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Some properties of a Hamiltonian are unique to its spectrum and are basis-independent. For example, I think whether the Hamiltonian's gap remains constant as $n$ goes to infinity, or whether the Hamiltonian obeys an area law, are independent of the basis in which the Hamiltonian is written.

But it seems to me that some other computationally interesting properties of the $k$-local Hamiltonian problem may be dependent on the basis in which the Hamiltonian is written.   Am I correct in saying that the answer to each of the following may depend on the basis in which the Hamiltonian is written - either the standard (computational) basis, or else the Hadamard basis?

  • Whether or not the $k$-local Hamiltonian is frustration-free, meaning that the global ground state of the full Hamiltonian corresponds to the lowest energy state of each local term?
  • Whether or not each term commutes with one another?
  • Whether the Hamiltonian is stoquastic - that is, whether, as a Hermitian matrix, it only includes real, non-positive off-diagonal elements?
  • Whether it is $k$-local to begin with, or what its $k$-value is?

Or does it even make sense to ask the above?

Here, I mean locality in the computational sense (and not geometric sense) - i.e., $H$ can be written as the sum of a polynomial number of local terms each acting solely on up to $k$ qudits (tensored with identities elsewhere) - but these qudits need not be geometrically near each other.


This is born out of me rowing various confusion boats - entanglement is clearly basis independent, and generally highly entangled states are computationally harder to work with than otherwise, but I've also believed that being non-stoquastic, or being non-frustration free, or non-commuting, or being a highly non-local Hamiltonian, corresponds to being among the most difficult to find ground states thereof.

I'm also learning slowly about IQP circuits in view of the recent experiments from Quera - hence the question about non-commuting Hamiltonians.

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  • $\begingroup$ Do you mind defining "basis-dependent": Which basis transformations do you allow for? And by frustrated, do you mean the opposite of "frustration free", or geometric frustration? What is your definition of stoquastic? -- Note that if you allow for any basis transformation, there is no notion of an area law, or even any kind of spatial locality, except in a completely arbitrary way. And if you allow for local transformations only, I would answer the questions all with "no". (Ok, for "stoquastic" it will depend on your definition.) $\endgroup$ Commented Dec 26, 2023 at 22:16
  • $\begingroup$ @NorbertSchuch thanks for your feedback (and your presumed downvote! :). Part of this was driven by your comments about an early question of mine on the Kagame lattice - in the computational basis it is a highly frustrated Hamiltonian and I had asked earlier whether that meant it must have a highly entangled ground state. But from yours (and DaftWullie's) commentary I realized that entanglement and frustration are not necessarily as related as I had intuited. $\endgroup$ Commented Dec 27, 2023 at 1:55
  • $\begingroup$ I had seen a really good presentation from Irani touching briefly on the Kagame lattice - which is highly frustrated - and she hinted that studies with classical algorithms such as DMRG suggested that the ground state was highly entangled. But if being frustration-free is basis-dependent and being highly entangled is basis-independent then that relationship frustration-freeness and entangled ground states is illusory maybe? $\endgroup$ Commented Dec 27, 2023 at 17:11
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    $\begingroup$ To iterate: By "basis-dependent", you mean local rotation on each single qubit/-dit? I would say that both being frustrated and being frustration-free (different concepts!) are unchanged under local basis rotations. And under arbitrary rotations, only the spectrum is preserved. --- (One way to define "frustrated" is as a property of the lattice (plus the fact that the interaction is antiferromagnetic).) $\endgroup$ Commented Dec 27, 2023 at 17:21
  • $\begingroup$ Ah, with the most recent edit (didn't check properly before), what you call "frustration-free" has nothing to do with the frustration observed e.g. for the kagome lattice. $\endgroup$ Commented Dec 27, 2023 at 17:27

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