notation,

![Cartoon of a spin spiral where the magnetization (arrows) rotates uniformly in the $x$-$z$ plane of a fixed coordinate system where $\hat{\bf M}(x) \cdot \hat{\bf z}=\cos\theta$. The inset shows a local coordinate system where ${\bf M}(x)$ always points along the $z'$ axis. []{data-label="fig:spinspiral"}](fig3.eps){width="8cm"}

$$\varepsilon_{\pm}({\bf k}) 
    = {\hbar^2\over 2m}{\left( k^2+{{1\over 4}}p^2 \pm\sqrt{(k_x p)^2+k_B^4 } \right)},
    \label{eqn:spiralenergy}$$

and $$\psi_\pm({\bf k,r})
    = e^{i{\bf k\cdot r}} e^{-i\sigma_y\theta/2} e^{-i\sigma_x\alpha/2}\eta_\pm,
    \label{eqn:Neelstate}$$ where $$\label{alpha}
\sin\alpha = {k_x p\over \sqrt{(k_x p)^2+k_B^4}},$$ and $$\eta_+ = {\left(  \begin{array}[c]{c} 1 \\ 0 \end{array}  \right)},\hspace{20pt}
    \eta_- = {\left(  \begin{array}[c]{c} 0 \\ 1 \end{array}  \right)}. \vspace{0.5em}$$ From these results, it is easy to compute the single-electron spin densities defined in Eq. (\[eqn:sesd\]). In the local $(x',y',z')$ frame, $$\label{eqn:spirald}
{\bf s'}_\pm(x,k_x)=\pm(0, \sin \alpha, \cos\alpha).$$

The corresponding calculation of the single-electron spin current densities Eq. (\[eqn:scscd\]) is straightforward but tedious and not very illuminating. Therefore, we pass directly to the total spin current density calculated by summing over all electrons as indicated in the second line of Eq. (\[eqn:spinandcurrent\]). Again in the local $(x',y',z')$ frame, $$\label{eqn:spiralcd}
{\bf Q'}(x)=A(p,k_{\rm B})(0,0,1).$$ where $A(p,k_{\rm B})$ is a constant. This shows that ${\bf
Q}(x)\propto {\bf M}(x)$, [*i.e.*]{}, the spin current density for a free electron spin spiral is 