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  State of Matrix Product States

+ 16 like - 0 dislike
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What is a good summary of the results about the correspondence between matrix product states (MPS) or projected entangled pair states (PEPS) and the ground states of local Hamiltonians? Specifically, what "if and only if" type of equality/approximation hold?

I know about this review-like paper by Verstraete, Cirac, and Murg but I feel some of its results are superseded by new ones...

This post has been migrated from (A51.SE)
asked Nov 1, 2011 in Theoretical Physics by Kaveh_kh (120 points) [ no revision ]
retagged Mar 7, 2014 by dimension10

1 Answer

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A short summary of the relation between MPS/PEPS and ground states local Hamiltonians:

First the direction from MPS/PEPS to Hamiltonians:

  • Every MPS/PEPS naturally appears as the exact ground state of a frustration free local Hamiltonian. ("parent Hamiltonian")

  • For generic MPS/PEPS, this ground state will be unique.

  • There is a number of cases beyond the generic one where one can make statements about the ground state degeneracy. In particular, for translational invariant MPS the ground state degeneracy is always constant.

  • For translational invariant MPS, there is always a spectral gap above the ground state manifold; for PEPS, this holds only in certain cases.

Conversely, from Hamiltonians to MPS/PEPS:

  • Given a gapped local 1D Hamiltonian, its ground state is well approximated by an MPS. (cf. [Hastings '07] for the scaling)

  • Given a local 2D Hamiltonian where the density of states does not grow too quickly, its ground state (as well as thermal states) is well described by a PEPS (cf. [Hastings '05],[Hastings '07] for the scaling)

This concerns only the analytical relations between MPS/PEPS and Hamiltonians. In practice, better bounds concerning approximability etc. will typically hold.

Concerning "if and only if" relations, I think all these results only hold rigorously in one direction (though typically they might be "if and only if", e.g., I would think that typical gapless Hamiltonians will not have MPS ground states).

If you are looking for something more concrete, please let us know.

This post has been migrated from (A51.SE)
answered Nov 1, 2011 by Norbert Schuch (290 points) [ no revision ]
Thanks for the summary and the links. I was thinking along similar lines: that is typically (but perhaps not an "almost for all" type) the correspondence is solid but degeneracies affect the Hamiltonians and the MPSs in slightly different ways. I am actually not looking at anything more concrete and this came up as a discussion with my friend but it is a favorite subject of mine. I will happily get a rain check for when concrete things come up (as they do!).

This post has been migrated from (A51.SE)

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