At inﬁnite temperature, where iniKal state is sum over all eigenstates: Transverse Field Ising; L = 14 1 0.9 0.7 h z z 1 (0) 1 (t)i 0.8 0.6 0.5 0.4 h = 0.2 h = 0.4 h = 0.6 0 2 4 6 8 Time t ⇣ ⌘ 1 J 10 12 14 16 18 At longer Kmes, quantum revivals! Transverse Field Ising; h = 0.4 1 0.8 0.6 h z z 1 (0) 1 (t)i 0.4 0.2 0 -0.2 -0.4 -0.6 L=8 L = 10 L = 12 L = 14 -0.8 -1 10 2 100 102 104 t 6 10 ⇣ ⌘ 108 1 J Note log on Kme axis! 1010 1012 1014 Similar for XYZ ✓ Plateau values are 1 XYZ; X = 0.2; L = 14 Jx2 Jz2 1 0.95 0.9 h 1z (0) 1z (t)i 0.85 0.8 0.75 0.7 0.65 Y Y Y Y Y 0.6 0.55 0 5 t 10 ⌘ ⇣ 1 Jz X=J_x/J_z, Y=J_y/J_z 15 = = = = = 0.1 0.2 0.3 0.4 0.5 20 ◆ 1 Jy2 Jz2 ! No revivals in XYZ – even though it’s integrable, the energy diﬀerences are incoherent Plateau value here for X=.2, Y=.3 is .87 XYZ; X = 0.2; Y = 0.3 Finite-‐size eﬀects from exponenKal splirng in levels 1 0.8 h 1z (0) 1z (t)i 0.6 0.4 0.2 0 -0.2 10 L=8 L = 10 L = 12 L = 14 2 10 1 100 101 t 102 ⌘ ⇣ 103 104 105 106 1 Jz Log on Kme axis What if integrability is broken but no disorder? Add interacKon term to the Ising Hamiltonian: L−1 L−1 j=1 j=1 K ∑σ xj σ xj+1 = −K ∑ψ 2 j−1ψ 2 jψ 2 j+1ψ 2 j+2 f=K=J/2 L=8 Appears as if the strong zero mode might live! F (-‐1) = 1 -‐1 An “almost” strong zero mode • Start with an integrable model with a strong zero mode, and include an interacKon that breaks the integrability, e.g. in Ising L X2 z z J2 j j+2 j=1 • Then start with = 1z + . . . and iterate to ﬁnd correcKons resulKng from the new interacKon. • In this case, the procedure seems to work, can ﬁnd the correcKons ( ⇠ 100000 terms at 8th order). Work in progress with Jack Kemp, Norm Yao and Chris Laumann = z 1 f J22 1 J2 f J22 1 x 1 x 1 z 2 Poles appear! z 3 They correspond to processes in perturbaKon theory that at e.g. J 2 = J require no energy to ﬂip the boundary spin. 2J22 f 2 x z x z z J26 11J24 + 19J22 9 1 2 3 4 5 iJ2 f 2 z x z x z 4iJ24 5iJ22 + i 1 1 2 2 3 iJ2 f 2 x x z 4iJ24 5iJ22 + i 1 2 4 2J22 f 2 z x z x z J26 11J24 + 19J22 9 1 1 3 3 5 ✓ iJ23 f 2 3iJ2 f 2 i J 6 11J24 + 19J22 9 J26 11J24 + 19J22 ✓ 2 iJ23 f 2 3iJ2 f 2 i J26 11J24 + 19J22 9 J26 11J24 + 19J22 6J2 f 2 z x z x z J26 11J24 + 19J22 9 1 1 3 3 4 2J22 f 2 z x z x z 1 1 2 2 4 4 2 4J2 5J2 + 1 ✓ ◆ 2J22 f 2 f2 x x z + 1 2 3 4J24 5J22 + 1 4J24 5J22 + 1 ✓ iJ23 f 2 3iJ2 f 2 i J26 11J24 + 19J22 9 J26 11J24 + 19J22 2J22 f 2 z x z z x z z J26 11J24 + 19J22 9 1 1 2 3 3 4 5 J2 f 2 z z z J24 2J22 + 1 1 2 3 ✓ J24 f 2 7J22 f 2 J26 11J24 + 19J22 9 J26 11J24 + 19J22 9 9 9 9 ◆ x 1 x 3 z 4 ◆ z 1 x 1 z 2 ◆ x 1 z 2 x 3 ◆ x 1 x 3 z 5 J=1, h =f z 3 x 3 Norm is only parKally cooperaKng 2 • Literally thousands of cancellaKons indicate that is proporKonal to idenKty operator! 2 = 1 + c2 (J2 ) 2 + c4 (J2 ) 4 + ... • However in this normalizaKon the coeﬃcients c 2j seem to diverge as j ! 1 . • So what is the eﬀect of this “almost” strong zero mode on the boundary magneKzaKon? The boundary magneKzaKon does decay, but is very long-‐lived. Ising; h = 0.05; L = 14 1 0.8 Dependence on J_2 is not monotonic! 0.4 h z z 1 (0) 1 (t)i 0.6 All All All All All All All All All All 0.2 0 -0.2 10 2 J2 J2 J2 J2 J2 J2 J2 J2 J2 J2 Paired Paired Paired Paired Paired Paired Paired Paired Paired Paired = 0.0 = 0.1 = 0.2 = 0.3 = 0.4 = 0.5 = 0.6 = 0.7 = 0.8 = 0.9 100 102 J2 J2 J2 J2 J2 J2 J2 J2 J2 J2 = 0.0 = 0.1 = 0.2 = 0.3 = 0.4 = 0.5 = 0.6 = 0.7 = 0.8 = 0.9 104 t 6 10 ⇣ ⌘ 1 J 108 1010 1012 1014 Log on Kme axis Due to the almost strong zero mode, the boundary magneKzaKon is long-‐lived because 1) the pairing of a macroscopic number of states survives at smallish system sizes. 2) The corresponding energy diﬀerence is small. H= X i J z z i i+1 + J2 X i z z i i+2 + X i h x i The pairing persists away from the poles: The mean overlap as a funcKon of J2 Ising; h = 0.05 1 Boundary Magnetisation 0.9 0.8 0.7 0.6 0.5 0.4 0.3 L=8 L=10 L=12 -4 -3 -2 -1 0 1 2 3 4 J2 Note independent of the sign of J 2 -‐-‐ frustraKon doesn’t mater! ED indicates the pairing between the states persists for smallish sizes, but not all energy diﬀerences in the pair shrink with size! 50 Histogram of energy diﬀerence for each pair 40 h = .05 L=12 L=10 L=8 J2 = .9 0 10 20 counts 30 J =1 10-15 10-10 10-5 Energy Di↵erence 100 The average variance in the energy splirng between paired states First Order Pole J₂ = J Second Order Pole J₂ = J/2 3rd Order J₂ = J/3 Second Order Pole J₂ = 3J Third Order Pole J₂ = 3J/2 Fourth Order Pole J₂ = J/4 ? Machine Precision Thus the physics depends on more than the strength of integrability breaking. No deﬁniKve conclusion yet • Many intermediate possibiliKes (good evidence that almost strong zero mode implies exponenKal splirng for all low-‐lying states) • Strong zero modes in the entanglement spectrum? • Parafermions seem to give an almost strong zero mode even in the integrable case. Is the reason the poles? • Does any of the non-‐integrable behaviour survive as L ! 1 ? ETH violaKon?