Spin-Wave Dispersion and Magnons

Derivation

Spin-Wave Dispersion and Magnons

D-264 Home PU-303 Threads waves · matter · symmetry Depends on Weiss Mean-Field Theory of Ferromagnetism, harmonic-oscillator-ladder-operators, bose-einstein-distribution

Statement

For the nearest-neighbour Heisenberg ferromagnet \(\hat H=-J\sum_{\langle ij\rangle}\hat{\mathbf S}_i\cdot\hat{\mathbf S}_j\) with \(J>0\) and spin \(S\), the Holstein–Primakoff transformation maps low-lying excitations onto free bosons (magnons) with dispersion \(\hbar\omega_{\mathbf k}=2JS z\,[1-\gamma_{\mathbf k}]\), where \(z\) is the coordination number and \(\gamma_{\mathbf k}=\frac{1}{z}\sum_{\boldsymbol\delta}e^{i\mathbf k\cdot\boldsymbol\delta}\). In the long-wavelength limit \(\hbar\omega_{\mathbf k}=D k^2\) with stiffness \(D=2JSa^2\) (simple cubic), and the resulting thermal magnon population reduces the spontaneous magnetisation as \(\Delta M/M_0\propto T^{3/2}\): the Bloch \(T^{3/2}\) law.

Why it matters

Spin waves are the elementary excitations of an ordered magnet, the magnetic analogue of phonons. They set the low-temperature thermodynamics of ferromagnets, and their quantisation into magnons underpins magnonics, spin-transport, and inelastic-neutron-scattering spectroscopy of magnetic materials.

The derivation is also a template for how a strongly interacting quantum many-body system — spins that do not commute and live in a finite Hilbert space — becomes, at low excitation density, a gas of nearly free bosonic quasiparticles. The Bloch law was one of the first quantitative triumphs of that picture and remains the textbook demonstration that broken continuous symmetry produces gapless Goldstone modes.

Assumptions

Ferromagnetic exchange \(J>0\), nearest-neighbour only.Antiferromagnetic or frustrated couplings change the ground state; the HP expansion about a fully polarised state is then illegitimate and one must expand about a Néel or spiral state instead.

Fully aligned ground state \(|0\rangle\) with every site at \(S^z=+S\).If the true ground state is not the fully polarised state (it is exact for the Heisenberg ferromagnet but not the antiferromagnet), the vacuum of the boson operators is wrong and the linear spin-wave spectrum is unphysical.

Low excitation density, \(\langle \hat a_i^\dagger \hat a_i\rangle\ll 2S\).The square-root in the HP map cannot be Taylor-expanded; magnon–magnon interactions and kinematic constraints (hard-core-like corrections) dominate and the free-boson dispersion fails.

Broken spin-rotation symmetry with a continuous order parameter.Without a continuous symmetry there is no Goldstone theorem; the mode acquires a gap and the \(k^2\) gapless dispersion — hence the \(T^{3/2}\) power — is lost.

Dimensionality \(d>2\) for finite-temperature order.In \(d\le2\) the magnon integral \(\int d^dk\,/(e^{\beta D k^2}-1)\) diverges at small \(k\): the Mermin–Wagner theorem forbids long-range order at \(T>0\) and the Bloch law does not apply.

Derivation

\[ \hat H=-J\sum_{\langle ij\rangle}\hat{\mathbf S}_i\cdot\hat{\mathbf S}_j =-J\sum_{\langle ij\rangle}\left[\hat S_i^z\hat S_j^z+\tfrac{1}{2}\left(\hat S_i^{+}\hat S_j^{-}+\hat S_i^{-}\hat S_j^{+}\right)\right] \]

Rewrite the dot product using \(\hat S^\pm=\hat S^x\pm i\hat S^y\), so \(\hat S_i^x\hat S_j^x+\hat S_i^y\hat S_j^y=\tfrac12(\hat S_i^+\hat S_j^-+\hat S_i^-\hat S_j^+)\). Exact operator identity. A

\[ \hat S_i^{z}=S-\hat a_i^\dagger \hat a_i,\qquad \hat S_i^{+}=\sqrt{2S}\,\sqrt{1-\frac{\hat a_i^\dagger \hat a_i}{2S}}\;\hat a_i,\qquad \hat S_i^{-}=\sqrt{2S}\,\hat a_i^\dagger\sqrt{1-\frac{\hat a_i^\dagger \hat a_i}{2S}} \]

Holstein–Primakoff map: bosons \([\hat a_i,\hat a_j^\dagger]=\delta_{ij}\) count spin deviations \(n_i=S-S_i^z\). It reproduces \([\hat S^z,\hat S^\pm]=\pm\hat S^\pm\) and \([\hat S^+,\hat S^-]=2\hat S^z\) exactly; the square root truncates the boson space at \(n_i=2S\). B

\[ \hat S_i^{+}\simeq\sqrt{2S}\,\hat a_i,\qquad \hat S_i^{-}\simeq\sqrt{2S}\,\hat a_i^\dagger,\qquad \hat S_i^{z}=S-\hat a_i^\dagger \hat a_i \]

Linear spin-wave theory: expand \(\sqrt{1-\hat n/2S}=1-\hat n/4S+\cdots\) and keep leading order, valid when \(\langle\hat n\rangle\ll 2S\). Dropped terms are quartic in bosons — the magnon–magnon interaction, treated perturbatively later. B

\[ \hat H\simeq -J\sum_{\langle ij\rangle}\left[(S-\hat a_i^\dagger \hat a_i)(S-\hat a_j^\dagger \hat a_j)+S\left(\hat a_i\hat a_j^\dagger+\hat a_i^\dagger \hat a_j\right)\right] \]

Substitute step 3 into step 1. The transverse term \(\tfrac12(\hat S^+\hat S^-+\hat S^-\hat S^+)\to S(\hat a_i\hat a_j^\dagger+\hat a_i^\dagger\hat a_j)\) hops a spin deviation between neighbours. A

\[ \hat H\simeq -J S^2 N_b + JS\sum_{\langle ij\rangle}\left(\hat a_i^\dagger \hat a_i+\hat a_j^\dagger \hat a_j-\hat a_i^\dagger \hat a_j-\hat a_j^\dagger \hat a_i\right) \]

Keep terms to quadratic order in bosons: \((S-\hat n_i)(S-\hat n_j)\to S^2-S(\hat n_i+\hat n_j)\), discarding the quartic \(\hat n_i\hat n_j\). Here \(N_b\) is the number of bonds; \(-JS^2N_b=E_0\) is the classical ground-state energy. B

\[ \hat H-E_0\simeq JS\sum_{i}\sum_{\boldsymbol\delta}\left(\hat a_i^\dagger \hat a_i-\hat a_i^\dagger \hat a_{i+\boldsymbol\delta}\right) \]

Convert the bond sum to a site sum over the \(z\) nearest-neighbour vectors \(\boldsymbol\delta\); each site contributes \(z\) diagonal terms and \(z\) hopping terms. Translational invariance makes every site equivalent. A

\[ \hat a_i=\frac{1}{\sqrt N}\sum_{\mathbf k}e^{i\mathbf k\cdot\mathbf r_i}\,\hat a_{\mathbf k},\qquad \hat a_i^\dagger=\frac{1}{\sqrt N}\sum_{\mathbf k}e^{-i\mathbf k\cdot\mathbf r_i}\,\hat a_{\mathbf k}^\dagger \]

Fourier transform to momentum space; \([\hat a_{\mathbf k},\hat a_{\mathbf k'}^\dagger]=\delta_{\mathbf k\mathbf k'}\) is preserved by the unitary lattice transform. Diagonalises the translationally invariant quadratic form. B

\[ \hat H-E_0\simeq \sum_{\mathbf k}\hbar\omega_{\mathbf k}\,\hat a_{\mathbf k}^\dagger \hat a_{\mathbf k},\qquad \hbar\omega_{\mathbf k}=JS\sum_{\boldsymbol\delta}\left(1-e^{i\mathbf k\cdot\boldsymbol\delta}\right)=2JSz\,(1-\gamma_{\mathbf k}) \]

Insert step 7; the site sum gives \(\sum_i e^{i(\mathbf k-\mathbf k')\cdot\mathbf r_i}=N\delta_{\mathbf k\mathbf k'}\). Using inversion symmetry of the lattice (\(\boldsymbol\delta\to-\boldsymbol\delta\)) the imaginary parts cancel, leaving \(1-\gamma_{\mathbf k}\) with \(\gamma_{\mathbf k}=\frac1z\sum_{\boldsymbol\delta}\cos(\mathbf k\cdot\boldsymbol\delta)\). Free-boson (magnon) Hamiltonian. B

\[ \gamma_{\mathbf k}=1-\tfrac12\,\overline{(\mathbf k\cdot\boldsymbol\delta)^2}+\cdots \;\Longrightarrow\; \hbar\omega_{\mathbf k}\xrightarrow{k\to0} 2JS\,a^2 k^2\equiv D k^2 \quad(\text{simple cubic},\ z=6) \]

Small-\(k\) expansion of the cosine; the linear term vanishes by inversion symmetry so the mode is gapless and quadratic. For a simple-cubic lattice \(\sum_{\boldsymbol\delta}(\mathbf k\cdot\boldsymbol\delta)^2=2a^2k^2\), giving stiffness \(D=2JSa^2\). This gaplessness is the Goldstone mode of broken spin-rotation symmetry. C

\[ \langle \hat n_{\mathbf k}\rangle=\frac{1}{e^{\hbar\omega_{\mathbf k}/k_BT}-1},\qquad \frac{\Delta M}{g\mu_B}=\frac1V\sum_{\mathbf k}\langle \hat n_{\mathbf k}\rangle=\frac{1}{(2\pi)^3}\int \frac{d^3k}{e^{Dk^2/k_BT}-1} \]

Magnons are bosons carrying \(\Delta S^z=-1\) (\(\Delta n=\sum_{\mathbf k}\hat n_{\mathbf k}\)), so each thermally excited magnon lowers \(M\) by \(g\mu_B\). Use the Bose–Einstein distribution (prior result) and pass \(\frac1N\sum_{\mathbf k}\to\frac{v_0}{(2\pi)^3}\int d^3k\). B

\[ \Delta M=\frac{g\mu_B}{(2\pi)^3}\left(\frac{k_BT}{D}\right)^{3/2}\!\!\int\frac{d^3x}{e^{x^2}-1} =\frac{g\mu_B}{(2\pi)^3}\,4\pi\,\Gamma\!\left(\tfrac32\right)\zeta\!\left(\tfrac32\right)\left(\frac{k_BT}{D}\right)^{3/2} \]

Substitute \(x=k\sqrt{D/k_BT}\) to extract the temperature scaling, then evaluate \(\int_0^\infty \frac{x^2\,dx}{e^{x^2}-1}=\tfrac{\sqrt\pi}{4}\zeta(3/2)\,\Gamma(3/2)\cdot\!\frac{4}{\sqrt\pi}\) via the standard Bose integral \(\int_0^\infty \frac{x^{s-1}dx}{e^{x}-1}=\Gamma(s)\zeta(s)\) with \(s=3/2\). Convergence at \(k\to0\) requires \(d=3\). C

Result

\[ \boxed{\;\hbar\omega_{\mathbf k}=2JSz\,(1-\gamma_{\mathbf k})\xrightarrow{k\to0}Dk^2,\quad D=2JSa^2;\qquad \frac{\Delta M}{M_0}=\frac{\zeta(3/2)}{S}\left(\frac{k_BT}{4\pi D}\right)^{3/2}\propto T^{3/2}\;}\]

Reading. Low-energy excitations of a ferromagnet are gapless, quadratically dispersing magnons — collective spin precessions. Populating them thermally tilts spins away from full alignment, so the spontaneous magnetisation falls from its \(T=0\) value \(M_0=g\mu_B S/v_0\) as \(T^{3/2}\), not exponentially and not linearly. The prefactor is fixed by the spin stiffness \(D\), i.e. by the exchange \(J\).

Units check. \([JS]=\)energy, so \(\hbar\omega_{\mathbf k}=2JSz(1-\gamma_{\mathbf k})\) is an energy (\(\gamma_{\mathbf k}\) dimensionless). \([D]=[JSa^2]=\)energy\(\times\)length\(^2\), so \(Dk^2\) is an energy. In \(\Delta M/M_0\), \(k_BT/(4\pi D)\) has units (energy)/(energy·length\(^2\))\(=\)length\(^{-2}\); raised to \(3/2\) gives length\(^{-3}\), matching \(\zeta(3/2)/S\) being dimensionless times a number density — consistent with \(\Delta M/M_0\) dimensionless after dividing by the site density \(1/v_0\).

Limiting cases

\(k\to0\): \(\hbar\omega_{\mathbf k}\to Dk^2\to0\) — gapless Goldstone mode, the uniform precession (\(\mathbf k=0\)) costs zero energy because a global spin rotation is a symmetry.
\(\mathbf k\) at zone boundary: \(\gamma_{\mathbf k}\to-1\) (bipartite), \(\hbar\omega_{\max}=4JSz/2=2JSz\cdot\!(1-\gamma)\) reaches its band top \(\sim 4JSz\)-scale, of order the exchange energy per bond.
\(S\to\infty\): quantum corrections \(\propto1/S\) vanish; linear spin-wave theory becomes exact and \(\Delta M/M_0\propto1/S\to0\) — the classical limit.
\(T\to0\): \(\Delta M/M_0\to0\) as \(T^{3/2}\); exponentially few magnons if a gap were present, but here power-law because gapless.
Large \(D\) (stiff magnet, big \(J\)): magnons are expensive, \(\Delta M\) suppressed, high Curie temperature.

Breaks when

Two or fewer dimensions. The magnon occupation integral \(\int d^dk/(e^{Dk^2/k_BT}-1)\sim\int k^{d-1}dk/k^2\) diverges at \(k\to0\) for \(d\le2\): infinitely many long-wavelength magnons destroy order at any \(T>0\) (Mermin–Wagner). No spontaneous magnetisation, no Bloch law.
Elevated temperature / near \(T_C\). \(\langle\hat n\rangle\) is no longer \(\ll2S\); the neglected quartic magnon–magnon interactions and the HP square-root corrections matter, the free-boson spectrum breaks down, and \(M(T)\) crosses over to critical behaviour \(\propto(T_C-T)^\beta\) rather than \(T^{3/2}\).
Antiferromagnet / frustrated exchange. The fully polarised state is not the ground state; expanding HP about it gives imaginary frequencies. One must expand about the Néel state and Bogoliubov-diagonalise, yielding linear \(\omega\propto k\) dispersion and a \(T^{3}\) (3D) magnetisation correction instead.
Long-range dipolar or anisotropy terms. These open a gap \(\Delta\) at \(k=0\); for \(T\ll\Delta/k_B\) the population is exponentially suppressed \(\sim e^{-\Delta/k_BT}\), replacing the \(T^{3/2}\) power law.

Failure modes

Sign/normal-ordering slip: writing \(\hat S^z=S+\hat a^\dagger\hat a\) instead of \(S-\hat a^\dagger\hat a\); magnons must remove alignment, so the deviation number carries a minus sign.
Keeping the quartic term in linear theory: retaining \(\hat n_i\hat n_j\) at "quadratic order" — it is quartic in bosons and belongs to the interaction, not the free spectrum.
Dropping the diagonal hopping: forgetting the \(+\hat a_i^\dagger\hat a_i\) piece from \(\hat S_i^z\hat S_j^z\) so that \(\omega_{\mathbf k}\propto-\gamma_{\mathbf k}\) alone — this gives a spurious gap at \(k=0\) and violates Goldstone.
Using \(T^{3/2}\) for an antiferromagnet whose magnons disperse linearly (\(T^3\) law) — confusing the two universality classes.
Treating magnons as fermions: applying Fermi–Dirac; magnons are bosons (\(\Delta S^z=\pm1\) integer), Bose–Einstein is mandatory.
Forgetting the \(1/z\) in \(\gamma_{\mathbf k}\) or miscounting neighbours, which mis-scales the stiffness \(D\).

Discussion

The physical content of the Holstein–Primakoff map is that a spin deviation — one unit of \(S^z\) lost — behaves like a boson. Because spins on different sites commute, these deviations can pile up independently, and at low density they scarcely notice each other: the magnet becomes a dilute Bose gas. The dispersion \(\hbar\omega_{\mathbf k}\propto1-\gamma_{\mathbf k}\) is exactly the lattice structure factor of the exchange, so measuring \(\omega_{\mathbf k}\) by inelastic neutron scattering directly maps out \(J\) and the range of the interaction.

The gapless quadratic form is not an accident. Rotating every spin uniformly is a symmetry of \(\hat H\), so the \(\mathbf k=0\) mode costs nothing — this is Goldstone's theorem in action. The quadratic (rather than linear) dispersion is special to the ferromagnet: because the order parameter (total \(S^z\)) is itself a conserved quantity, the two would-be Goldstone modes combine into a single mode with \(\omega\propto k^2\), unlike the antiferromagnet where \(\omega\propto k\). This quadratic law is precisely what feeds the density of states \(g(\omega)\propto\omega^{1/2}\) and produces the \(T^{3/2}\) exponent (contrast \(T^3\) for linear magnons or Debye phonons in 3D).

Beyond linear theory, the \(1/S\) expansion generates magnon–magnon interactions from both the quartic \(\hat n_i\hat n_j\) term and the HP square-root corrections. Dyson showed that for the ideal Heisenberg ferromagnet these produce no correction to the leading \(T^{3/2}\) term; the first interaction correction enters at order \(T^{5/2}\) (kinematic and dynamical parts partly cancel — the celebrated Dyson result). The full low-temperature series is \(\Delta M/M_0=a_{3/2}T^{3/2}+a_{5/2}T^{5/2}+\cdots\), and modern spin-wave theory recovers it systematically via a Dyson–Maleev or careful Holstein–Primakoff bookkeeping. That the awkward, non-polynomial square-root operator nonetheless yields a controlled asymptotic expansion is one of the subtler points of the subject.

Common misconceptions. (i) The Bloch law is not mean-field: Weiss theory predicts \(\Delta M/M_0\propto e^{-\Delta/k_BT}\) or a \(T^2\) form and misses the true \(T^{3/2}\), because it ignores the collective low-energy magnons. (ii) A magnon is not "one flipped spin" localised on a site; it is a coherent superposition — a wave of small precession spread over the whole lattice, carrying a single quantum \(\Delta S^z=-1\) collectively. (iii) The \(T^{3/2}\) law describes the approach to saturation at low \(T\), not the behaviour near \(T_C\), where critical fluctuations take over.

Worked examples

Magnon energy at a given wavevector in EuO.

EuO: simple-cubic-like fcc Eu sublattice, take \(S=7/2\), \(J/k_B\approx0.6\ \text{K}\), lattice constant \(a=5.14\ \text{Å}\). Find \(\hbar\omega\) at \(k=\tfrac{\pi}{2a}\) along a cubic axis using the small-\(k\) form. B

\[ D=2JSa^2 \]

Use the simple-cubic stiffness as an estimate. A

\[ D=2\,(0.6\,k_B)\,(3.5)\,(5.14\times10^{-10}\,\text{m})^2 \]

Insert numbers; \(J=0.6\,k_B=0.6\times1.381\times10^{-23}\,\text{J}=8.29\times10^{-24}\,\text{J}\). A

\[ D=2(8.29\times10^{-24})(3.5)(2.64\times10^{-19})=1.53\times10^{-41}\ \text{J·m}^2 \]

Arithmetic: \(a^2=2.64\times10^{-19}\,\text{m}^2\). A

\[ k=\frac{\pi}{2a}=\frac{\pi}{2(5.14\times10^{-10})}=3.06\times10^{9}\ \text{m}^{-1},\quad k^2=9.34\times10^{18}\ \text{m}^{-2} \]

Compute wavevector magnitude. A

\[ \hbar\omega=Dk^2=1.53\times10^{-41}\times9.34\times10^{18}=1.43\times10^{-22}\ \text{J} \]

Multiply; convert: \(1.43\times10^{-22}/1.381\times10^{-23}=10.4\ \text{K}\), or \(0.89\ \text{meV}\). A

\[ \hbar\omega\big|_{k=\pi/2a}\approx1.4\times10^{-22}\ \text{J}\approx0.9\ \text{meV}\approx10\ \text{K} \]

Reading. A meV-scale magnon — squarely in the range probed by cold-neutron spectrometers. The quadratic estimate slightly overshoots the true zone-interior value, where \(1-\gamma_{\mathbf k}\) bends below \(k^2\).

Units check. J·m\(^2\times\)m\(^{-2}=\)J. Good.

Fractional loss of magnetisation from the Bloch law.

Using the EuO stiffness from Example 1, estimate \(\Delta M/M_0\) at \(T=5\ \text{K}\) with \(S=7/2\). B

\[ \frac{\Delta M}{M_0}=\frac{\zeta(3/2)}{S}\left(\frac{k_BT}{4\pi D}\right)^{3/2},\qquad \zeta(3/2)=2.612 \]

Boxed result; per-spin normalisation \(M_0=g\mu_BS/v_0\), \(v_0=a^3\) folded in via the density of states used in step 11. B

\[ k_BT=1.381\times10^{-23}\times5=6.91\times10^{-23}\ \text{J} \]

Thermal energy at 5 K. A

\[ \frac{k_BT}{4\pi D}=\frac{6.91\times10^{-23}}{4\pi(1.53\times10^{-41})}=3.59\times10^{17}\ \text{m}^{-2} \]

Denominator \(4\pi D=1.92\times10^{-40}\). A

\[ \left(3.59\times10^{17}\right)^{3/2}=2.15\times10^{26}\ \text{m}^{-3} \]

Raise to 3/2: \((3.59\times10^{17})^{1/2}=5.99\times10^{8}\), times \(3.59\times10^{17}\). This is a magnon number density; multiply by \(v_0=a^3\) to get deviations per site. B

\[ v_0=a^3=(5.14\times10^{-10})^3=1.36\times10^{-28}\ \text{m}^3,\quad n_{\text{mag}}v_0=2.15\times10^{26}\times1.36\times10^{-28}=0.0292 \]

Deviations per site \(=\zeta(3/2)\,(n\,v_0\,\text{factor})\); combine: \(\frac{\Delta M}{M_0}=\frac{2.612}{3.5}\times0.0292\). B

\[ \frac{\Delta M}{M_0}\approx\frac{2.612}{3.5}\times0.029\approx0.022\ \ (\approx2\%) \]

Reading. At 5 K, roughly 2% of the saturation magnetisation is lost to thermal magnons — consistent with EuO's \(T_C\approx69\ \text{K}\), so 5 K is deep in the ordered, Bloch-law regime. Doubling \(T\) would raise the loss by \(2^{3/2}\approx2.83\).

Units check. \((\text{m}^{-2})^{3/2}=\text{m}^{-3}\), times \(\text{m}^3\) gives a pure number; \(\zeta(3/2)/S\) dimensionless. \(\Delta M/M_0\) dimensionless. Good.

Problems

Show explicitly that the linearised HP operators \(\hat S^+=\sqrt{2S}\,\hat a\), \(\hat S^-=\sqrt{2S}\,\hat a^\dagger\), \(\hat S^z=S-\hat a^\dagger\hat a\) reproduce \([\hat S^+,\hat S^-]=2\hat S^z\) only up to \(O(1/S)\) corrections, and identify the neglected term.
Solution
\([\hat S^+,\hat S^-]=2S[\hat a,\hat a^\dagger]=2S\). The exact requirement is \(2\hat S^z=2S-2\hat a^\dagger\hat a\). So the linear map gives \(2S\) instead of \(2S-2\hat a^\dagger\hat a\); the missing \(-2\hat a^\dagger\hat a\) is exactly the term restored by the square-root factor \(\sqrt{1-\hat n/2S}\). It is negligible when \(\langle\hat n\rangle\ll S\), i.e. relative error \(O(\langle\hat n\rangle/S)\). The full HP operators satisfy the algebra exactly.
For a body-centred-cubic lattice (\(z=8\), neighbours at \((\pm1,\pm1,\pm1)a/2\)), compute \(\gamma_{\mathbf k}\) and the small-\(k\) stiffness \(D\).
Solution
\(\gamma_{\mathbf k}=\frac18\sum_{\pm\pm\pm}e^{i(\pm k_x\pm k_y\pm k_z)a/2}=\cos\frac{k_xa}{2}\cos\frac{k_ya}{2}\cos\frac{k_za}{2}\). Small \(k\): \(\gamma_{\mathbf k}\approx1-\frac{a^2}{8}(k_x^2+k_y^2+k_z^2)=1-\frac{a^2k^2}{8}\). Then \(\hbar\omega=2JSz(1-\gamma)=2JS\cdot8\cdot\frac{a^2k^2}{8}=2JSa^2k^2\), so \(D=2JSa^2\) — same functional form as simple cubic, but with the bcc \(a\) and \(J\).
The magnon specific heat. From \(U=\sum_{\mathbf k}\hbar\omega_{\mathbf k}\langle\hat n_{\mathbf k}\rangle\) with \(\hbar\omega=Dk^2\), show \(C_V\propto T^{3/2}\) and find the exponent's origin.
Solution
\(U=\frac{V}{(2\pi)^3}\int d^3k\,\frac{Dk^2}{e^{Dk^2/k_BT}-1}\). Sub \(x=k\sqrt{D/k_BT}\): \(U\propto D\left(\frac{k_BT}{D}\right)^{5/2}\int\frac{x^4dx}{e^{x^2}-1}=A\,T^{5/2}\). Then \(C_V=\partial U/\partial T=\frac52 A\,T^{3/2}\propto T^{3/2}\). The exponent comes from the density of states \(g(\omega)\propto\omega^{1/2}\) of a 3D quadratic dispersion: \(g(\omega)d\omega\sim k^2dk\sim\omega^{1/2}d\omega\). (Contrast phonons: \(\omega\propto k\) gives \(g\propto\omega^2\) and \(C_V\propto T^3\).)
Estimate the spin-wave contribution to \(\Delta M/M_0\) at \(T=10\ \text{K}\) for a hypothetical simple-cubic ferromagnet with \(S=1\), \(J/k_B=10\ \text{K}\), \(a=3\ \text{Å}\).
Solution
\(D=2JSa^2=2(10\,k_B)(1)(3\times10^{-10})^2=2\times1.381\times10^{-22}\times9\times10^{-20}=2.49\times10^{-41}\,\text{J·m}^2\). \(k_BT=1.381\times10^{-22}\,\text{J}\). \(\frac{k_BT}{4\pi D}=\frac{1.381\times10^{-22}}{3.13\times10^{-40}}=4.41\times10^{17}\,\text{m}^{-2}\). \((4.41\times10^{17})^{3/2}=2.93\times10^{26}\,\text{m}^{-3}\). Times \(a^3=2.7\times10^{-29}\,\text{m}^3\): \(=7.9\times10^{-3}\). Then \(\frac{\Delta M}{M_0}=\frac{\zeta(3/2)}{S}\times7.9\times10^{-3}=2.612\times7.9\times10^{-3}\approx0.021\), about 2%.
Explain quantitatively why the Bloch \(T^{3/2}\) law fails in two dimensions by examining the convergence of \(\Delta M=\int d^2k/(e^{Dk^2/k_BT}-1)\) at small \(k\).
Solution
In 2D, \(\Delta M\propto\int_0^\Lambda \frac{2\pi k\,dk}{e^{Dk^2/k_BT}-1}\). As \(k\to0\), \(e^{Dk^2/k_BT}-1\approx Dk^2/k_BT\), so the integrand \(\sim\frac{2\pi k}{Dk^2/k_BT}=\frac{2\pi k_BT}{Dk}\), giving \(\int\frac{dk}{k}\), a logarithmic divergence at the lower limit. The number of thermally excited magnons is infinite for any \(T>0\), so no finite magnetisation survives — long-range order is destroyed. This is the Mermin–Wagner theorem: a continuous symmetry cannot be spontaneously broken at \(T>0\) in \(d\le2\) with short-range interactions. In 3D the extra factor of \(k\) in \(d^3k=4\pi k^2dk\) renders \(\int k^2dk/k^2=\int dk\) convergent, rescuing the \(T^{3/2}\) law.