Spin Hall effect emerging from a noncollinear magnetic lattice without spin–orbit coupling

The existence of the SHE and AHE in metals is determined by symmetry (in insulators apart from symmetry also the topology of the electronic structure is important). The symmetry of magnetic systems is normally described in terms of magnetic space groups, which contain, apart from the spatial symmetry operations, also the time-reversal symmetry operation. In absence of SOC (or other terms in the Hamiltonian that couple the magnetic moments to the lattice such as the shape anisotropy), however, the symmetry of the magnetic systems is higher than that contained in the magnetic space groups since the spins can be rotated independently of the lattice. This can be illustrated by considering the following minimal Hamiltonian

$$\begin{eqnarray}&&H=t\displaystyle \sum _{\langle {ij}\rangle \alpha }{c}_{i\alpha }^{\dagger }{c}_{{j}_{\alpha }}-J\displaystyle \sum _{i\alpha \beta }{({\boldsymbol{\sigma }}\cdot {{\boldsymbol{n}}}_{i})}_{\alpha \beta }{c}_{i\alpha }^{\dagger }{c}_{i\beta },\end{eqnarray} \tag{ 1 }$$

known as the double-exchange model (s–d model) [24–26] which describes itinerant s electrons interacting with local d magnetic moments. We assume that magnetic moments are only contributed by the spin degrees of freedom. Here, α and β stand for spin up and spin down, respectively. The first term is the nearest neighbor hopping term with $\langle {ij}\rangle$ denoting the nearest neighbor lattice sites. In the second term, J is the Hund's coupling strength between the conduction electron and the on-site spin moment, ${\boldsymbol{\sigma }}$ is the vector of Pauli matrices, and ${{\boldsymbol{n}}}_{i}$ is the spin magnetic moment on-site i. The magnetic texture is defined by the pattern of ${{\boldsymbol{n}}}_{i}$. In such a Hamiltonian, spin rotation only rotates the magnetic moments n_i. The corresponding symmetry groups are thus formed by combining the magnetic space groups with spin rotations [27, 28]. Such symmetry groups are referred to as spin-space groups. They apply generally for non-interacting Hamiltonians in absence of SOC.

We focus here only on the intrinsic contribution to the AHE and SHE, however, the other (extrinsic) contributions have the same symmetry and thus the symmetry discussion in the following is general. The intrinsic AHE and intrinsic SHE are well characterized via the Berry curvature formalism [1, 5, 10, 13]. The anomalous Hall conductivity (AHC) σ_αβ can be evaluated by the integral of the Berry curvature ${{\rm{\Omega }}}_{\alpha \beta }^{n}({\bf{k}})$ over the Brillouin zone (BZ) for all the occupied bands, where n is the band index. It should be noted that this method can also be applied to the THE, although it is commonly interpreted using the real space spin texture. Here, σ_αβ corresponds to a 3 × 3 matrix and indicates a transverse Hall current j_α generated by a longitudinal electric field E, which satisfies ${J}_{\alpha }={\sigma }_{\alpha \beta }{E}_{\beta }$. Within a linear response, Berry curvature can be expressed as

$$\begin{eqnarray}&&{{\rm{\Omega }}}_{n,\alpha \beta }(\overrightarrow{k})=2{\rm{i}}{{\hslash }}^{2}\displaystyle \sum _{m\ne n}\displaystyle \frac{\langle {\psi }_{n{\bf{k}}}| {\hat{v}}_{\alpha }| {\psi }_{m{\bf{k}}}\rangle \langle {\psi }_{m{\bf{k}}}| {\hat{v}}_{\beta }| {\psi }_{n{\bf{k}}}\rangle }{{({E}_{n}(\overrightarrow{k})-{E}_{m}(\overrightarrow{k}))}^{2}},\end{eqnarray} \tag{ 2 }$$

where n and m are band indices, and ${\psi }_{n{\bf{k}}}$ and ${E}_{n{\bf{k}}}$ denote the Bloch wave functions and eigenvalues, respectively, and $\hat{{\bf{v}}}$ is the velocity operator. Replacing the velocity operator with the spin current operator ${\hat{J}}_{\alpha }^{\gamma }=\displaystyle \frac{1}{2}\{{\hat{v}}_{\alpha },{\hat{s}}_{\gamma }\}$, where ${\hat{s}}_{\gamma }$ is the spin operator, we obtain the spin Berry curvature and corresponding spin Hall conductivity (SHC)

$$\begin{eqnarray}{\sigma }_{\alpha \beta }^{\gamma } & = & \displaystyle \frac{e}{{\hslash }}\displaystyle \sum _{n}{\displaystyle \int }_{{\rm{BZ}}}\displaystyle \frac{{{\rm{d}}}^{3}{\bf{k}}}{{(2\pi )}^{3}}{f}_{n}({\bf{k}}){{\rm{\Omega }}}_{n,\alpha \beta }^{\gamma }({\bf{k}}),\\ {{\rm{\Omega }}}_{n,\alpha \beta }^{\gamma }({\bf{k}}) & = & 2{\rm{i}}{{\hslash }}^{2}\displaystyle \sum _{m\ne n}\displaystyle \frac{\langle {\psi }_{n{\bf{k}}}| {\hat{J}}_{\alpha }^{\gamma }| {\psi }_{m{\bf{k}}}\rangle \langle {\psi }_{m{\bf{k}}}| {\hat{v}}_{\beta }| {\psi }_{n{\bf{k}}}\rangle }{{({E}_{n{\bf{k}}}-{E}_{m{\bf{k}}})}^{2}}.\end{eqnarray} \tag{ 3 }$$

The SHC (${\sigma }_{\alpha \beta }^{\gamma };$ α, β, γ = x, y, z) is a third-order tensor (3 × 3 × 3) and represents the spin current ${J}_{s,\alpha }^{\gamma }$ generated by an electric field E via ${J}_{s,\alpha }^{\gamma }={\sigma }_{\alpha \beta }^{\gamma }{E}_{\beta }$, where ${J}_{s,\alpha }^{\gamma }$ is a spin current flowing along the α-direction with the spin-polarization along the γ-direction, and f_n(k) is the temperature dependent Fermi–Dirac distribution.

We know that AHE vanishes while SHE remains if the time-reversal symmetry (operator T) exists in the system. In equation (2), T reverses the velocities ${\hat{v}}_{\alpha ,\beta }$ and brings an additional '−' sign by the complex conjugate. Thus, σ_αβ = 0 owing to ${{\rm{\Omega }}}_{n,\alpha \beta }(\overrightarrow{k})=-{{\rm{\Omega }}}_{n,\alpha \beta }(\overrightarrow{-k})$. In contrast, In equation (3), T generates one more '−' sign by reversing the spin in ${J}_{\alpha }^{\gamma }$. Then, ${\sigma }_{\alpha \beta }^{\gamma }$ can be nonzero since ${{\rm{\Omega }}}_{n,\alpha \beta }^{\gamma }(\overrightarrow{k})$ is even in k-space. In a magnetic system without SOC, T is broken, but a combination of T and a spin rotation (operator S) can still be a symmetry. For example, a coplanar magnetic system shows a TS symmetry, in which S rotates all spins by 180° around the axis perpendicular to the plane. Since S does not act on ${\hat{v}}_{\alpha ,\beta }$, TS causes vanishing σ_αβ just as T alone. In a general non-coplanar magnetic lattice, the TS symmetry is naturally broken, because one cannot find a common axis about which all spins can be rotated 180° at the same time, and thus the AHE can exist without SOC.

The situation is different for the SHE since ${J}_{\alpha }^{\gamma }$ in equation (3) contains an additional spin operator. As a consequence, (assuming that S is a rotation around the z axis) TS forces ${{\rm{\Omega }}}_{n,\alpha \beta }^{x/y}(\overrightarrow{k})$ to be odd where spin ${\hat{s}}_{x}$ or ${\hat{s}}_{y}$ is reversed by TS, but ${{\rm{\Omega }}}_{n,\alpha \beta }^{z}(\overrightarrow{k})$ to be even where ${\hat{s}}_{z}$ is unchanged by TS. Then, one can obtain zero ${\sigma }_{\alpha \beta }^{x/y}$ but nonzero ${\sigma }_{\alpha \beta }^{z}$. In a collinear magnetic lattice there exists more than one spin rotation S such that TS is a symmetry of the system and thus all of the ${\sigma }_{\alpha \beta }^{\gamma }$ components have to vanish. Therefore, we can argue that SHE can exist without SOC in general noncollinear magnetic lattices, regardless of FM, AFM, or the scalar spin chirality (coplanar or non-coplanar). In contrast, the AHC is zero for a coplanar magnetic lattice (zero scalar spin chirality), since TS acts as T alone in equation (2).

When SOC is included, SHE is allowed by symmetry in any crystal [29], while the AHE on the other hand can be present in magnetic systems that are not symmetric under time reversal combined with a translation or inversion (for example, a conventional collinear AFM). We summarize the necessary conditions for the existence of AHE and SHE in systems with and without SOC in figure 1.

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To demonstrate that the SHE can indeed be nonzero without SOC, we consider the s–d Hamiltonian (1) projected on a kagome lattice with the so-called q = 0 magnetic order, shown in figure 2(a). Such a coplanar AFM order is well studied in theory [14, 30] and appears in many realistic materials even at room temperature, for example Mn₃X (X = Ir, Ga, Ge, and Sn) [22, 23, 31–34] as we discuss in the following. For comparison, the SOC effect is also considered by adding to H in equation (1)

$$\begin{eqnarray}&&{H}_{\mathrm{so}}={{\rm{i}}{t}}_{2}\displaystyle \sum _{\langle {ij}\rangle \alpha \beta }{v}_{{ij}}{({\boldsymbol{\sigma }}\cdot {{\boldsymbol{n}}}_{{ij}})}_{\alpha \beta }{c}_{i\alpha }^{\dagger }{c}_{j\beta },\end{eqnarray} \tag{ 4 }$$

where v_ij is the antisymmetric Levi-Civita symbol and n_ij are a set of coplanar vectors anticlockwise perpendicular to the lattice vector R_ij, as defined in [30] and t₂ is the SOC strength.

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We first analyze the symmetry of the SHC tensor for the q = 0 magnetic order. Note that we use the Cartesian coordinate systems defined in figure 2. As discussed above, the existence of the TS symmetry leaves only ${\sigma }_{\alpha \beta }^{z}$ terms in the absence of SOC. Further, the combined symmetry ${{TM}}_{x}$, in which M_x is the mirror reflection along x and flips ${\hat{s}}_{z}$ and ${\hat{v}}_{x}$ in equation (3), leads to ${\sigma }_{{xx}}^{z}={\sigma }_{{yy}}^{z}=0$. We further obtain only two nonzero SHC tensor element ${\sigma }_{{xy}}^{z}=-{\sigma }_{{yx}}^{z}$ by considering the three-fold rotation around z. The magnetic order shown in figure 2(b) will also be relevant for the discussion in the following. This magnetic configuration differs from the q = 0 case only by a two-fold spin rotation around the y-axis and thus, without SOC its symmetry is exactly the same as that of the q = 0 case.

Setting the Hund coupling constant $J=1.7t$ and SOC strength t₂ = 0, we calculate the spin Berry curvature via equation (3). As expected, we find nonzero SHC ${\sigma }_{{xy}}^{z}$ fully in agreement with the symmetry considerations. Figures 3(a) and (b) show the band structures with t₂ = 0 and ${t}_{2}=0.2t$, respectively. One can see that SOC modifies slightly the band structure by gaping some band crossing points such as the BZ corners (K). Without SOC, we already observe nonzero ${\sigma }_{{xy}}^{z}$, while adding SOC reduces ${\sigma }_{{xy}}^{z}$ slightly at the Fermi energy that is set between the first and second bands at about −2.7 eV. We plot corresponding spin Berry curvature ${{\rm{\Omega }}}_{{xy}}^{z}$ in the hexagonal BZ In figures 3(d) and (e). Large ${{\rm{\Omega }}}_{{xy}}^{z}$ appears in the BZ without SOC, leading to net ${\sigma }_{{xy}}^{z}$. The SOC simply brings an extra contribution to ${\sigma }_{{xy}}^{z}$ at the band anti-crossing region around the K point.

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After proving the principle, we now identify materials that show strong SHE with negligible contribution from SOC. We naturally consider Mn₃X (X = Ga, Ge, Sn, and Ir) compounds, since they exhibit non:w collinear AFM order at room temperature (the AFM Néel temperature is over 365 K). Our recent ab initio calculations showed a sizable intrinsic SHE by including SOC [34]. Here, we further point out that SHE still presents without SOC and SOC actually plays a negligible effect for SHE in these materials.

The primitive unit cell of Mn₃Ga, Mn₃Ge and Mn₃Sn (space group $P{6}_{3}/{mmc}$, No. 194) includes two Mn₃X planes that are stacked along the c-axis according to a '–AB–AB–' sequence. Inside each plane, Mn atoms form a kagome-type lattice with Ga, Ge, or Sn lying at the center of the hexagon formed by Mn. Both the ab initio calculation [31] and neutron diffraction measurements [35–37] show that the Mn magnetic moments exhibit noncollinear AFM order, where the neighboring moments are aligned at an angle of 120°, as in figure 2(b). Large AHE in room temperature has recently been reported in Mn₃Ge and Mn₃Sn [22, 23]. These materials also exhibit other exotic phenomena including the Weyl semimetal phase [38], magneto-optical Kerr effect [39], anomalous Nernst effect [40], and topological defects [41]. Distinct from hexagonal Mn₃X compounds, the Mn₃Ir (space group ${Pm}\bar{3}m$, No. 221) crystallizes in a face-centered cubic structure with Mn atoms in the [111] planes forming a kagome lattice with the q = 0 magnetic order.

The symmetry of the SHE without SOC in these materials can be understood using a similar approach as we used for the 2D kagome lattice. The hexagonal Ga, Ge, and Sn materials can be viewed as stacking versions of the kagome lattice and thus we find that the symmetry of SHE is the same as the 2D kagome lattice, i.e. only ${\sigma }_{{xy}}^{z}=-{\sigma }_{{yx}}^{z}$ is nonzero. However, we find that SHE must vanish in Mn₃Ir without SOC, which is imposed by the higher symmetry of the cubic magnetic lattice. For completeness, we list the tensor matrices without and with SOC for all these compounds in the supplementary material available online at stacks.iop.org/njp/20/073028/mmedia.

Since the SHC tensor shape imposed by the symmetry has been systematically investigated for these materials in [34], we only discuss one of the largest SHC tensor elements ${\sigma }_{{xy}}^{z}$ based on the ab initio calculations [42] of the SHC. For comparison, we show the SHC without and together with SOC in figure 4. In the absence of SOC, Ga, Ge, and Sn compounds indeed exhibit nonzero SHC ${\sigma }_{{xy}}^{z}=-613,115,$ and $90({\hslash }/e){({\rm{\Omega }}\mathrm{cm})}^{-1}$, respectively, at the Fermi energy. One can see that SOC induces very few changes in the band structure and thereafter modifies the SHC weakly, especially at the Fermi energy for Ga, Ge, and Sn compounds. It is intuitive to observe comparable ${\sigma }_{{xy}}^{z}$ values for Ge and Sn compounds, despite the fact that Sn exhibits much larger SOC than Ge. These facts further verifies that the noncollinear magnetic structure, rather than SOC, is dominant for the SHE. The Ga compound shows an opposite sign in SHC compared to the Ge/Sn compound, since Ga has one valence electron fewer than Ge/Sn and the Fermi energy is lower in Mn₃Ga than in Mn₃Ge/Mn₃Sn.

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