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I am reading about short range and long range entanglement. I know that topological insulator is short range entangled (SRE), STP state as we can disentangle it by breaking the symmetry using magnetic impurity. It's also given that FQHE is LRE (long range entanglement) state.

I have done a numerical simulation for IQHE on a square lattice and have introduced a magnetic impurity in it, the edge states are still protected. Now I wish to ascertain this conclusion is correct and understand why. If it is correct then IQHE is LRE, otherwise it's SRE. Please explain, as I am uncertain.

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The IQH state cannot be changed to a product state via local unitary transformation (ie cannot be smoothly deformed into a product state without phase transition), thus, according to the original definition of long-range-entanglement in http://arxiv.org/abs/1004.3835, IQH state is a long-range-entangled state (ie has topological order). Later Kitaev proposed another definition where only states with non-trivial topological excitations are called long-range-entangled, under which IQH state is a "short-range-entangled" state.

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  • $\begingroup$ Professor Wen - Would IQHE be better understood as having "invertible" topological order? From my understanding, these are states that slip between the traditional definitions of SPT and SET phases, since they are long range entangled but do not have fractionalised quasi-particles? $\endgroup$
    – Aegon
    Commented Jul 30, 2016 at 2:49
  • $\begingroup$ Yes. IQH states have invertible topological orders. Traditionally (since 1990), IQH states were defined to have topological orders due to their robust gapless edge excitations $\endgroup$ Commented Jul 30, 2016 at 21:19

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