# Log-law entanglement with large central charge contradict bounds on the entanglement entropy

I am trying to learn more about entanglement entropy in large but finite-size systems at critical points. I am still relatively new to conformal field theory, so it is not unlikely I have misunderstood related results.

For ease, let's restrict to the von Neumann entropy for a $$(1+1)d$$ translationally invariant chain with periodic boundary conditions and a length of $$L$$. I typically have in mind a spin-$$1/2$$ chain or a lattice fermion model. A recurring, oft-cited statement is that critical points described by $$(1+1)d$$ conformal field theories with central charge $$c$$ have the following von Neumann entropy for $$l\leq \frac{L}{2}$$:

$$S(l) = \frac{c}{3}\ln \left(\frac{L}{\pi} \sin{\frac{\pi l}{L} }\right) +C$$

Here, $$C$$ is a non-universal constant. Such a scaling can be verified for spin-chains like the transverse field Ising model with $$c=1/2$$, as noted in a question on this site.

However, my confusion is that the von Neumann entropy for a chain with a local Hilbert space dimension of $$d$$ satisfies the inequality $$0 \leq S(l) \leq l \ln(d)$$. I am relatively new to understanding conformal field theories, but I believe there exist theories with extremely large central charges. Then naively, it would look as though the equation for $$S(l)$$ could exceed $$l \ln(d)$$ if one tunes $$c$$ large enough!

Perhaps the resolution would be that the non-universal constant $$C$$ will be very large and negative at large $$c$$. That is, to avoid violating $$S(l) \leq l \ln(d)$$, we need $$C \leq -\max_l \left( \frac{c}{3}\ln \left(\frac{L}{\pi} \sin{\frac{\pi l}{L}}\right) - l\ln(d) \right)$$. However, this does not seem quite right, since if $$C$$ is negative with too large a magnitude, then there will exist values of $$l$$ where $$S(l) < 0$$, which is forbidden.

Given that the resolution above doesn't seem to work, I'm tempted that there must be another finite-size entanglement entropy formula that handles the large $$c$$-case better, or that I am misunderstanding the range of applicability of the finite-size entanglement entropy formula.

What is the resolution of the bound $$0 \leq S(l) \leq l \ln(d)$$ on the entanglement entropy, the formula $$S(l) = \frac{c}{3}\ln \left(\frac{L}{\pi} \sin{\frac{\pi l}{L} }\right) +C$$ and the existence of theories with arbitrarily large central charge $$c$$? Is there a different formula for $$S$$ that captures the finite-size entropy better when $$c$$ is very large?

• The heuristic interpretation of the central charge is that it measures the number of degrees of freedom in your system. So the central charge is likely to be related to the dimension $d$ of the local Hilbert space, and you probably cannot get large $c$ while keeping $d$ fixed. Dec 7, 2021 at 8:40
• @SylvainRibault That is a very interesting idea. One thing that gives me pause is that I think the following quadratic fermion models $I_n = -\sum_j \left( c_j^\dagger c_{j+n} + c_{j+n}^\dagger c_j \right)$ have central charge $c=n$, and they should be Jordan-Wigner transformable to a $1d$ spin-$1/2$ chain with local Hilbert space dimension $d=2$. I think these models are briefly discussed in this paper. Thus, I feel even with $d=2$ we can get arbitrarily large central charge $c$. Dec 7, 2021 at 8:48
• @SylvainRibault Thinking more on this, maybe part of the resolution is that $c$ is bounded by some function of both the dimension $d$ of the local Hilbert space and $r$ the range of interaction. Then I could imagine that in the quantum chain in this problem perhaps the formulae break down for $l<r$. It reminds me of transfer matrices in $1d$ classical models where finite but longer-range interactions can be accounted for by merging sites/ increasing the local degrees of freedom to reduce the interaction to nearest-neighbor on the new degrees of freedom. Dec 7, 2021 at 22:52

For a critical spin chain of length $$L$$, the formula for the entanglement entropy of an interval $$S(l) = \frac{c}{3}\ln \left(\frac{L}{\pi} \sin{\frac{\pi l}{L} }\right) +C$$ for an interval of length $$l$$ is only valid for large intervals. It's simply inaccurate* (though see below) at small $$l$$, even when you include the constant. And at large $$l$$ where the formula holds, the constraint $$S(l) \leq l \log(d)$$ is easily satisfied, so there's no problem.

General correspondence between lattice model and CFT

Stepping back, it's good to remember that when we say a critical spin chain is described by a CFT, usually we just mean that the long-distance properties of the spin chain match those of the CFT.

For instance, when we say a spin chain correlation function $$\langle O(x) O(y)\rangle$$ (for a lattice operator $$O$$) matches a CFT correlation function $$\langle \phi(x) \phi(y)\rangle$$ (for some CFT field $$\phi$$), this relation only holds at long distance $$|x-y|$$. Here's one reason we can't expect a match at short distances: the CFT correlation function $$\langle \phi(x) \phi(y)\rangle$$ will generally blow up as $$x \to y$$, whereas on a spin chain, the correlation function $$\langle O(x) O(y)\rangle$$ is bounded even as $$x \to y$$. Instead, they match at long distance, where they have the same power-law decay. One way to express this is $$\langle \phi(0) \phi(r)\rangle \propto \lim_{\lambda \to \infty} \langle O(0) O(\lambda r) \rangle \lambda^{2\Delta}$$ for some scaling dimension $$\Delta>0$$. On the other hand, when $$|x-y|$$ is just a few lattice spacings, the spin chain correlation function $$\langle O(x) O(y)\rangle$$ and the CFT correlation function $$\langle \phi(x) \phi(y)\rangle$$ may bear no particular relation.

Just like a correlation function, the function $$S(l)$$ at small $$l$$ will depend on details of the lattice model. For instance, you could have two different critical spin chain Hamiltonians described by the same CFT in the IR, and their correlation functions and entropies at small length scales could be very different, even though they have the same universal behavior at long distance.

Addendum 1: Accuracy of CFT formula?

[*] I said the formula $$S(l) = \frac{c}{3}\ln \left(\frac{L}{\pi} \sin{\frac{\pi l}{L} }\right) +C$$ is simply wrong for small $$l$$. That might seem surprising if you look at the figure in the question you linked, where the entropies are calculated for a spin chain of length 12, and even for regions of size $$l=1,2,\dots$$ the fit looks perfect. Apparently, for this particular critical spin chain Hamiltonian, the long-distance scaling of entropy already provides a very accurate approximation at short distances. See Fig. 1 of Calabrese et al. where they plot the error between an exact spin chain calculation of $$S(l)$$ and the CFT formula above; in their particular case they also find a very good match at short distances. Maybe that's somehow related to the exact solvability of these particular lattice Hamiltonians?

If you chose a more generic lattice Hamiltonian that exhibited the same behavior in the IR, you might not be so lucky to have a high-accuracy match with the $$S(l)$$ formula at short distances.

Addendum 2: Upper bound on central charge?

There was a suggestion in the comments for another possible resolution: maybe there's a universal upper bound for the central charge $$c$$ of any CFT that can be realized by range-$$r$$ Hamiltonians on spin chains of local dimension $$d$$, with the upper bound depending only on ($$r,d$$). Well, such an upper bound is no longer needed to resolve the original question, but I will try to comment anyway.
First, any such bound would need to depend on both $$r$$ and $$d$$, since you can always coarse-grain the qudits so that they only have $$r=2$$ nearest-neighbor interactions, at the price of increasing $$d$$. Second, even if you restricted to e.g. nearest-neighbor qubit interactions, I think you would need to impose translation-invariance (under single-qubit translations). Otherwise, I think there are tricks to "simulate" more general 1D Hamiltonians (with longer-range interactions, or higher-dimensional qudits) using only nearest-neighbor qubit spin chains. (Such tricks are discussed by Piddock et al. in the 2D case here.) A more complicated CFT might then appear in the IR of a simple nearest-neighbor qubit chain.
However, restricting finally to nearest-neighbor translation-invariant qubit chains, there may well be an upper bound on the possible central charge of CFTs in the IR. The parameter space of such Hamiltonians (up to additive and multiplicative scaling, and some other redundancies, see e.g. below Eq. 5.8 here) is just a compact 9-dimensional real manifold. Imagine the set of critical spin chains (say we mean: gapless spin chains corresponding to some CFT in the IR) as a subset of this compact parameter space. If there were indeed nearest-neighbor qubit chains corresponding to CFTs with arbitrarily large central charge, then there would be a sequence of critical spin chains with central charges $$c_1,c_2,\ldots$$ diverging to infinity, and the corresponding sequence in the compact parameter space would need to have an accumulation point.
• The quick version of the answer becomes very clear for fixed $l/L$ and fixed $c,d$. In particular, maximum entropy predicted by the CFT formula occurs for $l=L/2$, where $S(l)\propto \mathrm{ln}$, which clearly grows slower with $L$ than the bound $S_\mathrm{max}=l \mathrm{ln}(d) = \frac{L}{2} \mathrm{ln}(2)$. This seems to imply that large central $c$ (at fixed $d$) require larger $L$ for their CFTs to be good low-energy effective field theories, which would be natural if there indeed is an $(r,d)$-dependent bound for $c$. Dec 25, 2021 at 17:58
• For addendum 2, nice argument about translationally-invariant Hamiltonians likely having an $(r,d)$ bound on $c$. The paper you linked by Cubitt, Montanaro, and Piddock is also very nice, especially their schematic figure 2. I hadn't appreciated how powerful breaking translation invariance could be. Dec 29, 2021 at 2:01