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asked by (400 points)

I have been doing some simple experiments with the transverse field Ising model and its Jordan Wigner transformed version, the Kitaev chain, in the Julia version of ITensor. I am trying to understand how DMRG deals with ground state degeneracies/ topological degeneracies. I have a few questions about this:

  1. When the transverse field is zero or small, DMRG ends up picking one of the two ferromagnetically ordered states at random. Why does it not pick linear combinations? My guess would be: Linear combinations of the ordered states have long range entanglement. This cannot be captured by DMRG because it arrives at the ground state by optimizing local bonds. Is that correct?

  2. Also, the MPS representation of these linear combinations would have a tiny bond dimension of 2. So what is the signature of long range entanglement in the MPS representation if it is not a large bond dimension?

  3. Is there any way though to force the DMRG to pick up one of the cat states?

  4. One possibility is to Jordan Wigner transform the Ising model to a Kitaev chain. When I compute ground states for the Kitaev chain with open boundary conditions in the topological phase, DMRG always picks up a uniform superposition of the even and odd parity states. This would correspond to the ordered states in the spin language. Again, why does DMRG do this? Does the long range entanglement argument translate into the fermion language as well?

  5. A side comment/ question about this: Even though DMRG picks up the "ordered" state, when I compute the expectation value of the Jordan Wigner spin sz, I get zero instead of +1 or -1 which is expected of the ordered state.

  6. Upon adding periodic or anti-periodic boundary terms, DMRG ends up picking the even or odd parity state (and this would correspond to picking up a cat state in the spin language). My question about this is the following: For the DMRG to successfully pick a definite parity state, I need a large enough bond dimension for the initial state psi0 (the number of sweeps does not matter so much). Why is this the case when the final bond dimension of the cat state is just 2? And how should the bond dimension of the initial state scale with system size to enable the DMRG to converge to a definite parity ground state?

Thank you!

1 Answer

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answered by (70.1k points)
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Hi Navya,
That's a lot of questions :^) But they are good ones. I will answer the ones I'm most confident about, then please comment below to remind me to try to answer the others as soon as I can or if my answer here is enough already.

  1. Yes this is partly right. But a more key and simple reason this happens is that the energy of the cat state is only a little smaller than a non-cat state, and this difference goes exponentially to zero with system size. You can think of DMRG as always making a trade-off between energy and entanglement, so it's as if DMRG doesn't "think it's worth it" to spend a whole lot more entanglement to gain only a very tiny bit of energy. More rigorously, I think what is happening is that DMRG is technically slightly stuck in a (barely) excited state and just can't 'find' the cat state which is the true ground state.

  2. The last part of my answer to (1) is the answer to this. The part about spending entanglement is more of a non-rigorous intuition. Really DMRG is a dynamical process where the MPS "environment" gets built from some initial state and then is used to project the Hamiltonian into a small part of the Hilbert space. This projected Hamiltonian and not the full one is used in the core step of DMRG to update the previous MPS basis on one site and then slide to the next site. So if one does not already start with an initial MPS that is a cat state, then the projection built from the non-cat initial state probably keeps DMRG from "seeing" that a cat solution exists. I think even a small deviation from a perfect cat initial state is probably unstable to a non-cat final state of DMRG for related reasons (how the MPS basis gets built and updated dynamically within DMRG).

  3. I don't think there's a simple way to force DMRG to find cat states other than to maybe start already in a perfect cat state and carefully design a flavor of DMRG to preserve that property. Maybe there is some kind of internal MPS symmetry or gauge that can be enforced but it's not immediately obvious to me.

  4. I have not found that DMRG makes a superposition of even and odd parity states for the Kitaev chain, and I studied a closely related system with DMRG in a paper of mine. I would say that in some philosophical sense you aren't really studying fermions if you don't enforce parity quantum number conservation. Or at least you can always choose to conserve this for a closed fermionic system. So I would make sure to use ITensor's QN DMRG mode and enforce fermion parity. Then you can just start off in a state of definite parity and our DMRG code will rigorously always stay in that parity sector. Then by how the mapping of the Kitaev chain works relative to the Ising chain, you will necessarily be in a cat state because the fermion parity conservation will enforce that.

For 5 & 6 I would suggest revisiting these questions in light of the idea of using QN conservation to enforce even or odd parity as discussed above.


commented by (70.1k points)
I advertised this question on our Twitter account since I thought it was an interesting one. Juan Carrasquilla helpfully pointed out the following:

for the case of the Ising chain there is a nice and simple way to get DMRG to find cat states: you can use the QN conservation system of ITensor to force DMRG to conserve Z2 parity in the spin language (conserving total Sz mod 2). To use this trick, I believe you have to set up the Ising Hamiltonian to be of the form -J XX - h Z where X and Z stand for the Pauli spin operators.

Then for at least the case of h=0 where it's simplest to think about, the all up plus all down cat state in the X basis is the same as the superposition of all even-parity states in the Z basis, and the all up minus all down is a superposition of all odd-parity states in the Z basis.

Please let me know if you want to try this and need help setting it up in the code.
commented by (400 points)
Hi Miles,

Thank you so much for the detailed answers, and I am sorry about the barrage of questions! :)

Your answers help a lot. I think I understand the spin version of things. I had some follow up questions about the Kitaev chain.

1. As you mentioned, the correct way to do fermion DMRG is by imposing fermion parity conservation. Thus, I could obtain the fermionic ground states in both the even and odd parity sectors. And then if the energies of both these ground states turn out to be the same (up to some DMRG resolution errors which I would need to estimate), I could conclude that the Hamiltonian is in a topologically non-trivial phase.

2. But I am also interested in another way of diagnosing the topological characteristics of the Hamiltonian. This second method involves adding appropriate tunneling terms which break the ground state degeneracy. One example of such a term would be the bond term connecting the first and last site which imposes periodic/ anti-periodic boundary conditions. I am interested in this method because it allows me to identify the operators which represent edge modes. However, the caveat is that I cannot impose fermion parity conservation anymore. Because if I do, then the imposed fermion parity conservation (and not the tunneling term) dictates which ground state is picked. So, I have a question about how to make this second method work.

Here is what I have noticed so far: When I turn off fermion parity conservation and add appropriate tunneling terms, DMRG picks up a definite parity ground state only if the bond dimension of the initial state psi0 is large enough (and the size of the required bond dimension grows with system size). For example, for a Kitaev chain with 100 sites,  I observe that psi0 should have a bond dimension of at least 50. If the bond dimension is any smaller, then DMRG picks up a uniform superposition of the ground states with different parities (even if I increase the number of sweeps/ maximum bond dimension in intermediate sweeps). I don't understand why DMRG chooses ground states in this manner or how this required bond dimension scales with system size. Basically, how is DMRG dealing with these tunneling terms, and how should I set up the DMRG to ensure that the algorithm registers these terms?

Also, thanks for letting me know about Juan's suggestion. I will give it a shot and get back to you in case I face difficulties.

Thank you!
commented by (70.1k points)
Hi Navya,
Yes #2 is a good question. We ran into this when studying the edge modes of Kitaev chain physics, though in the context of a more complicated model of electrons with spin which exhibits Kitaev chain physics in the low-energy limit. Here is the link to that paper which you may find useful: https://arxiv.org/pdf/1104.5493.pdf

The technique we came up with to see edge states turned out to be very beautiful and nice. It was to do the following:
(1) compute both ground states by running DMRG twice to get both ground states psi0 and psi1 (where the 0,1 refers to even and odd total fermion parity). This can be done by putting in initial states which have an even or odd number of particles initially.
(2) compute the 'overlap' or matrix element of the C operator between these two ground states, meaning the quantity phi_j = <psi0|C_j|psi1> . As discussed in Eqs. (9)-(13) in the paper linked above, one can use phi_j and a related quantity involving Cdag_j to extract the "wavefunction" of the edge states. I put wavefunction in quotes because this is an exact statement for a non-interacting system, which the Kitaev chain is, but only has an approximate or more complicated interpretation as a wavefunction for an interacting system as far as I know.

The question about what DMRG does with the parity conservation turned off is kind of a complicated one. I would not rely on the behavior of DMRG in ITensor or any other package for stabilizing these sort of things if possible, because it's a question about the (artificial) dynamics of the algorithm, initial state used, details of the implementation and code, etc. It's therefore hard to give a simple answer to questions like how DMRG is dealing with various terms or choosing ground states. I did give a partial answer in my answers numbers 1. and 2. above so I hope those help and they are about all I know about that topic.

Yes please do think about Juan's idea if you want to get cat states in the spin model!

Best regards,
commented by (400 points)
Hi Miles,

Thank you for the reference and for explaining the technique to numerically deduce edge states! It sounds very useful and interesting. I am going to give this a try and tell you how it goes. Appreciate your help!

Warm regards,
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