an electron is $$\varphi(x,E) = {\left(  \begin{array}[c]{c} a~e^{ik_+ x} \\ b~e^{ik_- x} \end{array}  \right)},$$ where $$\label{eqn:kapdef}
k_\pm^2=2mE/\hbar^2\pm k_B^2.$$ We compute the spin density ${\bf s}(x,E)$ and the spin current density ${\bf Q}(x,E)$ for this electron using the right sides of Eq. (\[eqn:sesd\]) and Eq. (\[eqn:sescd\]), respectively, with $\psi_\pm \to \varphi$.

It is straightforward to check that the components of these densities transverse to the magnetic field satisfy the semi-classical relations $$\label{eqn:scQs}
Q_x = s_x {\langle v \rangle}~~~~~~~~~~{\rm and}~~~~~~~~~~Q_y=s_y {\langle v \rangle},$$ where ${\langle v \rangle}$ is the velocity $$\label{eqn:vdef}
{\langle v \rangle}={\hbar\over m}{k_++k_-\over 2}={\hbar \over m}{\langle k \rangle}.$$ Moreover, the transverse components of the spin density satisfy $$\begin{aligned}
\label{eqn:perppiece}
    {\langle v \rangle}{ds_x\over dx} &=& -{\hbar k_B^2\over m}s_y,{\nonumber\\}{\langle v \rangle}{ds_y\over dx} &=& -{\hbar k_B^2\over m}s_x.\end{aligned}$$ These equations are the components of the vector equation $$\label{eqn:LLGa}
    {d{\bf s}\over dx} = -{k_B^2\over {\langle k \rangle}}{\bf s}\times\hat{\bf B}_{\rm ex}$$ where $\hat{\bf B}_{\rm ex}=\hat{\bf z}$.

We now make the [*ansatz*]{} that all three Cartesian components of the semi-classical majority and minority spin densities ${\bf
s}_\pm(x,k_x)$ satisfy Eq. (\[eqn:LLGa\]) when the direction of the magnetic field varies in space. Specifically, if ${\bf B}_{\rm
ex}(x)=B_{\rm ex}\hat{\bf B}_{\rm ex}(x)$, we suppose that $$\label{eqn:LLG}
    {d{\bf s}_\pm(x,k_x)\over dx} = -{k_B^2\over 