2
$\begingroup$

There is a paper (PhysRevB.95.014435) in which the dispersion relation for some Heisenberg model on the honeycomb lattice is derived from the Landau-Lifshitz equation: \begin{align} \frac{d S_i}{dt} = - S_i \times \mathcal H_{\rm eff} \end{align} Their attempt from Eq. 2 to Eq.4 is pretty simple and I'll try the same for the 2D triangular Heisenberg antiferromagnet (THAF) (in xy-plane), which has a much simpler Hamiltonian: \begin{align} \mathcal H = \sum_{\langle {ij}\rangle } J S_i S_j,\quad \mathcal H_{\rm eff} = J \sum_j S_j \end{align} where $\langle {ij}\rangle$ sums over all nearest neighbors. There are some papers out there (for example PhysRevB.74.180403) which have derived the dispersion to be \begin{align} \omega_{\bf k} = \sqrt{(1- \gamma_{\bf k} ) ( 1+ 2 \gamma_{\bf k} ) } \label{eq:thaf_disp} \end{align} with \begin{align} \gamma_{\bf k} = \frac{1}{z} \sum_{j} \mathrm{e}^{i \bf{k}( \bf{R}_i - \bf{R}_j )} = \frac{1}{3}\left(\cos k_{x}+2 \cos \frac{k_{x}}{2} \cos \frac{\sqrt{3}}{2} k_{y}\right) \, . \end{align} The ground-state of the THAF is the $120^{\circ}$-Neel order. My idea is similar to the derivation in Linear Spin Wave Theory and I'm starting by some rotation of spin vectors \begin{align} S_{i \in A} &= (\delta m_i^{x}, \delta m_i^{y}, 1) \\ S_{i \in B } &= ( \sqrt{3}/2 \delta m_i^{y} - 1/2 \delta m_i^{x}, -\sqrt{3}/2 \delta m_i^{x} - 1/2 \delta m_i^{y}, 1) \\ S_{i \in C} &= ( -\sqrt{3}/2 \delta m_i^{y} - 1/2 \delta m_i^{x}, \sqrt{3}/2 \delta m_i^{x} - 1/2 \delta m_i^{y}, 1) \end{align} where A,B,C are the three sublattices of the ground-state and $\delta m \ll 1$ . Then I tried to solve the Landau-Lifshitz equation: \begin{align*} \frac{d S_{i \in A}}{dt} &=- \begin{pmatrix} \delta m_i^{x} \\ \delta m_i^{y} \\ 1 \end{pmatrix} \times \left(\sum_j J S_{j\in B} + J S_{j \in C}\right) =- \sum_j J \begin{pmatrix} \delta m_i^{x} \\ \delta m_i^{y} \\ 1 \end{pmatrix} \times \begin{pmatrix} - \delta m_j^{x} \\ - \delta m_j^{y} \\ 2 \end{pmatrix} \approx - \sum_jJ \begin{pmatrix} \delta m_j^{y} + 2 \delta m_i^{y} \\ - \delta m_j^{x} - 2 \delta m_i^{x} \\ 0 \end{pmatrix} \\ \frac{d S_{i \in B}}{d t} &= -\begin{pmatrix} \frac{\sqrt{3}}{2} \delta m_i^{y} - \frac{1}{2}\delta m_i^{x} \\ -\frac{\sqrt{3}}{2} \delta m_i^{x} - \frac{1}{2} \delta m_i^{y} \\ 1 \end{pmatrix} \times \left(\sum_j J S_{j \in A} + J S_{j \in C} \right) \\ &= - \sum_j J \begin{pmatrix} \frac{\sqrt{3}}{2} \delta m_i^{y} - \frac{1}{2} \delta m_i^{x} \\ -\frac{\sqrt{3}}{2} \delta m_i^{x} - \frac{1}{2} \delta m_i^{y} \\ 1 \end{pmatrix} \times \begin{pmatrix} \frac{1}{2} \delta m_j^{x} - \frac{\sqrt{3}}{2} \delta m_j^{y} \\ \frac{\sqrt{3}}{2} \delta m_j^{x} + \frac{1}{2} \delta m_j^{y} \\ 2 \end{pmatrix} \approx - \sum_j J \begin{pmatrix} -(\sqrt{3} \delta m_i^{x} + \delta m_i^{y}) - ( \frac{\sqrt{3}}{2} \delta m_j^{x} + \frac{1}{2} \delta m_j^{y} ) \\ \frac{1}{2} \delta m_j^{x} - \frac{\sqrt{3}}{2} \delta m_j^{y} - (\sqrt{3} \delta m_i^{y} - \delta m_i^{x}) \\ 0 \end{pmatrix} \\ &=\sum_j J\begin{pmatrix} \frac{\sqrt{3}}{2} (2 \delta m_i^{x} + \delta m_j^{x} ) + \frac{1}{2}(2 \delta m_i^{y} +\delta m_j^{y} ) \\ \frac{\sqrt{3}}{2} (2\delta m_i^{y} + \delta m_j^{y} ) -\frac{1}{2} (2\delta m_i^{x} + \delta m_j^{x} ) \\ 0 \end{pmatrix} \\ \frac{d S_{i \in C}}{d t} &= - \sum_j \begin{pmatrix} -\frac{\sqrt{3}}{2} \delta m_i^{y} - \frac{1}{2} \delta m_i^{x} \\ \frac{\sqrt{3}}{2} \delta m_i^{x} - \frac{1}{2} \delta m_i^{y} \\ 1 \end{pmatrix} \times \begin{pmatrix} \frac{\sqrt{3}}{2} \delta m_j^{y} + \frac{1}{2} \delta m_j^{x} \\ -\frac{\sqrt{3}}{2} \delta m_j^{x} + \frac{1}{2} \delta m_j^{y} \\ 2 \end{pmatrix} \approx - \sum_j J \begin{pmatrix} \sqrt{3} \delta m_i^{x} - \delta m_i^{y} - (-\frac{\sqrt{3}}{2} \delta m_j^{x} + \frac{1}{2} \delta m_j^{y}) \\ (\frac{\sqrt{3}}{2} \delta m_j^{y} + \frac{1}{2} \delta m_j^{x}) + \sqrt{3} \delta m_i^{y} + \delta m_i^{x} \\ 0 \end{pmatrix} \\ &= \sum_j J \begin{pmatrix} \frac{1}{2} (2\delta m_i^{y} + \delta m_j^{y}) - \frac{\sqrt{3}}{2} (2 \delta m_i^{x} + \delta m_j^{x}) \\ - \frac{\sqrt{3}}{2} (2\delta m_i^{y} + \delta m_j^{y}) - \frac{1}{2} (2\delta m_i^{x} + \delta m_j^{x}) \\ 0 \end{pmatrix} \end{align*}

By using Bloch-Theorem: \begin{align} \delta m_i^{x} = X \exp(i \left( \bf{k} \bf{R}_i - \omega t \right) ), \quad \delta m_i^{y} = Y \exp(i \left( \bf{k} \bf{R}_i - \omega t \right) ) \end{align} Since I only have now one sublattice I don't need $X_A$, $X_B$ and $X_C$ etc. like in the paper. If you compare left-hand and right-hand side of the those equations of motions all do have the same structure. This structure looks like

\begin{align} i \omega \begin{pmatrix} X \\ Y \end{pmatrix} \mathrm{e}^{i (\bf{k} \bf{R}_i - \omega t)} = \sum_j J \begin{pmatrix} - 2 Y \\ 2X \end{pmatrix}\mathrm{e}^{i (\bf{k} \bf{R}_i - \omega t)} + \sum_j J\begin{pmatrix} -Y \\ X \end{pmatrix} \mathrm{e}^{i (\bf{k} \bf{R}_j - \omega t)} \end{align} where the Bloch theorem is already used. This would then lead to the following matrix \begin{align} i \omega \begin{pmatrix} X \\ Y \end{pmatrix} = J \begin{pmatrix} 0 & -2 - \gamma_k \\ 2 + \gamma_k & 0 \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} = H \begin{pmatrix} X \\ Y \end{pmatrix} \end{align} The paper sugested using $\psi^{\pm} = (X\pm iY)/\sqrt{2}$. This can be achieved by the Matrix \begin{align} U = \begin{pmatrix} 1 & i \\ 1 & -i \end{pmatrix} \end{align} and by calculating $i/2 \sigma_z UHU^{-1}$ I ended up with an hermitian matrix which uses $\psi^{\pm}$ as the amplitudes like sugested in the paper above: \begin{align} \begin{pmatrix} - \gamma_k - 2 & 0 \\ 0 & \gamma_k + 2 \end{pmatrix} \end{align} which would lead to $\omega_k = \pm \sqrt{(\gamma_k + 2)^2}$ which is obviously wrong but I cannot figure out where my mistake is or where I'm thinking wrong.

$\endgroup$

1 Answer 1

1
$\begingroup$

I see two possible problems in your consideration.

  1. You've investigated perturbations of ferromagnetic ground state. When spin variations $\delta m$ are zeros, spins on three sublattices are the same: $$ S_i = (0, 0, 1),\quad \forall i. $$

  2. The Landau-Lifshitz equation is a nonlinear one. Effective field ${\cal H}_{i,{\rm eff}}$ depends on neighboring spins. Hence you need to take into account variations of effective field: $$ \frac{d \delta S_i}{dt} = -\delta S_i \times {\cal H}_{i,{\rm eff}} - S_i \times \delta {\cal H}_{i,{\rm eff}}. $$

I didn't analyze your application of the Bloch theorem. I think there also could be problems. Neel state on triangular lattice is invariant under translation of states of triangular cells of spins, not of individual spins.

$\endgroup$
6
  • $\begingroup$ 1. Ok I understand that point. 2. How do you define $S_i$, $\delta S_i$, $\delta H_{i,\rm eff}$? Shoudn't it be $H_{j,\rm eff}$ on the right-hand side? $\endgroup$
    – Leviathan
    Nov 3, 2020 at 15:16
  • $\begingroup$ 2. $S_i$ - spins in the ground state, they do not depens on time; $\delta S_i$ - spins variations due to spin waves, they depend on time, you express them through $\delta m$; ${\cal H}{i,{\rm eff}} = J \sum_{j\ nbr\ i} S_j \rightarrow \delta {\cal H}{i,{\rm eff}} = J \sum_{j\ nbr\ i} \delta S_j$. $\endgroup$
    – Gec
    Nov 3, 2020 at 16:23
  • $\begingroup$ @Leviathan It might be possible that you are taking into account effective field variations due to the presence of $\delta m$ in your $S$. Then my second point might be not relevant to your problem. $\endgroup$
    – Gec
    Nov 3, 2020 at 17:17
  • $\begingroup$ If I understand you correctly it is $\delta S_{i \in A} = (\delta m_i^x,\delta m_i^y,1)$, $\delta S_{i \in B} = (\delta m_i^x, \frac{1}{2} + \delta m_i^y,-\frac{\sqrt{3}}{2})$, $\delta S_{i \in C} = (\delta m_i^x,-\frac{1}{2} + \delta m_i^y, - \frac{\sqrt{3}}{2})$, where I have used $\delta S_i = S_i + \delta S_i'$, where $\delta S_i'$ is the spin variation due to spin waves. Right? $\endgroup$
    – Leviathan
    Nov 3, 2020 at 17:22
  • $\begingroup$ No. In my formulas I mean $S_{i\in A} = (0, 0, 1)$, $\delta S_{i \in A} = (\delta m_i^x, \delta m_i^y, 0)$ and so on. I am ready to admit that my notation might be inconvenient. But it is a usual practice in considering perturbations to divide quantities onto a constant main part and a small perturbation. $\endgroup$
    – Gec
    Nov 3, 2020 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.