The ground state of the spin-$1$ AKLT model gives an example of a $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry-protected topological (SPT) phase, the Haldane phase. This state is a nice example of the Haldane phase because it has an exact matrix product states (MPS) representation. This state has exponentially decaying connected correlation functions.

Is there an example of a spin-$1$ state that's in the Haldane phase but has zero correlation length?

In particular, I'd want the protecting symmetry to be the same $\mathbb{Z}_2 \times \mathbb{Z}_2$ that protects the AKLT state (i.e. $\pi$ rotations about $x$, $y$, and $z$ axes), or the inversion symmetry that protects the AKLT state. I'm currently learning about SPT phases, so it's possible I'm missing or misunderstanding something trivial.

For example, in $1$d, there's the spin-$1/2$ cluster state, which is again an exact matrix product state protected by a $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry. The cluster state is even nicer than the AKLT state, in a sense, in that its correlation functions vanish identically after a finite distance. That is, unlike AKLT, its correlation length is zero.

My motivation is that I'm curious about whether there exists something similar to the cluster state but for spin-$1$ particles, and still in the Haldane phase and protected by the same symmetries (or a subset of them) as the AKLT state. I found a paper that mentions a generalization of cluster states to qudits, but it looks like the symmetry group there is larger, $\mathbb{Z}_3 \times \mathbb{Z}_3$ for spin-$1$. Further, it's not clear to me that breaking that symmetry down to some $\mathbb{Z}_2 \times \mathbb{Z}_2$ would yield the same $\mathbb{Z}_2 \times \mathbb{Z}_2$ ($\pi$ rotations about $x$, $y$, and $z$) that protects the AKLT state.

  • $\begingroup$ I guess you also insist on translational invariance? Note that the cluster state has a 2-site unit cell with regard to the symmetry action, so it really has 4 levels, not 2. $\endgroup$ Oct 23, 2022 at 12:41
  • $\begingroup$ And when you say "spin-1 state", do you require it to be SO(3)-invariant, or you just mean a 3-level system? $\endgroup$ Oct 23, 2022 at 12:48
  • $\begingroup$ @NorbertSchuch I'm happy with any chain of three-level systems, so long as it gives an SPT protected via the same operator implementation of the $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry protecting the AKLT state. For example, for an even number $L$ spin-$1$ particles, the symmetry for the $\pi$ rotation about $x$ would be $e^{i \pi \sum_i S^{x}_i} = \otimes_i \begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{pmatrix}_i$ and for the $\pi$ rotation about $y$ would be $e^{i \pi \sum_i S^{y}_i} = \otimes_i \begin{pmatrix} 0 & 0 & 1\\ 0 & -1 & 0\\ 1 & 0 & 0 \end{pmatrix}_i$, and $\endgroup$
    – user196574
    Oct 23, 2022 at 18:06
  • $\begingroup$ for $z$ would be $e^{i \pi \sum_i S^{z}_i} = \otimes_i \begin{pmatrix} -1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & -1 \end{pmatrix}_i$. The state would only need to be in the same phase as AKLT, protected by the symmetries above. I'm OK if the state itself is not $SO(3)$ invariant nor translationally invariant, but the $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetries should be implemented as above. $\endgroup$
    – user196574
    Oct 23, 2022 at 18:06
  • $\begingroup$ I should add that if instead the state's SPT nature was protected by inversion or time-reversal symmetries, those symmetries would also have to be implemented the same way they are for the AKLT state; i.e. with the same operators. $\endgroup$
    – user196574
    Oct 23, 2022 at 18:25

2 Answers 2


The following gives a spin-1 chain which is $\mathrm{SO(3)}$ invariant, with a 2-site unit cell (i.e. breaking translation symmetry to translation by two sites):

  1. Start with two spin-$\tfrac12$ at each site.

  2. Place each of the spin-$\tfrac12$ in a maximally entangled singlet state with one of the spin-$\tfrac12$ at the adjacent site.

  3. The resulting state is $\mathrm{SO(3)}$ invariant, and in the Haldane phase, for the representation $\tfrac12\otimes \tfrac12=0\oplus1$ acting at each site.

  4. This representation can be naturally embedded in $1\otimes 1 = 0\oplus 1 \oplus 2$. Thus, the chain above can also be seen as an $\mathrm{SO(3)}$ invariant chain with representation $1\otimes 1$ per site, which is in the Haldane chain.

  5. We can now split the two spin-$1$s. This gives a chain with a spin 1 per site, but which breaks translational symmetry (it is translational invariant under translation by two sites).

  6. By construction, the chain (i) is in the Haldane phase and (ii) has correlation length 0.

One way to look at the resulting state is to note that within the 2-site unit cell, the resulting state cannot have spin 2, which is precisely the property the AKLT Hamiltonian enforces. Thus, on 2 sites, the state can be obtained by applying the AKLT construction on two sites to two dangling spin-1/2 (i.e., put a singlet inbetween the two spin-1/2, and project each pair onto spin 1). This map is then applied to a chain of singlets, but before that, one needs to apply the map which makes the 2-site AKLT construction an isometry. This map is precisely $\rho^{-1/2}$, where $\rho$ is the boundary state of the 2-site AKLT construction.

This should also provide the most tangible way to construct an MPS representation of the state.

In fact, this way one can construct a family which interpolates from the given state to the AKLT model: Note that $\rho$ must be a linear combination of the symmetric and antisymmetric projector, giving a one-parameter family which contains, among others, the AKLT state and the given model.

  • $\begingroup$ One can also construct an explicit MPS tensor from this construction, but I doubt this is more instructive. $\endgroup$ Oct 24, 2022 at 16:15
  • $\begingroup$ +1, thank you for the construction. I'm checking this numerically now, but is it possible the natural embedding step induces correlations? I'm taking that step to be of the form $\frac{1}{\sqrt{6}}(|1,-1\rangle-|0,0\rangle+|-1, 1\rangle)(\langle \uparrow \downarrow|-\langle \downarrow \uparrow|)$, $\frac{1}{\sqrt{2}}(|+0\rangle-|0+\rangle)(\langle \uparrow \uparrow|)$, and so on $\endgroup$
    – user196574
    Oct 24, 2022 at 16:51
  • $\begingroup$ @user196574 The embedding is really the canconical identity map from $0\oplus1$ into $0\oplus1\oplus 2$. As such, it should not induce any correlations. (More generally, it is an isometry, and as such, should not introduce any correlations.) $\endgroup$ Oct 24, 2022 at 17:55
  • $\begingroup$ I think I see. I guess I would have thought that a generic isometry acting on the doubled site could induce correlations, similarly to how a generic two-body unitary acting on the doubled site, with each spin-$1/2$ in a different singlet pair, could induce correlations. However, you're definitely right; the resulting state numerically doesn't have correlations induced. It's looking like the resulting state after the procedure is finished is actually just replacing the spin-$1/2$ singlet pairs with spin-$1$ singlet pairs despite the isometry acting on pairs shifted relative to the singlets. $\endgroup$
    – user196574
    Oct 24, 2022 at 18:03
  • $\begingroup$ @user196574 It certainly should not replace the spin-1/2 across unit cells (first, this is impossible, and second, it would change the SPT). A fun fact is, however, that the state within the unit cell is indeed in the ground space of the 2-site AKLT Hamiltonian -- and thus, it can be understood as projecting two spin 1/2's at each site, just that the state underneath is more complicated. $\endgroup$ Oct 24, 2022 at 18:09

Here, I add some details based on Norbert Schuch's construction involving isometries replacing the AKLT projectors. Following a similar set of steps to that in Appendix A of this paper on AKLT scars, we immediately get that the two-site MPS matrices are

\begin{equation} M^{[1,1]} = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \end{equation}

\begin{equation} M^{[1,0]} = \frac{-1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{equation}

\begin{equation} M^{[1,-1]} = \frac{-1}{\sqrt{2}}\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \end{equation}

\begin{equation} M^{[0,0]} = \frac{-1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{equation}

where the superscripts stand for total spin $S$ and total spin-$z$ $S^z$ of the pair of spin-$1$s. The resulting transfer matrix is of the from $v v^\dagger$, where $v = \frac{1}{\sqrt{2}}(1,0,0,1)$.

This implies that two-spin connected correlation functions vanish as soon as the spins are in two different unit cells separated by at least one unit cell; such connected correlation functions vanish after the distance between the spins is $4$. Thus the correlation length is truly zero.

How about whether this state is an SPT protected by the correct $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry; i.e. how does this state transform under $e^{i \pi S^x_{tot}}$ and $e^{i \pi S^z_{tot}}$?

Under $e^{i \pi S^x_{tot}}$, we have $|1,1\rangle \leftrightarrow -|1,-1\rangle$, while $|1,0\rangle \rightarrow -|1,0\rangle$ and $|0,0\rangle \rightarrow |0,0\rangle$. By inspection, this is realized through the virtual spin transformation of $M \rightarrow \sigma^x M \sigma^x$.

Under $e^{i \pi S^z_{tot}}$, we have $|1, \pm 1\rangle \rightarrow -|1,\pm 1\rangle$, while $|1,0\rangle \rightarrow |1,0\rangle$ and $|0,0\rangle \rightarrow |0,0\rangle$. By inspection, this is realized through the virtual spin transformation of $M \rightarrow \sigma^z M \sigma^z$.

This is important, as we have a minus sign on the right hand side of $\sigma^x \sigma^z = - \sigma^z \sigma^x$, which, by the arguments in this paper on SPT states, implies we have an entanglement spectrum degeneracy protected by the $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry above, the hallmark of an SPT.


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