1 Hamiltonian of the spin system in the adiabatic approximation

Within the rigid-spin approximation the adiabatic dynamics of local spin moments is described by the Landau-Lifshitz equation.

$$\mathrm{M}_{ri}\mathbf{\dot{S}}_{ri}=-\frac{2\mu_{B}}{\hbar}\frac{\delta\mathcal{F}}{\delta\mathbf{S}_{ri}}\times\mathbf{S}_{ri}$$ (125)

where $\mathbf{S}_{ri}$ is a unit vector pointing along the spin-quantization axis in the atomic cell at the site i of the layer r, $\mathrm{M}_{ri}$ is the magnitude of the spin moment and $\mathcal{F}$ is the free energy of the system. If we are working in spherical coordinates then we can write $\mathbf{S}_{ri}$ as a function of the polar and azimuthal angles $\theta_{ri}$ and $\varphi_{ri}$ respectively.

$$\mathbf{S}_{ri}=(\sin\theta_{ri}\cos\varphi_{ri}, \sin\theta_{ri}\sin\varphi_{ri},\cos\theta_{ri})$$ (126)

Rewriting the dynamical equation into spherical coordinates, the equations of motion for the angles $\theta_{ri}$ and $\varphi_{ri}$ are given by:

$$M_{r}\dot\varphi_{ri}\sin\theta_{ri} = \frac{2\mu_{B}}{\hbar}\frac{\delta\mathcal{F}}{\delta\theta_{ri}}$$ (127)

$$-M_{r}\dot\theta_{ri}\sin\theta_{ri} = \frac{2\mu_{B}}{\hbar}\frac{\delta\mathcal{F}}{\delta\varphi_{ri}}\:,$$ (128)

implicitly in these equations we supposed that the magnitude of the spin moment $\mathrm{M}_{ri}$ depends only on the atomic layer for the spins i.

If we choose the polar axis of the reference system (Z) perpendicular to the magnetization in the ferromagnetic state, the system of equations (A.3 and A.4) can be linearized

$$M_{r}\dot\varphi_{ri} = \frac{2\mu_{B}}{\hbar}\frac{\delta\mathcal{F}}{\delta\theta_{ri}}\Bigr\vert _{\theta=\frac{\pi}{2},\varphi=0}$$ (129)

$$-M_{r}\dot\theta_{ri} = \frac{2\mu_{B}}{\hbar}\frac{\delta\mathcal{F}}{\delta\varphi_{ri}}\Bigr\vert _{\theta=\frac{\pi}{2},\varphi=0}$$ (130)

These equations are the canonical equations for the generalized coordinates $q_{ri}$ and momenta $p_{ri}$ which are defined as:

$$q_{ri} = \Bigl( \frac{M_{r}}{\mu_{B}}\Bigr)^{1/2}\varphi_{ri}$$ (131)

$$p_{ri} = \Bigl( \frac{M_{r}}{\mu_{B}}\Bigr)^{1/2}\theta_{ri}$$ (132)

Therefore doing an expansion up to the second order of the free energy in the angular variables, the corresponding Hamiltonian can be written as:

$$\mathcal{H}=\frac{1}{\hbar}\sum_{ri,sj}\bigl(q_{ri}A_{ri,sj}q_{sj}+ q_{ri}B_{ri,sj}p_{sj} + p_{ri}B_{sj,ri}q_{sj}+ p_{ri}C_{ri,sj}p_{sj}\bigr),$$ (133)

with

$$A_{ri,sj} = \Bigl( \frac{M_{r}}{\mu_{B}}\Bigr)^{1/2} \frac{\delta^{2}\mathcal... ...\vert _{\theta=\frac{\pi}{2},\varphi=0}\Bigl( \frac{M_{s}}{\mu_{B}}\Bigr)^{1/2}$$ (134)

$$B_{ri,sj} = \Bigl( \frac{M_{r}}{\mu_{B}}\Bigr)^{1/2} \frac{\delta^{2}\mathcal... ...\vert _{\theta=\frac{\pi}{2},\varphi=0}\Bigl( \frac{M_{s}}{\mu_{B}}\Bigr)^{1/2}$$ (135)

$$C_{ri,sj} = \Bigl( \frac{M_{r}}{\mu_{B}}\Bigr)^{1/2} \frac{\delta^{2}\mathcal... ...\vert _{\theta=\frac{\pi}{2},\varphi=0}\Bigl( \frac{M_{s}}{\mu_{B}}\Bigr)^{1/2}$$ (136)