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I have been struggling recently with a comprehensive problem on the relationship between topological superconductor and topological order. My question originates from reading a work conducted by Prof. Wen http://arxiv.org/abs/1412.5985 ,also a previous answered question Do topological superconductors exhibit symmetry-enriched topological order? ,in which Prof. Wen's comment also deals a lot with my questions presented below.

It has been claimed that 1+1d topological superconductor is a case of 1+1d fermionic topological order (or more strictlly speaking, a symmetry-enriched topological order).

My understanding on topological order is that it is characterized by topological degeneracy, which is degenrate ground states roubust to any local perturbation. And in the paper 1412.5985, it is argued that 1+1d topological superconductor is such a fermionic topological order with fermion-parity $Z_2^f$ symmetry breaking. So far I have learned about one-dimensional (or maybe it's quasi-one-dimensional as you like) topological superconductor, the first case where we have detected the appearance of Majorana zero modes is 1d D-class topological superconductor. And the terminology on D-class is defined from a classification working with BdG symmetry class of Hamiltonian. And D-class means such 1d superconductor possesses only particle-hole symmetry (PHS: $\Xi^2=+1$), and based on Kitaev's K-theory jobs the topological invatiant is $\mathbb{Z}_2$. In non-trivial topological phase labeled with topological number $\nu=-1$, a Majorana zero mode appears at each end of 1d D-class topological superconductor. And under my comprehension, the two Majorana zero modes form a two-fold degenracte ground states $|0\rangle$,$|1\rangle$ with different fermion-parity, and in a fermionic system like superconductor, the Hamiltonian has a fermion-parity $Z_2^f$ symmetry, hence the actual ground state configuration just don't have such symmetry from Hamiltonian and therefore spontaneous breaks it.

However why it should be claimed that such $Z_2^f$ breaking just directly leads to the appearance of 1d fermionic topological order in this 1d topological superconductor? (Regarding to the original words in the paper "two-fold topological degeneracy is nothing but the two-fold degeneracy of the $Z_2^f$ symmetry breaking"). And Prof. Wen in that previous posted question also said that this kind of 1d fermionic topological order state is therefore the result of long-range entanglement(LRE), so Majorana chain is indeed LRE and Kitaev has just used another way with out local unitary transformation definition(LU) to describe topological order. So I want ask how should one treat such open-line 1d Majorana chain as a long-range entanglement quantum state? And what about describing it with LU definition? How should people be supposed to demonstrate that?

And the last but also the most important question which makes me quite confused is that, it seems like all 1d topological superconductors, regardless of what class or what symmetry they have, are 1d fermionic topological order state due to the appearance of topological degeneracy, which so far in here considered as Majorana zero modes. Well, at least in the paper mentioned above, it is about a two-fold topological degeneracy. And in 1d D-class topological superconductor, it is toplogical robust to what perturbation ? It has no TRS, nor chiral symmetry at all. I don't know how to discuss the TRS breaking influence on this two-fold degeneracy. And fine, I assume it is indeed robust to any local perturbation. Then let's consider another case, the 1d DIII-class topological superconductor with TRS $\Theta^2=-1$, and according to Kramer's theorem, each end has a pair of time-reverse counterpart of Majorana zero modes, i.e. Majorana doublet, which forms a four-fold degenracy. After we introduce TRS-breaking perturbation e.g. Zeeman field, the Majorana doublet is lifted and splits a finite energy level-------I have read the references PhysRevB.88.214514 as well as PhysRevLett.111.056402 , and I don't know whether I have got it correctly or not. So in these works, I found words claim that such four-fold degenracy is TRS topological protected. Does that imply this kind of 1d toplogical superconductor has actually no topological degenracy with Majorana zero modes since it needs symmetry protection and therefore should belongs to SPT rather then SET ? If it is a yes, then does it mean not all 1d topological superconductors are topological order but just those with two-fold degenarcy may belong to topological order ? If not, then where did I actually get wrong ?

I'd appreciate every helpful comments and replys. Thank you.

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The key question here is that how to define/describe 1+1D fermionic topological order (ie with only fermion-number-parity symmetry $Z_2^f$) for interacting system? Kitaev's approach does not apply since that is only for non-interacting system. In other word, we like to ask "given a ground state of a strongly interacting 1+1D fermion system, how do we know it is topologically ordered or trivially ordered?"

There is a lot of discussions about Majorana chain and topological superconductor, but many of them are only for non-interacting systems. The key question here is how to define/describe those concepts for strongly interacting fermion systems?

A lot of the descriptions in the question are based on the picture of non-interacting fermions, while our 1D chain paper is for strongly interacting fermion systems. Using the picture of non-interacting fermions to view our 1D chain paper can lead to a lot of confusion. Note that there is not even single-particle energy levels in strongly interacting fermion systems, and there is no Majorana zero modes in strongly interacting fermion systems. In this case, how do we understand 1D topological superconductor?

With this background, the point we are trying to make in our 1D chain paper is that:

1) strongly interacting fermion chain that has only $Z_2^f$ symmetry can have a gapped state that correspond to the spontaneous symmetry breaking state of $Z_2^f$.

2) Such a symmetry breaking state is the 1+1D fermionic topologically ordered state (ie is in the same phase as 1D p-wave topological superconductor).

In other words, the 1+1D fermionic topologically ordered state on a chain can be formally viewed as SSB state of $Z_2^f$ (after we bosonize the fermion using Jordan-Wigner transformation). Unlike many other pictures, this picture works for interacting systems.

Also topological order is defined as the equivalent class of local unitary transformations. It is incorrect to say that topological order is characterized by topological degeneracy, since there are topological orders (ie invertible topological orders) that are not characterized by topological degeneracy.

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  • $\begingroup$ Dear Prof. Wen, why do you say there are no Majorana zero modes in a strongly interacting system? I believe even in the strongly interacting case there is still an operator satisfying the Majorana algebra localized near the boundary and commuting with the ground state projector, is there not? $\endgroup$ – Dominic Else Sep 24 '16 at 5:08
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    $\begingroup$ Majorana zero mode is single particle energy level at zero energy in a superconductor. Such a concept is only for non-interacting fermions. For strongly interacting fermions, the corresponding concept is topological degeneracy. There are always operators that connect the degenerate states. There are topological degeneracies that do not correspond to Majorana zero mode in interacting systems. $\endgroup$ – Xiao-Gang Wen Sep 27 '16 at 20:51
  • $\begingroup$ Thank you very much Prof. Wen for this detail answer, and one more quick question below. $\endgroup$ – Tom Gao Nov 5 '16 at 17:43
  • $\begingroup$ By comparing your clearification on a previous question about (1+1)D TSC as symmetry-enriched topological ordered phase and the one here, could I put it this way: Kiteav chain is characterized by non-LU transformation exhibiting topological two-fold degenracy, which are non-abelian anyons, and therefore it is a case of topological ordered state in (1+1)D non-interacting fermion system. However after putting on some sysmetry, like the one I've mentioned is DIII-TSC, it can be regarded as SET state in (1+1)D, as one knows if we break TR-symmetry, it decomposes into two copies of Kiteav chains. $\endgroup$ – Tom Gao Nov 5 '16 at 17:52
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So many questions! First of all, it's a bit misleading to describe a Kitaev superconductor as spontaneously breaking $Z_2^f$. In fact, if you put a Kitaev superconductor on a system without boundary (i.e. periodic boundary conditions) there is no ground state degeneracy; the two-fold degenerate ground state for open boundary conditions really should be thought of a boundary modes.

The reason why people sometimes say this is that you can make a non-local transformation (Jordan-Wigner transformation) which relates the Kitaev superconductor to a quantum Ising chain, which does spontaneously break the $Z_2$ symmetry. But the order parameter for the quantum Ising chain doesn't map to anything local in the Kitaev superconductor. Therefore, from a conceptual point of view I think it's better to think of the Kitaev chain as entirely new topological phase and not try to understand it in terms of spontaneous symmetry breaking.

How do you see it's a topologically ordered state? Well, with open boundary conditions it has those topologically protected boundary modes. There is no local term you can add to the Hamiltonian that gaps out those boundary modes. This is not really related to any symmetry; it's just that that those two states are totally indistinguishable unless you look at the whole system so no local interaction could assign them different energies. It is possible to re-interpret this fact in terms of the non-local mapping to the quantum Ising model, but maybe a bit confusing.

To make connection with other definitions of topological order, this topologically protected degeneracy actually implies that you can't smoothly connect the Hamiltonian $H_K$ of the Kitaev superconductor to that of a trivial superconductor, $H_T$, without closing the bulk gap. To see this, we make use of a really useful mathematical trick called quasi-adiabatic continuation [1], which tells you that if $H_K$ and $H_T$ could be smoothly connected without closing the gap, there would exist a local unitary $\mathcal{U}$ relating the ground-state subspaces of $H_K$ and $H_T$. But $H_K$ has a two-fold degeneracy that cannot be closed by a local perturbation, whereas $H_T$ is trivial and therefore there exists a local perturbation $h$ which lifts the degeneracy. But then $\mathcal{U} h \mathcal{U}^{\dagger}$ is a local perturbation which lifts the degeneracy of $H_K$ which is a contradiction.

With regard to your last question, the Kitaev superconductor in 1-D (D-class) is protected without requiring to any symmetry. On the other hand, the class DIII does require time-reversal symmetry. If you ignore the symmetry a class-DIII superconductor actually looks like two copies of the Kitaev superconductor, which is supposed to be trivial (because D class has a $Z_2$ classification). In other words, there are actually two Majorana zero modes on the edge, which is equivalent to a regular complex fermion and can be gapped out. But, the term which gaps out the edge is not allowed by time-reversal symmetry. Thus, the class D superconductor is topologically ordered, the class DIII superconductor is only SPT.

[1] https://arxiv.org/abs/1008.5137

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  • $\begingroup$ Great answer! Relatedly, I've been a bit puzzled by Wen calling the Kitaev chain an example of spontaneous symmetry breaking: indeed as you suggest one way of interpreting that statement is as a claim about the Jordan-Wigner dual spin chain. However it seems to me that Wen is really talking about the fermionic chain itself, whereas I agree with you that there is actually no spontaneous symmetry breaking in this case (i.e. the physical ground state is still symmetric under fermionic parity $P$). [cont] $\endgroup$ – Ruben Verresen Aug 22 '16 at 23:27
  • $\begingroup$ Trying to make sense of Wen's statement, I suspect the following: by looking at spin chains, Wen equates spontaneous symmetry breaking to long-range entanglement in the (exact) energy eigenstates. This is indeed correct for spin chains, e.g. taking the Ising model, we know the (exact) ground states are those of the form $|\textrm{spins up}\rangle \pm |\textrm{spins down}\rangle$. This means that if there is one lone spin in the universe that couples to one spin of our system, our ground state decoheres into all $|\textrm{spins up}\rangle$ or $|\textrm{spins down}\rangle$ [cont] $\endgroup$ – Ruben Verresen Aug 22 '16 at 23:28
  • $\begingroup$ i.e. any physical ground state will have spontaneously broken the symmetry. Generally: for spin chains long range entanglement implies spontaneous symmetry breaking. This is different for the Kitaev chain: similarly the ground states $|N_f\textrm{ odd}\rangle \pm |N_f\textrm{ even}\rangle $ (with $N_f$ the number of fermions) have long range entanglement, but this time one cannot write down any local operator that could decohere this superposition and as such the long range entanglement seems stable and so does not imply spontaneous symmetry breaking. I wonder if someone disagrees with this? $\endgroup$ – Ruben Verresen Aug 22 '16 at 23:29
  • $\begingroup$ Hi, thank you for your reply Else, and also thanks for @Ruben Verresen 's comments. First I would like to point out, usually as we know in superconductor system, there is SSB from $U(1)\to Z_2$ with certain ground state configuration however normal superconductor preserve fermion-parity $Z_2^f$ because ground states are just a Cooper pairs condensate and it allows various fermion number occupation configurations without modifying total fermion-parity. $\endgroup$ – Tom Gao Aug 23 '16 at 15:42
  • $\begingroup$ However here, we discuss on an open line segment 1d spinless p wave superconductor (well in experimental realization, one always add strong Zeeman field so as to obtain fully polarized electrons and at low energy effective scale we view them as spinless electrons), the Majorana zero modes exist at two end form two-fold degeneracy indeed cause the previous normal superconductor ground states become two degenerate sectors which differ with their parity: $|GS\rangle_\text{even}=|0\rangle_f\prod_j^{N-1}|0\rangle_j$, $|GS\rangle_\text{odd}=|1\rangle_f\prod_j^{N-1}|0\rangle_j$. $\endgroup$ – Tom Gao Aug 23 '16 at 15:42

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