I have a question about entanglement in condensed matter physics. It seems that topological order origins from long range entanglement, but what is long range entanglement? It is the same as long range correlation? I am interested in this issue and I am happy to have any discussion.

  • $\begingroup$ Long-range entanglement is not the same as long-range correlation, the latter is characteristic of "Landau-type" symmetry breaking order. For topological order, it is typically important you don't have long-range correlations. For a discussion of long/short-range entanglement see arxiv.org/abs/1004.3835 section II. I will not answer this question since I think Prof. @Xiao-Gang Wen can give a much more qualified answer. $\endgroup$ – Heidar Sep 20 '12 at 13:46
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    $\begingroup$ @Heidar, you are correct, but it makes me wonder -- is the long-range (classical) correlation of Landau-type order incompatible with topological order? Or does it just prevent us from seeing topological order because we think we know what we're looking at? Maybe Prof. Wen can weigh in on this as well. $\endgroup$ – wsc Sep 20 '12 at 14:32
  • $\begingroup$ I'd like to second wsc's question about the exact relation between correlation and entanglement. After all, the two definitely occur together in many cases (not least models of criticality, e.g. see the numerical work on MERA). My impression is that entanglement is somehow a "finer" view of correlation. $\endgroup$ – genneth Oct 1 '12 at 13:25
  • $\begingroup$ @genneth: I am not sure how these things can occur together at criticality (where the system is gapless). Maybe you are thinking more generally about long-range entanglement, while I have more specifically "topological order" in mind? $\endgroup$ – Heidar Oct 1 '12 at 14:30
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    $\begingroup$ @wsc: My naive reason to think that you can't have both long-range correlations and topological order, is that the latter is gapped by definition. While having long-range correlations (I think) always indicate you have gappless excitations. 'it is however not completely unusual to have gapless excitations at the boundary of the system, as in FQHE, but the bulk remains gapped. $\endgroup$ – Heidar Oct 1 '12 at 14:32

Long range entanglements are defined through local unitary transformations which is discussed in arXiv:1004.3835 Local unitary transformation, long-range quantum entanglement, wave function renormalization, and topological order by Chen, Gu, and Wen.

Basically, long range entangled states are states which are very different from product states and cannot be changed to product states through local unitary transformations.

  • $\begingroup$ Thanks. But is the definition of long/short entanglement through LU transformation equal to the definition from entanglement entropy, which says short range entanglement state gives zero entropy in large scale? $\endgroup$ – Brioschi Oct 1 '12 at 13:09
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    $\begingroup$ No. There are long entangled states (as defined by LU transformation) which have zero topological entanglement entropy. The integer fermionic QH states and $E_8$ bosonic QH states are examples. $\endgroup$ – Xiao-Gang Wen Oct 1 '12 at 21:31

I think I can give a more technical and detailed explanation of "long range entanglement". I felt it puzzling some time before and still puzzle me for some situations.

For generic topological states, the entanglement of ground states scales as $S=\alpha L-ln(D)$ ($D$ is the total quantum dimension of the system. For topologically nontrivial system, $D>1$) for the reduced density matrix $\rho_A$ whose perimeter is $L$. Good references of this point should be Preskill & Kitaev's paper and Levin & Wen's paper. If we start and keep on coarse-graining the wave function, i.e., block several lattice sites into one, we still end up with an entangled state. The reason is the following: Since we need $S>0$ all the time during our coarse-graining, $\alpha$ cannot be zero. Therefore, our state is still entangled no matter how you do the coarse-graining. The "long range" is in the sense of coarse-graining. For a topologically trivial state, $S=\alpha L$, the above reasoning does not apply here. So we will "probably" end up with a product state with $\alpha=0$ after coarse-graining.

Hope it helps!!


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