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I'm currently learning about symmetry-protected topological phases in one dimension. The ground state of the AKLT model provides one such example. In particular, the AKLT state for any length $L$ cannot be approximated to a fixed error $\epsilon$ with a local quantum circuit of depth independent of $L$ when the unitary gates respect appropriate symmetries. Here, the error is measured in terms of expectation values of local observables, and the unitary circuit is acting on a product state. The circuits I'm imagining in my head are brickwork circuits. By AKLT state, I'm referring to the unique ground state of the AKLT chain with periodic boundary conditions.

However, from what I've read, the AKLT state can be approximated with a finite-depth circuit acting on a product state when the unitary gates don't obey symmetries.

Could one perfectly reach the AKLT state of length $L$ with such a family of finite-depth circuits? By that, I mean whether for each $L$ there exists some finite depth local circuit $U$ (depth independent of $L$) where $\langle \text{AKLT}| U|\text{product}\rangle = 1$. Here $|\text{product}\rangle$ is some product state, like the fully-polarized state $|++++...+\rangle$.

Here is some background. The Affleck-Kennedy-Lieb-Tasaki (AKLT) model is a quantum chaotic spin-1 chain that nevertheless has an exactly solvable ground state for all $L$. This exactly solvable ground state is an example of a symmetry protected topological (SPT) state. From what I gather, SPT states are short-range entangled, which means that they can be (approximately) disentangled to a product state with a local quantum circuit with depth independent of $L$.

What sets SPT states apart from other short-range entangled states is that if the local unitaries are constrained to be symmetric under appropriate symmetries, one can't find a local circuit with depth independent of $L$ to disentangle the state. This is only tangential to my problem at hand; it means that the short-depth circuit will necessarily have unitaries breaking the invariance under $\pi$ rotations of at least two of the $x,y,z$ axes.

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    $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic ♦
    Commented Aug 16, 2022 at 21:01
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    $\begingroup$ If you only want an example of how to go with non-symmetric circuits from a trivial to an SPT state - not necessarily the AKLT state - the answer becomes "it is possible", if you go to a zero-correlation length state (e.g. the "AKLT without the projector", i.e. singlets between adjacent sites, with local Hilbert space 1/2 $\times$ 1/2 = 0$\oplus$1. $\endgroup$
    – Norbert Schuch
    Commented Aug 23, 2022 at 19:29

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You can not get to the AKLT state exactly from a product state with a finite-depth circuit. The reason is that AKLT has a finite correlation length, while any state obtained from a product state with finite-depth circuit (regardless of symmetry) must have zero correlation length.

Update: in fact, one can prove that in a chain of spin-1's, there does not exist a globally SO(3) symmetric finite-depth circuit that maps a trivial phase to a SPT phase.

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  • $\begingroup$ +1 This is an interesting answer to read. I want to give a friendly challenge to it: on page 47 of a PhD thesis asks the question "How much symmetry needs to be broken to be able to map the ground state of a non-trivial SPT phase to a product state using an FDUC [Finite Depth Unitary Circuit]?" and claims that for AKLT, breaking $Z_2 \times Z_2$ to $Z_2$ suffices. Page 49 gives an explicit symmetry breaking for an AKLT-like model, but I can't convert it to the full AKLT chain. $\endgroup$
    – user196574
    Commented Aug 17, 2022 at 6:51
  • $\begingroup$ Another friendly challenge - "SPT phases are also called states with short-range entanglement or short-range entangled (SRE) states. To be more precise, SRE states are states that can be transformed, by applying a finite-depth local unitary quantum circuit, into a product state." This is from page 53 of Chiu et al.'s review article. If the AKLT state is an SPT state, it's an SRE state that by definition can be transformed with a finite-depth local unitary circuit into a product state. How do you reconcile this with your answer? $\endgroup$
    – user196574
    Commented Aug 17, 2022 at 7:11
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    $\begingroup$ For your second question, this is a well-known defect of the proposal (that SPT can be transformed by FDLUC to a product state), for the same reason. We either have to allow errors, or replace FDLUC with locally generated unitary. $\endgroup$
    – Meng Cheng
    Commented Aug 17, 2022 at 12:34
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    $\begingroup$ The explicit symmetry breaking transformation on 49 is from a product state to a product state. $\endgroup$
    – Meng Cheng
    Commented Aug 17, 2022 at 12:38
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    $\begingroup$ @NorbertSchuch One thing I should clarify is that my original question asked about both exact constructions and approximations (and I commented on Meng Cheng's answer given that), but because I realized that Meng Cheng's answer firmly answered the existence of exact constructions in the negative, and that approximations was probably another can of worms, I edited my question to remove the approximations part so that I could accept Meng Cheng's answer. I'll probably post a separate question on approximations once I've clarified some things in my head. $\endgroup$
    – user196574
    Commented Aug 24, 2022 at 2:40

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