2010
DOI: 10.1088/1742-5468/2010/10/p10007
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ASEP on a ring conditioned on enhanced flux

Abstract: We show that in the asymmetric simple exclusion process (ASEP) on a ring, conditioned on carrying a large flux, the particle experience an effective longrange potential which in the limit of very large flux takes the simple form U = −2 i =j log | sin π(n i /L − n j /L)|, where n 1 , n 2 , . . . n N are the particle positions, similar to the effective potential between the eigenvalues of the circular unitary ensemble in random matrices. Effective hopping rates and various quasistationary probabilities under suc… Show more

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Cited by 97 publications
(176 citation statements)
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References 32 publications
(86 reference statements)
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“…is not a priori satisfied, since J T is only proportional to Σ T in the limit T → ∞ because of the potential boundary terms in (29). For V 0 = 0, however, J T is exactly proportional to Σ T for all T , so the operator symmetry (38) holds, as can be verified from the expression (14) of L k .…”
Section: Fluctuation Relation and Upper Boundsmentioning
confidence: 99%
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“…is not a priori satisfied, since J T is only proportional to Σ T in the limit T → ∞ because of the potential boundary terms in (29). For V 0 = 0, however, J T is exactly proportional to Σ T for all T , so the operator symmetry (38) holds, as can be verified from the expression (14) of L k .…”
Section: Fluctuation Relation and Upper Boundsmentioning
confidence: 99%
“…which implies, by applying the change of variables (29) and by neglecting the potential boundary terms [62],…”
Section: Fluctuation Relation and Upper Boundsmentioning
confidence: 99%
See 1 more Smart Citation
“…In this way one readily obtains σ x n = σ y n = j x = j y = 0. In the isotropic model with a twist the parity symmetry is broken, but its place is taken by another symmetry, which we specify here for twisting angles θ R = −θ L = π/2: It involves left-right reflection R(A⊗B ⊗...⊗C) = (C ⊗....⊗B ⊗A)R, global rotation in the XY -plane U rot = diag(1, i) ⊗N and Σ x = (σ x ) ⊗N and reads ρ = V ρV † , where V = Σ x U rot R [20]. It is straightforward to check that neither of the set of observables σ α n , j α , changes sign under the V symmetry, and therefore they are generically nonzero.…”
Section: The Stationary Lindblad Equation (5) Can Thus Be Split Into mentioning
confidence: 99%
“…We imagine that the noise force db in (19) comes from friction between the particle and a surrounding solvent, but now imagine that the solvent is moving with constant velocity v. We refer to this as a system with advection (of the particle, by the solvent). In this case the equations of motion are obtained by applying a Galilean transformation to (19), which yields dq t = p t dt, dp t = db…”
Section: Example System: Comparisons Between the Auxiliary Process Anmentioning
confidence: 99%