leon balents- frustrated magnets (2) materials survey

8/3/2019 Leon Balents- Frustrated Magnets (2) Materials Survey

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AB2X4 spinels

One of the most

common mineralstructures

Common valence:

A2+

,B3+

,X2-

X=O,S,Se

A

X

B

cubic Fd3m


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Deconstructing the

spinel

A atoms: diamond lattice

Bipartite: notgeometrically

frustrated A


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Deconstructing the spinel

B atoms: pyrochlore

decorate the plaquettes

of the diamond lattice

B


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ACr2O4 spinelspyrochlore lattice

S=3/2 Isotropicmoment

X=O spinels: B-Bdistance close enough

for direct overlap

dominant AFnearest-neighborexchange


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H=0 Susceptibility

Frustration:

T


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DegeneracyHeisenberg model

Ground state constraint: total spin 0 per

tetrahedron

Quantum mechanically: not possible

H =

ij

Si Sj =

1

2

t

it

Si

2

+ const.


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Classical spin liquid

No LRO (Reimers)


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No LRO (Reimers)

Dipolar correlations

Si = b

ab


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No LRO (Reimers)

Dipolar correlations

Si = b

ab

a

bi


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Unusual ringcorrelations seen inCdCr2O4 related

Y2Ru2O7: J. van Duijn

et al, 2007

2

-

-

-

- -

Broholm et al


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Ordering

Many perturbations

important forordering:

Spin-latticecoupling

Further exchangeSpin-orbit effects

Quantumcorrections

ZnCr2O4

CdCr2O4

HgCr2O4

S.H. Lee + many others


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Magnetization Plateaus

Classically: M = Ms H/Hs

Plateau indicates 3:1structure

H. Ueda at al, 2005/6


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Magnetization PlateausPlateau mechanism:

spin-lattice couplingfavors collinearity

Order on plateau maybe selected by

spin-lattice

quantum effects

R state observedin neutrons

Matsuda et al


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A-site spinelsSpectrum of materials

1 900

FeSc2S4

10 205

CoAl2O4

MnSc2S4

MnAl2O4

CoRh2O4 Co3O4

V. Fritsch et al. PRL 92, 116401 (2004); N. Tristan et al. PRB 72, 174404 (2005); T. Suzuki et al. (2006)

s = 5/2

s = 3/2

Orbitaldegeneracy

s = 2

Naively unfrustrated

f=|CW |

TN


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Why frustration?

Roth, 1964: 2nd and

3rd neighborexchange notnecessarily small

Exchange paths:A-X-B-X-B

comparableMinimal modelJ1-J2 exchange

J1J2


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Ground state evolutionCoplanar spirals

q0 1/8

Neel

J2/J1

J2/J1 = 0.2 J2/J1 = 0.4 J2/J1 = 0.85 J2/J1 = 20

Spiral surfaces:


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Monte Carlo

f = 11 atJ2/J1 = 0.85

MnSc2S4


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Phase Diagram0

Spiral spinliquid

paramagnet

MnSc2S4

J3

CWTN

ObD

ObJ3

Entropy and J3compete to determineordered state

Spiral spin liquidregime has intensity

over entire spiralsurface


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Diffuse scattering

Ordered state

(qq0) spiral

Specific heat?

Comparison to Expt.Expt. Theory

agrees withtheory for FM J1


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Cs2CuCl4Spatially anisotropictriangular lattice

Cu2+ spin-1/2 spins

couplings:

H=1

2

ij

Jij

Si Sj

Dij Si

Sj

J=0.37meV

J=0.3JD=0.05J

D = Da


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Neutron scatteringColdea et al, 2001/03: a 2d spin liquid?

Very broad spectrumsimilar to 1d (in somedirections of k space).Roughly fits power law.

Fit of peak dispersion tospin wave theory requiresadjustment of J,J by 40%- in opposite directions!


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Dimensional reduction?Frustration of interchain coupling makes itless relevant

First order energy correction vanishes

Leading effects are in fact O[(J)4/J3]!


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Dimensional reduction?Frustration of interchain coupling makes itless relevant

First order energy correction vanishes.Numerics: J/J < 0.7 is weak

Weng et al,2006

Very different fromspin wave theory

Very weak inter-chaincorrelations


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Build 2d excitations from 1d spinons

Exchange:

Expect spinon binding to lower inter-chainkinetic energy

Use 2-spinon Schroedinger equation

ExcitationsJ

2

S+

i S

j + S

i S+

j


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Broad lineshape: free spinonsPower law fits well to free spinon resultFit determines normalization

J(k)=0 here


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Bound stateCompare spectra at J(k)0:

Curves: 2-spinon theory w/ experimental resolution Curves: 4-spinon RPA w/ experimental resolution


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Transverse dispersion

Bound state andresonance

Solid symbols: experiment

Note peak (blue diamonds) coincides

with bottom edge only for J(k)


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Spectral asymmetry

Comparison:

Vertical lines: J(k)=0.

leon balents- frustrated magnets (2) materials survey

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