I'm confused about the valence bond solid (VBS) in condensed matter literature. The idea is a lattice is covered by spin singlets and thus spin rotational invariant. It seems that it's commonly accepted there are four ground states by rotating the whole lattice by $\frac{\pi}{2}$. I have roughly two questions:

1) how does the boundary condition change these states (possibly reduce the degeneracy?)it seems one at least needs period boundary condition or open boundary condition to have these four-fold degeneracy.

2) why is the order parameter a complex number? I don't quite understand this.


The model I had in mind is 2D spin-1/2 AFM and so the four degenerate states are singlets oriented up/down/right/left (hence rotating the whole lattice by $\frac{\pi}{2}$). But I suppose other examples are also welcome

  • $\begingroup$ What is the geometry you have in mind? 1D, 2D? The fourfold degeneracy is typical of 1D open chains $\endgroup$ – lcv Jun 2 at 1:13

I believe the name valence bond solid comes in opposition with resonating valence bond (an idea due to Anderson) where there is a coherent mixture of bonds and so they "resonate". Instead in the VBS state they are fixed. However I would say that a more precise characterization of VBS is as an equivalent form of matrix product states (MPS). Looking at an MPS as a VBS can give some physical insight but is usually more cumbersome to do computation. There is a famous old review by Werner Fannes and probably also Nachtergaele that discusses neatly this equivalence.

In any case the phenomenon you mention in 1) is typical of the 1D spin 1 Heisenberg antiferromagnetic chain and is exact for the AKLT model. The idea is the following. Substitute each spin 1 with two spin 1/2 (thus enlarging the Hilbert space to 4 dimensional). Build singlets between neighboring spins. Then project down onto the original Hilbert space (the triplet sector for the pair of neighboring spins).

If you do this and you have open boundary conditions, at the end you will have one free spin 1/2 remaining at each end. This gives rise to an exact 4-fold degeneracy for the AKLT model with open end. The degeneracy is lifted but exponentially small for the Heisenberg AFM. Of course if you close the chain (changing boundary conditions) the degeneracy disappears.

As for 2) the string order parameter can be complex simply because it's not the expectation value of an hermitian operator. However you will have to be a bit more specific herw as there are several string order parameters (and some are indeed real)

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  • $\begingroup$ This does not quite address the OP's question. (S)he is thinking of a case where translational/rotational symmetry is spontaneously broken, giving rise to degeneracy (which is there also for periodic boundary conditions). In your case, there is no symmetry-breaking but rather symmetry-protected topological order, giving degeneracy with open boundaries due to different (arguably more subtle) reasons. $\endgroup$ – Ruben Verresen Jun 2 at 17:01
  • $\begingroup$ Where did you read that the OP thinks of a case where translational symmetry is spontaneously broken? I agree that the question is vague at that point. I explicitly asked but received no response. And yes what I described here is an example of SPT. $\endgroup$ – lcv Jun 2 at 17:19
  • $\begingroup$ The OP specified that the degenerate ground states are related by rotating the lattice over pi/2. $\endgroup$ – Ruben Verresen Jun 2 at 17:22
  • $\begingroup$ Thanks for Icv's answer and Ruben's comment. I was indeed thinking about 2D spin-1/2 AFM model. It's interesting to know the AKLT example also, but it doesn't seem obvious to me that these two systems are related even though they have the same degeneracy in open boundary condition. I see we can choose to define a string order parameter for AKLT. I'd appreciate if you don't mind pointing me to a source explicitly giving a "string" order parameter for 2D AFM model $\endgroup$ – Histoscienology Jun 2 at 18:16
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    $\begingroup$ Histoscienology concerning your last question: Precisely, that's what RVBs are. @RubenVerresen your interpretation was correct :) $\endgroup$ – lcv Jun 2 at 18:57

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