1
$\begingroup$

I've only recently, and still only haphazardly and rather poorly, begun to understand Ising models with local interactions. I'm interested in particular in the simple one-dimensional Ising model with nearest and next-nearest neighbor interactions, which have been referred to in the literature as ANNNI Hamiltonians, or anisotropic, next-nearest neighbor interactions (with none, either, or both a transverse and longitudinal external magnetic field).

Depending on the strengths of the nearest-neighbor interactions relative to the next-nearest neighbor interactions (and also to the external magnetic fields) there could be very lovely and dynamic frustration going on - the nearest neighbor might favor parallel spins $\mid\uparrow\uparrow\rangle$ or $\mid\downarrow\downarrow\rangle$ but the next-nearest neighbor interactions might force antiparallel spins $\mid\downarrow\uparrow\rangle$ or $\mid\uparrow\downarrow\rangle$, or vice-versa.

  • Can we say anything about if and when frustration leads to an entangled ground state?

The answer might be related to so-called area laws, which consider the amount of entanglement relative to the dimension of the chain. For a one-dimensional chain an area law suggests that there may be little entanglement, but is there any additional entanglement borne out of frustration?


This was also partly inspired by Sandy Irani's presentation at the Israeli Institute for Advanced Studies here, where she mentions frustration in the context of a Kagome lattice.

Kagome Lattice

I gathered from her talk that the Kagome lattice puts some stress on naive implementations of classical many-body algorithms like DMRG, but nonetheless from clever DMRG simulations there's some evidence that the lattice does indeed have a uniquely quantum (non-degenerate and highly entangled) ground state.

$\endgroup$
1
  • 1
    $\begingroup$ The question seems to broad. Clearly, Ising models have a basis of the ground space consisting exclusively of unentangled states; and conversely, unfrustruated models such as the Heisenberg model on bipartite lattices have entangled ground states. $\endgroup$ Jun 10, 2023 at 21:54

1 Answer 1

3
$\begingroup$

In the Ising model, I don't think frustration is the key feature.

Let's start with an arbitrary Ising model with no transversal field. (I'll let you have arbitrary interactions, and arbitrary longitudinal fields). The Hamiltonian is diagonal. The eigenstates are basis states. There is no entanglement. If the ground state is degenerate, then any superposition of the degenerate states is also a ground state, and will generally be entangled, but that's kind of artificial.

On the other hand, as soon as you apply an arbitrarily weak transversal field, you can apply perturbation theory to describe the ground state and see that it involves a superposition which, generically, will be entangled. (I'm not saying it's absolutely impossible to construct counter-examples. For instance, if I have a NN Ising model on a bipartite lattice, and only apply the transverse field on one partition, the superposition is a separable state.)

$\endgroup$
4
  • $\begingroup$ Thanks! I’ll have to study this more. I’m trying to grok more about frustration and degeneracy w.r.t. entanglement and area laws. It seems so pretty and vibrant! $\endgroup$ Jun 12, 2023 at 15:50
  • $\begingroup$ We have that frustration is not necessary for entanglement (as the nondegenerate transverse field model explains), nor is frustration sufficient for entanglement, as @Norbert’s Heisenberg model illustrates. $\endgroup$ Jun 12, 2023 at 15:53
  • $\begingroup$ OK, but maybe don't bring the Heisenberg model into it yet (it's definitely not what your question was about!) It just complicates things which are already complicated enough! $\endgroup$
    – DaftWullie
    Jun 12, 2023 at 16:18
  • 1
    $\begingroup$ @Mark You got "sufficient" and "necessary" wrong. $\endgroup$ Jun 14, 2023 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.