A half-metallic half-Heusler alloy having the largest atomic-like magnetic moment at optimized lattice constant

For half-Heusler alloys, the general formula is XYZ, where X can be a transition or alkali metal element, Y is another transition metal element, typically Mn or Cr, and Z is a group IV element or a pnicitide. The atomic arrangements within a unit-cell show three configurations. Before this study, most of the predictions of half-metallic properties of half-Heusler alloys at the lattice constants differing from their optimized lattice constant. Based on the electropositivity of X and electronegativity of Z for half-Heusler alloys, we found that one of the configurations of LiCrS exhibits half-metallic properties at its optimized lattice constant of 5.803A, and has the maximum atomic-like magnetic moment of 5μB. The challenges of its growth and the effects of the spin-orbit effect in this alloy will be discussed.

For half-Heusler alloys, the general formula is XYZ, where X can be a transition or alkali metal element, Y is another transition metal element, typically Mn or Cr, and Z is a group IV element or a pnicitide.The atomic arrangements within a unit-cell show three configurations.Before this study, most of the predictions of half-metallic properties of half-Heusler alloys at the lattice constants differing from their optimized lattice constant.Based on the electropositivity of X and electronegativity of Z for half-Heusler alloys, we found that one of the configurations of LiCrS exhibits half-metallic properties at its optimized lattice constant of 5.803Å, and has the maximum atomic-like magnetic moment of 5µ B .The challenges of its growth and the effects of the spin-orbit effect in this alloy will be discussed.© 2016 Author(s).All article content, except where otherwise noted, is licensed under a Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).[http://dx.doi.org/10.1063/1.4967365]Since the prediction of half-metallic properties in the half-Heusler of NiMnSb by de Groot et al., 1 much attention has been paid exploring half-metallic half-Heusler alloys for device applications. 2,3n half-metals (HM), one spin channel is metallic while the other channel is insulating.Thus, the density of states (DOS) of a HM at the Fermi energy, E F , is contributed entirely by states of one spin channel.Consequently, the spin polarization can achieve 100%.This characteristic offers exciting and unique opportunities for the occurrence of infinite magnetoresistance as predicted by the Julliére formula. 4eusler alloys have been well studied.The formula characterizing half-Heusler alloys is XYZ where X can be an alkali metal element or a transition metal element, TME, such as Ni in NiMnSb.Y is another TME and Z is either a group IV or a group V element.These alloys have the C1 b crystal structure, which is similar to the L2 1 structure of a full-Heusler alloy (X 2 YZ) except missing one X to form a vacancy. 5There are three possible atomic arrangements for the XYZ in a unit cell, i.e.ZXY, YZX, and XYZ, and we call them α, β, and γ configurations, respectively.The positions of the three atoms, X, Y and Z for each configuration are listed in Table I according to the notation defined by Wyckoff. 6Fig. 1 shows the C1 b structure.Many of half-Heusler alloys have been predicted to be half-metals 7,8 with lattice constants deviating away (mostly larger) from the optimized lattice constants.For example, Damewood et al. examined half-metallic properties and stability of lithiated half-Heusler alloys, LiMnZ (Z=N, P, Si). 8 They found that the studied alloy exhibit half-metallic properties at lattice constants larger than its' optimized lattice constant.In fact, many zinc-blende (ZB) compounds, such as MnZ, are not half-metallic at their equilibrium lattice constants but can also a To whom correspondence should be addressed.E-mail address: rlzhang@nju.edu.cn.be HMs at larger lattice constants. 9Akinaga et al. argued that strained materials are tolerable because it is possible to grow them in thin film forms on substrates, such as CrAs grown on GaAs. 10 However, it is a common sense to expect that devices using strained materials will not have a long life-time.
In order to have a long lived device, the material should at least be metastable.For half-Heusler alloys, the better ones used for device materials should at least be at the optimized lattice constants corresponding to their respective configurations.Furthermore, a large magnetic moment (>3 µ B ) can lead to the Curie temperature, T C , above room temperature. 11Consequently, can the half-metallic Heusler alloys exhibit the largest or even atomic magnetic moments contributed by the 3d-states of the TME?
In this paper, we address the following issue: is it possible to find a configuration of a half-metallic half-Heusler alloy possessing the largest atomic-like magnetic moment, 5 µ B , at an optimized lattice constant so that devices made of the alloy not only can have high T C but also do not suffer large strains for making a long life time spintronic device?Instead of carrying out experimental search, we use the common practice of utilizing spin polarized first principles calculations to predict the possibility.
Shaughnessy et al. 12 suggested a point of view in half-Heusler alloys unifying the d-p and d-d bonding based on the 'electronegativity'.In these alloys, Z is the most electronegative among the three atoms.So, it should be possible for Z to form a d-p bonding with its nearest neighbor (nn) and determine the primary gap: the bonding-antibonding gap.If there are two TME in an alloy, weak bonds between the two TME form the d-d bonding states and contribute to the states near E F resulting in a much smaller gap.By combining the concept of "electropositivity", results of Li-based half-Heusler alloys in β-LiMnZ with pinictide Z, the nearest neighbor to both Li and Mn, are understood.Li is the most electropositive among the three atoms and easily gives up its outermost electron to its neighbor.The most electronegative Z atom gains an electron from Li.To fill its p-shell, Z takes additional two s-electrons from Mn since they are nn pair.The five remaining d-electrons at the Mn form local magnetic moment of 5 µ B per Mn.If Z is a group IV element, then the moment is 4 µ B per Mn. 8 However, the lattice constants are consistently larger than the optimized values.This may be caused by the fact that the pnictides and Si do not have a sufficiently strong enough electronegativity.The residual covalent character between the bonds formed by the TME and Z do not provide enough of a large volume in order to align their spins.If this explanation is true, we should try to use elements at the right side of the pnictides in the periodic  at one column right of the pnictides in the periodic table needs only two electrons to fill its outer valence shell.Mn should be replaced by TME of valence 6 for achieving the largest magnetic moment -the d-shell is half filled.In the following, we report on first principles results of LiCrS meeting the expectation.
The spin-polarized version of VASP 13,14 was used for the calculations.The VASP package provides projector-augmented-wave (PAW) pseudopotentials 15 for Li, Cr, and S that were constructed by using the generalized gradient approximation (GGA) of Perdew et al. (PBE). 16The GGA was also used to treat the exchange-correlation between electrons in the crystal calculations.We used the basis of plane waves with a 1000 eV kinetic energy cutoff for all calculations.The Monkhorst-Pack mesh of (15, 15, 15) was adopted to calculate the total charge density.
In Table II, the lattice constants of the primitive cell, the total energies, and the magnetic moments of α , β , and γ-LiCrS are presented.The α configuration has the moment of 5 µ B at its optimized lattice constant.It is the first case having this unique feature and in fact it has a range of the lattice constants (5.8 -5.83 Å) to have the total energy of -15.650 eV and moment of 5 µ B .We present the total density of states (DOS) for αand β-LiCrS at their optimized lattice constants 5.803Å and 5.795Å in Figs.2(a) and 2(c), respectively.The DOS for the γ configuration is not shown because it exhibits a very similar result to the β configuration, because in both cases, Cr and S are nearest neighbors (nn), and the position of Li atom does not drastically alter the DOS.As shown in Figs.2(a) and 2(c), the gap of spin down channel for α-LiCrS is well defined, while there are states of spin down channel for β-LiCrS overlapping with the states of spin up channel at the Fermi level.Therefore, α-LiCrS demonstrates the half-metal (HM) property, while β configuration is not a half metal at its optimized lattice constant.Based on Julliére formula, 4 the spin polarization of α-LiCrS should be 100% at the Fermi level.However, the actual situation including the surface reconstruction, finite-temperature effects, etc. may lead to the deviation from the theoretical predictions.
In order to understand the states contributing to the gap in α-LiCrS, we plot the partial DOS of the so-called e g (doubly degenerate d z In the β configuration, there is a large gap for the spin down channel even though the E F doesn't fall within it [Fig.2(d)].The valence band is contributed by the t 2g states of Cr and by significant p states of S. They form the bonding part of the bonding-antibonding gap due to hybridization of the t 2g states of Cr and sp 3 states of S under the tetrahedral environment.However, the α configuration is significantly different compared to the β configuration.From Fig. 2(b), the gap of the minority spin channel appears between the p states of S and the d states of Cr.The valence band is contributed primarily by the p states of S. In the α configuration, Cr and S are second neighbors.It's hard to form a bonding-antibonding gap coming from the hybridization of the d states of Cr and the sp 3 states of S. We attribute the gap of minority spin channel to the large electronegativity of the S atom to hold closer the transferred electrons from the nn Li and the second neighbor Cr leading to a large enough volume around the Cr to align its remaining five d-electrons governed by the Pauli principle.Consequently, the largest atomic-like magnetic moment 5 µ B around Cr in α-LiCrS is obtained.
To confirm our attribution, we demonstrated the spin up (ρ ↑ ), spin down (ρ ↓ ), total (ρ ↑ + ρ ↓ ), and spin polarized (ρ ↑ − ρ ↓ ) charge density distributions for α-LiCrS at the optimized lattice constant in Figs.3(a), 3(b), 3(c), and 3(d), respectively.It is found that the charge density approaches zero around Li, and presents small amplitude around S both for spin up and spin down electrons [Figs.3(a) and 3(b)].However, the charge density around Cr demonstrates large amplitude for spin up electrons [Fig.3(a)], and is close to zero for spin down electrons [Fig.3(b)].As a result, the total charge density is strong around Cr and S, and near zero around Li [Fig.3(c)], while the spin-polarized charge density is strong around Cr, and approximately zero around S and Li [Fig.3(d)].This feature comes from the fact that Li gives up its electron to the most electronegative element S.However, S needs another electron to fill its 3p orbit, and it captured the electron from Cr.For Cr, the normal electronic configurations are (3d) 5 (4s) 1 .It becomes (3d) 5 when it losses an electron.The five electrons align the same direction due to the intra-atomic exchange interaction.
It is worthwhile to compare the bonding features of α and β configurations.As to why the β-LiCrS is not a half-metal at its optimized lattice constant.We suggest the following explanations.The β configuration, in principle, would be the best candidate to resolve our issue because it has the lowest total energy.However, the Cr and S form nn pair.Their separation is smaller than the corresponding one in the α configuration.Thus, the S atom takes one electron each from its two nn.It does not leave enough space of the remaining five electrons at Cr to align their spins.
Table II indicates that the lattice constants of α and β configurations are very close.This can possibly present a challenge for the growth of α configuration.One may use a quenching scheme to grow the α configuration at a lattice constant larger than 5.79 Å based on the fact that it has a range of lattice constants (5.803 to 5.83 Å) with the total energy between -15.649 and -15.652 eV.Another issue is the spin-orbit effect for the α configuration.We found that with the spin-orbit interaction, the moment is changed to 4.9993 from .The spin-orbit effect is not negligible.This issue needs further study because there are s-and p-states at E F and the gap states from the S atom.The situation differs from the ones studied in Ref. 17 where the states at E F are d-like.
In conclusion, we predict using first-principles calculations that the α configuration of LiCrS exhibits a favorable property for long life time spintronic applications.It shows a large gap (3.26 eV), the largest atomic-like magnetic moment at its optimized lattice constant.There are remaining challenging issues, such as, the growth and the effect of the spin-orbit interaction.In our system, we only studied the collinear magnetization.The noncollinear magnetization can affect the demonstration of half-metallicity.Generally, the compounds that include Mn may exhibit noncollinear magnetic behaviour.Noncollinear magnetism can also originate from the geometric frustration or magnetic anisotropy.One can try to keep away from the above conditions to avoid the emergence of noncollinear magnetization.
FIG.2.The charge densities of LiCrS in the three configurations at their optimized lattice constants, respectively.

Figs. 4
FIG. 3. Charge density distribution of α-LiCrS at the optimized lattice constant.The letters Cr, S, and Li are used to help reader find the atoms.

FIG. 4 .
FIG.4.The contours of charge density of (a) α-LiCrS, (b) β-LiCrS at their optimized lattice constants, respectively.In the red regions, the amplitude of charge density is larger than the other place, while in the olive regions the amplitude of charge density is smaller than the other place.Arrows (i) and (ii) point to bound charges near the S atoms.The letters Cr, S, and Li are used to help reader find the atoms.

TABLE I .
The positions of the three atoms, X, Y and Z, and the vacancy in terms of the Wyckoff notation: table.A new physical condition appears.Any element FIG. 1. Diagrammatic sketch of C1 b structure.

TABLE II .
The optimized lattice constants, total energies, and magnetic moments for three configurations of LiSCr alloys.