“avoided” by going beyond the critical point Liquid/gas are “essentially indistinguishable” Figure from Sonntag R E, Borgnakke C, Van Wylen G J, “Fundamentals of Thermodynamics”
between solid/liquid at, e.g. higher pressures? NO! in solid, translation symmetry is spontaneously broken, while it is not in liquid/gas Spontaneous Symmetry Breakings (SSB) clearly distinguish different phases, implying existence of phase transitions
characterize each phase Ferromagnet: magnetic order Superfluid (3D): off-diagonal long-range order (order of U(1) phase of wavefunctions) etc. “order” ⊇ Spontaneous Symmetry Breaking
many quantum phases that are beyond understanding in terms of conventional orders/spontaneous symmetry breaking “topological phases” how to define them? how to distinguish different phases?
connected adiabatically to a trivial (product) state it belongs to a topologically ordered phase (with long-range entanglement) e.g.: Fractional Quantum Hall states Z2 topological phase (RVB spin liquid/Kitaev’s toric code)
a local 1D Hamiltonian is connected to a trivial state adiabatically Absence of (genuine) topologically ordered phase in 1D! However, there can be more variety of phases if some symmetries are imposed Verstraete et al. (2005), Hastings (2007), Chen-Gu-Wen (2011)
1) the ground state is in a trivial phase, 2) the symmetry is spontaneously broken in the ground state (SSB phase), OR 3) the symmetry is unbroken, but the ground state cannot be connected to a trivial state by any local unitary evolution respecting the symmetry
to a trivial state adiabatically respecting the symmetry The ground state belongs to a Symmetry-Protected Topological (SPT) Phase SPT phases exist in 1 dimension, as well as in higher dimensions
phases was developed after the discovery of Topological Insulators [Kane-Mele (2005)] e.g. 2D Topological Insulators (Quantum Spin Hall Insulators) can be understood in terms of two quantum Hall states with up- and down- spins, and opposite magnetic field
with opposite magnetic fields seems artificial (and indeed is unrealistic) However, presence of the edge states is protected by Kramers degeneracy Topological Insulator is a SPT phase protected by Time-Reversal Symmetry
(for periodic boundary condition) Correlation function of any local operator decays exponentially There is no local order parameter; no symmetry is broken spontaneously No order, that’s it? 21
each other. Effective coupling: Je e L/ Consider a chain with open boundary condition 2x2=4 groundstates below the Haldane gap (nearly degenerate) 23 Kennedy (1990)
basis, + and - alternate, with 0’s in between No long-range order w.r.t. local observables, but a hidden (topological) order measurable by the “string order parameter” Ostr lim |j k|⇥⇤ ⇤Sj ei⇥ P k 1 l=j Sl Sk ⌅ Den Nijs & Rommelse (1989) 25
= J j ⇤ Sj · ⇤ Sj+1 ˜ H = UHU 1 = J j ⇥ Sx j ei Sx j+1 Sx j+1 + Sy j ei (Sz j +Sx j+1)Sy j+1 + Sz j ei Sz j Sz j+1 ⇤ non-local unitary transformation Global discrete symmetry (π-rotation about x, y, z axes = Z2 x Z2) U = j<k ei Sz j Sx k [simple expression by M.O. (1992)] [well-defined only for open b.c.] 26
˜ H 4-fold groundstate degeneracy for H only with the open b.c.! = edge states Ferromagnetic order for ˜ H String order for H UHU 1 = ˜ H U Sz j ei Pk 1 l=j Sz l Sz k ⇥ U 1 = Sz j Sz k 28
“Haldane phase” and understood many interesting things on it. However, (as far as I know) nobody asked “when does it work?” A simple analysis reveals that, a certain symmetry is needed for the Haldane phase to be distinct from a trivial phase. Now we expect this from the general argument, which was unknown at that time. Nevertheless, we could have (rather easily) reached the notion of SPT phases back then!
81, 063349 (2010) Frank Pollmann (MPIPKS Dresden) Erez Berg (Weizmann Institute) Ari Turner (Johns Hopkins) In 2009, I came back to the old problem and started a collaboration with
nonlocal -- if the transformed Hamiltonian is nonlocal, the argument does not work. Because the transformation is self-dual, for to be local, the original Hamiltonian must have global D2 = Z2 x Z2 symmetry (π-rotation about x, y, z axes) Pollmann, Berg, Turner, M.O. 2009- One of the symmetries needed for the Haldane phase as a SPT phase
the edge state survive in more general models? Consider perturbations to AKLT model Generic perturbations will lift the edge degeneracy! However, if the perturbation respect time reversal, it should keep the “Kramers degeneracy” of S=1/2 edge state i.e. time reversal symmetry also protects the Haldane phase cf.) edge state of topological insulator
(π-rotation about x,y,z axes): lost string order does not work as an order parameter Time reversal: lost edge state does not characterize the Haldane phase Nevertheless, Haldane phase is still distinct from other phases by QPTs Protected by inversion symmetry! Bx ˠ Dˠ
I-invariance. The adiabatically connected state remains I-odd. On the other hand, a trivial groundstate is I-even. Any adiabatic evolution of the trivial state is also I-even as long as I-invariance is kept. | ⇥ = j | ⇥j There must be a phase transition between the two groundstates (robustness of Haldane phase protected by the inversion symmetry) 35
generalized to general integer S U = j<k ei Sz j Sx k What did I find back in 1992? The hidden Z2xZ2 symmetry is unbroken in S=2,4,6,8,.... AKLT state while broken in S=1,3,5,7,..... AKLT state! “even-odd effect” 36
in the (uniform) S=2 AKLT state. Q (1992): Is it indistinguishable from a trivial state, or are we just unaware of appropriate hidden order/symmetry? 37 remained open until 2009…
S=1 (3-fold deg.) The degeneracy will be lifted by perturbations, and generically no degeneracy remains! (no Kramers degeneracy) 38 This suggests a rather surprising conclusion that S=2 “Haldane phase” is essentially indistinguishable from a trivial state! (although it could be argued back in 1992)
adiabatically connected to a trivial state?! Is this really the case? Yes! There exists a 1-parameter family of Matrix Product State (and corresponding Hamiltonian) interpolating S=2 AKLT state and large-D state Pollmann, Berg, Turner, M.O. 2009
presence of the S=1 edge state makes the system distinct from trivial states? In general, the answer is NO. The “S=1 edge states” of the “S=2 Haldane phase” are killed by introducing anisotropies
K = 0: 2 x (S=1/2 edge spin) K>0 : rung singlet (trivial), no edge state K<0 : S=1 edge spin (ʙS=2 Haldane phase) NO phase transition at K=0 !! (S=1 Haldane gap at K=0)
change by level crossing at the edge (w/o bulk transition) Sb=0 vs. 1/2 doublet singlet Kramers theorem requires all the edge levels be doubly degenerate! The degeneracy can be only removed by bulk phase transition. 44 cf.) Todo et al. (2001)
matrices NA x NB diagonal matrix Entanglement Spectrum Entanglement Entropy Entanglement spectrum contains more information than entanglement entropy! Schmidt decomposition 45
| B µ B The entire entanglement spectrum has exact double degeneracy in the Haldane phase! 1 = 2, 3 = 4, 5 = 6, . . . This degeneracy is protected by any one of the three symmetries. A B Minimal entanglement entropy log(2) when 1 = 2 = 1/ ⇥ 2, = 0( 3) 46
in the presence of D2(=Z2 x Z2) symmetry [π-rotation about x,y, and z axes] (II) Kramers degeneracy of edge spins, robust in the presence of time-reversal S=1 Haldane phase is “protected” by ANY one of 49 (III) Space Inversion symmetry about a bond center (Gu-Wen/Pollmann-Berg-Turner-M.O.)
application of field theory and topology to condensed matter physics - a prediction which sounded like crazy at the time but eventually is supported by many analytical/numerical/experimental evidences but also - an inspiration for many modern concepts including Matrix-Product/Tensor-Network States and Symmetry-Protected Topological Phases