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Integrable Matrix Product States from boundary integrability
by B. Pozsgay, L. Piroli, E. Vernier
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Submission summary
| Authors (as registered SciPost users): | Balázs Pozsgay · Eric Vernier |
| Submission information | |
|---|---|
| Preprint Link: | https://arxiv.org/abs/1812.11094v3 (pdf) |
| Date submitted: | 2019-04-25 02:00 |
| Submitted by: | Pozsgay, Balázs |
| Submitted to: | SciPost Physics |
| Ontological classification | |
|---|---|
| Academic field: | Physics |
| Specialties: |
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| Approach: | Theoretical |
Abstract
We consider integrable Matrix Product States (MPS) in integrable spin chains and show that they correspond to "operator valued" solutions of the so-called twisted Boundary Yang-Baxter (or reflection) equation. We argue that the integrability condition is equivalent to a new linear intertwiner relation, which we call the "square root relation", because it involves half of the steps of the reflection equation. It is then shown that the square root relation leads to the full Boundary Yang-Baxter equations. We provide explicit solutions in a number of cases characterized by special symmetries. These correspond to the "symmetric pairs" $(SU(N),SO(N))$ and $(SO(N),SO(D)\otimes SO(N-D))$, where in each pair the first and second elements are the symmetry groups of the spin chain and the integrable state, respectively. These solutions can be considered as explicit representations of the corresponding twisted Yangians, that are new in a number of cases. Examples include certain concrete MPS relevant for the computation of one-point functions in defect AdS/CFT.
Author comments upon resubmission
List of changes
Referee 1.
1. We included a diagram at the Conclusions, with proper referencing. We hope that the relations are clear now.
2. We are especially thankful for this question, because we did not consider it before, and it is actually nice. Now we added a footnote about this: The later theorem in the text can be used to show that indeed all such states are of this form (they enjoy local group invariance), if the MPS is group invariant in any volume, and it is completely reducible. We did not include a full detailed proof of this claim. On the one hand side it is not really important, on the other hand the proof would use material which is presented only later in the text. So we just added this footnote.
3. There can be cases with non-trivial F parts, but we did not treat such cases here. We added some comments about this.
4. We clarified this.
5. At the moment we have not yet investigated the XYZ case, so we don't know the answer to this question. The present paper only deals with the polynomial cases. Perhaps later research can clarify this.
Referee 2
1. Yes and we explained it.
2. Indeed there was a mismatch in the notations, we tried to clarify it.
3. Indeed, and we modified that part of the sentence.
4. Here G stands for the matrix in the defining representation, we added the explanation in the text.
5. We explained this now, the G' representation is inherited from the defining rep. of the full original group.
6. Corrected.
7. Corrected.
8. This is a point, where we hoped that our manuscript was clear, but the comments of the referee show that it was not clear. So we have now added a few more explanations. The $\psi(u)$ is a four-legged object, so it has 4 indices. This is not written out explicitly, we just state that $\psi_{ab}(u)$ are matrices acting on $V_A$. And we regard $\psi(u)$ as an element of $V_1\otimes V_2\otimes End(V_A)$. We explained in the text now, we hope it is clear. Also, in the Appendix we added a detailed derivation with the indices spelled out, that the two forms of the sq.r.r. are the same.
9. Corrected.
10. See point 8. In the Appendix there is one example for this.
11. We replaced the notation here.
12. See above.
13. See point 8 and the new Appendix.
14. Corrected.
15. Here $L^*$ is just some threshold value.
Current status:
Reports on this Submission
Report #2 by Anonymous (Referee 4) on 2019-5-12 (Invited Report)
- Cite as: Anonymous, Report on arXiv:1812.11094v3, delivered 2019-05-12, doi: 10.21468/SciPost.Report.947
Report
The authors have addressed all the issues I have raised and added an additional Appendix explaining the derivation of (3.17) which they have now explicitly stated in (3.16). However I still find it difficult to understand equation (3.17). First, I believe there should be no tensor product in the r.h.s. of (3.16). This follows from (A.11): matrices $\psi_{ab}(u)$ and $\omega_k$ act on the same auxiliary space $V_A$. Second, tensor products $\omega \otimes \psi(u)$ and $\psi(u) \otimes \omega$ of collections of matrices $\omega_k$ and $\psi_{ab}(u)$ in (3.17) act on a tensor product $V_A \otimes V_A$ contradicting to what is stated right below, and so (3.17) is not equivalent to (3.16). Please correct this issue.
Requested changes
1) p.8, line 2, and p.28, line 5 from the bottom: $V_A\otimes V_0$ should be $V_0\otimes V_A$.
2) a tensor product symbol is missing in (3.8) and (A.3).
3) p.28, line 5 from the bottom: concernes should be concerns.