But can the strong zero mode occur in an interacKng system without disorder? Yes, in the pure XYZ spin chain/coupled Majorana wires This is a strongly interacKng model; in fermion language it contains a four-‐fermion interacKon term. It is integrable. My proof is by explicit construcKon. The XYZ spin chain is an integrable model of coupled Majorana chains: L−1 x y H = ∑ (J xσ jxσ j+1 + J yσ jyσ j+1 + J zσ zj σ zj+1 ) j=1 L−1 = ∑ (i J xψ 2 j−1ψ 2 j+2 + i J yψ 2 jψ 2 j+1 − J z ψ 2 j−1ψ 2 jψ 2 j+1ψ 2 j+1 ) j=1 The model is gapped and ordered unless two of the couplings are equal and less than or equal to the other. ordered gapless Jz = 1 / 2 Jz = 4 XXZ limit Jx = Jy = 1 L=8 S z =0 F (-‐1) = 1 -‐1 1 -‐1 It’s not diﬃcult to iterate the zeroth order result to get the ﬁrst few terms. For XXZ: Zeroth order First correcKon 1 1 x x y y z Ψ = σ − (σ 1 σ 2 + σ 1 σ 2 )σ 3 − 2 (σ 1xσ 3x + σ 1yσ 3y )σ 4z Jz Jz 1 1 z x x y y x x y y z + 2 (σ 1 σ 2 + σ 1 σ 2 )(σ 3σ 4 + σ 3σ 4 )σ 5 + 2 σ 2 + ... Jz Jz z 1 Jx = Jy = 1 It gets ugly, but much less nasty than you might expect, and much much less nasty than the disordered case. Presumably the integrability is the reason. The exact strong zero mode: ∞ Ψ=∑ k=0 Jx Jy J z2 b−1 ( ) z 〈a a 〉〈a a 〉…〈a a 〉 σ ∑ 12 34 2 k−1 2 k b {a j },b Sum over all integers obeying a1 < a2 < · · · < a2k < b where 〈ab〉 ≡ () Jy Jz a−b J (1− J 2 z 2 x () Jx x x )σ aσ b + J z \ a−b J z2 y y (1− 2 )σ aσ b Jy Norm is good: Ψ = 1 2 J 1− J ( 2 x 2 z J 1− J )( 2 y 2 z ) Ψ is a very complicated operator, but Ψ is proporKonal to the idenKty! It toggles back and forth between the sectors. 2 The blowing up at J x = J z or J y = J z means that the strong zero mode transiKon occurs at the same coupling as the transiKon out of the ordered phase. Is the strong zero mode a ﬂuke of integrability? The MBL arguments imply it occurs at strong enough disorder. However, XYZ/coupled chains are not integrable if translaKon invariance is broken. So as one increases the disorder, are there two eigenstate transiKons or none? A very conjectural phase diagram gap Strong zero mode ``Almost’’ strong zero mode Strong zero mode Eigenstate phase transiKons Weak only disorder The sizes of the weak/almost regions depend on interacKon strength. What good is a strong zero mode? Fix the boundary spin in Ising to be up at Kme t=0. Then let it evolve in Kme. Of course it ﬂips – there’s a transverse ﬁeld. However, for all subsequent Kmes, it is more likely to be up, even if iniKal state is at inﬁnite temperature! This is easy to see using the strong zero mode. z Recall that 1 = 1 and that the normalized strong zero mode in Ising is ✓ ◆ 2 1/2 = h J2 1 1 + ... By inverKng the standard calculaKon of Ising raising/lowering operators, z 1 = ✓ 2 1 h J2 ◆1/2 + X ak (k) k Strong zero mode Raising/lowering operators InserKng this into inﬁnite-‐temperature correlator gives: X X 1 2L s hs| z z (0) 1 1 (t)|si 1 = L 2 =1 ei(Es0 s,s0 2 Es )t hs| z 0 0 z |s ihs | 1 |si 1 h + oscillating terms 2 J