Magnetoelastics in the frustrated spinel ZnCr 2 O 4. Roderich Moessner CNRS and ENS Paris

Transcription

1 Magnetoelastics in the frustrated spinel ZnCr 2 O 4 Roderich Moessner CNRS and ENS Paris CEA Saclay, June 2005

2 Overview Neutron scattering experiments on ZnCr 2 O 4 Modelling magnetism on the B sublattice of the spinels degeneracy and instability Frustrated magnetoelastics theory for single tetrahedron Landau theory for phase transition dynamics: string modes Collaborators Oleg Tchernyshyov (Johns Hopkins); Shivaji Sondhi (Princeton); John Chalker (Oxford) Collin Broholm (Johns Hopkins); S.-H. Lee (NIST)

3 Experiments on the spinel ZnCr 2 O 4 I S.-H. Lee et al. Neutron scattering: lattice parameters and density of states

4 Experiments on the spinel ZnCr 2 O 4 II S.-H. Lee et al.

5 The simple spinel oxides AB 2 O 4 (after Takagi) d 0.5 d 1.5 d 2.5 d 3.5 LiTi 2 O 4 LiV 2 O 4 AlV 2 O 4 LiMn 2 O 4 BCS SC heavy Fermion charge-ordered d 1 d 2 d 3 d 4 MgTi 2 O 4 {Zn,Mg,Cd}V 2 O 4 {Zn,Mg,Cd}Cr 2 O 4 ZnMn 2 O 4 valence spin+orbital spin+structural bond solid ordering phase transition ions on B-sublattice form pyrochlore lattice properties tunable by varying ions on A, B sublattices many more compounds exist: ZnCr 2 O 4 is one of many... need to study elastic degrees of freedom

6 Modelling ZnCr 2 O 4 Zn, O nonmagnetic nominal valence of Cr: d 3 (half-filled t 2g orbitals) isotropic S = 3/2 Cr sublattice: pyrochlore structure corner-sharing tetrahedra highly frustrated classical Heisenberg model on pyrochlore lattice Q: Interplay of elastic degrees of freedom and frustration? Yamashita+Ueda

7 Magnetoelastic Hamiltonian H tot = H m + H me + H e magnetic exchange H m = J ij S i S j magnetoelastic coupling (x a... displacements) H me = aij dj ij dx a (S i S j )x a elastic energy H e = ab k abx a x b (k ab... elastic constants)

8 Unfrustrated magnetoelastics: chain in d = 1 S i S j = c nn is uniform for nearest neighbours Simplest case: dj ij /dx a = J δ a,i : H me + H e = a J c nn x a +kx 2 a minimised by x a = J c nn /(2k) = E min = (J c nn ) 2 /(4k) grows with c nn H m minimised by extremal c nn = S i S j = S 2 global minimum of H tot : only uniform contraction! quantum S = 1/2 chain: S i S j cannot independently extremised modulated S i S j modulated distortion dimerisation

9 Frustrated magnetoelastics in a nutshell Frustration provides alternative route to modulation Frustration degeneracy of ground states Degenerate states not symmetry equivalent S i S j can be non-uniform Distortions (strengthen)weaken (un)frustrated bonds Energy balance: distortions generally present at low T magnetic energy: linear gain (S i S j ) x elastic energy: quadratic cost kx 2 Remainder of talk: details of degeneracies and distortions

10 Magnetic frustration: Ising spins Consider Ising spins σ i = ±1 with antiferromagnetic J > 0: H = J ij σ i σ j? Not all terms in H can simultaneously be minimised But we can rewrite H: ( q ) H = J 2 σ i + const 2 i=1 Number of ground states: N gs = ( 4 2) = 6 for one tetrahedron Degeneracy is hallmark of frustration (cf. spin ice)

11 Magnetic frustration: Heisenberg spins Consider Heisenberg spins S i with S 2 i = 1. For a single tetrahedron: φ H = J 2 ( 4 i=1 S i ) 2 J 2 L2 L = 0 α Two continuous internal d.o.f. (α, φ) in ground state Q: Degeneracy for full pyrochlore lattice?

12 Degeneracy of pyrochlore Heisenberg magnet Hamiltonian consists of sum over individual tetrahedra, α: H = J 2 α L 2 α = L α = i α S i = 0 tetrahedra Constraint counting: Degrees of freedom (per tetr.): D = (4/2) (3 1) = 4 Ground-state constraints: K = 3 Ground-state dimensionality: F = D K = 1 F is extensive! many interesting instabilities many different compounds ρ(ω) Νδ(0) J ω

13 triplet T 2 Elastic degrees of freedom: single tetrahedron Symmetry group: permutations of four objects, T d Distortions of tetrahedron classified by irreducible representations: singlet A, doublet E, triplet T 2 uniform A doublet E

14 Coupling of distortions to spins singlet A gives uniform distortion disregard triplet T 2 does not couple to ground states: opposing bonds are weakened/strengthened, but L = 0 S 1 S 2 = S 3 S 4 = no net energy gain only need to consider doublet E with x = (x 1,x 2 ) doublet E couples to E irrep provided by bonds S i S j

15 The ground states of a single tetrahedron f (f 1,f 2 ) with f 1 = [3S 1 S 2 + 1]/ 3; f 2 = S 1 S 3 S 2 S 4 collinear (RGB) f 2 coplanar (CYM) α f f 1 white

16 Biquadratic effective interaction Denoting elastic constants by J,k gives: E = Em 0 J f x + kx 2 /2 = C 0 C 4 (S i S j ) 2 ij C 4 > 0 collinear state C 4 < 0 white state Adding 4-spin exchange (S 1 S 2 )(S 3 S 4 ) + (S 1 S 3 )(S 2 S 4 ) (S 1 S 4 )(S 2 S 3 ) allows changing sign of C 4 selecting colplanar state

17 Undistorted state: zero-energy excitations Ground-state constraint L = 0 precludes rearranging single spin alternating mode around hexagonal loop remains within L = 0 manifold structure factor measured by S.-H. Lee (top); theory (bottom) + + _+

18 Fate of string modes under distortion dn/dω Strings of parallel spins continue to sustain eigenmodes: δs = ( 1) x S 1/JS 1/2JS J 1 J 2 J 1 Energy non-zero in distorted state: E δj Peak in d.o.s. shifted away from ω = 0 Different estimates for δj do not agree 0 JS 2JS 3JS 4JS ω Details of spin ordering pattern unknown!

19 Landau theory for phase transition Pyrochlore lattice point group contains inversion: O h = T d I Irreps now have additional even/odd label g, u for inversion. have two order parameters, g,u, with Landau free energy F : F = a g g 2 + b g g 3 cos 3θ g + c g g 4 f 2 f 2 f 2 + a u u 2 + c u u 4 + d u u 6 cos 6θ u + b u u 2 g cos (2θ u + θ g ) +... E g f 1 E g f 1 E g+e u f 1 Many scenarios (a) f 2 (b) f 2 (c) f 2 Phonons away from q = 0: even more potential order parameters E g+e u (d) f 1 E u +E g (e) f 1 E u +E g (f) f 1

20 Resulting Néel states In Landau theory, spin S i and bond S i S j orders distinct Nonetheless, one constrains the other q = 0 phonon can give q 0 Néel state E g phonon E u phonon E g phonon

21 Summary Analysis of magnetoelastic properties of ZnCr 2 O 4 frustrated magnets highly unstable towards distortions elastic theory based on (point) groups T d,o h First-order phenomenon simpler than normal frustrated bonds weakened satisfied bonds strengthened (Hexagonal) string modes dispersionless excitations Landau theory for phase transition: (too) many scenarios Open questions which phonon goes soft? spin state?

22 Summary Analysis of magnetoelastic properties of ZnCr 2 O 4 frustrated magnets highly unstable towards distortions elastic theory based on (point) groups T d,o h First-order phenomenon simpler than normal frustrated bonds weakened satisfied bonds strengthened (Hexagonal) string modes dispersionless excitations Landau theory for phase transition: (too) many scenarios Open questions which phonon goes soft? spin state? Thank you for your attention!