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For the purposes of this question, let us define a quantum integrable model as one solvable by the Bethe Ansatz. That structure endows the model with a set of conserved charges $\{H^{(n)}\}$ whose number is extensive in the system size $N$. My question is about how local are these charges.

Working in the thermodynamic limit $N \to \infty$, let us say that an operator $O$ is local if it is supported on an $\mathcal{O}(1)$ number of sites, and that it is quasilocal if its support decays sufficiently quickly with distance (say faster than $1/r^{d}$ in $d$ dimensions). We might also reasonably apply this terminology to an operator which can be written as a sum of terms with these support properties.

My question was prompted by reading this paper. There the authors study the spin-$\frac{1}{2}$ Heisenberg chain

$$H = \dfrac{J}{4} \sum_{j=1}^{N} \left[\sigma_{j}^{x} \sigma_{j+1}^{x} + \sigma_{j}^{y} \sigma_{j+1}^{y} + \Delta (\sigma_{j}^{z} \sigma_{j+1}^{z} -1) \right],$$

where $J>0$, $\sigma_{j}^{\alpha}$, $\alpha=x,y,z$ are the Pauli matrices, and they consider the regime $\Delta \geq 1$.

At the bottom of p2, the authors identify an infinite set of conserved charges $\{H^{(n)}\}$, which they dub ultra-local. They justify this terminology with the sentence

These charges are ultra-local in the sense that they can be written as $H^{(m)} = \sum_{j=1}^{N} h_{j}^{(m)}$, where the operators $h_{j}^{(m)}$ act nontrivially on a block of at most $m$ sites adjacent to $j$.

Here is where I am confused. If I understand correctly, the index $m$ can be extensive in the system size $N$, so the operators $h_{j}^{(m)}$ could in principle have $\mathcal{O}(N)$ support. If we assume the definition of locality I gave above, which seems fairly reasonable, how then can the authors call these conserved charges ultra-local?

Qmechanic
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anon1802
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1 Answers1

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By $H^{(m)}$ the authors denote the usual conserved charge, i.e. the $m$th logarithmic derivative of the transfer matrix at a suitable value of the spectral parameter. For example, for $m=1$ this gives the XXZ spin chain Hamiltonian: a sum of terms $h^{(1)}_j$ that each act at blocks of at most one site adjacent to site $j$, i.e. at nearest neighbours $j,j+1$. This is true for any system size $N$.

If you similarly compute the next ultralocal charge $H^{(2)}$ you will get a sum of terms that act at sites $j,j+1,j+2$. (You might want do this explicitly as an exercise. The result can be found e.g. in my lecture notes on the arXiv.) Again this form is the same for any $N$.

In general, for fixed $m$ the summands act nontrivially at at most $m+1$ adjacent sites, independently of $N$. There's nothing extensive about these summands when you fix $m$.

The only thing that changes with $N$ is how many summands there are (the sum runs up to $N$), and how many such charges you have. Indeed, up to a possible common factor (depending on the normalisation of the R-matrix) the entries of the transfer matrix are polynomials in the spectral parameter of degree at most $N$, so you can in principle compute nontrivial charges for $0\leq m\leq N$. (Here $m=0$ corresponds to the momentum operator, i.e. the log of the translation = cyclic shift operator.)

Jules Lamers
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  • Thanks for your answer, Jules --- you confirmed my initial understanding of the paper. I think I do still find it a bit of a strange notion of locality though. Sure, for a fixed $m$ the support of $h_{j}^{(m)}$ is independent of $N$. But in the thermodynamic limit $N \to \infty$, we could still have conserved charges with arbitrarily large support. That does not seem very local to me! – anon1802 Oct 02 '21 at 07:48
  • @anon1802 No, but the point is that each charge separately is local. That you get more and more as $N$ increases is part of the notion of quantum integrability. These charges should be independent, so it's perhaps not that surprising that the charges become less and less local as $m$ increases – Jules Lamers Oct 02 '21 at 08:11
  • I agree that a characteristic feature of quantum integrability is a number of conserved charges that is extensive in $N$. But it does not seem like this is inherently at odds with locality. As a trivial example, you could have $N$ charges, each localized on a different site of the lattice. Actually something like this happens in many-body localized systems, which you might call "quasi-integrable". But from what you are saying in your answer, it seems like in more typical integrable models, the charges can be much less local than this. – anon1802 Oct 02 '21 at 09:17
  • @anon1802 By translation symmetry you can't have a charge that acts nontrivially at only one site as part of the family of commuting operators – Jules Lamers Oct 02 '21 at 09:33
  • Ah, sorry, I should have been clearer: the $H^{(m)}$ have to be translationally invariant if that is a symmetry of the model, but the $h_{j}^{(m)}$ could still be 1-local, right? – anon1802 Oct 02 '21 at 10:00
  • @anon1802 There are only very few independent operators acting at a single site. Once you know how a 1-local operator acts at a single site, I think that translational symmetry fixes its action at all other sites – Jules Lamers Oct 02 '21 at 11:48
  • Yes, fair enough. So, to conclude, as $N \to \infty$, the conserved charges become increasingly non-local, and in translationally-invariant systems that is in some sense a constraint imposed by that symmetry. Thanks for the discussion! – anon1802 Oct 02 '21 at 12:03
  • Yes: the charge for any fixed $m$ is local as $N$ grows, but, indeed, as $N$ increases charges with larger and larger 'support' are part of the family of commuting symmetries. (And I would expect -- or, at least, am not surprised -- that the 'less and less local' might be related to translational invariance, but I haven't thought about this in great detail) – Jules Lamers Oct 02 '21 at 12:11