# How to understand the entanglement in a lattice fermion system?

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Topological insulator is a fermion system with only short-ranged entanglement, what does the entanglement mean here?

For example, the Hilbert space $V_s$ of a lattice $N$ spin-1/2 system is $V_s=V_1\otimes V_2\otimes...\otimes V_N$, where $V_i$ is the Hilbert space of the spin on site $i$. And the meaning of an entanglement state belongs to $V_s$ is clear — a state which can not be written as a direct tensor product of the $N$ single spin states.

Now consider a spinless fermion system lives on the same lattice as spin-1/2, in the 2nd quantization framework, the fermion operators $c_i,c_j$ on different lattices $i,j$ do not commute with each other and the Hilbert space $V_f$ of the fermion system can not be written as a direct product of $N$ single fermion Hilbert spaces. Thus, how to understand the entanglement in this fermion system?

Mathematically, we can make a natural linear bijective map between $V_f$ and $V_s$, simply say, just let $\mid 0\rangle=\mid \downarrow\rangle,\mid 1\rangle=\mid \uparrow\rangle$. Thus, can we understand the entanglement of a fermion state in $V_f$ through its corresponding spin state in $V_s$?

This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user K-boy

edited Apr 6, 2015
Good question. The issue of entanglement in fermionic Fock space has been studied in a number of works. As far as I know there is no unique definition of entanglement; see this nice review for more details. The map between the fermion and spin picture can be achieved using a Jordan-Wigner transformation, which is a bit more subtle than the simple idenfication you alluded to. When going between spin and fermion pictures, local operators get transformed to highly non-local operators in general.

This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user Mark Mitchison
@ Mark Mitchison Thanks for your suggested review.

This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user K-boy

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Mark Mitchison is right. The concept of entanglement in systems of indistinguishable particles is more controversial than it is in the case of systems composed of distinguishable subsystems. You need to define first what do you mean by it when it comes, for example, to fermions. Do you mean entanglement between particles (connected with single Slater determinants), modes, pairing of states or whether a given state can be written as a convex combination of Gaussian states or sth completely else. You also should specify do you want to consider fermionic state with a ﬁxed number of fermions (and then use the criteria from here) or just to ﬁx the parity of the fermionic state and not the number of fermions, obtaining e.g. Gaussian states. This is also important, because even though physical states have a fixed number of fermions, Gaussian fermionic states are important approximations to physically non-trivial states, such as the superconducting BCS state. Of course then, the super-selection rule should also play a role somehow.

And about your question, you can find a nice definition of short-ranged entanglement in topological insulators in Sec. II of http://arxiv.org/pdf/1004.3835v2.pdf

This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user Piotr Ćwikliński
answered Nov 15, 2013 by (70 points)
@ Piotr Ćwikliński Thanks for your wonderful explanation.

This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user K-boy
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It is mentioned that "the Hilbert space $V_f$ of the fermion system can not be written as a direct product of $N$ single fermion Hilbert spaces." This is not correct.

Let $V_n$ be the total Hilbert space of $n$ spinless fermions on a lattice, and $N$ is the number of lattice sites. Then we have $V_f=\oplus_{n=0}^N V_n =\otimes_i V_i$, where $V_i$ is the two dimensional Hilbert space in site-$i$, describing an empty site $|0>$ and an occupied site $|1>$. So if we include states with different fermion numbers, the total fermion Hilbert space $V_f$ CAN be written as a direct product of N single fermion Hilbert spaces, just like your spin example.

The entanglement in a lattice fermion (or boson) system is many-body entanglement. To understand patterns/equivalent-classes of many-body entanglement is to understand topological order. However, we cannot understand the patterns/equivalent-classes of many-body entanglement of a fermion state in $V_f$ through its corresponding spin state in $V_s$. The reason is that the the local unitary transformations that define the  equivalence relation between different many-body enatanglement (or between different many-body states) are different for bosonic and fermionic systems. See http://arxiv.org/abs/1010.1517 for details. So although bosonic and fermionic systems can have exactly the same Hilbert space (as described by your natural linear bijective map), the meaning of short-range and long-range many-body entanglement are different for bosonic and fermionic systems. The classification of topological orders and SPT orders for bosonic and fermionic systems are different, due to the different definitions of  local unitary transformation.

answered Apr 4, 2014 by (3,485 points)
edited Aug 7, 2015

@Xiao-Gang Wen Thank you Prof.Wen. Can I understand that the meaning of "local" in the "local perturbation" or "local operator" is different for bosonic and fermionic systems? To be specific, take your recent paper(http://arxiv.org/abs/1412.5985) or the 1D Kitaev chain(http://arxiv.org/abs/cond-mat/0010440) as an example, the fermionic topological order (2-fold degeneracy) is due to the spontaneous fermion-parity ($Z_2^f$) symmetry breaking, and consider two kinds of perturbations to the fermionic Hamiltonian:(1)fermion operator $c_j+c_j^\dagger$;(2)spin-flip operator $S_j^x$ expressed by fermion operators via Jordan-Wigner transformation, both of which break the $Z_2^f$ symmetry.

@Xiao-Gang Wen My confusions are: Are the above two kinds of perturbations "local operators" ? The spin-flip operator $S_j^x$ seems to be "local" at least for the bosonic spin system, so is it also "local" for the fermionic system? The perturbations (1) and (2) seem to be able to lift the 2-fold degeneracy since they break the $Z_2^f$ symmetry, so if they indeed represent "local perturbations", then this implies that the fermionic topological degeneracy can be lifted by "local perturbation" (1) or (2). So why we still call it "topological degeneracy"? Thank you very much!

$c_j+c_j^\dagger$ is not a local perturbation. Anything you can add to the Hamiltonian of a fermionic system must preserve fermion parity. $S_j^x$ is local for spins, but when you do Jordan-Wigner transformation it becomes a non-local operator for fermions (roughly like $(c_j+c_j^\dagger)\prod_{k<j}(1-2n_k)$. So neither is local.

@Meng Thanks for your comments. But I still feel a little subtle for the "non-locality" of $S_j^x$ for fermions. Since $S_j^x$ can only change the local fermion Hilbert space at site $j$ (i.e., only change the local fermion number at site $j$) while does not affect other local fermion Hilbert spaces, as long as we have the map between $V_f$ and $V_s$ mentioned in the last paragraph in the original question. So, does the meaning of "local" or "non-local" not only depend on the Hilbert space itself, but also depend on the physical variables like fermion operators $c_j$ or spin operators $S_j$?

$S_i^x$ only "changes" the local fermion occupation, which is true, but its matrix element depends on the infinite string of fermions. So it is non-local.

Topological degeneracy is NOT robust to non-local perturbations. This is why locality is so important for the definition of topological phase.

Yes, and it is nothing but the operator $i\gamma_1\gamma_2$ where $\gamma_{1,2}$ are the two zero modes localized on the two ends of the superconductor.
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