advanced32 min studyLesson 524 in Physics

Ferromagnetic spin-wave dispersion

Derive the single-magnon band with explicit bond counting and separate exact one-particle results from dilute spin-wave theory.

Before you begin

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The question for today

What you’ll understand

Obtain a magnon band from a translated single-spin deviation.
Identify its long-wavelength stiffness and field gap.
Count branches and state the limits of a noninteracting magnon description.

A spin wave is a collective change in orientation spread across many sites, and its quantum is a magnon. In an isotropic ferromagnet the fully polarized state makes the one-magnon problem unusually simple: a local spin deviation hops between neighbors without creating additional deviations. Translation symmetry then labels its exact eigenstates by wavevector. The resulting band explains why long twists cost little energy, while also showing where a single-particle picture ceases to describe the full many-spin system.

Start from the correct polarized vacuum

Write a nearest-neighbor chain as $H=-J_F\sum_i\mathbf s_i\cdot\mathbf s_{i+1}+b\sum_i s_i^z$ , with $J_F>0$ and $b=g\mu_BB\geq0$ . Every unordered bond appears once. This is the module convention J negative, rewritten with a positive ferromagnetic magnitude. In a positive field the electron-like magnetic moments point along positive z, so the angular-momentum vacuum has all projections minus S. At zero field any globally rotated fully polarized state has the same exchange energy.

One application of a spin raising operator creates a local deviation from that vacuum. It increases physical angular momentum along z by ħ and decreases the field-directed magnetic moment by $g\mu_B$ . A coherent sum of such local deviations with lattice phases is a one-magnon momentum state. Because isotropic exchange conserves total z projection, the subspace with exactly one raised projection is closed. Its diagonalization does not require an uncontrolled replacement of the full many-body Hilbert space by independent particles.

|k\rangle=\frac1{\sqrt{2SN}}\sum_j e^{ikja}s_j^+|F\rangle,\qquad s_j^z|F\rangle=-S|F\rangle,\qquad E(k)=b+2J_FS[1-\cos(ka)]

N is the number of sites of a periodic chain, a is lattice spacing, and E is excitation energy above the polarized ground state. k takes allowed reciprocal wavevectors. The band is exact in the one-magnon sector of this nearest-neighbor Hamiltonian.↔ Scroll wide equations horizontally.

A lattice difference becomes a quadratic dispersion

A deviation changes the exchange energy on its two neighboring bonds and can move through transverse exchange. The resulting discrete difference gives a cosine band. For one spin per primitive cell there is one positive-frequency magnon branch. The two transverse components of a precessing ferromagnetic orientation form a conjugate pair; counting each component as an independent positive-frequency branch would double the states. Lattices with several magnetic sites per primitive cell can instead support additional bands.

Expanding the cosine near zero wavevector yields a quadratic energy above the Zeeman gap. The stiffness D has units energy times length squared and should not be confused with a velocity. Angular frequency is E divided by ħ, so group velocity contains an additional inverse ħ. At zero field the zero-wavevector mode corresponds to changing the direction of the total spin and costs no exchange energy. Anisotropy and dipolar interactions can modify this low-energy behavior and are absent from the present derivation.

E(k)=b+Dk^2+O(k^4a^4J_FS),\qquad D=J_FSa^2,\qquad v_g=\frac1\hbar\frac{dE}{dk}=\frac{2J_FSa}{\hbar}\sin(ka)

The expansion requires |ka| much smaller than one. D is the energy-dispersion stiffness for this one-dimensional nearest-neighbor model, and vg has velocity units. The exact lattice expression remains periodic in k.↔ Scroll wide equations horizontally.

One exact magnon does not make all magnons free

A Holstein–Primakoff representation around the negative-z vacuum has $s_i^z=-S+a_i^\dagger a_i$ . Keeping only quadratic boson terms reproduces the one-magnon band and permits multiple independent occupations. Higher-order terms encode interactions and the finite local spin Hilbert space. The quadratic truncation is justified for a dilute density of spin deviations compared with S, not for arbitrary temperatures or a magnetization reduced nearly to zero.

Using a chain makes the algebra transparent, but the infinite short-range isotropic chain has no spontaneous ferromagnetic order at positive temperature. The band remains a valid zero-temperature one-magnon statement and can describe suitable finite-size excitations. In three-dimensional ordered ferromagnets, dilute quadratic magnons lead to the familiar low-temperature three-halves magnetization correction under additional bulk and gapless assumptions. That thermal power law should not be carried over unchanged to a one-dimensional chain or to a gapped anisotropic magnet.

Work it through

Compare the cosine band with its small-k limit

For a zero-field chain take JF=5 meV, S=1, and a=0.30 nm. Find the energy at ka=π/3, the stiffness D, and the small-k estimate at ka=0.10. Compare the last estimate with the cosine expression.

At $ka=\pi/3$ , the cosine is one half. Thus $E=2(5)(1)(1-1/2)=5\,\mathrm{meV}$ . This point is evaluated with the full lattice dispersion, not a small-k expansion.
The stiffness is $D=5(1)(0.30)^2=0.45\,\mathrm{meV\,nm^2}$ . Its length-squared factor is essential when k is expressed in inverse nanometres.
At ka=0.10 the approximation gives $E\simeq J_FS(ka)^2=0.0500\,\mathrm{meV}$ . Using cos(0.10)≈0.995004 gives 0.049958 meV, so the quadratic estimate is slightly high and already quite accurate.

The result

The lattice energy is 5 meV at π/3; D=0.45 meV nm², and the long-wavelength estimate is 0.0500 meV versus approximately 0.049958 meV exactly.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Inspect an energy proportional to wavevector squared

For each independent comparison, Reset and set BOTH Coefficient a=0.45 and Position x. Compare x=0.50 and x=1.00. Predict y=0.1125 and y=0.45, with tangent slopes 0.45 and 0.90. Interpret x as a dimensionless wavevector and y as a rescaled low-k energy. This y=ax² graph has no lattice periodicity, Zeeman gap, quantum occupation, or magnon interactions.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA quadratic dispersion has a wavevector-dependent slope and a fourfold energy increase when a small wavevector doubles.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The transverse components are conjugate coordinates of a single ferromagnetic mode. Additional magnetic sites in the primitive cell can create extra branches, but they are absent here.

Your turn

Try it yourself

A zero-field spin-half chain has one-magnon zone-edge energy E(π/a)=12 meV and lattice spacing a=0.25 nm. Determine JF and D, then predict the small-k energy at k=0.40 nm−1. State the relevant expansion parameter.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

At the zone edge, one minus the cosine equals two. Infer JF from the exact band before using the long-wavelength expansion.

Compare with the solution

The edge energy is $4J_FS=2J_F$ , giving $J_F=6\,\mathrm{meV}$ . Then $D=6(1/2)(0.25)^2=0.1875\,\mathrm{meV\,nm^2}$ . At the specified k, the quadratic energy is $Dk^2=0.1875(0.40)^2=0.0300\,\mathrm{meV}$ . The dimensionless expansion parameter is $ka=0.40(0.25)=0.10$ , so the small-k estimate is justified. This calculation characterizes the zero-temperature excitation band; it does not assert a nonzero thermal ordering temperature for an isolated chain.

Keep the idea

Take it with you

The exact one-magnon cosine band follows from a closed single-deviation subspace.
At small k the zero-field energy is Dk², with D carrying energy times length squared.
Multiple magnons interact beyond the dilute quadratic approximation, and dimensionality constrains thermal order.

Read further

Original sources and deeper explanations to follow at your own pace.

Kusminskiy: Quantum Magnetism, Spin Waves, and Light, §4.1 and §§4.3–4.4, PDF pp. 48–52 and 56–59 — exact one-magnon states and the spin-wave truncation

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