TITLE: How undecidable is the spectral gap?
QUESTION [47 upvotes]: Nature just published a paper by Cubitt, Perez-Garcia and Wolf titled Undecidability of the Spectral Gap, there is an extended version on arxiv which is 146 pages long. Here is from the abstract:"Many challenging open problems, such as the Haldane conjecture, the question of the existence of gapped topological spin liquid phases, and the Yang–Mills gap conjecture, concern spectral gaps. These and other problems are particular cases of the general spectral gap problem: given the Hamiltonian of a quantum many-body system, is it gapped or gapless? Here we prove that this is an undecidable problem. Specifically, we construct families of quantum spin systems on a two-dimensional lattice with translationally invariant, nearest-neighbour interactions, for which the spectral gap problem is undecidable".
I am curious about the undecidable part. The abstract says "our result implies that there exists no algorithm to determine whether an arbitrary model is gapped or gapless, and that there exist models for which the presence or absence of a spectral gap is independent of the axioms of mathematics". "Axioms of mathematics" is kind of vague, so in the extended version they phrase it in a Gödelian manner:"Our results imply that for any consistent, recursive axiomatisation of mathematics, there exist specific Hamiltonians for which the presence or absence of a spectral gap is independent of the axioms". But still, axiomatization of which mathematics, how much mathematics do they need to construct their Hamiltonians. Is it real analysis? ZF? ZFC? I can't figure it out even from their theorem statements.

Is this a mathematical proof or something at the "physical level of rigor"? If so, does it produce "concrete" undecidable statements, or are these Hamiltonians as obscure as "I am unprovable"? Does it represent a new way of proving independence results compared to forcing, etc.? In other words, is it an advance on Gödel sentences and the continuum hypothesis? 

EDIT: Cubitt gave an interview where he commented on the nature of the result informally:"It's possible for particular cases of a problem to be solvable even when the general problem is undecidable, so someone may yet win the coveted $1m prize... The reason this problem is impossible to solve in general is because models at this level exhibit extremely bizarre behaviour that essentially defeats any attempt to analyse them... For example, our results show that adding even a single particle to a lump of matter, however large, could in principle dramatically change its properties".

REPLY [20 votes]: I interpret your question to be asking about the transition from
computable undecidability to Gödelian or logical
undecidability, and furthermore about the extent to which this
logical undecidability might depend on which axioms of mathematics
we have adopted.
The answer is that one may quite generally deduce that there are
concrete instance of logical undecidability, whenever one has a
computably undecidable problem, no matter what theory you have
adopted as your axioms of mathematics, whether it is PA or ZFC or
ZFC plus large cardinals or what have you. The two kinds of
undecidability — computable undecidability and logical
undecidability — are tightly intertwined, and every
computable undecidable problem is saturated with instances of
logical independence with respect to every (computably
axiomatizable, sound) theory.
To see this, suppose that $A$ is a computably undecidable decision
problem, such as the halting problem or the tiling problem or,
now, the spectral gap problem, and suppose that $T$ is your
favorite foundational theory, the axioms of mathematics. We
require only that $T$ has a computably enumerable list of axioms,
that $T$ is strong enough to express the decision problem $A$, and
that $T$ is sufficiently sound.
Since $A$ is computably undecidable, there is no computable
procedure to determine on input $n$ whether or not $n\in A$.
Consider the algorithm that on input $n$ searches for a proof from
$T$ that $n\in A$ or a proof from $T$ that $n\notin A$. Since the
membership problem for $A$ was computably undecidable, this
algorithm cannot correctly settle all instances of the problem,
since otherwise it would be a computable decision procedure. Since
$T$ is sound, however, the algorithm is correct about the
instances of the problem that it does settle. So there must be a
specific instance of the problem $n$ (in fact, infinitely many
such $n$) for which $T$ does not prove $n\in A$ and $T$ does not
prove $n\notin A$. Thus, for this specific $n$ and this specific
$T$, the question of whether $n\in A$ is independent of $T$.
This argument works even as you strengthen your axioms, and so it
doesn't matter whether you use PA or ZFC or ZFC plus large
cardinals or what have you. A stronger theory may settle some of
the concrete instances that were not settled by the weaker
theories, but every theory will admit concrete instances that it
doesn't settle, for otherwise the proof-searching algorithm would
be a computational procedure deciding $A$, contrary to the
assumption that $A$ is undecidable.
See also John Pardon's question, Are the two meanings of “undecidable” related?