Last week a bomb landed on the arXiv: Ji et al. posted a proof that MIP*=RE which implies, among other goodies, a solution to Tsirelson’s problem. I’m speaking here as if their proof actually holds; I can’t check that myself, as the paper is 165 pages of serious complexity theory, and I’m not a computer scientist. What I could understand made sense, though, and the authors have a good track record of actually proving what they claim, so I’ll just be positive and assume that MIP* is in fact equal to RE.

My first reaction to the result was ill-informed and overly dramatic (which of course a journalist took from my comment on Scott Aaronson’s blog to the Nature News article about it), but a few days after that, I still find the result disturbing, and perhaps you should too.

To explain why, first I need to explain what Tsirelson’s problem is: it asks whether the supremum of the violation of a Bell inequality we can achieve with tensor product strategies, that is, with observables of the form $A_i = A_i’\otimes \mathbb{I}$ and $B_j = \mathbb{I} \otimes B_j’$, is the same as the one we can achieve with commuting strategies, where we only require that $[A_i,B_j] = 0$ for all $i,j$.

It is easy to show that these two values must coincide for finite-dimensional systems, and furthermore that the tensor product value can be approximated arbitrarily well with finite-dimensional systems, which opens up a terrifying possibility: if there exists a Bell inequality for which these two values differ, it would mean that the commuting value can only be achieved by infinite-dimensional systems, and that it can’t even be approximated by finite-dimensional ones! It would make it possible for an experiment to exists that would prove the existence of literal, unapproachable infinity in Nature. Well, Ji et al. proved that there exists a Bell inequality for which the tensor product value is at most $1/2$, whereas the commuting value is $1$. Do you feel the drama now?

Now for the soothing part: Ji et al. cannot show what this Bell inequality is, they cannot even say how many inputs and outputs it would need and, more importantly, they cannot show what the commuting strategy achieving the value $1$ is. It turns out that even showing a purely mathematical commuting strategy that can do this is really hard.

We need more than that, though, if we’re talking about an actual experiment to demonstrate infinity: we need the commuting strategy to be physical. That’s the part I was ill-informed about in my initial reaction: I thought it was natural for QFTs to only commute across spacelike separated regions, and not be separated by a tensor product, but this is not the case: not a single example is known of a non-pathological QFT that is not separated by a tensor product, at least when considering bounded spacetime regions that are spacelike separated and are more than some distance apart. Even that wouldn’t be enough, as a QFT that only commutes might be residually finite, which would mean that it can only achieve the tensor product value in a Bell inequality.

So, I’m not shitting bricks anymore, but I’m still disturbed. With this result the infinity experiment went from being mathematically impossible to merely phisically impossible.

UPDATE: Perhaps it would be useful to show their example of a nonlocal game that has a commuting Tsirelson bound larger than the tensor product Tsirelson bound. For that we will need the key construction in the paper, a nonlocal game $G_\mathcal{M}$ that has a tensor product Tsirelson bound of 1 if the Turing machine $\mathcal{M}$ halts, and a tensor product Tsirelson bound at most $1/2$ if the Turing machine $\mathcal{M}$ never halts.

Consider now $\mathcal{M}(G’)$ to be the Turing machine that computes the NPA hierarchy for the nonlocal game $G’$, and halts if at some level the NPA hierarchy gives an upper bound strictly less than 1 for the commuting Tsirelson bound.

The nonlocal game $G_{\mathcal{M}(G’)}$ will then have a tensor product Tsirelson bound of 1 if the nonlocal game $G’$ has a commuting Tsirelson bound strictly less than 1, and $G_{\mathcal{M}(G’)}$ will have a tensor product Tsirelson bound at most $1/2$ if the nonlocal game $G’$ has a commuting Tsirelson bound equal to 1 (as in this case the Turing machine $\mathcal{M}(G’)$ never halts).

The nonlocal game we need will then be the fixed point $G^*$ such that $G_{\mathcal{M}(G*)} = G^*$. It cannot have a commuting Tsirelson bound strictly less than 1, because then it would need to have a tensor product Tsirelson bound equal to 1, a contradiction. Therefore it must have a commuting Tsirelson bound equal to 1, which also implies that it must have a tensor product Tsirelson bound at most 1/2.