\text{Non-Collinear Magnons}

\begin{align*} &\Psi^{+}_p=\psi^+_p e^{i(kr_p-\omega t)}&\\ &\Psi^{-}_p=\psi^-_p e^{i(kr_p-\omega t)}& \end{align*}

\text{notice how different are $\Psi$ and $\psi$}

(\Psi^+_p)^\dagger \neq \Psi^-_p

(\psi^+_p)^\dagger = (\psi^-_p)

\text{whereas}

\begin{equation*} \begin{align*} i \frac{d \Psi_p^{+}}{d t} & =+\delta \mathbf{H}_{\mathbf{p}} \cdot \xi_{\mathbf{p}}-\Psi_{\mathbf{p}}^{+} \mathbf{H}_{\mathbf{p}}^0 \cdot \mathbf{e}_{\mathbf{r}_{\mathbf{p}}} \\ i \frac{d \Psi_p^{-}}{d t} & =-\delta \mathbf{H}_{\mathbf{p}} \cdot \xi_{\mathbf{p}}^{\star}+\Psi_{\mathbf{p}}^{-} \mathbf{H}_{\mathbf{p}}^0 \cdot \mathbf{e}_{\mathbf{r}_{\mathbf{p}}} \end{align*} \end{equation*}

\begin{equation*} \delta \mathbf{m}_{\mathbf{q}}= \Psi_{\mathbf{q}}^{+} \xi_{\mathbf{q}}^{\star}+\Psi_{\mathbf{q}}^{-} \xi_{\mathbf{q}} \end{equation*}

\text{let us suppose that we have a time dependent magnetic field with a parallel component $h_0$ and perpendicular small }\\ \text{component $\delta h$}.\text{ We can write it as:}

h_p(t)=h^0 e_{r_p}+h^+\xi_p^\star+h^-\xi_p

\begin{equation*} \begin{align*} i \frac{d \Psi_p^{+}}{d t} & =+\delta \mathbf{H}_{\mathbf{p}} \cdot \xi_{\mathbf{p}}-\Psi_{\mathbf{p}}^{+} \mathbf{H}_{\mathbf{p}}^0 \cdot \mathbf{e}_{\mathbf{r}_{\mathbf{p}}} +(\delta \mathbf{h}_{\mathbf{p}} \cdot \xi_{\mathbf{p}}-\Psi_{\mathbf{p}}^{+} \mathbf{h}_{\mathbf{p}}^0 \cdot \mathbf{e}_{\mathbf{r}_{\mathbf{p}}}) \\ i \frac{d \Psi_p^{-}}{d t} & =-\delta \mathbf{H}_{\mathbf{p}} \cdot \xi_{\mathbf{p}}^{\star}+\Psi_{\mathbf{p}}^{-} \mathbf{H}_{\mathbf{p}}^0 \cdot \mathbf{e}_{\mathbf{r}_{\mathbf{p}}}-(\delta \mathbf{h}_{\mathbf{p}} \cdot \xi_{\mathbf{p}}^\star-\Psi_{\mathbf{p}}^{-} \mathbf{h}_{\mathbf{p}}^0 \cdot \mathbf{e}_{\mathbf{r}_{\mathbf{p}}}) \\ \end{align*} \end{equation*}

\text{including the magnetic field will lead to this equation}

\text{it will be interesting to investigate the case where $h^0=0$ (and replace $\delta h$ )}

\begin{equation*} \begin{align*} i \frac{d \Psi_p^{+}}{d t} & =+\delta \mathbf{H}_{\mathbf{p}} \cdot \xi_{\mathbf{p}}-\Psi_{\mathbf{p}}^{+} \mathbf{H}_{\mathbf{p}}^0 \cdot \mathbf{e}_{\mathbf{r}_{\mathbf{p}}} +h^+_p \\ i \frac{d \Psi_p^{-}}{d t} & =-\delta \mathbf{H}_{\mathbf{p}} \cdot \xi_{\mathbf{p}}^{\star}+\Psi_{\mathbf{p}}^{-} \mathbf{H}_{\mathbf{p}}^0 \cdot \mathbf{e}_{\mathbf{r}_{\mathbf{p}}}-h^-_p \\ \end{align*} \end{equation*}

\text{by using the Fourier transform, we find out}

\omega \begin{pmatrix} \vdots \\ \hat{\psi}_p^+\\ \vdots\\ \hat{\psi}_p^-\\ \vdots \\ \end{pmatrix} =\mathcal{H(k)} \begin{pmatrix} \vdots \\ \hat{\psi}_p^+\\ \vdots\\ \hat{\psi}_p^-\\ \vdots \\ \end{pmatrix} + \begin{pmatrix} \vdots \\ \hat{h}_p^+\\ \vdots\\ \hat{h}_p^-\\ \vdots \\ \end{pmatrix}

\begin{pmatrix} \vdots \\ \hat{\psi}_p^+\\ \vdots\\ \hat{\psi}_p^-\\ \vdots \\ \end{pmatrix} =\frac{1}{\omega-\mathcal{H(k)}} \begin{pmatrix} \vdots \\ \hat{h}_p^+\\ \vdots\\ \hat{h}_p^-\\ \vdots \\ \end{pmatrix}

\text{So, by doing a Fourier transform, we get the band structure as the lignes of high intensity}