Strong-coupling phases of frustrated bosons on a two-leg ladder with ring exchange

Physical Review B - Condensed Matter and Materials Physics

Volumn 78, Issue 5, 2008, Pages

  Sheng, D N ^a   Motrunich, Olexei I ^b   Trebst, Simon ^c   Gull, Emanuel ^d   Fisher, Matthew P A ^c  

^a CALIFORNIA STATE UNIVERSITY   (United States)

^b California Institute of Technology   (United States)

^c UNIVERSITY OF CALIFORNIA   (United States)

^d ETH ZURICH   (Switzerland)

Author keywords

[No Author keywords available]

Indexed keywords

EID: 50849103387     PISSN: 10980121     EISSN: 1550235X     Source Type: Journal    
DOI: 10.1103/PhysRevB.78.054520     Document Type: Article

References (48)

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35. In this small system with eight bosons (Fig. 7), there is also an abrupt change inside the DBL phase where the ground-state momentum jumps from (kx, ky) = (0,0) to (kx, ky) = (2π/3,π) at around K 2.7J. We can understand these quantum numbers from specific orbital occupations in the DBL wave-function approach, and the change in the total ground-state momentum arises from a change in orbital occupations in the d2 bands. Near K 2 VMC finds the optimal DBL state to be det1 [N1 (0) =8, N1 (π) =0] abc × det2 [N2 (0) =6, N2 (π) =2] abc. The label "abc" stands for antiperiodic boundary conditions when specifying the mean-field orbitals. The total bosonic wave function satisfies periodic boundary conditions and carries momentum (0,0). For the larger K, VMC finds the optimal DBL state det1 [N1 (0) =8, N1 (π) =0] × det2 [N2 (0) =5, N2 (π) =3], where now the orbitals are specified with periodic boundary conditions. This state carries momentum (kx, ky) = (2π/3,π).
36. Note that ΔE in Fig. 11 maintains a positive sign throughout the DBL phase, which can be understood roughly as follows: A boson propagates via a bonding mode (with wave vector qx (0)) in one half and via an antibonding mode (qx (π)) in the other half. The total phase accumulated by going around the system is (qx (0) + qx (π)) Lx /2=πρ Lx =0mod2π for ρ=1/3 and the shown Lx which are multiples of 6. Thus, the zero twist boundary condition provides the best (crude) matching for all K. When we tried different Lx (not shown), the ΔE could be of either sign.
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