left and right moving electrons. Thus we have a Landauer-type formula [@5] $$\label{landau}
{\bf j}_{\rm s}=\frac{a}{2\pi}\sum_{{\bf k }_{\parallel}}
\left\{\int_{k>0}{\rm d}k\left[\langle S_k|{\bf
  j}_{n-1}|S_k\rangle f(\omega-\mu_L)+\langle S_{-k}|{\bf
  j}_{n-1}|S_{-k}\rangle f(\omega-\mu_{R})\right]\right\}$$ where $|S_k\rangle$, $|S_{-k}\rangle$ are wave-functions in the layer considered corresponding to electrons incident from left and right, respectively. Here $\omega$, the energy of the Bloch wave $k$, is given by the tight-binding formula $$\label{omega}
\omega=u_{{\bf k}_{\parallel}}+2t\cos ka$$ where $u_{{\bf k}_{\parallel}}=2t(\cos k_xa+\cos k_za)$. We take $t<0$ so that positive $k$ corresponds to positive velocity $\hbar^{-1}\partial\omega/\partial k$ as we have assumed. The current ${\bf j}_{\rm s}$ in layer $S$ calculated by Eq. (\[landau\]) does not depend on the particular planes $n-1$, $n$ between which it is calculated. On changing the integration variable in Eq. (\[landau\]) we find $$\label{landau2} {\bf j}_{\rm s}=\frac{1}{2\pi} \sum_{{\bf
k}_{\parallel}}\int{\rm d}\omega\left[{\bf J}_{+}f(\omega-\mu_{\rm
  L})+{\bf J}_{-}f(\omega-\mu_{\rm R})
\right]$$ where $$\label{landau3}
{\bf J}_{\pm}=\frac{\langle S_{\pm k}|{\bf j}_{n-1}|S_{\pm
    k}\rangle}{-2t\sin ka}.$$ Here $k=k(\omega,{\bf k}_{\parallel})$ is the positive root of Eq. (\[omega\]). Eq. (\[landau2\]) may be written as $$\label{landau33}
{\bf j}_{\rm s}=\frac{1}{4\pi}\sum_{{\bf k }_{\parallel}}
\int{\rm d}\omega\left\{
\left({\bf J}_{+}+{\bf J}_{-}\right)
\left[f(\omega-\mu_{\rm L})+f(\omega-\mu_{\rm R})\right]+
\left({\bf J}_{+}-{\bf J}_{-}\right)
\left[f(\omega-\mu_{\rm L})-f(\omega-\mu_{\rm  R})\right]
\right\}.$$

Before discussing this spin current we briefly consider the charge current $j^{\rm c}$, and we denote the analogues of ${\bf J}_{\pm}$ 