{\langle k \rangle}}{\bf s}_\pm(x,k_x)\times\hat{\bf B}_{\rm ex}(x),$$ where $k_+$ and $k_-$ for ${\bf s}_+(x,k_x)$ are defined by Eq. (\[eqn:kapdef\]) with $E=\hbar^2(k_x^2-k^2_{\rm
B})/2m$.[@KMEvan] Similarly, $k_+$ and $k_-$ for ${\bf
s}_-(x,k_x)$ are defined by Eq. (\[eqn:kapdef\]) with $E=\hbar^2(k_x^2+k^2_{\rm B})/2m$. With suitable boundary conditions, we solve the differential equation Eq. (\[eqn:LLG\]) to determine the semi-classical, one-electron spin densities. The total, spin-resolved, spin densities follow by inserting these one-electron quantities into $$\label{scm}
{\bf s}_\pm(x) 
    = \int{\left[ f_\pm({\bf k}-{eE\tau\over\hbar}\bm{\hat{x}})-f_\pm({\bf k}) \right]}{\bf s}_\pm(x,k_x){k_x\over {\langle k \rangle}}d^3{\bf k}.$$ This equation differs from Eq. (\[eqn:spinandcurrent\]) by the weighting factor $k_x/{\langle k \rangle}$.[@ksign] This factor guarantees that the flux carried by each electron is proportional to its velocity (see Appendix I). This is confirmed by Fig. \[fig:LLGcheck\](b) which shows quantitative agreement between a fully quantum calculation of ${\bf s}(x)$ using Eq. (\[eqn:Schrodinger\]), Eq. (\[eqn:sesd\]) and Eq. (\[eqn:spinandcurrent\]) and a semi-classical calculation using Eq. (\[eqn:LLG\]) and Eq. (\[scm\]).

![(a) Cartesian components of an imposed magnetization ${\bf
M}(x)$ used in the other panels; (b) comparison of quantum (lines) to semi-classical (solid dots) calculations for the Cartesian components of the spin density ${\bf s}(x)$; (c) same comparison for the Cartesian components of the spin current density ${\bf Q}(x)$. []{data-label="fig:LLGcheck"}](fig2.eps){width="8cm"}

In light of the 