by $J_{\pm}^{\rm c}$. Since the charge current is conserved throughout the structure $J_{+}^{\rm c}$ and $J_{-}^{\rm c}$ can be calculated in different ways, [*e.g.*]{} in the lead for $J_{+}^{\rm c}$ and in the polariser for $J_{-}^{\rm c}$. Since $T=\left|F\right|^2=\left|F^{\prime}\right|^2$ we find $J_{+}^{\rm
  c}+J_{-}^{\rm c}=0$ and for small bias $eV_{b}=\mu_{\rm L}-\mu_{\rm R}$ the charge current is given by $$\label{current}
j^{\rm c}=\frac{2e^2V_{\rm b}}{h}\sum_{{\bf
k}_{\parallel}}T$$ where the transmission coefficient $T$ is given by Eq. (\[T\]) with $k=k(\mu,{\bf k}_{\parallel})$, $\mu$ being the common chemical potential as $V_{\rm B}\rightarrow 0$. This is the well-known Landauer formula [@5].

The spin transfer torque on the switching magnet is given by $$\label{torque}
{\bf T}^{\rm s-t}=\langle{\bf j}_{\rm spacer}\rangle-
\langle{\bf j}_{\rm lead}\rangle,$$ where $\langle{\bf j}_{\rm spacer}\rangle$ and $\langle{\bf j}_{\rm
lead}\rangle$ are spin currents in the spacer and lead respectively. For zero bias ($\mu_{\rm L}=\mu_{\rm
  R}$) there is clearly no charge current in the structure and straight-forward calculation shows that all components of spin current in the spacer and the lead vanish, except for a non-zero $y$-spin current in the spacer. There is therefore a non-zero $y$ component of spin-transfer torque acting on the switching magnet for zero bias, and its dependence on the angle $\psi$ between the magnetisations is found 