Frustrated magnetism on Hollandite lattice

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Frustrated magnetism on Hollandite lattice Saptarshi Mandal (ICTP, Trieste, Italy) Acknowledgment: A. Andreanov(MPIKS, Dresden) Y. Crespo and N. Seriani(ICTP, Italy) Workshop on Current Trends in Frustrated Magnetism, JNU, New Delhi February 13, 2015

Plan of the Talk Introduction Hollandite lattice and αmno 2 Experimental Magnetic properties Model Phase diagram Effect of external magnetic field Recent interest

Frustrated Magnetism: AFM interaction or FM + AFM interaction Geometrical Frustration Triangular lattice Kagome lattice Pyrochlore lattice

MnO 2 and its various compounds Wide applications Catalyst for oxygen reduction reaction. W. Xiao et al, J. Phys. Chem.C 114 1694 (2010) Microbial fuell cell. RSC Adv. 3 7902 (2013) Electrode materials for Li-ion batteries, Lithiam-air batteries. Supercapacitor G. -R. Li et al Langmuir 26, 2209 (2010) αmno 2 compounds, ex. BaMn 8 O 16, KMn 8 O 16. Comes in hollandite and ramsdellite lattice structures. βmno 2 appears in rutile structures. γmno 2 a combination of ramsdellite αmno 2 and rutile βmno 2 domains. So far the best material for battary use.

αmno 2 and its properties BaMn 8 O 16, KMn 8 O 16, α Mn0 2 ṅh 2 0 a = 2.86Å, b = 2.91Å, c = 3.44Å Diameters of pores 4.6Å Spin moment of Mn is 3 2

Previous experimental finding. AFM state for K <0.7 MnO 2 (synthesized with hydrothermal technique), N. Yamamoto et al Jpn. J. Appl. Phys. 13, 723 (1974). AFM transition for K 0.16 MnO 2 at T N = 18 K Strobel et al J. Sol. State Chem. 55, 67 (1984) A helical magnetic structure was also suggested for K 0.15 MnO 2. H. Sato et al J. Alloys Comp. 262263, 443 (1997). FM state for 52K to 20K for K 1.5 (H 3 O) x Mn 8 O 16 Below 20 K spatial anisotropic susceptibilities indicate a helical ground state. H. Sato et al Phys. Rev. B 59, 12836 (1999). Spin glass behaviour for K x MnO 2 (0.087 < x 0.125). J. Luo et al J. Phys. Chem. C 114, 8782 (2010). J. Luo et al J. Appl. Phys. 105, 093925 (2009), X.-F. Shen et al J. Am. Chem. Soc. 127, 6166 (2005).

Recent interest Spin-glass behaviour with Ising model. Y. Crespo et al Phys. Rev. B 88, 014202 (2013) Electronic and magnetic properties..ab initio calculations. Y. Crespo et al Phys. Rev. B 88, 0144428 (2013) other works.. Hollandite as a new class of multiferroics, Scientific Reports 4, 6203 (2014)

Hollandite lattice: lattice of αmno 2

Modelling αmno 2 Mn-O-Mn angle varies 100 to 130 Goodenough- Kanamori-Anderson rule DFT insights Expeimental insight H = J 1 Si. S j + J 2 Si. S j + J 3 Si. S j (1) ij ij 2 ij 3

Method.. Interaction matrix method, S i,α = k ei K. R i Sk,α H = S k,α H α,β (k) S k,β, H α,β (k) S k,β = λ kmin Sk,β Numerical Simulation, inhomogenous meanfield method. H = ij J ij S i. S j, H = i h i. S j, hi = j J ij S j bravais vs non-bravais lattice, λ kmin = E site 2 4 1 6 3 I 5 2 4 5 6 II 1 3 2 1 5 4 3 IV 6 5 2 III 4 6 3 1 H = H α,β (k) Sk,α c Sk,β c, α, β 1, 4; E = J 1 cos2φ ( J 2 + 2 J 3 )cosφ E = J 1 ( J 2 +2 J 3 ) 2 8J 1, cos φ = J 2 +2 J 3 4J 1 E φ = 2.125, φ = 138.66

Ground states of Hollandite lattice ( S.M et. al. Phys. Rev. B 90, 104420, (2014) J 1 AFM FM J 2 J 3 J 2 J 1 AFM FM J 2 J 3 J 2

s n α s n β = cos(2nφ)ˆx + sin(2nφ)ẑ, = cos((2n + 1)φ)ˆx + sin((2n + 1)φ)ẑ.

s n α = cos(2nφ)ˆx + sin(2nφ)ẑ, s α 1 n s n β = cos((2n + 1)φ)ˆx + sin((2n + 1)φ)ẑ, s β 1 n = s n α = s β n.

phase diagram-i C2 AFM (0,2) C AFM J 2 J1 C2 H C H ( 4,0) F H A2 H (4,0) FM (0, 2) A2 AFM J 3 J 1 E col = J 1 J 2 2 J 3, E hel = J 1 ( J 2 +2 J 3 ) 2 8 θ = 2φ Each helical state is continually connected with co-linear phase.

Comparision with Ising model degeneracy!! E GFP = J 1, E hel = J 1 ( J 2 +2 J 3 ) 2 8J 1 Area of the GFP is smaller than the area of helical phase. Boundary between GPF is discontinuous but the boundary between helical and colinear phase is continuous. Macroscopic degeneracies of GPF is absent in helical phase.

Chirality and degeneracy Definition: C J1,J 2(3) = s 1 s 2 + s 2 s 3 + s 3 s 1. = ±(2 sin φ + sin 2φ)ẑ For J 1 Ferromagnetic the system is not frustrated and simple colinear magnetism is observed.

Neutron diffraction pattarn Magnetic structure factor F( Q) = 1 Nt Nt l=1 ei Q. r l S( rl ) = ( N 5,7 uc k=1 j 1,3 s k,je i Q. d j + 6,8 j 2,4 s k,je i Q. d j )e i Q. R k. 2 1.5 N y 0.5 C H C2 H A2 H F H 0.0 0.0 0.5 N x 1.5 2 1.0 0.8 0.6 0.4 0.2 0.0 _ 0.2 The position of peaks and the value of F M ( Q) / F M ( Q max ) is different for each phase.

Ground state magnetisation and susceptibility F M ( Q) / F M ( Q max ) Phase (0,2φ-7) (1,2φ-7) (2,2φ-7) (0,2φ-6) (1,2φ-6) (2,2φ-6) C-H 0.5618 0 0.0597 1 0 0.1063 C2-H 0 0.5618 0 0 1 0 A2-H 0 1 0 0 0.5933 0 F-H 1 0 0.1063 0.6566 0 0.0698 Magnetisation: m µ = i s i,µ = 0 Non zero susceptbility tensor: χ µ,λ = 1 ( s i,µ s j,λ s i,µ s j,λ ) = χδ µ,λ, µ, λ x, z N i,j = 0.5 (2) H. Sato, et. al. J. Alloys Comp. 262263, 443 (1997).

Effect of magnetic field: 00 11 00 11 000 111 00 000 111 000 111 11 000 111 00 11 000 111 00 11 C H 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 H. Sato et al. Phys. Rev. B 59, 12836,(1999) Unknown FM state between T 2 and T 3 We consider T=0. H is the magnetic field applied perpendicular to the plane of polarization H is the magnetic field applied along the plane of polarization

Effect of perpendicular magnetic field H = H 0 + h Nt i=1 s y,i, s i = (1 2 i ) s 0,i i ŷ (1 2 j ) self-consistent Eq: ( ) j i J ij s (1 2 0,i. s 0,j + J ij j = h i ) s i = (1 2 ) s 0,i ŷ, = h 2(J 1 +J 2 +2J 3 E GS ) Ground state energy: E = E GS m = h 2 4(J 1 +J 2 +2J 3 E GS ) h 2(J 1 +J 2 +2J 3 E GS ), χ = 1 2(J 1 +J 2 +2J 3 E GS ) > 0. Critical Magnetic field h c y = 2(J 1 + J 2 + 2J 3 E GS ).

Effect of parallel magnetic field, H = H 0 + h Nt i=1 s x,i. s α n = cos(θ α n + δθ α n )ˆx + sin(θ α n + δθ α n )ẑ, α β. δθ (α,β) n = h sin θ n (α,β) ). 2 (J 1 cos 2 (2φ)+(J 2 +2J 3 ) cos 2 φ E GS E = E GS Magnetisation: m x = h 2 8(J 1 cos 2 (2φ)+(J 2 +2J 3 ) cos 2 φ E GS ) h 4(J 1 cos 2 2φ+(J 2 +2J 3 ) cos 2 φ E GS ). Susceptibility: χ = 1 4(J 1 cos 2 (2φ)+(J 2 +2J 3 ) cos 2 φ E GS ) No critical field, For strong parallel field, the spins are canted perpenicular to the field.

Susceptibility 2 1 J 3 J 1 1 1 0.8 0.6 0.4 0.2 0.0 2 4 2 J 2 4 2 J 1 temperature dependence of Susceptibilities? χ > χ, χ χ = 0.87 χ, χ, φ, from neutron diffraction may help to determine J 1, J 2, J 3

Scientific Reports 4, 6203 (2014) a minimal model includes 4 different nearest neighbour coupling, J 1, J 2, J 3, J 4. Could be frustrating. The transition temperature 50 o K is identical to that of α-mno 2, indicating that dynamics along Mn Mn ladder is the main factor for finite T mechanism.

A. M. Larson et. al, Inducing Ferrimagnetism in Insulating Hollandite Ba 1.2 Mn 8 O 16,Chem of Mat. Ba x Mn 8 O 16 from a complex AFM with (T N ) =25 K to a ferrimagnet with Curie temperature (T C )=180 K via partial Co sustitution for Mn.

Thank you