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asked by (160 points)

The Haldane phase of spin-1 Heisenberg chain belongs to symmetry protected topological phase. In PhysRevB.81.064439 and many other works it is said that the entanglement spectrum has two-fold degeneracy and it correspond to gapless edge modes. In this work they use iTEBD method. I suppose finite-size DMRG method will also get the double degeneracy of entanglement spectrum if the system is long enough.
In fact in the ITensor calculation of spin-1 chain, I get an entanglement spectrum like
0.48469
0.48469
0.00741
0.00741
0.00741
0.00741
4.00531E-4
4.00531E-4
3.32937E-5
3.32937E-5
3.32937E-5
3.32937E-5
They are almost double degenerate. But their value are not same as the iTEBD result in PhysRevB.81.064439. I just wonder if this method is efficient to detect entanglement spectrum and topological properties in general. Thank you very much!

1 Answer

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answered by (70.1k points)
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Best answer

Yes, you can use finite system DMRG to analyze the entanglement spectrum of SPT phases. These are systems with a finite correlation length, so if the system size exceeds the correlation length then the bulk behavior should generally be the same as for an infinite system. (Of course there can be exceptions, such as if the system breaks a symmetry yet the state is a superposition of two or more symmetry broken ground states.)

I would not be surprised if the precise values you get do have some finite-size and finite-sweep effects that make them not exactly identical to other reported values. The key thing of course is that you should see agreement with converged infinite values in the limit of taking larger and larger system sizes and converging your calculation to a more accurate level.

Best regards,
Miles

P.S. I didn't see any numerical values reported in the PRB article you mentioned? Where do they appear in the article?

commented by (160 points)
Hi Miles,
Thank you for your answer. In that famous paper by F. Pollmann et al., they show the entanglement spectrum in Fig. 2.
Now I know why I'm wrong. In fact they list the Schmidt eigenvalues in the figure, but I use the eigenvalues of the reduced density matrix calculated in ITensor, which is the square of Schmidt eigenvalues. Now everything is OK. Thank you!

Best regards,
Jie Hou
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