Ballistic spin transport through a metallic system of two junctions with strong spin–orbit coupling

Abstract

Inversion asymmetry at an interface separating two different metals can produce a strong interfacial Rashba spin–orbit coupling (ISOC) that induces various charge and spin transport phenomena. For the sake of analyzing such emergent phenomena in a ballistic approximation, we consider a simple model of a metallic system of two interfaces with strong spin–orbit coupling. We show that the ISOC would be responsible for inducing a z-dependent and z-independent secondary lateral charge currents from the primary spin current injected in the system (spin-charge current conversion), where the z-direction corresponds to the direction of flow of the primary spin current. We explain the fundamental background behind this induction in terms of quantum interference effects. Furthermore, we show that the presence of a tunneling barrier between the two interfaces increases spin current attenuation, but reduces it once the width of the tunneling barrier becomes sufficiently large.

Appendix A: Obtaining the scattering coefficients

We have assumed that V(z) is equal to $V\Big (\theta (z+a)-\theta (z-a)\Big )$ (see Eq. (3)). Hence, $V'(z)=V(\delta (z+a)-\delta (z-a))$.

We can let

$$\begin{aligned} V_1(z)=V(z), \end{aligned}$$

(A1)

and

$$\begin{aligned} V_\mathrm{RSOC}(z)=\lambda z^{-1}\dfrac{\partial V(z)}{\partial z}(\vec {z} \times \vec {p})\cdot \vec {\sigma }. \end{aligned}$$

(A2)

It can be noticed from the Hamiltonian in Eq. (1) that the scattering problem is composed of two problems, one that corresponds to the potential $V_{1}(z)$ which is the step potential problem and other that corresponds to the potential $V_\mathrm{RSOC}(z)$ which is the dirac-delta potential localized at the interfaces. To obtain the energy eigenfunctions of ${\hat{H}}$, we first solve the common step potential problem, which gives us the eigenfunctions in Sect. 2.2, but by modifying and adding an additional terms to the scattering coefficients obtained in case $V_\mathrm{RSOC}=0 (r_0, t_0, r_1, t_1).$

In our case, the additional terms corresponding to $r_0$ and $t_0$ are $\vec {{\mathbf {r}}} \cdot \vec {\sigma }$ and $\vec {{\mathbf {t}}} \cdot \vec {\sigma }$, respectively. While the additional terms corresponding to $r_1$ and $t_1$ are $\vec {{\mathbf {r}}}_1 \cdot \vec {\sigma }$ and $\vec {{\mathbf {t}}}_1 \cdot \vec {\sigma }$, respectively. The exact form of the scattering coefficients $({\hat{r}}_k, {\hat{t}}_k, {\hat{t}}_{1k}$ and ${\hat{r}}_{1k})$ in terms of the wavevector k of the electron is obtained by making the energy eigenfunctions obey to the boundary conditions imposed by the dirac-delta potential term $V_\mathrm{RSOC}(z)$.

The boundary conditions are:

• 1. $\phi _{k\sigma }(z)$ is continuous along z.

• 2. $\dfrac{d\phi _{k\sigma }(z)}{dz}\Big |_{z=\pm a}=-2m\Gamma \phi _{k}(\pm a),$

where, $V_\mathrm{RSOC}=-\Gamma V(\delta (z+a)-\delta (z-a))$.

By handling the boundary conditions above, the x and y terms of the SOC-part of the transmission and reflection coefficients can be written in the following form:

$$\begin{aligned} r_{x/y}= & {} t_{1x/y}e^{-i(k+k')a}+r_{1x/y}e^{i(k'-k)a} \end{aligned}$$

(A3)

$$\begin{aligned} t_{x/y}= & {} t_{1x/y}e^{i(k'-k)a}+r_{1x/y}e^{-i(k+k')a} \end{aligned}$$

(A4)

The functions $\chi _{1/2}, {\mathcal {M}}, {\mathcal {F}}, {\mathcal {A}}, {\mathcal {B}}$ and ${\mathcal {C}}$ are just formulated for the sake of simplifying Eq. (7) and they have the following form:

$$\begin{aligned} \chi _1= & {} \dfrac{\mathcal {AB}+2k}{{\mathcal {C}}}, \quad \chi _2=\dfrac{2{\mathcal {A}}k+{\mathcal {B}}}{{\mathcal {C}}}. \end{aligned}$$

(A5)

$$\begin{aligned} {\mathcal {M}}= & {} (\alpha \beta ^2p)^2(\chi _1+\chi _2)+k'(1+{\mathcal {A}}). \end{aligned}$$

(A6)

$$\begin{aligned} {\mathcal {F}}= & {} ({\mathcal {A}}+1)e^{-ika}\Big (e^{(2ik'a)}+e^{(-2ik'a)}\Big ). \end{aligned}$$

(A7)

$$\begin{aligned} {\mathcal {A}}= & {} \dfrac{\big [{\mathcal {B}}(k-k')+{\mathcal {C}}\big ]e^{ik'a}-2k(k+k')e^{-ik'a}}{\big [{\mathcal {B}}(k+k')-{\mathcal {C}}\big ]e^{-ik'a}-2k(k-k')e^{ik'a}}. \end{aligned}$$

(A8)

$$\begin{aligned} {\mathcal {B}}= & {} (k+k')e^{-2ik'a}+(k-k')e^{2ik'a}. \end{aligned}$$

(A9)

$$\begin{aligned} {\mathcal {C}}= & {} (k+k')^2e^{-2ik'a}-(k-k')^2e^{2ik'a}. \end{aligned}$$

(A10)