In the following I have in mind antiferromagnetic spin chains in periodic boundary conditions on a chain of even length $L$.

Consider the spin-$S$ spin chain $$H = J \sum_{i=1}^L (S^x_i S^x_{i+1} + S^y_iS^y_{i+1})$$

It appears to gapless for all $S$, meaning that the energy gap between the ground state and the first excited state vanishes as $L\to \infty$.

I'm most interested in the Hamiltonian for $J>0$, when it is antiferromagnetic. (On the other hand, note that the Hamiltonians with $J$ and $-J$ are related by a simple unitary similarity transformation of $\pi$ rotations about the $z$-axis on every other site, so the Hamiltonians with $\pm J$ have the same eigenvalues. That is, the ferromagnetic and antiferromagnetic versions of the model have the same spectra. Potentially the ferromagnet might give a more natural proof.)

How can I show that $H$ is always gapless for every spin $S$?

By the Lieb-Schultz-Mattis theorem, for half-integer spins $\frac{1}{2}, \frac{3}{2},... $, the first excited state is within energy $O(1/L)$ of the ground state and hence $H$ is gapless. However, the usual version of the argument I know doesn't hold for integer spin $S$. The argument I know constructs a variational state with low energy that is orthogonal to the ground state, but the orthogonality is only guaranteed for half-integer spins.

That is, gaplessness is relatively straightforward to prove for half-integer spins $\frac{1}{2}, \frac{3}{2},... $, but it's not clear how to extend the proof to integer spins $S$. Importantly, the addition of a $\Delta \sum_i S^z_i S^z_{i+1}$ term with $\Delta >0$ can open a gap in integer spin chains, so it might be that a very different style of proof from Lieb-Schultz-Mattis will be needed.

  • $\begingroup$ “…and hence H is gapless.” A basic question about language: by gapless do you also include the possibility of a gapped degenerate ground state? If my understanding is correct, the LSM construction cannot distinguish the two possibilities - having nontrivial ground state degeneracy vs being gapless. $\endgroup$ Commented Mar 28 at 14:32
  • $\begingroup$ @NandagopalManoj That's a very good point. For example, when we consider a spin-1/2 $XXZ$ chain, LSM applies everywhere despite being gapless for $|\Delta| \leq 1$ and $\mathbb{Z}_2$-symmetry breaking and gapped elsewhere. The splitting between the two dressed cat states in the symmetry broken phase is exponentially small in system size with an O(1) gap above. Typically I would say the $|\Delta|>1$ regime in the spin-1/2 $XXZ$ chain is not gapless. $\endgroup$
    – user196574
    Commented Mar 28 at 17:36
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    $\begingroup$ @NandagopalManoj However, I think I'm willing to assume that if a result tells us we are either gapless or have nontrivial ground state degeneracy, then we are gapless in the case of $H = J \sum_{i=1}^L (S^x_i S^x_{i+1} + S^y_iS^y_{i+1})$. That is, for this Hamiltonian, I will accept the weaker result that there is at least one eigenstate with energy within $1/L^\alpha$ of the ground state. At the very least, because of the periodic boundary conditions, this will rule out the Haldane phase. $\endgroup$
    – user196574
    Commented Mar 28 at 17:37
  • $\begingroup$ Do you know for a fact that the spin-S Hamiltonian of the form you wrote is gapless for all S? Is this from your own numerics or is there a reference? $\endgroup$ Commented Mar 29 at 0:23
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    $\begingroup$ @NandagopalManoj A good one is arxiv.org/abs/1212.6255. In the integer-$S$ chains in periodic boundary conditions, the gapped phase with a unique ground state (the trivial "even Haldane" and the SPT "odd Haldane" phases) rapidly occupy an extremely tiny portion of the phase diagram. I was surprised by how tiny that portion is (see table II) already for spin-2 and 3. $\endgroup$
    – user196574
    Commented Mar 29 at 6:40

1 Answer 1


I have a hand-wavy answer for this. These are not rigorous results.

Takeaway: The ferromagnetic XX chain is in the superfluid phase of the 1d Bose-Hubbard model (BHM). Particle number in the BHM is analogous to $S^z$ in the XX chain. This is a phase with quasi-long range order in the $S^-$ order parameter, a gapless spectrum with $\Delta \sim L^{-1}$, and power law correlations.

Bose-Hubbard model

This is an interacting model of bosons $$ H = \sum_i\left(-t (b^\dagger_i b_{i+1} + b^\dagger_{i+1}b_i) - \mu b^\dagger_i b_i + \frac{U}{2} b^\dagger_ib^\dagger_i b_i b_i \right) $$ Here is the intuition for the Bose-Hubbard model phase diagram (figure 3 in the paper). We start at certain values of $\mu/U$ and look at the behavior while increasing $t/U$. If $\mu /U$ is at a value such that having $N$ or $N+1$ particles at the same site have equal energy (for some $N$), then an arbitrarily small amount of hopping would drive a transition into a superfluid phase, because the ground state would prefer to have superpositions of $N$ and $N+1$ bosons on each site to gain hopping energy. This is a Bose-condensed phase which spontaneously breaks the particle number conservation symmetry.

For generic $\mu/U$ which prefers to have $N$ particles at a site, you need some finite hopping strength before the bosons condense, since there is a gap to having $N\pm 1$ particles at a site. For more discussion on BHM, refer these notes. Note that what I have described is a mean-field picture of what happens, there are some important details which are different in 1 dimension.

Spin-$1/2$ XX chain

It is straightforward to map the spin-1/2 XX chain to a BHM. A spin-$1/2$ degree of freedom is equivalent to a hard-core boson (a bosonic site with $U \to \infty$ such that it can only hold up to one particle). Therefore, the ferromagnetic XX chain becomes a BHM with $U \to \infty$, $\mu = 0$ and $t = J$. Note that this satisfies the condition that $N$ and $N+1$ bosons at one site have the same energy when $N=0$. Therefore, an arbitrarily small hopping drives a transition into a gapless superfluid phase with $\Delta \sim L^{-1}$, consistent with the exact solution from the Jordan-Wigner transformation.

Spin-$S$ XX chain

How do we generalize this to spin $S$? There is no choice of $U$ for the BHM that recovers the spin-$S$ XX chain. But, intuitively, it is believable that the idea is the same. At one site, the XX chain does not care if $S^z = -S, -S+1, \dots, S-1,S$. So an arbitrarily small amount of "hopping" $$ S^x_{i}S^x_{i+1} + S^y_{i}S^y_{i+1} = S^+_{i}S^-_{i+1} + S^-_{i}S^+_{i+1} $$ makes the particle number want to fluctuate and drives a transition into a superfluid.

Can we make this more believable? The best I could do is to map this to a problem of $2 S$ spin-$1/2$ degrees of freedom per site, with the mapping of operators to Pauli matrices $$ S_i^{\mu} = \frac{1}{2}\sum_{\alpha=1}^{2S} \sigma_{i,\alpha}^\mu $$ This satisfies all the spin-$S$ commutation relations, so is a valid representation of the original problem (note that this is a redundant description as we are introducing new degrees of freedom -- my intuition is that this is okay for the low-energy physics of the ferromagnetic case, but maybe not for the anti-ferromagnetic case).

In this language, the Hamiltonian becomes $$ H = \frac{-J}{4}\sum_{i=1}^L \sum_{\alpha,\beta=1}^{2S} \left( \sigma^x_{i,\alpha} \sigma^x_{i+1,\beta} + \sigma^y_{i,\alpha} \sigma^y_{i+1,\beta} \right) $$ which can be reduced (following the discussion for the spin-$1/2$ case) to a Bose-Hubbard model with $U \to \infty$, albeit with a non-standard geometry but that is not consequential for any of our arguments. This is further proof that the spin-$S$ XX chain is in the superfluid phase of the Bose-Hubbard model.

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    $\begingroup$ +1 These are some really interesting ideas! I'm especially curious about splitting the spins $S$ into $2S$ spin-$1/2$ particles (similar in spirit to AKLT). One idea is maybe being able to construct/prove the existence of an eigenstate in the many spin-$1/2$ model that is close in energy to the ground state, and then projecting into the spin-$S$ space by way of a rectangular "projector" onto the completely symmetric space of each site. Hopefully there are conditions under which such a projected state will remain orthogonal to the ground state. $\endgroup$
    – user196574
    Commented Apr 19 at 19:07
  • $\begingroup$ Indeed! The nice thing here (unlike the usual AKLT construction) is that the on-site projector commutes with the Hamiltonian (and the ground state, one could assume). So the projection should be trivial if the ground state is in the right sector. I expect it should be because you dont want the 2S qubits to project into a smaller spin (<S) sector because you want maximum "ferromagnetic" coupling. Sorry if I'm rambling.. hopefully it makes sense. $\endgroup$ Commented Apr 19 at 22:30

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