Spin Waves and Magnons

At zero temperature a Heisenberg ferromagnet has all its spins aligned. The cheapest way to excite it is not to reverse one spin completely — that would cost a full exchange energy $\sim 2JSz$ against all $z$ neighbors — but to tip every spin by a tiny angle and let the tilt precess with a phase that advances from site to site. This collective mode is a spin wave; its quantum is a magnon. Spin waves are to magnetic order what phonons are to the crystal lattice: the low-energy, long-wavelength excitations that carry away order as the temperature rises.

The classical spin wave

Take a one-dimensional chain of spins ${\vec{S}}_n$ with nearest-neighbor ferromagnetic coupling, described by the Heisenberg Hamiltonian $\mathcal{H}=-2J{\sum }_n{\vec{S}}_n\cdot {\vec{S}}_{n+1}$ with $J>0$. Treating each spin as a classical vector, its motion is the torque from the exchange field of its neighbors,

In the ground state every ${\vec{S}}_n=S\hat{z}$ and the right side vanishes. For a small-amplitude wave, write $S_n^z\approx S$ and let the transverse components $S_n^x,S_n^y$ be small. Linearizing and looking for a travelling-wave solution $S_n^{x,y}\propto e^{i(kna-\omega t)}$ gives the spin-wave dispersion

Each spin precesses on a cone about $\hat{z}$; the phase of the precession advances by $ka$ from one site to the next, so the tips trace a helix frozen in a snapshot and rotating in time.

At long wavelength ($ka\ll 1$) the dispersion becomes quadratic,

with $D=2JS{a}^2$ the spin-wave stiffness. The quadratic law $\omega \propto k^2$ is the signature of a ferromagnetic spin wave and contrasts sharply with the linear $\omega \propto k$ of an acoustic phonon. Its origin is the conservation of total spin: a uniform tilt ($k=0$) is a rotation of the whole magnetization and costs no energy, so ${\omega }_k\to 0$ as $k\to 0$, and because the ground state already extremizes the energy the leading correction is second order in $k$.

Magnons as quantized excitations

Quantizing the transverse precession turns the classical spin wave into a set of independent harmonic oscillators, one per wavevector $k$, exactly as for phonons. Their quanta are magnons. A single magnon of wavevector $k$:

• carries energy $\hslash {\omega }_k$, given by the dispersion above;
• lowers the total spin $S_{\text{tot}}^z$ of the crystal by exactly one unit of $\hslash$ — the tilt is shared over all $N$ spins, each reduced by $1/N$, but the total reduction is one full quantum;
• is a boson: any number of magnons can occupy the same mode, and their thermal occupation follows the Bose–Einstein distribution

That one magnon reduces $S_{\text{tot}}^z$ by a single unit — rather than by the $2S$ of a fully reversed spin — is why the spin wave is the cheap excitation. It spreads the cost of one unit of demagnetization over the whole crystal.

The Bloch law

The number of magnons excited at temperature $T$ fixes how much the magnetization has fallen from its zero-temperature value. Each magnon removes one unit of spin, so the fractional magnetization deficit is

Converting the sum to an integral over the Brillouin zone and using the long-wavelength dispersion $\hslash {\omega }_k=D{k}^2$ (only the low-energy modes are populated at low $T$),

The three-dimensional phase space ${\mathbf{d}}^{3}k\propto k^2\mathbf{d}k$ combined with the quadratic dispersion produces the exponent $\frac{3}{2}$. This is the Bloch $T^{3/2}$ law:

with $T_0$ set by $D$. It describes the low-temperature magnetization of iron, cobalt, and nickel far more accurately than the mean-field Brillouin curve, which predicts an exponentially small deficit and misses the data badly. Mean-field theory treats each spin in an average field and knows nothing of the soft collective modes; the Bloch law counts exactly those modes.

The same magnon gas carries a heat capacity. The internal energy is $U={\sum }_k\hslash {\omega }_k⟨n_k⟩$, and the identical integral gives $U\propto T^{5/2}$, so the magnon contribution to the specific heat is

In a magnetic insulator at low temperature this $T^{3/2}$ term adds to the phonon $T^3$ term (from the Debye model), and their different exponents let a plot of $C$ versus $T$ separate the two.

Antiferromagnetic magnons

An antiferromagnet also has spin-wave excitations, but their dispersion differs fundamentally. The two-sublattice ground state is not an eigenstate of the Heisenberg Hamiltonian (unlike the ferromagnet), and the linearized equations couple the precessions of the two sublattices. The result is a linear long-wavelength dispersion,

with $c$ a spin-wave velocity — the same form as an acoustic phonon or a photon, not the quadratic ferromagnetic law. The linear dispersion changes the thermodynamics: the low-temperature magnon heat capacity of an antiferromagnet goes as $T^3$, matching the phonon exponent rather than the ferromagnetic $T^{3/2}$.

Measuring magnons

Magnon dispersion curves are measured by inelastic neutron scattering, the same technique used for phonons. A neutron carries both a magnetic moment, which couples to the electron spins, and a wavelength comparable to the lattice spacing. When it creates or absorbs a magnon of wavevector $\vec{k}$ and energy $\hslash {\omega }_k$, its own energy and momentum change to conserve both,

where $\vec{G}$ is a reciprocal-lattice vector. Scanning the scattered neutron energy at fixed momentum transfer maps out ${\omega }_k$ directly and confirms the $k^2$ ferromagnetic and $k$ antiferromagnetic laws, together with the stiffness $D$ that the Bloch law needs. This closes the magnetism module: the exchange interaction sets the ordered ground state, mean-field theory gives the transition and the sublattice structure, domains and hysteresis govern the macroscopic magnet, and magnons are the excitations that erode the order as temperature rises.