_{\ell}\) runs over \(-3,-2,-1,0,+1,+2,+3\): seven orbitals, holding fourteen electrons when full. With seven electrons the shell is exactly half full.

Rule 1: all seven spins parallel, so

\begin{equation*} S = 7 \times \tfrac{1}{2} = \tfrac{7}{2}. \end{equation*}
Rule 2: with one electron in each orbital there is no freedom left, and
\begin{equation*} L = \sum m_{\ell} = -3-2-1+0+1+2+3 = 0 . \end{equation*}
Rule 3: half-filled, so \(J = S = \tfrac{7}{2}\). The term symbol is \(^{8}S_{7/2}\).

The Landé factor follows from the relation above with \(L=0\), \(J=S\):

\begin{equation*} g_{J} = 1 + \frac{J(J+1)+J(J+1)-0}{2J(J+1)} = 1 + 1 = 2 , \end{equation*}
as it must be for a pure-spin ion. Hence
\begin{equation*} p = g_{J}\sqrt{J(J+1)} = 2\sqrt{\tfrac{7}{2}\cdot\tfrac{9}{2}} = 2\cdot\tfrac{\sqrt{63}}{2} = \sqrt{63} = 3\sqrt{7} = 7.94 . \end{equation*}
Gd\(^{3+}\) is the largest free-ion moment among the rare earths, which is why gadolinium compounds dominate magnetic refrigeration and MRI contrast agents.

The Brillouin function and Curie's law

Place \(N\) such ions per unit volume in a field \(B\). The component of \(\vb{J}\) along the field is quantised, \(J_{z} = M_{J}\hbar\) with \(M_{J} = -J,\dots,+J\), so the energy levels are

\begin{equation} E_{M_{J}} = g_{J}\mu_{B}B\,M_{J}, \qquad M_{J} = -J,\,-J+1,\,\dots,\,J . \end{equation}
Evaluating the Boltzmann average of \(\mu_{z}\) over these \(2J+1\) levels gives
\begin{equation} M = N g_{J}\mu_{B} J\, B_{J}(x), \qquad x = \frac{g_{J}\mu_{B} J B}{k_{B}T}, \end{equation}
where \(B_{J}\) is the Brillouin function
\begin{equation} B_{J}(x) = \frac{2J+1}{2J}\coth\!\left(\frac{2J+1}{2J}x\right) - \frac{1}{2J}\coth\!\left(\frac{x}{2J}\right). \end{equation}

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