We know that the string order of a spin chain is defined as $$\mathcal{O}^\alpha=\lim_{i-j\to\infty}\left\langle S_i^\alpha\prod_{k=i+1}^{j-1}\exp(i\pi S_k^\alpha)\ S_j^\alpha \right\rangle$$ now consider a spin-1 chain and the state to be $|+1,+1,\cdots,+1\rangle$. Then the string order should be $\langle1\cdot(-1)^{j-i-1}\cdot1\rangle$, then this number is not defined.

I think the way to avoid this is by consider that any state cannot be so pure so that as $i-j\to\infty$ this number could have exponential decay or something like that, making the string order turns into 0. I'm highly likely to be wrong, but what the correct way to explain this?

Edit: as suggested, the string order might only be able to study the ground state topological properties, and for systems (Hamiltonian) with different class of symmetry, the string order parameter is different. However, what if one want to study an eigenstate's topological properties in a non-equilibrium scenario?


Your mistake is in claiming that that is 'the' string order. There is no such unique string order; rather, it depends on the phase of matter one wants to probe.

Let me illustrate the point with an example that is more familiar: suppose someone tells you that 'the' symmetry-breaking order parameter is $$\lim_{|i-j| \to \infty}\langle S^x_i S^x_j \rangle. $$ And then suppose that same person complains that this is not well-defined for an anti-ferromagnet, where the above quantity would have an alternating sign on every other site.

Your answer to that person would be simple: firstly, for different symmetry-breaking orders you could need different order parameters (in general one tries to find some operator $\mathcal O$ which does not commute with the symmetry yet which would obtain a finite expectation value in a symmetry-broken ground state, and then one is interested in the long-range two-point correlator of that operator). Secondly, it is understood that one tries to peel off short-distance phenomena, as they are irrelevant and often easy to avoid given the physics at hand. In particular, if one expects an anti-ferromagnetic phase, one would instead propose $\lim_{|i-j| \to \infty} (-1)^{i-j} \langle S^x_i S^x_j \rangle$.

Having cleared that issue up, the natural question you might then wonder is "but how does one in general figure out the correct string order parameter for a given symmetry-protected topological (SPT) phase?" That is a good question, and its answer has been figured out (at least for a huge class of SPT phases). I can for example recommend https://arxiv.org/abs/1204.0704 for further details on this.

  • $\begingroup$ Thank you for answering. I got a little bit confused here: the string order parameter, of course I know the form should be related to the symmetry of Hamiltonian, and the form I gave is only a $\mathbb{Z}_2\times\mathbb{Z}_2$ special one -- should the string order parameter only trace over the ground state of one Hamiltonian? What if I want to study the topological properties of arbitrary eigenstate (say, when study non-equilibrium physics such as MBL)? $\endgroup$ – RoderickLee Jul 31 '18 at 4:57
  • $\begingroup$ @RoderickLee You can also use that same string order parameter for excited states in the context of MBL. The main point is whether the string order decays to zero or not --- not whether it has a well-defined limit as such. To avoid such issues, one could for example look at $\lim_{|i-j| \to \infty} \sup |\langle \textrm{string operator}_{i,j} \rangle |$. Again, it might be conceptually useful to realize that your question is not particular to SPTs: you could ask the equivalent question of whether you can use the usual symmetry-breaking order parameter for excited states. The answer is: yes. $\endgroup$ – Ruben Verresen Jul 31 '18 at 18:43
  • $\begingroup$ thank you for the good comment. I'm only wondering, say in AKLT Hamiltonian where the string order parameter is defined the same form as I posted in question. Is it reasonable to ask about the parameter in state $|+1,+1,\cdots,+1\rangle$? As we can see, this is still an eigenstate of AKLT Hamiltonian. $\endgroup$ – RoderickLee Aug 2 '18 at 14:47
  • $\begingroup$ @RoderickLee Just because one choice of string order has a well-defined limit for one eigenstate, doesn't mean it has to be so for any eigenstate of the same Hamiltonian. Using the above 'limsup' definition works for all cases though, so doesn't that settle the issue? $\endgroup$ – Ruben Verresen Aug 2 '18 at 18:59
  • $\begingroup$ Thank you. One last question: if "The main point is whether the string order decays to zero or not", then is $|+1,+1,\cdots,+1\rangle$ a non-trivial state? Or just because it breaks the symmetry so it cannot be considered this way? If so, how about $\dfrac{1}{\sqrt{2}}\left(|+1,+1,\cdots,+1\rangle + |-1,-1,\cdots,-1\rangle\right)$? $\endgroup$ – RoderickLee Aug 2 '18 at 19:36

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