Spin waves

Spin waves (magnons) are propagating disturbances of an ordered magnetic lattice

The magnetic ordering arises due to the exchange interactions $J_{ij}$ between electrons on atomic sites $i$ and $j$.

The dispersion relation $\omega(\mathbf{q})$ of spin waves can be measured by inelastic neutron scattering, and directly depends on the exchange interactions

Spin waves are deviations from an ordered state, so in principle can only be seen in the ordered state

Linear Spin Wave Theory

We'll now derive the spin wave dispersion for a ferromagnet in linear spin wave theory

The steps involved in the derivation are:

1. Beginning with a spin Hamiltonian, express the spin operator vector $\mathbf{S}_i = (\hat{S}_i^x, \hat{S}_i^y, \hat{S}_i^z)$ as the ladder operators $(\hat{S}_i^+, \hat{S}_i^-, \hat{S}_i^z)$

2. Map the raising (lowering) operators to bosonic anihilation (creation) operators $a^{\dagger}$ ($a$) via the Holstein-Primakoff transformation.

3. The transformed Hamiltonian is a series expansion in powers of $a^{\dagger}, a$, but in linear spin wave theory we take only the terms which are linear in $a^{\dagger}a$.

4. Fourier transform the Hamiltonian.

5. Diagonalise the Hamiltonian, ensuring that the commutation relation of the bosonic operators are respected (for two-sublattice system such as antiferromagnetic, the Bogoliubov transformation can be used for this purpose).

Heisenberg Hamiltonian:

The Hamiltonian of an ordered magnet can be expressed in terms of the spin operators for site $i$, $\mathbf{S}_i$.

For example, the Heisenberg Hamiltonian is:

$$ \mathcal{H} = \sum_{mi, nj} J_{mi, nj} \mathbf{S}_i\cdot\mathbf{S}_j $$

where $J_{mi, nj}$ is the exchange interaction with the indices $i$ and $j$ running over magnetic atoms within a single magnetic unit cell, and $m$ and $n$ label different unit cells.

$\mathbf{S}_i=(\hat{S}_i^x, \hat{S}_i^y, \hat{S}_i^z)$ is the spin operator for site $i$, which can be re-expressed in terms of the ladder operators $\hat{S}_i^+, \hat{S}_i^-, \hat{S}_i^z$:

$$ \begin{align} \mathcal{H} &= \sum_{mi, nj} J_{mi, nj} (\hat{S}_i^x\hat{S}_j^x + \hat{S}_i^y\hat{S}_j^y + \hat{S}_i^z\hat{S}_j^z) \\ &= \sum_{mi, nj} J_{mi, nj} ((\hat{S}_i^+\hat{S}_j^- + \hat{S}_i^-\hat{S}_j^+)/4 + \hat{S}_i^z\hat{S}_j^z) \end{align} $$

since $\hat{S}^{\pm} = \hat{J}^x \pm i\hat{S}^y$ so $\hat{S}^{x,y} = \frac{\sqrt{\pm 1}}{2}(\hat{S}^+ \pm \hat{S}^-)$

Holstein-Primakoff transformation

To determine the magnons (magnetic normal modes), we map the excitations to a simple harmonic operator

The ordered ground state is the state with the maximum azimuthal quantum number $|m = S\rangle$, and the action of the ladder operator is:

$$ \hat{S}^{\pm}|m\rangle = \sqrt{(S\mp m)(S+1\pm m)} |m\pm 1\rangle $$

We now map the ladder operators to boson creation and anihilation operators $a^{\dagger}$ and $a$:

$$ a|n\rangle \sqrt{n} |n-1\rangle \qquad a^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle $$

mapping also the ordered state $|m=S\rangle$ to the vacuum state $|n=0\rangle$, such that the action of $a^{\dagger}|n\rangle$ is analogous to $\hat{S}^-|S\rangle$

So, substituting $m = S - n$ and re-arranging the ladder operator terms:

$$ \begin{align} \hat{S}^+|n\rangle &= \sqrt{(2Sn) (1 - \color{red}{\frac{n-1}{2S}})} |n-1\rangle \\ \hat{S}^-|n\rangle &= \sqrt{2S(n+1) (1 - \color{red}{\frac{n-1}{2S}})} |n+1\rangle \end{align} $$

In linear spin wave theory we ignore the terms in red

Using the identities $n = aa^{\dagger}$ and $S^{z}|m\rangle = m|m\rangle$ we have the following mappings

$$ \begin{align} \hat{S}_i^+ &= a_i\sqrt{2S_i} \\ \hat{S}_i^- &= a_i^{\dagger}\sqrt{2S_i} \\ \hat{S}_i^z &= S_i - aa^{\dagger} \end{align} $$

So the Heisenberg Hamiltonian (ignoring terms which are not linear in $a a^{\dagger}$) is now:

$$ \mathcal{H} = \sum_{mi, nj} J_{mi, nj} \left[ \frac{\sqrt{S_i S_j}}{2} \left( a_i a_j^{\dagger} + a_i^{\dagger}a_j \right) - S_i a_ia_i^{\dagger} - S_j a_ja_j^{\dagger} \right] $$

Fourier transformation

In the expression for the spin Hamiltonian we have a sum over all magnetic sites. This can be simplified to just sites $i$ and $j$ within the magnetic unit cell by Fourier transforming the Hamiltonian. The Fourier transform of the Holstein-Primakoff operator is:

$$ a^{(\dagger)} = \frac{1}{\sqrt{N}} \sum_{\mathbf{q}} \exp (-i\mathbf{q}\cdot\mathbf{r}) b_{\mathbf{q}}^{(\dagger)} $$

Let us take one term from the Hamiltonian, $\sum_{mi, nj} J_{mi, nj} a_i a_j^{\dagger}$; its Fourier transform is:

$$ \mathrm{FT}\left[ \sum_{mi, nj} J_{mi, nj} a_i a_j^{\dagger} \right] = \frac{1}{N} \sum_{mi, nj} \sum_{\mathbf{q}, \mathbf{q}'} J_{mi, nj} \exp(-i\mathbf{q}\cdot\mathbf{r}_m) \exp(-i\mathbf{q}'\cdot\mathbf{r}_n) b_{i,\mathbf{q}} b_{j,\mathbf{q}'}^{\dagger} $$

The periodicity of the crystal implies that the exchange interaction is translationally invariant such that $J_{mi, nj} = J_{ij}(\mathbf{d})$ where $\mathbf{d} = \mathbf{r}_m - \mathbf{r}_n$ so the Fourier transform becomes:

$$ \frac{1}{N} \sum_{\mathbf{q}, \mathbf{q}'} \left[ \sum_{ij} J_{ij}(\mathbf{d}) \exp(-i\mathbf{q}'\cdot\mathbf{d}) \right] \exp(-i(\mathbf{q}-\mathbf{q}')\cdot\mathbf{r}_{m}) b_{i,\mathbf{q}} b_{j,\mathbf{q}'}^{\dagger} $$

where the terms in the square brackets is the Fourier transform of the exchange interactions $J_{ij}(\mathbf{q})$ where the indices $ij$ label only sites within the magnetic unit cell. We now use the identity $(1/N) \sum_{\mathbf{q}, \mathbf{q}'} \exp (-i(\mathbf{q}-\mathbf{q}')\cdot\mathbf{r}) = \delta_{\mathbf{q}\mathbf{q}'}$ to obtain:

$$ \mathrm{FT}\left[ \sum_{mi, nj} J_{mi, nj} a_i a_j^{\dagger} \right] = \sum_{ij} J_{ij}(\mathbf{q}) b_{i,\mathbf{q}} b_{j,\mathbf{q}}^{\dagger} $$

So the Fourier transformed Hamiltonian becomes:

$$ \mathcal{H}_{\mathbf{q}} = \sum_{ij} J_{ij}(\mathbf{q}) \left[ \frac{\sqrt{S_i S_j}}{2} \left( b_{i,\mathbf{q}}b_{j,\mathbf{q}}^\dagger + b_{i,\mathbf{q}}^\dagger b_{j,\mathbf{q}} \right) - S_i b_{i,\mathbf{q}}b_{i,\mathbf{q}}^\dagger - S_j b_{j,\mathbf{q}}^\dagger b_{j,\mathbf{q}} \right] $$

Which can be expressed as a matrix equation:

$$ \mathcal{H}_{\mathbf{q}} = \left( \begin{array}{cccccc} b_1^{\dagger} & \dots & b_N^{\dagger} & b_1 & \dots & b_N \end{array} \right) \left( \begin{array}{cc} \left[\frac{\sqrt{S_i S_j}}{2} J_{ij}(\mathbf{q}) - \delta_{ij} S_i J_{ij}(\mathbf{q}=0) \right] & 0 \\ 0 & \left[\frac{\sqrt{S_i S_j}}{2} J_{ij}(\mathbf{q}) - \delta_{ij} S_i J_{ij}(\mathbf{q}=0) \right] \end{array} \right) \left( \begin{array}{c} b_1 \\ \vdots \\ b_N \\ b_1^{\dagger} \\ \vdots \\ b_N^{\dagger} \end{array} \right) $$

where the square brackets denote a block matrix and we have omitted the $\mathbf{q}$ indices on the bosonic operators.

The off-diagonal blocks are only zero in this case because we are using the Heisenberg Hamiltonian. For anisotropic Hamiltonians (e.g. $XXZ$ or Dzyaloshinskii-Moriya interactions), these blocks will not be zero

Setting:

$$ \mathbf{b} = \left( \begin{array}{c} b_1 \\ \vdots \\ b_N \\ b_1^{\dagger} \\ \vdots \\ b_N^{\dagger} \end{array} \right) $$

The Hamiltonian is:

$$ \mathcal{H}_{\mathbf{q}} = \mathbf{b}^{\top} h_{\mathbf{q}} \mathbf{b} $$

With the matrix $h_{\mathbf{q}}$ as shown previously and $\top$ indicates a complex conjugate transpose.

At each $\mathbf{q}$ the eigenvalues of the matrix $h_{\mathbf{q}}$ will be the magnon energies $\omega(\mathbf{q})$ and the eigenvalues can be used to calculate the spin-spin correlation function and hence the neutron scattering cross-section.

Heisenberg Ferromagnet

For a ferromagnet, there is only one magnetic atom in the unit cell, so the Hamiltonian is diagonal already:

$$ \mathcal{H}_{\mathbf{q}} = \left( \begin{array}{cc} b^{\dagger} & b \end{array} \right) \left( \begin{array}{cc} \frac{S}{2} J(\mathbf{q}) - SJ & 0 \\ 0 & \frac{S}{2} J(\mathbf{q}) - SJ \end{array} \right) \left( \begin{array}{c} b \\ b^{\dagger} \end{array} \right) $$

The Fourier transform is just:

$$ \begin{align} J(\mathbf{q}) &= J \left( \exp(-i\mathbf{q}\cdot\mathbf{d}) + \exp(i\mathbf{q}\cdot\mathbf{d}) \right) \\ &= 2 J \cos (\mathbf{q}\cdot\mathbf{d}) \end{align} $$

So we obtain $\omega(\mathbf{q}) = JS \left( \cos(\mathbf{q}\cdot\mathbf{d}) - 1 \right)$

In general, however, the Hamiltonian is not diagonal and so must be diagonalised.

This can be done using the Bogoliubov transformation for an antiferromagnet

SpinW uses a more general method due to Colpa, as explained in the SpinW paper

Now we have $\omega_\mathbf{q}$ and half the story. Remember for neutron scattering:

$$ \frac{d^2\sigma}{d\Omega dE} \propto F^2(\mathbf{Q}) \sum_{\alpha, \beta}(\delta_{\alpha\beta} - \hat{q}_\alpha\hat{q}_\beta ) S^{\alpha\beta} (\mathbf{q}, \omega) $$

With the correlation function:

$$ S^{\alpha \beta}(\mathbf{k},\omega) = \frac{1}{2\pi\hbar}\int dt e^{-i \omega t} \langle S^\alpha (\mathbf{q}, 0) S^\beta (-\mathbf{q}, t) \rangle $$

$\langle S^\alpha(\mathbf{q}, t)\rangle$ is related to the Fourier transform of the eigenvectors of the Hamiltonian at $\mathbf{q}$.

Summary

Linear spin wave theory calculations requires:

The exchange interactions
The magnetic structure

It expands the spin ladder operators as a power series in bosonic creation/anihilation operators, keeping only linear terms

For each momentum transfer $\mathbf{q}$ a Hamiltonian matrix has to be diagonalised to obtain the magnon energies and spin-spin correlation functions

The size of the matrix is proportional to the number of magnetic atoms in the unit cell

So, larger (or more complex) magnetic structures require more memory; and more $\mathbf{q}$-points require longer processing time

References

S. Toth and B. Lake, J. Phys.: Condens. Matter 27 166002 (2015) arxiv:1402.6069

S. Petit Collection SFN 12 105-121 (2011) 10.1051/sfn/201112006

Linear Spin Wave Theory Review