LPTM. Quantum-Monte-Carlo Approach to the Thermodynamics of Highly Frustrated Spin-½ Antiferromagnets. Andreas Honecker 1

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1 Quantum-Monte-Carlo Approach to the Thermodynamics of Highly Frustrated Spin-½ Antiferromagnets LPTM Laboratoire de Physique Théorique et Modélisation Andreas Honecker 1 Laboratoire de Physique Théorique et Modélisation, Université de Cergy-Pontoise, France Ido Niesen 2, Philippe Corboz 2, Bruce Normand 3, Frédéric Mila 4, Jonas Stapmanns 5, Stefan Wessel July 12, 218

2 Take-home message Efficient Quantum Monte-Carlo simulations can be performed for certain highly frustrated quantum magnets using a suitable computational basis. References: 1. A.H., S. Wessel, R. Kerkdyk, T. Pruschke, F. Mila, B. Normand, Phys. Rev. B 93, 5448 (216) 2. S. Wessel, B. Normand, F. Mila, A.H., SciPost Phys. 3, 5 (217) 3. J. Stapmanns, P. Corboz, F. Mila, A.H., B. Normand, S. Wessel, arxiv: S. Wessel, I. Niesen, J. Stapmanns, B. Normand, F. Mila, P. Corboz, A.H., arxiv:187.1???? See also: 5. F. Alet, K. Damle, S. Pujari, Phys. Rev. Lett. 117, (216) 6. K.-K. Ng, M.-F. Yang, Phys. Rev. B 95, (217)

3 Outline ❶ Motivation: thermodynamics of the spin-½ Shastry-Sutherland magnet SrCu2(BO3)2 ❷ Quantum-Monte Carlo: sign problem + dimer basis ❸ Application to fully frustrated bilayer ❹ Application to the 2D Shastry-Sutherland model ❺ Summary

4 Outline ❶ Motivation: thermodynamics of the spin-½ Shastry-Sutherland magnet SrCu2(BO3)2 ❷ Quantum-Monte Carlo: sign problem + dimer basis ❸ Application to fully frustrated bilayer ❹ Application to the 2D Shastry-Sutherland model ❺ Summary

5 Shastry-Sutherland model dimer product state J/J H = J S i S j + J S i S j square lattice dimers

6 Shastry-Sutherland model SrCu2(BO3)2 structure of a layer dimer product state J/J H = square lattice J S i S j + dimers J S i S j Cu, O, B, Sr

7 Shastry-Sutherland model SrCu2(BO3)2 structure of a layer dimer product state J/J S =1/2 Cu, O, B, Sr

8 Magnetic susceptibility Dimer expansions? High-temperature series? o experiment low-temperature region inaccessible Zheng Weihong, C. J. Hamer, J. Oitmaa. Phys. Rev. B 6, 668 (1999)

9 Exact diagonalization Magnetic susceptibility Specific heat experiment J /J=.62,J=77K, N=16 J /J=.64,J=88K, N=16 J /J=.66,J=14K, N=16 J /J=.635,J=85K, N=16 J /J=.635,J=85K, N=2 χ(emu/cu mol).3.2 C/T(mJ/K 2 mol) experiment J /J=.62,J=77K N=16 J /J=.64,J=88K, N=16 J /J=.66,J=14K, N=16 J /J=.635,J=85K, N=16 J /J=.635,J=85K, N= T(K) T(K) S. Miyahara, K. Ueda, J. Phys. Soc. Jpn. Suppl. B 69, 72 (2) apparently good agreement for J /J.635, but are N=16,, 2 spins really sufficient?

10 Outline ❶ Motivation: thermodynamics of the spin-½ Shastry-Sutherland magnet SrCu2(BO3)2 ❷ Quantum-Monte Carlo: sign problem + dimer basis ❸ Application to fully frustrated bilayer ❹ Application to the 2D Shastry-Sutherland model ❺ Summary

11 Stochastic Series Expansion A.W. Sandvik, J. Kurkijärvi, Phys. Rev. B 43, 595 (1991); A.W. Sandvik, Phys. Rev. B 59, 14157(R) (1999); O.F. Syljuåsen, A.W. Sandvik, Phys. Rev. E 66, 4671 (22); F. Alet, S. Wessel, M. Troyer, Phys. Rev. E 71, 3676 (25) Hamiltonian: H = Partition function: H b = bond b (b,t) bond term ( = 1/T ) H (b,t) H b = H (b,d) diagonal + H (b,o) o diagonal Z = Tr e H = n= n n! Tr ( H)n = n= n n! n = 1 (b n,t n )...(b 1,t 1 ) n i=1 i+1 H (bi,t i ) i basis i Sample (Monte Carlo) weights i+1 H (bi,t i ) i

12 Stochastic Series Expansion A.W. Sandvik, J. Kurkijärvi, Phys. Rev. B 43, 595 (1991); A.W. Sandvik, Phys. Rev. B 59, 14157(R) (1999); O.F. Syljuåsen, A.W. Sandvik, Phys. Rev. E 66, 4671 (22); F. Alet, S. Wessel, M. Troyer, Phys. Rev. E 71, 3676 (25) Hamiltonian: H = Partition function: H b = bond b (b,t) bond term ( = 1/T ) H (b,t) H b = H (b,d) diagonal + H (b,o) o diagonal Z = Tr e H = n= n n! Tr ( H)n = n= n n! n = 1 (b n,t n )...(b 1,t 1 ) n i=1 i+1 H (bi,t i ) i basis i Sample (Monte Carlo) weights i+1 H (bi,t i ) i sign problem

13 Sign problem off-diagonal terms may cause problems example: one triangle H (b,o) = J 2 S + b,1 S b,2 + S b,1 S+ b,2 h""# H (b3,o) H (b2,o) H (b1,o) ""#i sample with absolute values of weights ha signi P k W (C) A(C) sign(c) hai = = C P hsigni W (C) sign(c) k average sign can get exponentially small exponentially large error bars but this is basis-dependent C

14 Dimer basis eigenstates of a spin-½ dimer: (well-defined magnetization) Si = 1 p 2 ( "#i #"i) i = 1 p 2 ( "#i + #"i) +i = ""i i = ##i total spin operators ~T = ~ S 1 + ~ S 2 and spin-difference ~D = ~ S 1 ~ S2 ~T 2 T z T + T D z D + D Si i p 2 +i p 2 i i 2 p 2 +i p 2 i Si +i 2 1 p 2 i p 2 Si i 2 1 p 2 i p 2 Si

15 Outline ❶ Motivation: thermodynamics of the spin-½ Shastry-Sutherland magnet SrCu2(BO3)2 ❷ Quantum-Monte Carlo: sign problem + dimer basis ❸ Application to fully frustrated bilayer ❹ Application to the 2D Shastry-Sutherland model ❺ Summary

16 From the Shastry-Sutherland model to the fully frustrated bilayer J 2 J addition of J2 preserves the exact dimer ground state J2 = J : fully frustrated bilayer (materials: Ba2CoSi2O6Cl2 [H. Tanaka et al., J. Phys. Soc. Jpn. 83, 1371 (214)])

17 Ground-state phase diagram discontinuous level crossing transition E. Müller-Hartmann, R.R.P. Singh, C. Knetter, G.S. Uhrig, Phys. Rev. Lett. 84, 188 (2) J? J = J k dimer triplet antiferromagnet exact singlet product state (1) J? /J k What are the thermodynamic properties?

18 Hamiltonian in dimer basis in terms of dimer d: total ~T d = ~ S d,1 + ~ S d,2 and difference spin operators ~D d = ~ S d,1 ~ Sd,2 H = J? 2 X d + X hd,d i ~T 2 d 3 2 Jk + J 2 J? J J k ~T d ~T d + J k J 2 ~D d ~D d sign problem disappears for J = J k

19 Hamiltonian in dimer basis in terms of dimer d: total ~T d = ~ S d,1 + ~ S d,2 and difference spin operators ~D d = ~ S d,1 ~ Sd,2 H = J? 2 X d + X hd,d i ~T 2 d 3 2 Jk + J 2 J? J J k ~T d ~T d + J k J ~D d ~D d 2 sign problem disappears for J = J k

20 Magnetic susceptibility.12 L = 32.8 χ J T/J transition from gapped to ordered S = 1 state between J /J = 2.26 and 2.34 J /J = J /J = 2.26 J /J = 2.34 J /J = 2.5 S=1 square lattice J = J k

21 Scanning the transition region E/J Internal energy L = 32 T/J =.5 T/J =.54 T/J =.56 T/J = J /J ρ s Singlet density discontinuous behavior extends up to T.54J continuous crossover at higher T enhanced singlet-triplet fluctuations upon approaching the first-order transition line L = 32 T/J =.5 T/J =.54 T/J =.56 T/J =.6 T/J = J /J J = J k

22 Finite-T phase diagram 1 J = J k.8 T/J long-range ordererd S=1 AF J /J SU(2) symmetry Mermin-Wagner theorem forbids first-order line finite-temperature ordering transition singlet-product state

23 Finite-T phase diagram 1 J = J k.8 T/J triplet dominated long-range ordererd S=1 AF first-order line singlet dominated singlet-product state J /J binary variable on each dimer: singlet versus triplet (SU(2) symmetry equivalence of 3 components)

24 Finite-T phase diagram 1 J = J k.8 T/J triplet dominated long-range ordererd S=1 AF Ising critical point first-order line singlet dominated singlet-product state J /J binary variable on each dimer: singlet versus triplet (SU(2) symmetry equivalence of 3 components) Ising critical point

25 Logarithmic specific heat scaling 1 12 J = J k C L = 48 L = 36 L = 24 L = 16 L = 12 C max 8 4 J /J = 2.31 L = ln L T/J compatible with 2D Ising universality class

26 Outline ❶ Motivation: thermodynamics of the spin-½ Shastry-Sutherland magnet SrCu2(BO3)2 ❷ Quantum-Monte Carlo: sign problem + dimer basis ❸ Application to fully frustrated bilayer ❹ Application to the 2D Shastry-Sutherland model ❺ Summary

27 Ground-state phase diagram fully frustrated bilayer.8 1 ipeps J 2 J2/J.6 J J.4.2 Shastry- Sutherland J /J SrCu2(BO3)2

28 Average sign J 2 = 1 mild sign problem.8 recovery at low T (exact ground state) <sign> L = 1 J /J =.3 J /J =.4 J /J =.5 J /J =.525 J /J =.55 J /J = T/J

29 Example: J /J =.5 J 2 = Magnetic susceptibility Specific heat χ J C.4.1 ED, N = 2 QMC, N = 32 QMC, N = 2.2 ED, N = 2 QMC, N = 32 QMC, N = T/J T/J Quantum Monte Carlo (QMC) with N=2 spins Exact diagonalization (ED) for N=2 correct, except for persistent finite-size effects close to the maximum of C

30 Summary Thanks: 2D Shastry-Sutherland Model: progress with QMC, but SrCu2(BO3)2 remains open fully frustrated square lattice: detailed analysis of finite-t properties possible: first-order transition terminating in Ising critical point other possible applications: 3D bicubic dimer model planar pyrochlore (checkerboard) kagome, pyrochlore, J. Stapmanns, P. Corboz, F. Mila, A.H., B. Normand, S. Wessel, arxiv: Efficient Quantum Monte-Carlo simulations can be performed for certain highly frustrated quantum magnets using a suitable computational basis.