frustrated magnetism on hollandite lattice
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Frustrated magnetism on Hollandite lattice Saptarshi Mandal (ICTP, - PowerPoint PPT Presentation

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


  1. 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

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

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

  4. 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) ◮ α MnO 2 compounds, ex. BaMn 8 O 16 , KMn 8 O 16 . Comes in hollandite and ramsdellite lattice structures. ◮ β MnO 2 appears in rutile structures. ◮ γ MnO 2 a combination of ramsdellite α MnO 2 and rutile β MnO 2 domains. So far the best material for battary use.

  5. α MnO 2 and its properties ◮ BaMn 8 O 16 , KMn 8 O 16 , α − Mn 0 2 ˙ nH 2 0 ◮ a = 2 . 86˚ b = 2 . 91˚ c = 3 . 44˚ A , A , A ◮ Diameters of pores ∼ 4 . 6˚ A ◮ Spin moment of Mn is 3 2

  6. 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).

  7. 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 multi- ferroics, Scientific Reports 4, 6203 (2014)

  8. Hollandite lattice: lattice of α MnO 2

  9. Modelling α MnO 2 ◮ Mn-O-Mn angle varies 100 ◦ to 130 ◦ ◮ Goodenough- Kanamori-Anderson rule ◮ DFT insights ◮ Expeimental insight � � S i .� � � S i .� � � S i .� H = J 1 S j + J 2 S j + J 3 S j (1) � ij � � ij � 2 � ij � 3

  10. Method.. k e i � K .� ◮ Interaction matrix method, � R i � S i ,α = � S k ,α ◮ H = � S k ,α H α,β ( k ) � H α,β ( k ) � S k ,β = λ k min � S k ,β , S k ,β ◮ Numerical Simulation, inhomogenous meanfield method. ij J ij � S i .� i � h i .� � j J ij � ◮ H = � S j , H = � S j , h i = � S j ◮ bravais vs non-bravais lattice, λ k min = E site H = H α,β ( k ) � k ,α � S c S c k ,β , α, β ∈ 1 , 4; 2 1 4 6 3 1 I 5 2 3 E = J 1 cos 2 φ − ( | J 2 | + 2 | J 3 | ) cos φ 4 5 6 II 1 2 3 E = − J 1 − ( | J 2 | +2 | J 3 | ) 2 1 cos φ = − | J 2 | +2 | J 3 | 4 , 5 3 IV 6 8 J 1 4 J 1 5 2 III 4 E φ = − 2 . 125 , φ = 138 . 66 6

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

  12. s α � = cos(2 n φ )ˆ x + sin(2 n φ )ˆ z , n s β � = cos((2 n + 1) φ )ˆ x + sin((2 n + 1) φ )ˆ z . n

  13. s α s α 1 s α � = cos(2 n φ )ˆ x + sin(2 n φ )ˆ z , � = − � n n n s β s β 1 s β � = cos((2 n + 1) φ )ˆ x + sin((2 n + 1) φ )ˆ z , � = − � n . n n

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

  15. Comparision with Ising model degeneracy!! ◮ E GFP = − J 1 , E hel = − J 1 − ( | J 2 | +2 | J 3 | ) 2 8 J 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.

  16. Chirality and degeneracy Definition: C J 1 , J 2(3) = � s 1 × � s 2 + � s 2 × � s 3 + � s 3 × � s 1 . = ± (2 sin φ + sin 2 φ )ˆ z ◮ For J 1 Ferromagnetic the system is not frustrated and simple colinear magnetism is observed.

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

  18. Ground state magnetisation and susceptibility | � F M ( � Q ) | / | � F M ( � Q max ) | Phase (0,2 φ -7) (1,2 φ -7) (2,2 φ -7) (0,2 φ -6) (1,2 φ -6) (2,2 φ -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: 1 � � � χ µ,λ = � 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) .

  19. Effect of magnetic field: �� �� ��� ��� �� �� ��� ��� �� �� ��� ��� ��� ��� �� �� ��� ��� �� �� C−H �� �� �� �� �� �� �� �� �� �� �� �� �� �� �� �� �� �� 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

  20. Effect of perpendicular magnetic field � � N t ◮ H = H 0 + h ⊥ (1 − ∆ 2 i =1 s y , i , � s i = i ) � s 0 , i − ∆ i ˆ y � (1 − ∆ 2 j ) � � ◮ self-consistent Eq: � √ − J ij i ) � s 0 , i .� s 0 , j + J ij ∆ j = h ⊥ j ∈ i (1 − ∆ 2 h ⊥ � ◮ � (1 − ∆ 2 ) � s i = s 0 , i − ∆ˆ y , ∆ = 2( J 1 + J 2 +2 J 3 − E GS ) h 2 ◮ Ground state energy: E = E GS − ⊥ 4( J 1 + J 2 +2 J 3 − E GS ) h ⊥ 1 ◮ m ⊥ = 2( J 1 + J 2 +2 J 3 − E GS ) , χ ⊥ = 2( J 1 + J 2 +2 J 3 − E GS ) > 0 . ◮ Critical Magnetic field h c y = 2( J 1 + J 2 + 2 J 3 − E GS ) .

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

  22. Susceptibility 2 1 −1 0.8 0.6 J 3 J 1 0.4 1 0.2 0.0 −2 −4 −2 2 4 J 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

  23. Scientific Reports 4 , 6203 (2014) a minimal model in- cludes 4 different near- est neighbour coupling, J 1 , J 2 , J 3 , J 4 . Could be frustrating. The transition tempera- ture 50 o K is identical to that of α -MnO 2 , indicat- ing that dynamics along Mn − Mn ladder is the main factor for finite T mechanism.

  24. 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 temper- ature ( T C )=180 K via partial Co sustitution for Mn.

  25. Thank you

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