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  Do topological superconductors exhibit symmetry-enriched topological order?

+ 7 like - 0 dislike
2390 views

Gapped Hamiltonians with a ground-state having long-range entanglement (LRE), are said to have topological order (TO), while if the ground state is short-range entangled (SRE) they are in the trivial phase. The topological properties of a system with topological order (such as anyonic statistics, ground-state degeneracy etc.) are protected for any perturbations of the Hamiltonian that does not close the energy gap above the ground state. See more details here.

If we further require that the system is protected by a symmetry $G$, the LRE states split further into several classes called symmetry-enriched topological order (SET). The SRE state (which is trivial without symmetry protection) is also split into several classes called symmetry protected topological order (SPT). The added physical features of these systems (such as gappless edge-states) are only protected for perturbations which do not close the gap and do not break the symmetry $G$.

Topological insulators are know belong to SPT states, they are SRE and their topological properties are only protected by a symmetry. Related cousins to topological insulators are topological superconductors. In this context, one usually think of superconductors as insulators + particle-hole symmetry (which comes from the Bogoliubov de Gennes Hamiltonian + Nambu spinor notation). This might lead you to conclude that topological superconductors are also SPT states.

However, it is known (by some) that superconductors cannot be described by a local gauge-invariant order-parameter as usual symmetry-breaking phases of matter (but a non-local gauge invariant order parameter exist.) A s-wave superconductor is actually topologically ordered (and thus LRE), and exhibits anyonic statistics, ground-state degeneracy on higher-genus manifolds, low-energy topological quantum field theory (TQFT) description and so on. It is topologically in the same phase as the famous Kitaev Toric code, the $\mathbb Z_2$ topological order. See details here.

Now, my question is the following: is it wrong to consider topological superconductors (such as certain p-wave superconductors) as SPT states? Aren't they actually SET states?

This post imported from StackExchange Physics at 2014-04-05 04:35 (UCT), posted by SE-user Heidar
asked Jul 17, 2013 in Theoretical Physics by Heidar (855 points) [ no revision ]
This question was inspired by a recent question from Oaoa physics.stackexchange.com/questions/71151/… .

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Heidar
A beginner 's question : in your first reference, fig (3), page 6, it seems that SRE and LRE could be considered without symmetry or with symmetry, and in the symmetry-case, with symmetry-breaking (SB) or symmetry respected (SY). So it seems that SRE and LRE differ only by the fact that we could or not unentangle a state by local unitary operation , and does not depends on the behaviour relative to the symmetries. Is this correct?

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Trimok
In the same reference, page 7, the authors state that : "(C) Symmetry breaking and long range entanglement can appear together in a state, such as SB-LRE 1, SB-LRE 2, etc in Fig. 3(b). The topological superconducting states are examples of such phases".

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Trimok
@Trimok Yes exactly, if we do not declare any symmetry to be sacred then the difference between SRE states and LRE states is as you state it. Thus the topological properties of LRE states are protected only by the energy gap, you may break any symmetry you like just the gap is not closed (you may in principle hit the system with a hammer or shoot it with a gun, as long as the gap is protected). The quote about topological superconductors you just found seem to exactly answer my question! Thanks a lot! Not so much a beginner after all! ;)

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Heidar
I "might" have a definite answer for you Heidar. But only for the specific $p+ip$ case. I recently attended one of Kitaev's lectures and he said that $p+ip$ does NOT have topological order. I don't have a full understanding myself considering that the talk was very brief! Also, to my annoyance, his lecture was mainly based on unpublished work! But if it helps in any way, he drew this table on the board: i.stack.imgur.com/7mOry.jpg. Sorry I can't be of much help. But it might be a wise strategy to assume Kitaev is right and reverse-engineer his results!

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user NanoPhys
One should be careful because I'm not really sure how standardized this terminology us. "Topological order" has been attached to many things in trivial part of Gang Xiao Wen's classification. I'm particularly not sure how "most" people consider gauge-invariance in this discussion. Suppose you put your system in a pile of superconducting grains - is the topological order maintained? So is there maybe a physical way to re-phrase this question? Maybe in terms of adiabatic connectivity under particular conditions?

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user BebopButUnsteady
@NanoPhys Yes, the famous unpublished work of Kitaev which everything has been solved. I have heard a lot of rumors about its existence, but almost nothing about its content. :D I think there is a lot confusion in the field, partly because of (as BebopButUnsteady) says, non-equivalent uses of the word "topological order" and LRE. I am mainly using the definitions and conventions of Wen and collaborators. For example, Kitaev says that the $E_8$-state is not LRE. However I think it is LRE, under Wen's definition of LRE, at least I have heard him say that. (continued)

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Heidar
But I haven't understood why exactly (that could be another question I should ask). At least I think I understand why Kitaev puts the $E_8$-state under non-LRE. If I recall correctly, in this talk pirsa.org/displayFlash.php?id=13050050, John McGreevy talks about east coast (Wen) vs west coast (Kitaev) definitions of LRE (I cannot confirm this, I don't have audio here right now). The east cost one my above definition (using local unitary operations) and the west cost one requires that the topological entanglement entropy $\gamma$ is non-zero. (continued)

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Heidar
For the $E_8$ state it is zero $\gamma = 0$, and thus non-LRE under (what I think is) Kitaev's definition. I am however not really sure why that should be the case for $p+ip$ state. My vague understanding is that all superconductors (possibly except the ones with gapless points) are topologically ordered, with a Toric code type $\mathbb Z_2$ topological order. Where the vortices, charge excitations and their bound states give rise to an abelian anyon model. But when extra symmetry is assumed, such as time-reversal, there can be an extension of to ising anyons due to Majorana zero modes.

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Heidar
I am not sure whether my understanding is correct. At least, I am very intrigued to understand why Kitaev considers the $p+ip$ superconductor as SRE.

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Heidar
@BebopButUnsteady Thanks a lot for the correction of the title! My grammar in English is not what it should be!

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Heidar
This paper came out today: arxiv.org/pdf/1307.4403.pdf . It has a large discussion on the p+ip superconductor on the first few pages. It might be of interest to you.

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Olaf

2 Answers

+ 4 like - 0 dislike

Let me first answer your question "is it wrong to consider topological superconductors (such as certain p-wave superconductors) as SPT states? Aren't they actually SET states?"

(1) Topological superconductors, by definition, are free fermion states that have time-reversal symmetry but no U(1) symmetry (just like topological insulator always have time-reversal and U(1) symmetries by definition). Topological superconductor are not p+ip superconductors in 2+1D. But it can be p-wave superconductors in 1+1D.

(2) 1+1D topological superconductor is a SET state with a Majorana-zero-mode at chain end. But time reversal symmetry is not important. Even if we break the time reversal symmetry, the Majorana-zero-mode still appear at chain end. In higher dimensions, topological superconductors have no topological order. So they cannot be SET states.

(3) In higher dimensions, topological superconductors are SPT states.

The terminology is very fusing in literature:

(1) Topological insultor has trivial topological order, while topological superconductors have topological order in 1+1D and no topological order in higher dimensions.

(2) 3+1D s-wave superconductors (or text-book s-wave superconductors which do not have dynamical U(1) gauge field) have no topological order, while 3+1D real-life s-wave superconductors with dynamical U(1) gauge field have a Z2 topological order. So 3+1D real-life topological superconductors (with dynamical U(1) gauge field and time reversal symmetry) are SET states.

(3) p+ip BCS superconductor in 2+1D (without dynamical U(1) gauge field) has a non-trivial topological order (ie LRE) as defined by local unitary (LU) transformations. Even nu=1 IQH state has a non-trivial topological order (LRE) as defined by LU transformations. Majorana chain is also LRE (ie topologically ordered). Kitaev does not use LU transformation to define LRE, which leads to different definition of LRE.

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user Xiao-Gang Wen
answered Aug 21, 2013 by Xiao-Gang Wen (3,485 points) [ no revision ]

To my best knowledge, topological order is thermally unstable, e.g., arXiv:0709.2717; if a s-wave superconductor is a SET state, what is the manifestation of its topological order? Can those experimental signatures at finite temperature (but below Tc), such as Meissner effect and zero resistance be related to to topological order?

+ 0 like - 0 dislike

Let me shortly elaborate -- though not answering (at all? certainly :-) since all these notions are pretty new for me -- about what I know from my pedestrian work on superconductors.

I believe everything is much more complicated when you look at the details, as usual. I may also comment (and would love to be contradicted about that of course) that the Wen's quest for a beautiful definition about topological order somehow lets him going far away from realistic matter contrarieties.

$s$-wave superconductor may be seen as long-ranged topological order / entangled state, as in the review you refer to : Hansson, Oganesyan and Sondhi. But it is also symmetry protected. I think that's what you call a symmetry-enriched-topological-order. I'm really angry when people supposed the symmetry classification for granted. $s$-wave superconductors exhibit time-reversal symmetry in addition to the particle-hole one, and thus a chiral symmetry for free, so far so good. Suppose you add one magnetic impurity. Will the gap close ? Of course not ! The gap is robust up to a given amount of impurities (you can even adopt a handy argument saying that the total energy of the impurities should be of the same order than the energy gap in order to close the gap). You may even end up with gapless-superconductor... what's that beast ? Certainly not a topological ordered staff I presume. More details about that in the book by Abrikosv, Gor'kov and Dzyaloshinski. So my first remark would be : please don't trust too much the classification, but I think that was part of the message of your question, too. [Since we are dealing with the nasty details, there are also a lot of predicted re-entrant superconductivity when the gap close, and then reopens at higher magnetic field (say). Some of them have been seen experimentally. Does it mean the re-entrant pocket is topologically non-trivial ? I've no idea, since we are too far from the beautiful Wen's / east-coast definition I believe. But open -> close -> open gap is usually believe to give a topologically non trivial phase for condensed matter physicists.]

So my understanding about this point is: what does the symmetry do ?

  • If it creates the gap in the bulk an/or the gap closure at the edge, then it's not a good criterion. Most of the symmetry like this are generically written as chiral or sub-lattice symmetry (the $C\equiv PT$ in the classification). Some topological insulators only have this symmetry (grapheme in particular if I remember correctly), since the particle-hole ($P$) symmetry defines superconductors. According to Wen, this situation does not lead to topological order in the discussion we have previously.

  • If the symmetry reenforces the gap, then it protects you against perturbation. Once again, the time-reversal symmetry of the $s$-wave superconductor protects against any time-reversal perturbation (the Anderson theorem). But it is not responsible for the appearance of the gap ! I confess that's really a condensed-matter physicist point of view, which could be really annoying for those wanting beautiful mathematical description(s). But clearly the $T$-symmetry should change nothing about the long-ranged-entanglement for $s$-wave (according you believe in of course :-)

As for $p+ip$, it's also called polar phase in superfluid (recall there is no $p$-wave superconductor in nature at the moment, only in neutral superfluid). As Volovik discuss in his book, this phase is not stable (chapter 7 among others). It is sometimes referred as a weak topological phase, which makes no sense. It just corresponds to a fine tuning of the interaction(s) in superfluid. B-phase is robust, and fully gapped. So the pedestrian way is just a way to say that the $p+ip$ is not robust, and that you should have a structural transition (of the order parameter) to the more robust B-phase. NB: I may well confound the names of the subtle phases of superfluid, since it is a real jungle there :-(.

Finally, to try to answer your question: I would disagree. A topological-superconductor (in the condensed matter sense: a $p$-wave say) does not have $T$ symmetry. So it's hard for me to say it is symmetry-enriched the way I used it for $s$-wave. That's nevertheless the symmetry reason why $p$-wave (if it exists in materials) should be really weak with respect to impurities. [Still being in the nasty details: For some reasons I do not appreciate fully, the Majorana may be more robust than the gap itself, even if it's difficult to discuss this point since you need to discuss a dirty semiconductor with strong spin-orbit and paramagnetic effect in proximity with a $s$-wave superconductor, which is amazingly more complicated than a $p$-wave model with impurities, but this latter one should not exist, so...] Maybe I'm totally wrong about that. It sounds that words are not really helpful in topological studies. Better to refer to mathematical description. Say, if the low energy sector is described by Chern-Simons (CS), would we (or not ?) be in a topological order ? Then the next question would be: is this CS induced by symmetry or not ? (I have no answer for this, I'm still looking for mechanism(s) about the CS -- an idea for an other question, too).

Post-scriptum: I deeply apologize for this long answer which certainly helps nobody. I nevertheless have the secret hope that you start understanding my point of view about that staff: symmetry may help and have an issue in topological order, but they are certainly not responsible for the big breakthrough Wen did about long-ranged-entanglement. You need a local gauge theory for this, with global properties.

This post imported from StackExchange Physics at 2014-04-05 04:36 (UCT), posted by SE-user FraSchelle
answered Jul 17, 2013 by FraSchelle (390 points) [ no revision ]

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