# Why is $H = J \sum_i (S^x_i S^x_{i+1} + S^y_iS^y_{i+1})$ always gapless for any spin $S$?

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.

• “…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. Commented Mar 28 at 14:32
• @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. Commented Mar 28 at 17:36
• @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. Commented Mar 28 at 17:37
• 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? Commented Mar 29 at 0:23
• @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. Commented Mar 29 at 6:40

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.

• +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. Commented Apr 19 at 19:07
• 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. Commented Apr 19 at 22:30