Past exam of the condensed matter physics course of the University of Oxford / 2019 / 4

A material showing paramagnetism has an induced magnetization in the direction of a weak applied field, giving positive magnetic susceptibility, and no spontaneous magnetization once the field is removed in the paramagnetic phase. Alignment of permanent atomic or nuclear moments is one mechanism.

A material showing diamagnetism develops an induced moment opposing the applied field, giving negative magnetic susceptibility. Closed electronic shells supply a familiar orbital contribution; the response need not require permanent moments.

A ferromagnet has a phase with a spontaneous ordered magnetization below its Curie temperature, produced by cooperative alignment of moments. Its magnetic domains can cancel in a macroscopic zero-field sample, so a measured zero net moment alone does not show that the material lacks ferromagnetism.

The helium atom has a closed $1s^2$ electronic shell with total electronic spin and orbital angular momentum zero. The nucleus of helium-4 also has spin zero. At low temperature there is no Curie response from permanent atomic or nuclear moments; the leading atomic response is diamagnetism.

Let $n_4$ denote the number of atoms per volume and let $⟨r^2⟩$ be the mean squared radius of one electron. Use the symmetric magnetic vector potential $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. Expanding the electron kinetic Hamiltonian $(\mathbf{p}+e\mathbf{A}{)}^2/(2m_e)$ gives the quadratic field termwhere isotropy gives $⟨x^2+y^2⟩=2⟨r^2⟩/3$. The term linear in orbital angular momentum vanishes in the spherical electronic ground state. ThusFor weak susceptibility, $B\simeq {\mu }_{0}H$, so the Langevin diamagnetic susceptibility isIf the supplied density is a mass density ${\rho }_4$, use $n_4={\rho }_4/(4u)$, with $u$ the atomic mass unit. The result is approximately temperature independent at fixed density. The factor two for the two electrons is already included; $⟨r^2⟩$ has not been defined as their summed radius.

Let ${\mu }_a=2.13{\mu }_N$ be the magnitude of the saturated nuclear moment of helium-3, with ${\mu }_N=e\hslash /(2m_p)$. A spin-one-half nucleus has the two Zeeman energies $\mp {\mu }_{a}B$. For independent nuclei the one-site partition function and magnetization density areIn the weak-field limit ${\mu }_{a}B\ll k_{B}T$, expand the hyperbolic tangent:The nuclear moment's negative sign means that its moment points oppositely to its spin; it does not make the thermal magnetic susceptibility negative. The response is paramagnetism governed by Curie's law.

There is no extra factor $1/3$ here: the stated $2.13{\mu }_N$ is the moment of a fully polarized spin-one-half state. Equivalently, writing the moment operator as $g_N{\mu }_N\mathbf{I}$ requires $g_N=-4.26$, and $g_N^2{\mu }_N^{2}I(I+1)/3=(2.13{\mu }_N{)}^2$. The formula presumes independent localized nuclear moments; exchange correlations invalidate it close to the much lower ordering scale discussed in part (e).

Add the electronic diamagnetism of each helium atom to the nuclear magnetic susceptibility, using the same $n_3$ for both contributions in the helium-3 sample:The density cancels when the two terms balance. The sign-change temperature in the independent-atom approximation is thereforeNo numerical atomic radius was supplied, so the exact numerical value remains conditional on that electronic length scale. Taking $⟨r^2⟩\sim a_0^2$, where $a_0$ is the Bohr radius, gives $T_0\simeq 0.32\mathrm{K}$. A slightly more explicit estimate uses the helium variational effective charge $\zeta =27/16$, for which a hydrogenic $1s$ orbital has $⟨r^2⟩=3a_0^2/{\zeta }^2\simeq 1.05a_0^2$, givingAbove this scale the electronic contribution is negative and dominates; below it the positive nuclear contribution dominates within the independent-moment approximation. The estimate is not a precision prediction of a real interacting helium phase.

Let $s=⟨S_i^z⟩$ be uniform, with dimensionless spin operators $S=1/2$. The factor $1/2$ in the source Hamiltonian removes double counting of pairs. Decoupling ${\mathbf{S}}_i\cdot {\mathbf{S}}_j$ about a uniform polarization gives the one-site mean-field theory HamiltonianA body-centred-cubic site has eight nearest neighbors and six next-nearest neighbors, soFor a spin one half in this effective field, the mean-field self-consistency equation isLinearizing at the continuous onset gives $s=J_{0}s/(4k_{B}T)+O(s^3)$, henceEquivalently, the general spin-$S$ result is $k_B{T}_C^{\mathrm{MF}}=S(S+1)J_0/3$. The antiferromagnetic next-nearest-neighbor exchange reduces the uniform ordering scale rather than contributing its absolute magnitude. The negative nuclear g-factor does not change this exchange instability; aligning the spins also aligns their moments with one another.

For these particular competing couplings, the uniform mode is indeed the largest exchange eigenvalue. Writing $x=\cos (q_{x}a/2)$, $y=\cos (q_{y}a/2)$ and $z=\cos (q_{z}a/2)$ giveswith equality at $\mathbf{q}=0$ modulo reciprocal lattice vectors. To see the bound, replace a negative $xyz$ by its absolute value for an upper bound, let $t=(|xyz|{)}^{1/3}$, and use $x^2+y^2+z^2\ge 3t^2$. The expression is at most $8t^3-6t^2+3$, whose maximum on $0\le t\le 1$ is five. Thus another modulation does not preempt the stated ferromagnetic onset at this level of mean-field theory. Fluctuations can change the actual transition temperature.