Spin filtering in a Rashba–Dresselhaus–Aharonov–Bohm double-dot interferometer

To study the scattering of a spin-1/2 electron by the double-dot interferometer with arbitrary SOI and AB flux (figure 1), we model it as a square interferometer as shown in figure 2. Each QD/QN is replaced by a bond connecting two sites (a, b and c, d in figure 2) with the corresponding bonds subject to SOI. We emphasize that the solution is not limited to this model. One can model each QD/QN by an arbitrary number of sites M. This changes only the transmission of the spin-polarized electrons, with no significant effect on the spin filtering conditions and the polarization direction of the outgoing electrons [32].

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In the framework of the nearest neighbors tight-binding model, the Schrödinger equation for the spinor |ψ_v〉 at site v is written as

$$\begin{eqnarray} &&\left(\varepsilon-\varepsilon_{v}\right)|\psi_{v}\rangle=-\sum_{u}J_{uv}U_{uv}|\psi_{u}\rangle, \end{eqnarray} \tag{ 5 }$$

where ε_v is the site energy, J_uv is a real hopping amplitude and U_uv is a 2 × 2 unitary matrix which describes the AB and AC phases acquired by an electron moving from site u to site v. The sum in equation (5) is over the nearest neighbors u of v. At this stage we do not specify the details of these matrices and hopping amplitudes. Except for the sites 0, a, b, c, d and 1 (figure 2), the hopping amplitude along the leads is j and the site energies on the leads are set to zero. The leads are free of SOI. Thus, the dispersion relation of the leads is $\varepsilon =-2\mathrm {j}\cos \left (ka\right )$ with a the lattice constant. The tight-binding Schrödinger equations for the spinors at sites 0, a, b, c, d and 1 are

$$\begin{eqnarray} &&\begin{array}{l} \left(\varepsilon-\varepsilon_{0}\right)|\psi_{0}\rangle=-J_{0a}U^{\dagger}_{0a}|\psi_{a}\rangle-J_{0c}U^{\dagger}_{0c}|\psi_{c}\rangle-\mathrm{j}|\psi_{-1}\rangle,\\ \left(\varepsilon-\varepsilon_{a}\right)|\psi_{a}\rangle=-J_{ab}U^{\dagger}_{ab}|\psi_{b}\rangle-J_{0a}U_{0a}|\psi_{0}\rangle,\\ \left(\varepsilon-\varepsilon_{b}\right)|\psi_{b}\rangle=-J_{b1}U^{\dagger}_{b1}|\psi_{1}\rangle-J_{ab}U_{ab}|\psi_{a}\rangle,\\ \left(\varepsilon-\varepsilon_{c}\right)|\psi_{c}\rangle=-J_{cd}U^{\dagger}_{cd}|\psi_{d}\rangle-J_{0c}U_{0c}|\psi_{0}\rangle,\\ \left(\varepsilon-\varepsilon_{d}\right)|\psi_{d}\rangle=-J_{d1}U^{\dagger}_{d1}|\psi_{1}\rangle-J_{cd}U_{cd}|\psi_{c}\rangle,\\ \left(\varepsilon-\varepsilon_{1}\right)|\psi_{1}\rangle=-j|\psi_{2}\rangle-J_{b1}U_{b1}|\psi_{b}\rangle-J_{d1}U_{d1}|\psi_{d}\rangle. \end{array} \end{eqnarray} \tag{ 6 }$$

Eliminating |ψ_a〉, |ψ_b〉, |ψ_c〉 and |ψ_d〉 from equations (6) one ends up with the equations

$$\begin{eqnarray} &&\begin{array}{l} \left(\varepsilon-y_{0}\right)|\psi_{0}\rangle=\boldsymbol{W}^{\dagger}|\psi_{1}\rangle-\mathrm{j}|\psi_{-1}\rangle,\\ \left(\varepsilon-y_{1}\right)|\psi_{1}\rangle=-\mathrm{j}|\psi_{2}\rangle+\boldsymbol{W}|\psi_{0}\rangle, \end{array} \end{eqnarray} \tag{ 7 }$$

where

$$\begin{eqnarray} &&\begin{array}{l} y_{0}=\varepsilon_{0}+\displaystyle\frac{J^{2}_{0a}}{\varepsilon-\varepsilon_{a}-\frac{J^{2}_{ab}}{\varepsilon-\varepsilon_{b}}}+\displaystyle\frac{J^{2}_{0c}}{\varepsilon-\varepsilon_{c}-\frac{J^{2}_{cd}}{\varepsilon-\varepsilon_{d}}},\\ y_{1}=\varepsilon_{1}+\displaystyle\frac{J^{2}_{b1}}{\varepsilon-\varepsilon_{b}-\frac{J^{2}_{ab}}{\varepsilon-\varepsilon_{a}}}+\displaystyle\frac{J^{2}_{d1}}{\varepsilon-\varepsilon_{d}-\frac{J^{2}_{cd}}{\varepsilon-\varepsilon_{c}}}, \end{array} \end{eqnarray} \tag{ 8 }$$

and

$$\begin{eqnarray} \boldsymbol{W}&=\frac{J_{0a}J_{ab}J_{b1}}{J^{2}_{ab}-\left(\varepsilon-\varepsilon_{a}\right)\left(\varepsilon-\varepsilon_{b}\right)}U_{b1}U_{ab}U_{0a}+\frac{J_{0c}J_{cd}J_{d1}}{J^{2}_{cd}-\left(\varepsilon-\varepsilon_{c}\right)\left(\varepsilon-\varepsilon_{d}\right)}U_{d1}U_{cd}U_{0c}\nonumber\\ &\equiv\gamma_{{\rm upper}}U_{{\rm upper}}+\gamma_{{\rm lower}}U_{{\rm lower}}. \end{eqnarray} \tag{ 9 }$$

Here, the coefficients γ_upper and γ_lower are defined as

$$\begin{eqnarray} &&\begin{array}{l} \gamma_{{\rm upper}}=\displaystyle\frac{J_{0a}J_{ab}J_{b1}}{J^{2}_{ab}-\left(\varepsilon-\varepsilon_{a}\right)\left(\varepsilon-\varepsilon_{b}\right)},\\ \gamma_{{\rm lower}}=\displaystyle\frac{J_{0c}J_{cd}J_{d1}}{J^{2}_{cd}-\left(\varepsilon-\varepsilon_{c}\right)\left(\varepsilon-\varepsilon_{d}\right)} \end{array} \end{eqnarray} \tag{ 10 }$$

and U_lower = U_d1U_cdU_0c, U_upper = U_b1U_abU_0a are the unitary matrices corresponding to transitions through the lower and upper paths of the interferometer, respectively. Equations (7) describe the effective tight-binding equations for hopping between sites 0 and 1 and have the same form as in the diamond interferometer [30, 32]. All the details of the interferometer are embodied in the effective hopping matrix W and effective site energies y₀ and y₁.

We next consider the scattering of a wave coming from the left, i.e.

$$\begin{eqnarray} &&|\psi_{n}\rangle=|\chi_{{\rm in}}\rangle {\rm e}^{{\rm i}kna}+r|\chi_{{\rm r}}\rangle {\rm e}^{-{\rm i}kna}, \quad n\leqslant 0, \nonumber\\ &&|\psi_{n}\rangle=t|\chi_{{\rm t}}\rangle {\rm e}^{{\rm i}k\left(n-1\right)a}, \quad n\geqslant 1, \end{eqnarray} \tag{ 11 }$$

where |χ_in〉, |χ_r〉 and |χ_t〉 are the incoming, reflected and transmitted normalized spinors, respectively, with the corresponding reflection and transmission amplitudes r and t. Substituting equations (11) into (7), one finds

$$\begin{eqnarray} &&t|\chi_{{\rm t}}\rangle\equiv\mathcal{T}|\chi_{{\rm in}}\rangle, \quad r|\chi_{{\rm r}}\rangle\equiv\mathcal{R}|\chi_{{\rm in}}\rangle \end{eqnarray} \tag{ 12 }$$

with the 2 × 2 transmission and reflection amplitude matrices

$$\begin{eqnarray} &&\mathcal{T}=2\mathrm{ij}\sin\left(ka\right)\boldsymbol{W}\left(Y\boldsymbol{I}-\boldsymbol{W}^{\dagger}\boldsymbol{W}\right)^{-1}, \end{eqnarray} \tag{ 13 }$$

$$\begin{eqnarray} &&\mathcal{R}=-\boldsymbol{I}-2\mathrm{ij}\sin\left(ka\right)X_{1}\left(Y\boldsymbol{I}-\boldsymbol{W}^{\dagger}\boldsymbol{W}\right)^{-1}. \end{eqnarray} \tag{ 14 }$$

Here we define

$$\begin{eqnarray} &&X_{0,1}=y_{0,1}+\mathrm{j}\,{\rm e}^{-{\rm i}ka}, \quad Y=X_{0}X_{1}. \end{eqnarray} \tag{ 15 }$$

Using equation (9), the matrix W^†W involved in both $\mathcal {T}$ and $\mathcal {R}$, is found to be

$$\begin{eqnarray} &&\boldsymbol{W}^{\dagger}\boldsymbol{W}=\gamma^{2}_{{\rm upper}}+\gamma^{2}_{{\rm lower}}+\gamma_{{\rm upper}}\gamma_{{\rm lower}}\left(u+u^{\dagger}\right), \end{eqnarray} \tag{ 16 }$$

with u = U^†_upperU_lower the unitary matrix representing anticlockwise hopping from site 0 back to site 0 around the loop. Equation (4) then yields $u+u^{\dagger }=2\left (\cos \omega \cos \phi +\sin \omega \sin \phi \:\hat {\boldsymbol {\omega }}\cdot \boldsymbol {\sigma }\right )$ and equation (16) can thus be written as

$$\begin{eqnarray} &&\boldsymbol{W}^{\dagger}\boldsymbol{W}=A+\boldsymbol{B}\cdot\boldsymbol{\sigma} \end{eqnarray} \tag{ 17 }$$

with

$$\begin{eqnarray} &&\begin{array}{l} A=\gamma^{2}_{{\rm upper}}+\gamma^{2}_{{\rm lower}}+2\gamma_{{\rm upper}}\gamma_{{\rm lower}}\cos\omega\cos\phi, \\ \boldsymbol{B}=2\gamma_{{\rm upper}}\gamma_{{\rm lower}}\sin\omega\sin\phi\:\hat{\boldsymbol{n}}\equiv B\hat{\boldsymbol{n}}. \end{array} \end{eqnarray} \tag{ 18 }$$

Here, $\hat {\boldsymbol {n}}\equiv \hat {\boldsymbol {\omega }}$ is a real unit vector along the direction of ω. As shown in [30], the spin-dependent transmission of the interferometer is determined by the eigenvalues of the matrix W^†W. These eigenvalues are given by

$$\begin{eqnarray} &&\begin{array}{l} \boldsymbol{W}^{\dagger}\boldsymbol{W}|\pm\hat{\boldsymbol{n}}\rangle=\lambda_{\pm}|\pm\hat{\boldsymbol{n}}\rangle,\\ \lambda_{\pm}=A\pm B=\gamma^{2}_{{\rm lower}}+\gamma^{2}_{{\rm upper}}+2\gamma_{{\rm lower}}\gamma_{{\rm upper}}\cos\left(\phi\mp\omega\right), \end{array} \end{eqnarray} \tag{ 19 }$$

where $|\pm \hat {\boldsymbol {n}}\rangle$ are the eigenstates of the spin component along the unit vector $\hat {\boldsymbol {n}}$, i.e.  $\hat {\boldsymbol {n}}\cdot \boldsymbol {\sigma }|\pm \hat {\boldsymbol {n}}\rangle =\pm |\pm \hat {\boldsymbol {n}}\rangle$. For an incoming spinor $|\pm \hat {\boldsymbol {n}}\rangle$ the corresponding transmission amplitudes t_± are [30]

$$\begin{eqnarray} &&|t_{\pm}|=\frac{2\mathrm{j}\sin\left(ka\right)}{|Y-\lambda_{\pm}|}\sqrt{\lambda_{\pm}}, \end{eqnarray} \tag{ 20 }$$

and the outgoing electrons are polarized along a different direction $\pm \hat {\boldsymbol {n}}'$, i.e. their spinor is $|\chi ^{{\mathrm { out}}}_{\pm }\rangle =|\pm \hat {\boldsymbol {n}}'\rangle$. Hence, the transmission amplitude matrix (13) can be rewritten as

$$\begin{eqnarray} &&\mathcal{T}=t_{-}|-\hat{\boldsymbol{n}}'\rangle\langle-\hat{\boldsymbol{n}}|+t_{+}|\:\hat{\boldsymbol{n}}'\rangle\langle\hat{\boldsymbol{n}}|. \end{eqnarray} \tag{ 21 }$$

Here, $|\pm \hat {\boldsymbol {n}}'\rangle$ are the eigenstates of the matrix WW^† [30], namely

$$\begin{eqnarray} &&\boldsymbol{W}\boldsymbol{W}^{\dagger}|\pm\hat{\boldsymbol{n}}'\rangle=\lambda_{\pm}|\pm\hat{\boldsymbol{n}}'\rangle, \end{eqnarray} \tag{ 22 }$$

where the eigenvalues λ_± are given by equations (19). Using equation (9), the matrix WW^† is given by

$$\begin{eqnarray} &&\boldsymbol{W}\boldsymbol{W}^{\dagger}=\gamma^{2}_{{\rm upper}}+\gamma^{2}_{{\rm lower}}+\gamma_{{\rm upper0}}\gamma_{{\rm lower}}\left(u'+u'^{\dagger}\right) \end{eqnarray} \tag{ 23 }$$

with u' = U_upperU^†_lower the unitary matrix representing clockwise hopping from site 1 back to site 1 around the loop.

It is important to emphasize that the spinors $|\pm \hat {\boldsymbol {n}}\rangle$ and $|\pm \hat {\boldsymbol {n}}'\rangle$, being the eigenstates of u + u^† and u' + u'^†, respectively, are completely determined by the AB and AC phases and are independent of the electron energy ε. As we show below, this implies that spin filtering can be achieved independent of energy, with the spin polarization direction controlled solely by the external electric and magnetic fields (see below). In the next subsection we analyze the general conditions for spin filtering arising from the transmission amplitude matrix (21) with the transmission amplitudes (20).

The spin-polarized current (along $\hat {\boldsymbol {n}}'$) at the output of the interferometer is given by [37]

$$\begin{eqnarray} &&I=\frac{e}{\hbar}\int^{\infty}_{-\infty}\frac{\mathrm{d}\varepsilon}{2\pi}\left[f_{{\rm L}}(\varepsilon)-f_{{\rm R}}(\varepsilon)\right]P_{\hat{\boldsymbol{n}}'}(\varepsilon){\rm Tr}\left[\mathcal{T}^{\dagger}\mathcal{T}\right], \end{eqnarray} \tag{ 24 }$$

where f_L,R(ε) = [1 + e^{(ε−μ_L,R)/k_BT} ]⁻¹ is the Fermi distribution in the left (L) or right (R) lead with the corresponding chemical potential μ_L and μ_R, k_B is the Boltzmann constant and T is the temperature. The spin polarization $P_{\hat {\boldsymbol {n}}'}(\varepsilon )$ along $\hat {\boldsymbol {n}}'$ is defined as

$$\begin{eqnarray} &&P_{\hat{\boldsymbol{n}}'}(\varepsilon)\equiv\frac{{\rm Tr}\left[\mathcal{T}^{\dagger}\mathcal{T}\boldsymbol{\sigma}\cdot\hat{\boldsymbol{n}}'\right]}{{\rm Tr}\left[\mathcal{T}^{\dagger}\mathcal{T}\right]}=\frac{|t_{+}|^{2}-|t_{-}|^{2}}{|t_{+}|^{2}+|t_{-}|^{2}}, \end{eqnarray} \tag{ 25 }$$

where in the last step we used equation (21). For $P_{\hat {\boldsymbol {n}}'}(\varepsilon )=\pm 1$ the outgoing electrons with energy ε are fully polarized along $\pm \hat {\boldsymbol {n}}'$. This occurs if and only if |t_∓| = 0, or equivalently λ_∓ = 0 (equation (20)). Without loss of generality, let us consider the case λ₋ = 0, in which the outgoing electrons are polarized along $\hat {\boldsymbol {n}}'$. From equation (19) it follows that λ_± ⩾ 0 and the equality λ₋ = 0 occurs only if

$$\begin{eqnarray} &&\begin{array}{l} \gamma_{{\rm lower}}=\gamma_{{\rm upper}}\equiv\gamma,\\ \cos(\phi+\omega)=-1. \end{array} \end{eqnarray} \tag{ 26 }$$

It should be noted that the spin filtering conditions (26) are valid for an arbitrary two-path interferometer with SOI and AB flux. The details of a specific interferometer enter through the AC phase ω and the effective hopping amplitudes γ_lower and γ_upper. The first condition in equations (26) can be interpreted as a requirement for a symmetry relation between the two paths. The second condition in equations (26), namely ω = −ϕ + π, imposes a relation between the AB flux and the SOI strength.

To achieve an outgoing spin-polarized beam at finite temperature or bias voltage, the spin polarization $P_{\hat {\boldsymbol {n}}'}(\varepsilon )$ should be equal to unity in the relevant energies μ_R − (a few k_BT) < ε < μ_L + (a few k_BT) in which electron transport occurs. Therefore, it is desirable that the spin filtering conditions (26) will be satisfied independent of energy. Since the second condition in equations (26) is energy independent, we focus for the moment on the first condition. From equations (10), one readily sees that this condition holds independent of energy if

$$\begin{eqnarray} &&\begin{array}{l} J_{0a}J_{ab}J_{b1}=J_{0c}J_{cd}J_{d1},\\ \varepsilon_{a}+\varepsilon_{b}=\varepsilon_{c}+\varepsilon_{d},\\ J^{2}_{ab}-\varepsilon_{a}\varepsilon_{b}=J^{2}_{cd}-\varepsilon_{c}\varepsilon_{d}. \end{array} \end{eqnarray} \tag{ 27 }$$

Compared to the corresponding relations in the diamond interferometer [30, 32], equations (27) allow much more freedom for the values of the various parameters (see below). Therefore, spin filtering can be achieved even in a very asymmetric interferometer. Furthermore, as argued at the end of the previous subsection, the spinor of the outgoing electrons, $|\hat {\boldsymbol {n}}'\rangle$, is independent of energy. Hence, the spin filtering is energy independent provided that equations (27) hold.

To satisfy equations (27), one can adopt several approaches. One possibility is to use a single gate electrode for each branch of the interferometer, so that ε_a = ε_b ≡ ε_ab and ε_c = ε_d ≡ ε_cd. The conditions (27) then read

$$\begin{eqnarray} &&\begin{array}{l} J_{0a}J_{b1}=J_{0c}J_{d1},\\ \varepsilon_{ab}=\varepsilon_{cd},\\ J_{ab}=J_{cd}. \end{array} \end{eqnarray} \tag{ 28 }$$

The first condition in equations (28) can be satisfied by properly tuning the hopping amplitudes from the leads to the QDs/QNs, as shown in several experiments [38–41]. The second condition can be satisfied by tuning the gate electrodes. The third condition can be satisfied by controlling the potential barrier between sites a and b (and/or c and d). A further possibility to satisfy equations (27) is by using two gate electrodes on each branch of the interferometer [35]. Then, by tuning two site energies, say ε_a and ε_c, one can fulfill the second and the third conditions in equations (27). The first condition is again satisfied by tuning one of the hopping amplitudes from the leads to the QDs/QNs, say J_b1.

Alternatively, one can work at low temperatures in the linear-response regime, where all the electrons have the same energy, equal to the Fermi energy of the leads ε_F. The first condition in equations (26) should then be satisfied for a single specific energy ε = ε_F. Setting ε_F = 0 in equations (10), one has

$$\begin{eqnarray} &&\frac{J_{0a}J_{ab}J_{b1}}{J^{2}_{ab}-\varepsilon_{a}\varepsilon_{b}}=\frac{J_{0c}J_{cd}J_{d1}}{J^{2}_{cd}-\varepsilon_{c}\varepsilon_{d}}. \end{eqnarray} \tag{ 29 }$$

In this case one has to tune only a single site energy, e.g. ε_a. However, in experiment it may be easier to introduce a single gate electrode for each dot so that one has to tune a single site energy $\tilde {\varepsilon }=\varepsilon _{a}=\varepsilon _{b}$. In addition, one should tune the magnetic field or the electric field perpendicular to the plane in order to satisfy the second condition in equations (26).

Having fulfilled the spin filtering conditions (26), one would like to optimize the transmission T₊ = |t₊|² of the polarized electrons. Using equation (20), the transmission has the form [30]

$$\begin{eqnarray} &&T_{+}=|t_{+}|^{2}=\frac{4\mathrm{j}^{2}\sin^{2}\left(ka\right)\lambda_{+}}{P+Q\cos\left(ka\right)+R\cos\left(2ka\right)}, \end{eqnarray} \tag{ 30 }$$

where

$$\begin{eqnarray} &&\begin{array}{l} P=\left(y_{0}y_{1}-\lambda_{+}\right)^{2}+\left(y_{0}+y_{1}\right)^{2}\mathrm{j}^{2}+\mathrm{j}^{4},\\ Q=2\mathrm{j}\left(y_{0}y_{1}-\lambda_{+}+\mathrm{j}^{2}\right)\left(y_{0}+y_{1}\right),\\ R=2\mathrm{j}^{2}\left(y_{0}y_{1}-\lambda_{+}\right). \end{array} \end{eqnarray} \tag{ 31 }$$

The dependence of the transmission on the magnetic flux is only through λ₊. Substituting ω = −ϕ + π and γ_lower = γ_upper = γ into equations (19), one has λ₊ = 4γ²sin²ϕ. Since one does not expect the tight-binding model to be valid near the band edges, we confine ourselves to the center of the band, ε = 0 or ka = π/2, where the details of the model chosen are not so important. At the band center we have γ(ε = 0) ≡ γ₀ = J³₁/J²₂ with J³₁ ≡ J_0aJ_abJ_b1 = J_0cJ_cdJ_d1 and J²₂ ≡ J²_ab − ε_aε_b = J²_cd − ε_cε_d, as required by equations (27). The denominator in equation (30) becomes , which is minimal at ε₀ = −(J²_0aε_b + J²_0cε_d)γ₀/J³₁ and ε₁ = −(J²_b1ε_a + J²_d1ε_c)γ₀/J³₁. In this case the transmission is T₊ = 4j²λ₊/(λ₊ + j²)², and this has its maximal value of 1 at λ₊ = j². For a specific filter one would usually decide around which flux ϕ₀ one would like to work. We thus optimize the transmission for a specific flux ϕ = ϕ₀. One has a perfect transmission T₊(ε = 0,ϕ = ϕ₀) = 1 at a flux ϕ = ϕ₀ if one tunes the parameters so that $\gamma _{0}=J^{3}_{1}/J^{2}_{2}=\mathrm {j}/\left (2\sin \phi _{0}\right )$, $\varepsilon _{0}=-\left (J^{2}_{0a}\varepsilon _{b}+J^{2}_{0c}\varepsilon _{d}\right )\gamma _{0}/J^{3}_{1}$ and $\varepsilon _{1}=-\left (J^{2}_{b1}\varepsilon _{a}+J^{2}_{d1}\varepsilon _{c}\right )\gamma _{0}/J^{3}_{1}$. With these choices, the transmission T₊(ε = 0,ϕ) reads

$$\begin{eqnarray} &&T_{+}(\varepsilon=0,\phi)=\frac{4\sin^{2}\phi\sin^{2}\phi_{0}}{\left(\sin^{2}\phi+\sin^{2}\phi_{0}\right)^{2}}. \end{eqnarray} \tag{ 32 }$$

The transmission (32) is plotted in figure 3(a) as a function of ϕ for two values of ϕ₀. Figure 3(b) shows the transmission T₊ versus ka for the flux fixed at ϕ = ϕ₀ and for J_0a = J_b1 = J_0c = J_d1 = 2j, J_ab = J_cd = 8j and $\varepsilon _{a}=\varepsilon _{b}=\varepsilon _{c}=\varepsilon _{d}=\sqrt {J^{2}_{ab}-J^{3}_{1}/\gamma _{0}}$. These values correspond to a completely symmetric interferometer in which the hopping amplitudes and site energies of the lower and upper branches are identical. The transmission depends smoothly on energy and remains close to unity in a range around ka = π/2 which increases with increasing ϕ₀. As expected, the transmission in this case resembles the transmission of the symmetric diamond interferometer [30].

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Figures 3(c) and (d) are the same as figure 3(b) but for an asymmetric interferometer. In figure 3(c) we set J_0a = j, J_b1 = 4j, J_0c = J_d1 = 2j, J_ab = J_cd = 8j and $\varepsilon _{a}=\varepsilon _{b}=\varepsilon _{c}=\varepsilon _{d}=\sqrt {J^{2}_{ab}-J^{3}_{1}/\gamma _{0}}$. This choice corresponds to an asymmetric interferometer described by equations (28), in which the two branches have the same site energies ε_a = ε_b = ε_c = ε_d and hopping amplitudes J_ab = J_cd, but different hopping amplitudes from the leads, namely J_0a ≠ J_0c and J_b1 ≠ J_d1. Figure 3(d) shows the transmission of a completely asymmetric interferometer for which J_0a = J_b1 = J_d1 = 2j, J_0c = 1.6j, J_ab = 8j, J_cd = 10j and ε_a ≠ ε_b ≠ ε_c ≠ ε_d. We set ε_a = j and then the values of ε_b, ε_c and ε_d are determined from equations (27) and from the equation $\gamma _{0}=J^{3}_{1}/J^{2}_{2}=\mathrm {j}/\left (2\sin \phi _{0}\right )$. The calculated values are found to be ε_b = 44.22j, ε_c = 43.37j, ε_d = 1.85j and ε_b = 26.38j, ε_c = 24.87j, ε_d = 2.51j for ϕ₀ = 0.1π and 0.2π, respectively. A comparison between figures 3(b)–(d) reveals that the transmission peak at ka = π/2 gets narrower as the interferometer becomes more asymmetric. However, comparing the solid and dashed curves in figures 3(b)–(d), we see that the narrowing of the transmission peak can be circumvented by working at a higher flux ϕ = ϕ₀.

The results of this subsection thus suggest that perfect spin filtering, independent of energy, can be accomplished in an asymmetric interferometer. By tuning the hopping amplitudes, site energies and AB flux, one can simultaneously satisfy the conditions (27) and obtain an ideal transmission of the spin-polarized electrons. The polarization direction of these spin-polarized outgoing electrons is discussed in the next subsection.

Let us consider now the form of the unitary matrices U_uv = e^{iϕ_uv + iK_uvσ} in the presence of Rashba and Dresselhaus SOIs. Firstly, consider the AB phase $\phi _{uv}=-\frac {e}{\hbar }\int ^{v}_{u}\boldsymbol {A}\cdot \boldsymbol {dr}$. With the gauge $\boldsymbol {A}=-By'\hat {\boldsymbol {x}}'$ (figure 2), the AB phases are non-zero only for the bonds ab and cd, with the latter being ϕ_ab = −ϕ_cd = ϕ/2. To account for the Rashba and Dresselhaus SOIs, we denote the angle between the x-axis and the crystallographic (100) axis as ν (figure 2). With respect to the crystallographic axes, the unit vectors along the bonds ab and cd are then $\hat {\boldsymbol {g}}_{ab}=\hat {\boldsymbol {g}}_{cd}=(\cos \nu ,\sin \nu ,0)$. Using equation (3) and denoting

$$\begin{eqnarray} &&\alpha^{2}_{uv}=\alpha^{2}_{{\rm R},uv}+\alpha^{2}_{{\rm D},uv}, \quad \tan\theta_{uv}=\alpha_{{\rm D},uv}/\alpha_{{\rm R},uv} \quad (uv=ab,cd), \end{eqnarray} \tag{ 33 }$$

one ends up with the unitary matrices

$$\begin{eqnarray} &&\begin{array}{l} U_{ab}={\rm e}^{{\rm i}\phi/2+\mathrm{i}\alpha_{ab}\sigma_{ab}}, \quad U_{cd}={\rm e}^{-{\rm i}\phi/2+\mathrm{i}\alpha_{cd}\sigma_{cd}},\\ U_{0a}=U_{b1}=U_{0c}=U_{d1}=\boldsymbol{I}, \end{array} \end{eqnarray} \tag{ 34 }$$

where σ_uv = −sin ξ_uvσ_x + cos ψ_uvσ_y, with ξ_uv = θ_uv + ν and ψ_uv = θ_uv − ν (uv = ab,cd). Note that $\sigma ^{2}_{uv}=F^{2}_{uv}=1+\sin (2\nu)\sin (2\theta _{uv})$ and therefore e^iα_uvσ_uv = c_uv + is_uvσ_uv, with $c_{uv}=\cos (\alpha _{uv}F_{uv})$ and $s_{uv}=\sin (\alpha _{uv}F_{uv})/F_{uv}$. To identify the AC phase ω and the blocked and transmitted spin directions, $-\hat {\boldsymbol {n}}$ and $\hat {\boldsymbol {n}}'$, we calculate the matrices u = U^†_upperU_lower = U^†_abU_cd and u' = U_upperU^†_lower = U_abU^†_cd. Straightforward algebra yields

$$\begin{eqnarray} &&\begin{array}{l} u={\rm e}^{-{\rm i}\phi}\,{\rm e}^{-{\rm i}\alpha_{ab}\sigma_{ab}}\,{\rm e}^{{\rm i}\alpha_{cd}\sigma_{cd}}={\rm e}^{-{\rm i}\phi}\left(\delta+\mathrm{i}\boldsymbol{\tau}\cdot\boldsymbol{\sigma}\right),\\ u'={\rm e}^{{\rm i}\phi}\,{\rm e}^{{\rm i}\alpha_{ab}\sigma_{ab}}\,{\rm e}^{-{\rm i}\alpha_{cd}\sigma_{cd}}={\rm e}^{{\rm i}\phi}(\delta+\mathrm{i}\boldsymbol{\tau}'\cdot\boldsymbol{\sigma}), \end{array} \end{eqnarray} \tag{ 35 }$$

where

$$\begin{eqnarray} &&\begin{array}{l} \delta=c_{ab}c_{cd}+s_{ab}s_{cd}(\sin\xi_{ab}\sin\xi_{cd}+\cos\psi_{ab}\cos\psi_{cd}),\\ \tau_{x}=-\tau'_{x}=s_{ab}c_{cd}\sin\xi_{ab}-s_{cd}c_{ab}\sin\xi_{cd},\\ \tau_{y}=-\tau'_{y}=s_{cd}c_{ab}\cos\psi_{cd}-s_{ab}c_{cd}\cos\psi_{ab},\\ \tau_{z}=\tau'_{z}=s_{ab}s_{cd}(\sin\xi_{cd}\cos\psi_{ab}-\sin\xi_{ab}\cos\psi_{cd}), \end{array} \end{eqnarray} \tag{ 36 }$$

and δ² + |τ|² = 1 from unitarity. Comparing equations (35) with (16), (19), (22) and (23), one derives the following relations:

$$\begin{eqnarray} &&\cos\omega=\delta, \quad \hat{\boldsymbol{n}}=\hat{\boldsymbol{\tau}}, \quad \hat{\boldsymbol{n}}'=-\hat{\boldsymbol{\tau}}'. \end{eqnarray} \tag{ 37 }$$

Equations (36) and (37) show that the transmitted spin direction $\hat {\boldsymbol {n}}'$ differs from the blocked one $-\hat {\boldsymbol {n}}$ in that the components along the x and y axes are reversed.

Let us examine several special cases of equations (36) and (37). Firstly, suppose that one of the branches of the interferometer, say the lower one, is free of SOI. Substituting α_cd = 0 (and therefore c_cd = 1, s_cd = 0), equations (36) take the form

$$\begin{eqnarray} &&\begin{array}{l} \delta = c_{ab},\\ \tau_{x} = -\tau'_{x}=s_{ab}\sin\xi_{ab},\\ \tau_{y}=-\tau'_{y}=-s_{ab}\cos\psi_{ab},\\ \tau_{z}=\tau'_{z}=0. \end{array} \end{eqnarray} \tag{ 38 }$$

Hence, the AC phase in this case is

$$\begin{eqnarray} &&\omega=\alpha_{ab}F_{ab}=\alpha_{{\rm D},ab}\sqrt{1+2\sin(2\nu)\alpha_{{\rm R},ab}/\alpha_{{\rm D},ab}+(\alpha_{{\rm R},ab}/\alpha_{{\rm D},ab})^{2}}. \end{eqnarray} \tag{ 39 }$$

Secondly, if the Rashba mechanism is the dominant SOI, i.e. α_R,uv ≫ α_D,uv, then θ_uv ≈ 0 and equations (36) are reduced to

$$\begin{eqnarray} &&\begin{array}{l} \delta=\cos(\alpha_{{\rm R},ab}-\alpha_{{\rm R},cd}),\\ \tau_{x}=-\tau'_{x}=\sin(\alpha_{{\rm R},ab}-\alpha_{{\rm R},cd})\sin\nu,\\ \tau_{y}=-\tau'_{y}=-\sin(\alpha_{{\rm R},ab}-\alpha_{{\rm R},cd})\cos\nu,\\ \tau_{z}=\tau'_{z}=0. \end{array} \end{eqnarray} \tag{ 40 }$$

The AC phase is then simply ω = α_R,ab − α_R,cd. In the opposite limit where α_D,uv ≫ α_R,uv, one has θ_uv ≈ π/2 and equations (36) give

$$\begin{eqnarray} &&\begin{array}{l} \delta=\cos(\alpha_{{\rm D},ab}-\alpha_{{\rm D},cd}),\\ \tau_{x}=-\tau'_{x}=\sin(\alpha_{{\rm D},ab}-\alpha_{{\rm D},cd})\cos\nu,\\ \tau_{y}=-\tau'_{y}=-\sin(\alpha_{{\rm D},ab}-\alpha_{{\rm D},cd})\sin\nu,\\ \tau_{z}= \tau'_{z}=0. \end{array} \end{eqnarray} \tag{ 41 }$$

Note that in both limits α_R,uv ≫ α_D,uv and α_D,uv ≫ α_R,uv, the AC phase is ω = α_ab − α_cd. This is not surprising, since the Rashba and Dresselhaus interactions are related by a unitary transformation. Furthermore, equations (40) and (41) show that in both limits the polarization of the outgoing electrons is fixed and determined only by the orientation of the crystal axes. For α_R,uv ≫ α_D,uv the direction of spin polarization is $\hat {\boldsymbol {n}}'=-\hat {\boldsymbol {\tau }}'=s_{\omega }(\sin \nu ,-\cos \nu ,0)$ and for α_D,uv ≫ α_R,uv the direction is $\hat {\boldsymbol {n}}'=-\hat {\boldsymbol {\tau }}'=s_{\omega }(\cos \nu ,-\sin \nu ,0)$, where s_ω ≡ sign(sin ω). This is different from the diamond interferometer in which $\hat {\boldsymbol {n}}'$ is a non-trivial function of the SOI strength in both the Rashba and Dresselhaus limits. The origin of this difference is the geometry of the two interferometers. The double-dot interferometer consists of two parallel bonds while the diamond interferometer consists of four non-parallel bonds. Hence, one can have a fixed spin polarization at the output of the interferometer provided that Rashba SOI dominates over the Dresselhaus SOI, or vice versa.