# Spin wave theory

Spin wave theory
관련코스 현대광학
소분류 물리
선행 키워드
연관 키워드

The magnetization of ferromagnetic material can be described by LLG equation. Only considering exchange and dipole interaction and neglecting the damping, anisotropy and others, the equation becomes:

${\operatorname{d}\!\mathbf{M}\over\operatorname{d}\!t}=\gamma\mu_{0}\mathbf{M}\times\mathbf{H}_{eff}$
$\mathbf{H}=\mathbf{H}_{0}+\mathbf{H}_{0ex}+\mathbf{h}(t)+\mathbf{h}_{ex}(t)$
$\mathbf{M}=\mathbf{M}_{0}+\mathbf{m}(t)$

$\mathbf{H}_{0}$ is static external magnetic field and $\mathbf{h}(t)$ is external dynamic magnetic field. $\mathbf{H}_{0ex}$ and $\mathbf{h}_{ex}(t)$ is effective field by exchange interaction. Also, the exchange interaction is $\mathbf{H}_{ex}=\lambda_{ex}\nabla^{2}\mathbf{M}$ where $\lambda_{ex}$ is exchange stiffness constant. $\mathbf{M}_{0}$ and $\mathbf{m}(t)$ are static and dynamic magnetization. LLG equation can be written again as:

${\operatorname{d}\!\mathbf{m}\over\operatorname{d}\!t}=\gamma\mu_{0}[\mathbf{M}_{0}\times(\mathbf{H}_{0}+\mathbf{H}_{ex})+\mathbf{M}_{0}\times(\mathbf{h}_{0}+\mathbf{h}_{ex})+\mathbf{m}\times(\mathbf{H}_{0}+\mathbf{H}_{ex})+\mathbf{m}\times(\mathbf{h}_{0}+\mathbf{h}_{ex})]$

$\mathbf{M}_{0}$ and $\mathbf{H}_{0}$ are parallel. Since both dynamic magnetization and magnetic field are so small, outer product of two can be neglected and $M_{0}$ is large enough so that in the direction $\mathbf{m}$ is put 0. If the magnetization is taken to be $\mathbf{m}\propto e^{i(\mathbf{k}\cdot\mathbf{r}-{\omega}t)}$ and external magnetic field is in direction z:

$-i\omega\mathbf{m}=\widehat{\mathbf{z}}\times[-\omega_{M}\mathbf{h}+(\omega_{0}+\omega_{M}\lambda_{ex}k^{2})\mathbf{m}]$

where $\omega_{M}\equiv-\gamma\mu_{0}M_{s}$ and $\omega\equiv-\gamma\mu_{0}H_{0}$

From previous equation, susceptibility tensor can be found. The tensor is defined as:

$\mathbf{m}=\mathbf{\bar{\chi}}\cdot\mathbf{h}$

where the component is:

$\mathbf{\bar{\chi}}= \begin{bmatrix} \chi & -i\kappa \\ i\kappa & \chi \end{bmatrix}$

and

$\chi=\frac{(\omega_{0}+\omega_{M}\lambda_{ex}k^{2})\omega_M}{(\omega_{0}+\omega_{M}\lambda_{ex}k^{2})^{2}-\omega^{2}}$
$\kappa=\frac{\omega\omega_M}{(\omega_{0}+\omega_{M}\lambda_{ex}k^{2})^{2}-\omega^{2}}$

Since $\mathbf{b}$ field can be written using susceptibility tensor:

$\mathbf{b}=\mu_0(\mathbf{\bar{I}}+\mathbf{\bar{\chi}})$

Using magneto-static approximation boundary:

$\nabla\times\mathbf{h}=0$
$\nabla\cdot\mathbf{b}=0$

$\mathbf{h}$ can be put as $-\nabla\psi$, and the boundary condition becomes:

$\nabla\cdot(\mu_0(\mathbf{\bar{I}}+\mathbf{\bar{\chi}})\cdot(-\nabla\psi))=0$

therefore,

$(1+\chi)(k_{x}^{2}+k_{y}^{2})+k_{z}^{2}=0$

This is called Walker's equation.

And the dispersion relation can be written as:

$\omega=\sqrt{(\omega_{0}+\omega_{M}\lambda_{ex}k^{2})(\omega_{0}+\omega_{M}(\lambda_{ex}k^{2}+\sin^{2}\theta))}$