MIP*=RE

Another Update (Jan. 16): Yet another reason to be excited about this result—one that somehow hadn’t occurred to me—is that, as far as I know, it’s the first-ever fully convincing example of a non-relativizing computability result. See this comment for more.

Update: If you’re interested in the above topic, then you should probably stop reading this post right now, and switch to this better post by Thomas Vidick, one of the authors of the new breakthrough. (Or this by Boaz Barak or this by Lance Fortnow or this by Ken Regan.) (For background, also see Thomas Vidick’s excellent piece for the AMS Notices.)

Still here? Alright, alright…

Here’s the paper, which weighs in at 165 pages. The authors are Zhengfeng Ji, Anand Natarajan, my former postdoc Thomas Vidick, John Wright (who will be joining the CS faculty here at UT Austin this fall), and my wife Dana’s former student Henry Yuen. Rather than pretending that I can provide intelligent commentary on this opus in the space of a day, I’ll basically just open my comment section to discussion and quote the abstract:

We show that the class MIP* of languages that can be decided by a classical verifier interacting with multiple all-powerful quantum provers sharing entanglement is equal to the class RE of recursively enumerable languages. Our proof builds upon the quantum low-degree test of (Natarajan and Vidick, FOCS 2018) by integrating recent developments from (Natarajan and Wright, FOCS 2019) and combining them with the recursive compression framework of (Fitzsimons et al., STOC 2019).
An immediate byproduct of our result is that there is an efficient reduction from the Halting Problem to the problem of deciding whether a two-player nonlocal game has entangled value 1 or at most 1/2. Using a known connection, undecidability of the entangled value implies a negative answer to Tsirelson’s problem: we show, by providing an explicit example, that the closure Cqa of the set of quantum tensor product correlations is strictly included in the set Cqc of quantum commuting correlations. Following work of (Fritz, Rev. Math. Phys. 2012) and (Junge et al., J. Math. Phys. 2011) our results provide a refutation of Connes’ embedding conjecture from the theory of von Neumann algebras.

To say it differently (in response to a commenter’s request), some of the major implications are as follows.

(1) There is a protocol by which two entangled provers can convince a polynomial-time verifier of the answer to any computable problem whatsoever (!!), or indeed that a given Turing machine halts.

(2) There is a two-prover game, analogous to the Bell/CHSH game, for which Alice and Bob can do markedly better with a literally infinite amount of entanglement than they can with any finite amount of entanglement.

(3) There is no algorithm even to approximate the entangled value of a two-prover game (i.e., the probability that Alice and Bob win the game, if they use the best possible strategy and as much entanglement as they like). Instead, this problem is equivalent to the halting problem.

(4) There are types of correlations between Alice and Bob that can be produced using infinite entanglement, but that can’t even be approximated using any finite amount of entanglement.

(5) The Connes embedding conjecture, a central conjecture from the theory of operator algebras dating back to the 1970s, is false.

Note that all of these implications—including the ones for pure math and the foundations of quantum physics—were obtained using tools that originated in theoretical computer science, specifically the study of interactive proof systems.

I can remember when the class MIP* was first defined and studied, back around 2003, and people made the point that we didn’t know any reasonable upper bound on the class’s power—not NEXP, not NEEEEXP, not even the set of all computable languages. Back then, the joke was how far our proof techniques were from what was self-evidently the truth. I don’t remember a single person who seriously contemplated that two entangled provers could convince a polynomial-time verifier than an arbitrary Turing machine halts.

Still, ever since Natarajan and Wright’s NEEXP in MIP* breakthrough last year, all of us in quantum computing theory knew that MIP*=RE was a live possibility—and all through the summer and fall, I heard many hints that such a breakthrough was imminent.

It’s worth pointing out that, with only classical correlations between the provers, MIP gives “merely” the power of NEXP (Nondeterministic Exponential Time), while with arbitrary non-signalling correlations between the provers, the so-called MIPns gives the power of EXP (Deterministic Exponential Time). So it’s particularly striking that quantum entanglement, which is “intermediate” between classical correlations and arbitrary non-signalling correlations, yields such wildly greater computational power than either of those two.

The usual proviso applies: when I’ve blogged excitedly about preprints with amazing new results, most have stood, but at least two ended up being retracted. Still, assuming this one stands (as I’m guessing it will), I regard it as easily one of the biggest complexity-theoretic (and indeed computability-theoretic!) surprises so far in this century. Huge congratulations to the authors on what looks to be a historic achievement.

In unrelated news, for anyone for whom the 165-page MIP* paper is too heavy going (really??), please enjoy this CNBC video on quantum computing, which features several clips of yours truly speaking in front of a fake UT tower.

In other unrelated news, I’m also excited about this preprint by Avishay Tal, which sets a new record for the largest known separation between quantum query complexity and classical randomized query complexity, making substantial progress toward proving a conjecture by me and Andris Ambainis from 2015. (Not the “Aaronson-Ambainis Conjecture,” a different conjecture.)

161 Responses to “MIP*=RE”

  1. R. Crist Says:

    A solution to the Connes’ Embedding Conjecture solution is as big a deal to operator algebraists as the actual result is to complexity theory. It’s interesting that two of the biggest problems in operator algebras (CEC and the Paving Problem) were solved by using equivalences to problems in other areas.

  2. Joshua B Zelinsky Says:

    With the disclaimer that I’ve only looked at the paper very briefly, it seems like not only have they proven that MIP* = RE, they’ve also proven that the Connes’ Embedding Conjecture is false. That’s surprising, although not incredibly so, since if I understand correctly, it was already known that the Embedding Conjecture implied an upper bound of RE on MIP*, so there already was a connection between the two problems.

  3. fred Says:

    Is there a short explanation of the consequences of this that can be understood by a casual computational complexity enthusiast? Or 165 pages is the Kolmogorov complexity and I’ll just move on?

  4. Bob Hearn Says:

    Interesting. You write “I don’t remember a single person who seriously contemplated that two entangled provers could convince a polynomial-time verifier than an arbitrary Turing machine halts… I regard it as easily one of the biggest complexity-theoretic (and indeed computability-theoretic!) surprises so far in this century.”

    Can you briefly explain why this is so shocking? You might recall my result with Erik Demaine that team games of imperfect information are undecidable, even when the games involve a fixed amount of state (i.e. are played on a finite board). To me this has the same flavor of “but wait, where did the infinite TM tape go?” that MIP*=RE does. It certainly shocked me at the time, anyway. Is it that same basic sense you are talking about, or something else?

  5. fred Says:

    Hi Scott,

    maybe a question that’s too personal, but I’d like your opinion on this…

    Would you encourage experts in the same field to marry each other? The upsides are obvious, but are there downsides?
    Like, the kids are confused that mom and dad talk and/or possibly argue so much about the entangled value of non-local two player games? 😛

  6. Scott Says:

    fred #2: Did you try reading just the abstract and intro? Trust me, it will be a lot easier than the full 165 pages. 🙂

    Some of the major implications are:

    (1) There is a protocol by which two entangled provers can convince a polynomial-time verifier that a given Turing machine halts (or of the answer to any computable problem whatsoever).

    (2) There is a two-prover game, analogous to the Bell/CHSH game, for which Alice and Bob can do markedly better with a literally infinite amount of entanglement than they can with any finite amount of entanglement.

    (3) There is no algorithm even to approximate the entangled value of a two-prover game (i.e., the probability that Alice and Bob win the game, if they use the best possible strategy and as much entanglement as they like). Instead, this problem is equivalent to the halting problem.

    (4) The Connes embedding conjecture, a central conjecture from the theory of operator algebras dating back to the 1970s, is false.

  7. Scott Says:

    Bob Hearn #3: One reason why it’s shocking—at least relative to everyone’s thinking back in the day—is that the original expectation was that, if anything, entanglement would only decrease the power of multi-prover interactive proof systems. Indeed, there are well-known examples of proof systems that are sound with unentangled provers that become unsound when the provers were entangled. So it was a major advance when, in 2012, Ito and Vidick at least showed that the proof of MIP=NEXP could be “immunized” against entangled provers (yielding NEXP⊆MIP*)—with no thought at the time of getting even more power than you did classically! But now it turns out that, not only do you get the same vast power as classically (NEXP), not only do you get even a little bit more power (QMAEXP), not only do you get a lot more power (NEEXP), you go all the way up to the friggin’ halting problem!

    Was any similar progression followed in the example you mentioned, involving games with imperfect information?

  8. Scott Says:

    fred #4:

      Would you encourage experts in the same field to marry each other? The upsides are obvious, but are there downsides?
      Like, the kids are confused that mom and dad talk and/or possibly argue so much about the entangled value of non-local two player games?

    Uhhh … whatever might be the downsides of marrying someone in the same field as you, they don’t include your kids being confused by technical talk. Married couples don’t typically want or expect their kids to understand every last thing that they say to each other.

    Meanwhile, the upsides include never having to explain things like conference deadlines to your spouse. Also, being able to travel to many conferences with your spouse, and being able to tolerate your spouse’s work friends because many of them are also your work friends.

    (For context, Dana and I are not in exactly the same field—I do quantum while she does classical complexity theory, especially PCP and derandomization. But yeah, it’s close enough that we have many of the same friends, which indeed is how we met in the first place.)

  9. James Giammona Says:

    Hi Scott,
    In a previous comment to a different post (https://www.scottaaronson.com/blog/?p=3678#comment-1763790), you mentioned:
    “Connes’ very abstract-looking embedding conjecture for C* algebras is actually equivalent to the natural statement that any correlations allowed by quantum mechanics can be arbitrarily well approximated using a finite number of entangled qubits”.

    Does this conjecture being false imply that some QM correlations cannot be approximated by a finite number of entangled qubits? (Is that surprising or non-intuitive?) How is this different (or related) to the case that some Hamiltonians cannot be efficiently approximated?

  10. RandomOracle Says:

    To add to why this result is so surprising, if we go in either direction in terms of strengthening or weakening the correlations between the provers we get complexity classes that are strictly weaker than RE.

    Specifically, if the provers share classical correlations the resulting class is MIP which we know is equal to NEXP. If they share general non-signalling correlations we get MIP_ns which is contained in EXP.

    It’s really remarkable that in the case of quantum correlations the resulting class is ridiculously more powerful than either of these two.

  11. Scott Says:

    James Giammona #9: Yes, that’s also an implication. And yes, it’s also extremely surprising, at least relative to what the consensus had been until a year or two ago (when we started to get more and more indications that the consensus was wrong). I’ll add something about that to the main post.

    I’m not sure what result you were referring to with “some Hamiltonians can’t be efficiently approximated,” but whatever it is, it’s almost certainly very different. For one thing, note that with the new result we’re not talking about efficient approximation, but simply about approximation with any finite number of qubits!

  12. Gerard Says:

    Scott: #6

    “(1) There is a protocol by which two entangled provers can convince a polynomial-time verifier that a given Turing machine halts (or of the answer to any computable problem whatsoever).”

    For someone who is very uninformed about things like interactive proofs and quantum complexity theory, a naive reading of this would seem to mean something like “2 quantum computers working together with a classical computer can solve the halting problem”. I’m pretty sure that’s not what you mean to say here, so I’m hoping you can clarify/elaborate on this statement a bit.

    More generally do you think this this result will have practical implications or is it purely theoretical ?

  13. domotorp Says:

    Does this result have any implication about classical multi-prover protocols, like if there are two Provers who can communicate with each other, but only with certain constraints, then the class equals RE?

  14. Scott Says:

    RandomOracle #10: Thanks for the reminder! That’s an extremely relevant point that I’ll also add to the main post.

  15. Mateus Araújo Says:

    I’m shitting bricks here. I never thought I’d see this problem being solved in my lifetime, and I was sure that Tsirelson’s problem had a positive answer: that of course the closure of the set of finite dimensional correlations must be the set of infinite dimensional correlations, because Nature is in some vague sense fundamentally finite. Oh well, it’s still possible that there is a mistake somewhere 😉

  16. Scott Says:

    Gerard #12:

      For someone who is very uninformed about things like interactive proofs and quantum complexity theory, a naive reading of this would seem to mean something like “2 quantum computers working together with a classical computer can solve the halting problem”. I’m pretty sure that’s not what you mean to say here, so I’m hoping you can clarify/elaborate on this statement a bit.

    The fact that these are two provers convincing a verifier is essential here. You can’t simplify that away, as you’re trying to above, without completely destroying the meaning.

    The setting you should imagine is that there are two quantum computers with no bound whatsoever (not even “less than infinity”) on their power, how many qubits they can have, how long they can run for, etc. So then, you might say, it’s no surprise that they have superpowers like solving the halting problem!

    But crucially, the QCs are not trusted. They can say whatever they want, but why should a classical polynomial-time verifier believe them?

    The whole point of the new result is that, as long as the two QCs share an unlimited amount of entanglement, they can convince the verifier, in time polynomial in the length of a program, that they executed that program correctly—no matter how astronomical the program’s running time.

      More generally do you think this this result will have practical implications or is it purely theoretical ?

    As mentioned above, executing the protocol in interesting cases would require two astronomically-powerful QCs that share an astronomical amount of entanglement. For that reason alone, direct applications seem unlikely. As for indirect applications, who knows? When MIP=NEXP itself was first proved (by Babai, Fortnow, and Lund around 1990), it may have seemed comically remote from any applications. But it led directly to PCP Theorem, which became the central tool for making statements about the inapproximability of practical NP-hard problems.

  17. Scott Says:

    domotorp #13: If the provers were limited to communicating with each other in all and only the ways that perfectly simulated quantum entanglement, then yes, I suppose. 😀 But AFAIK the result is very specific to entangled provers.

  18. Mateus Araújo Says:

    Maybe it’s a good idea to pre-emptively clear up a misconception that always crops up when physicists discuss this subject: “How can it be undecidable to determine whether the Tsirelson bound of a bipartite Bell inequality is 1 or at most 1/2? We already have an algorithm for approximating it, the NPA hierarchy!”

    No, we don’t. The NPA hierarchy is not an algorithm, because algorithms need to halt after a finite number of steps, and the hierarchy does not, it goes on forever. Every fixed level of the hierarchy is an algorithm, and sometimes we know that the hierarchy has converged and we don’t need to go to higher levels, and often it gives us very good approximations, but this is not the case in general, and cannot be. In general we only know that the hierarchy converges (to the commuting set) in the limit of infinitely many levels.

    This result implies that there can’t be an algorithm for determining the Tsirelson bound, so in a sense the NPA hierarchy is the best we can hope for.

  19. Scott Says:

    Mateus #15:

      I was sure that Tsirelson’s problem had a positive answer: that of course the closure of the set of finite dimensional correlations must be the set of infinite dimensional correlations, because Nature is in some vague sense fundamentally finite.

    But you can still believe that “Nature is in some vague sense fundamentally finite”! The upshot is simply that, if Nature had involved infinite-dimensional Hilbert spaces, then that could in principle have been testable in a more direct way than you would’ve thought a priori. But it’s not like we know how to (attempt to) apply the infinite-dimensional unitaries anyway, in order to do the requisite experiment.

  20. Scott Says:

    Mateus #18: Yes, thanks for that!

  21. James Giammona Says:

    Scott # 11, thanks for answering my questions! Your answers were quite helpful.
    I was referring to the cost of simulating some quantum systems with a quantum computer. My understanding was that certain quantum systems could not be efficiently simulated on a quantum computer (i.e. would take exponential time or space resources) (perhaps that’s not a proven result?).
    Your explanation that this is about approximation in general and not about efficient approximation is a great clarification.

  22. Tobias Fritz Says:

    Scott #18:

    “But it’s not like we know how to (attempt to) apply the infinite-dimensional unitaries anyway, in order to do the requisite experiment.”

    Not only that, but it is also questionable whether there even exists any physical system that could realize the relevant commuting operators with spacelike separation even in principle.

    In fact, suppose that AQFT is a suitable mathematical framework for modelling physical reality. Then the split property, which is expected to hold for all sensible QFTs, implies that all spacelike commuting operators can always be realized on separate tensor factors while generating the same algebra. See Theorem 4.1 in this paper. This suggests that all physically realizable quantum correlations with spacelike separation are consistent with the tensor product paradigm.

    So while this paper is obviously an enormous achievement and a fantastic breakthrough for complexity theory and operator algebras, I think that there is reason to be skeptical about claims that it is relevant to the foundations of physics.

  23. Scott Says:

    James #21: Sure, there are quantum systems that can’t be efficiently simulated on a QC. But for that very reason, to whatever extent you believe in the “Quantum Extended Church-Turing Thesis” (QECT), those systems should be of limited relevance to Nature. And for all its revolutionary implications, the new breakthrough on MIP* has nothing to say one way or the other about the validity of the QECT.

  24. Mateus Araújo Says:

    Scott #19: That’s no consolation at all, having an actual experimental result to rub in my face how wrong I was. Maybe I deserve that for making fun of Miguel Navascués for believing that such an experiment must exist.

    It is still possible, of course, that the unitaries necessary to do the experiment are unphysical in a strong way, such as needing infinite energy or time to be applied. Until somebody does that I have to accept that as far as we know the experiment is possible.

  25. Mateus Araújo Says:

    Lol, I checked the paper (section 12.4), and their “explicit example” is clearly using mathematician values of “explicit”. We are still far from knowing whether the experiment is physically possible.

  26. Sebastian Oberhoff Says:

    One further point that I think is worth highlighting is that while it’s now possible to convince the verifier that a halting computation does indeed halt, it’s not the case that one can convince the verifier that a non-halting computation doesn’t halt.

    When I first heard of this result I thought: “Whoa, so by conversing with entangled, omniscient provers I could become convinced of the consistency of ZFC, effectively overcoming the Second Incompleteness Theorem!” Alas, this is not the case.

  27. Scott Says:

    Mateus #24: Does the comment of Tobias Fritz (#22) assuage your worries?

  28. Gerard Says:

    Scott #16

    “The setting you should imagine is that there are two quantum computers with no bound whatsoever (not even “less than infinity”) on their power, how many qubits they can have, how long they can run for, etc. So then, you might say, it’s no surprise that they have superpowers like solving the halting problem!”

    I still find that surprising. As far as I know, there are no bounds in the usual Turing Machine model either, yet no TM can solve the halting problem.

    “The fact that these are two provers convincing a verifier is essential here.”

    What is it about the provers that establishes their separateness ? I mean why couldn’t one simulate the entire configuration using just a single QTM ?

  29. Scott Says:

    Gerard #28: No, you’re still not getting it. The provers can even run for a literally infinite amount of time. Thus, if they didn’t also have to convince the verifier, there would be no issue whatsoever.

    Sure, whatever resources the provers end up needing, you can simulate them using similar resources running on a single machine. (With the proviso that the “resources” in question might be literally infinite!)

    But again, the relevant question here is why the verifier is convinced that the provers didn’t cheat. That’s where the assumption of no communication between the provers (only entanglement between them) becomes essential.

  30. Mateus Araújo Says:

    Scott #27: I’ll have to read the papers he linked in order to draw a conclusion (after all AQFT is not a suitable mathematical framework for modelling physical reality), but Fritz’s word is enough to let me sleep tonight.

    I’m wondering about a different problem, though. We know that the tensor product Tsirelson bound is decidable (as it can be approximated from below by just Tarskiing it, the most horrible algorithm I can imagine). Since now we know that the NPA hierarchy does not converge to it from above, but to the commuting Tsirelson bound, shouldn’t there be an algorithm that does converge to it from above? And since it is a decidable number, can’t we hope the algorithm to be something nicer than the NPA hierarchy? (the fact that the algorithm for approximating it from below is anything but nice maybe indicates that this is a misguided hope, or simply that nobody tried hard enough yet).

  31. Boaz Barak Says:

    Tobias Fritz #22 – I am not a physicists, so maybe completely off base, but isn’t your comment an example why the result could potentially have applications to physics?

    Not in the sense that it has such applications now, but that it opens some possibilities that would not have existed had the Connes embedding conjecture been true. Specifically, suppose you had two physical theories that disagreed with one another only in that one allowed for these more general types of operators. Before this result, it might have seemed that even in principle, one would not be able to design an experiment that could distinguish between the two theories, but now based on this results we could potentially design such an experiment. This is similar to how Bell’s inequalities can be used to design an experiment to rule out hidden local variables theories.

  32. William Slofstra Says:

    Mateus @30: While we can compute a sequence that converges to the tensor-product value from below, it’s not an algorithm for the same reason the NPA hierarchy isn’t an algorithm: we don’t have any way to know how close we are to the true value. This new result doesn’t just show that MIP*=RE, it shows that two-prover one-round MIP* = RE, which is equivalent to showing that the tensor-product value of a nonlocal game is not computable, in the sense that there’s no way to decide whether the tensor-product value of a game is equal to 1, or less than 1/2. So not only does the NPA hierarchy not converge to the tensor-product value, there is no computable sequence which converges to the tensor-product value from above.

    We don’t know yet whether it’s possible to have a computable sequence which converges to the commuting-operator value from below. If MIP* = RE, it seems likely that MIP^co, the equivalent class where the provers can use commuting operator strategies, could be equal to coRE. This is still open (see the open questions section of the paper), so it will be interesting to see if it can be done with the same techniques.

  33. Tobias Fritz Says:

    @Boaz Barak #22: that’s an interesting point. But if we have two physical theories that disagree with each other, and they are both concrete enough to make predictions, then they presumably will already make different predictions in much simpler situations. It follows that we will have easier ways to decide which one is wrong. More precisely, we could use both theories to model our proposed Bell test experiment, and if they give different predictions there, then of course they must already differ in the probabilities predicted for one particular choice of setting. Thus we don’t actually need to do the full Bell test: merely doing the relevant measurements with that particular choice of setting is enough.

    But what in case that the theories are not concrete enough to make predictions, but they are rather metatheoretical frameworks such as (A)QFT? In that case, it seems very unlikely that we’d be able to realize a Bell test of the relevant kind anyway; we need a concrete model of our apparatuses in order to determine how to implement the relevant measurements and state preparation. In other words, unless we get extremely lucky by doing something random, a pure metatheoretical framework is not enough to tell us what to do. So also in this case the Bell test doesn’t help.

    For a simpler version of the same argument, consider just a plain Bell test. Even without performing it, we know that either it will disprove local hidden variables, or it will disprove quantum theory. In case that it did the latter, we presumably would already have noticed a deviation from quantum theory in our modelling of the experiment’s individual components. Hence the only physical insight we can hope to obtain by actually conducting a Bell test is to exclude wacky theories in which physics conspires against us by suddenly changing its behaviour as soon as we conduct a Bell test.

  34. Greg Kuperberg Says:

    Joshua #2 – They have disproved the Connes embedding conjecture, but that’s not a separate result. MIP* with tensorable entanglement is in RE, while MIP* with commuting operator entanglement is in coRE. The Connes conjecture is (or was) that tensorable entanglement approximates all commuting operator entanglement. The Connes conjecture would imply that MIP* is in RE cap coRE = R, which is of course inconsistent with RE-completeness by Turing’s theorem.

  35. William Slofstra Says:

    Some sloppiness in my previous comment: showing MIP*(2,1) = RE is equivalent to showing that computing the tensor-product value is as hard as the halting problem. (Otherwise you might not get all of RE.)

  36. Scott Says:

    Boaz #31: Yes, the prospect of an “experimental test of the finite-dimensionality of the universe’s state space” was also one of the things that excited me about this line of research! But we should be clear that we don’t know, even in principle, how to apply the test in question (which would require specific infinite-dimensional unitary operators)—even assuming the case where the universe’s Hilbert space is infinite-dimensional. And that this is a genuine difference between this result and the Bell inequality. (A second difference, of course, is that the Bell inequality is easy to prove. 🙂 )

  37. Joshua B Zelinsky Says:

    Greg #34,

    Yes, that’s a really good point. The connection here is substantially more direct to previous work involving the conjecture than I had initially realized when I started reading it.

  38. Daniel Harlow Says:

    An aspect of this which is naively puzzling to me is as follows: for these astronomical complexity classes, why should quantum mechanics be helpful? After all BQP is contained PSPACE, so if we don’t even care how long it takes things to run then can’t we just simulate all quantum physics classically (or maybe classically plus a random number generator)?

  39. Boaz Barak Says:

    Tobias #33 Scot #31: Let me try to explain better what I mean.
    Bell’s Inequality experiments have 3 components as far as I can tell:

    1) You prove that a certain experiment will always succeed with probability at most p if the parties involved are modeled using classical hidden variables.

    2) You prove that quantum mechanics allows in principle a strategy that will succeed with probability p’ which is strictly larger than p

    3) You find a concrete implementation that manages to succeed in implementing this strategy.

    I think the formal Bell inequality is just steps 1 & 2, and then to make an actual experiment you need to do step 3 which is to actually show how you can realize the unitaries that are needed. Similarly, the new result can be thought of somewhat analogous to step 1 & 2 – it shows that there is an experiment that could be realized with commuting operators but not tensorizing operators (I hope I’m using the right notation).

    It doesn’t give step 3 – a concrete implementation of an experiment which actually achieves a value better than that achievable using tensorizing operators. Step 3 would inherently depend on the particular infinite-dimensional theory (this is a little bit like the question of efficiency of the prover in classical interactive proofs), but at least now there is a “fighting chance” in coming up with such an experiment.

    Perhaps the biggest difference between Bell’s inequality and this result is the conceptual case. In Bell’s inequality, we actually believe in the more general theory (i.e., quantum mechanics) as opposed to the more restricted one (i.e., hidden variables). In contrast, if you believe in some theories such as AQFT then (if I understood Tobias) you actually believe that the set of realizable distributions is more restricted than the one that is allowed by all commuting operators.

    Still, I do think it’s interesting because it says that if you believe in AQFT then you are actually positing a “law of nature”: it is a non trivial restriction on the set of realizable distributions. It might be interesting because if you are already positing this law of nature, you are already on a “slippery slope” and may as well take Aaronson’s position and not just rule out the distributions that could arise only via these infinite-dimension states but also the ones that could only arise using exponentially complex computation..

  40. Scott Says:

    Boaz #39: Yes, I don’t think we disagree about anything here! I would only add that, for me, saying that MIP*=RE is “missing only the concrete implementation aspect,” somewhat understates the magnitude of the difficulty. 🙂

    Let’s say it plainly: to implement the experiment we’re talking about would involve doing a specific quantum computation on a literally infinite number of qubits. So a skeptic might retort: if you actually had an infinite number of qubits available for acting on, then wouldn’t that fact alone already settle the infinite-dimensionality of the universe’s Hilbert space, with no need for further experiments?

    The reply to that is: well, maybe you don’t know how many qubits you have. Instead, you just have some black boxes with unknown (possibly infinite) numbers of qubits, together with finitely many knobs and measuring devices that act on the black boxes in ways that we assume to be consistent with the rules of QM. We can assume further that each knob has only finitely many settings, and each measuring device returns one of only finitely many outcomes.

    In such a case, yes: if the black boxes were of exactly the right sort, then the whole upshot of the MIP*=RE breakthrough is that the boxes’ observed behavior could certify the presence of infinitely many qubits inside them, to high statistical confidence. And this is conceptually amazing.

    From a “practical” standpoint, though, the trouble is that, to whatever extent it’s “physically plausible” to imagine worlds with infinitely many qubits at all, we almost certainly still wouldn’t be able to build the requisite black boxes in those worlds, at least not in any finite amount of time. The reason—physicists on the thread should tell me if I’m wrong!—is that accessing more and more qubits would almost certainly involve accessing higher and higher energy modes, but any doable experiment can marshal only finite energy.

    By contrast, as soon as Bell published his inequality, I think it was clear that, to whatever extent the world is described by quantum mechanics at all, in principle this experiment could be done. And that would still have been true, even if we were unlucky and the Bell experiment had turned out to be impractical for the next billion years, rather than doable in the next 20.

  41. Scott Says:

    Daniel Harlow #38:

      An aspect of this which is naively puzzling to me is as follows: for these astronomical complexity classes, why should quantum mechanics be helpful? After all BQP is contained PSPACE, so if we don’t even care how long it takes things to run then can’t we just simulate all quantum physics classically (or maybe classically plus a random number generator)?

    That’s an excellent question … the asking of which is precisely what led everyone for 17 years to the wrong intuition that MIP* should be relatively weak. 😀

    The key point is that, in the definition of MIP*, there’s no a-priori upper bound on how many entangled qubits the provers are allowed to use. And the more entangled qubits they have, the longer it could take to simulate their strategy.

    Naïvely, this seems like just a technical problem. Surely there’s some theorem showing that, without loss of generality, the provers will never need more than exp(n) entangled qubits (or however many), in order to approximate whatever they can do with arbitrarily many entangled qubits, at least as far as protocols running in poly(n) time are concerned?

    Seemed reasonable … to me, and to nearly everyone else.

    But the upshot of the new paper is that it’s wrong. More entangled qubits let the provers do more, without limit. And so, for example, if you hand the verifier an n-bit program that runs for T steps before halting, then just by running a poly(n)-time protocol, the verifier can check that the provers must have shared at least T entangled qubits. (Or even a literally infinite number of qubits, if we’re willing to go outside the world of tensor-product Hilbert spaces.) Insane.

  42. Sniffnoy Says:

    So, some basic questions here. The paper discusses a number of sets of correlations. There’s C_c, C_q, C_qs, C_qa, and C_qc. These are all distinct (with C_c < C_q < C_qs < C_qa being already known, and C_qa < C_qc being the negative resolution of Tsirelson’s problem that the paper proves). One could also add in another set of correlations to this chain, the set of all non-signalling correlations (let’s call it C_ns).

    Am I correct in inferring that you can define an MIP-analogue for any of these sets? Like, classical correlations C_c is MIP (which is equal NEXP). MIP* (now shown to be RE) I gather is C_qs? And if you use C_qa, is that just also MIP* since C_qa is just the closure of C_qs? And then above it’s been discussed how C_qc gets you MIP_co which is speculated to be coRE? And C_ns gets you MIP_ns which is equal to EXP? And I guess if you allowed all correlations then that would include signalling correlations and so you’d just get IP (which is equal to EXP)?

    So, questions:

    1. Do I have all this right? MIP* comes from C_qs, right? So does C_qa in fact get you the same thing? And what does using just C_q get you? Is that also MIP*?

    2. Has there been any study of this for weirder sets of correlations?

    3. So, a check of my understanding here… the paper shows that Tsirelson’s problem resolves in the negative, i.e., shows that C_qa < C_qc. But a bunch of the discussion above (like Scott #40) talks about distinguishing a finite number of qubits from an infinite number. Isn’t that C_q vs C_qs instead? Or what’s going on here? Is Scott’s informal characterization correct, but it’s not C_q vs C_qs, but rather some other thing that can informally be described the same way and indirectly leads to C_qa ≠ C_qc? What’s going on here?

  43. Mateus Araújo Says:

    Slofstra #32: I’m confused. What I have in mind is the Tarski algorithm (which is an actual algorithm). We parametrize the state and observables for some dimension d, and we’ll have a formula over the reals (the complex numbers drop out because the expectation values are always real) with a lot of parameters, but that boils down to \(\exists |\psi\rangle,\{A_i\},\{B_i\};v(G)=k\). We can then do quantifier elimination to solve this problem, and then do a binary search over k. This shows that for every fixed d the set is decidable.

    Ok, now I see the problem: we can always increase d, and we can’t know when to stop, so the “algorithm” doesn’t terminate, just like the NPA hierarchy. Damn.

    It is still valid question, though, whether we can have a tensor-product analogue of the NPA hierarchy, that converges to the tensor-product value from above. It wouldn’t be an algorithm, but it would still be better than what we have now, which is absolutely nothing.

    (I was writing under the assumption that MIP^* was defined for the commuting strategies, so it wouldn’t be a contradiction for the tensor-product Tsirelson bound to be decidable. It is not, though, as you noticed they used the “correct” definition for MIP^*. I’ll just be a pessimist and conjecture that the commuting Tsirelson bound is also undecidable)

  44. gentzen Says:

    In discussions triggered by the “MIP* contains NEEXP” paper asdf exclaimed: “Something still seems paradoxical.” I wonder whether seeing this result on par with other paradoxes like the Banach-Tarski paradox would be a valid perspective.

    An oracle for the halting problem could lie (by claiming that some non-halting computation would actually halt), and there is no protocol allowing “a mere Turing machine with infinite patience” to catch it lying. But for two oracles with infinite entanglement, there is a protocol that overcomes this barrier. This is sort of unbelievable.

    Joel David Hamkins once asked: What would it take in a book to convince a rational person that it had been written by or directly inspired by a god? I guess there is no way for the book to do that trick. But now imagine two books with infinite entanglement…

  45. Scott Says:

    gentzen #44: Yes, it’s “sort of unbelievable.” Two quibbles though:

    (1) A prover, which does whatever it needs to do to get you to accept, is a fundamentally different type of thing from an oracle, which just unfailingly answers a certain kind of question. If oracles are gods, maybe it would help to think of provers as trickster angels or devils? 🙂

    (2) Personally, I’d already be convinced of godlike powers if an entity could reliably solve NP-complete problems (eg, finding short proofs of theorems whenever they exist), or PSPACE-complete problems (which it could demonstrate via an interactive protocol). The halting problem would be massive overkill.

  46. Mateus Araújo Says:

    Sniffnoy #42: About your question 3, this is indeed confusing. For practical purposes, all that matters are the sets C_qa and C_qc, as C_qa is the closure of C_q (and also the closure of C_qs), and nobody wants to work with an open set.

    Now, despite C_qa explicitly coming from infinite-dimensional states, people still refer to it as “finite”, because you can approximate C_qa arbitrarily well with correlations produced from finite-dimensional states. C_qc, on the other hand, can not be approximated with correlations from finite-dimensional states (as we now know), so it is literal, unnaproachable infinity.

  47. Mateus Araújo Says:

    Scott #40: I actually find conceivable (although very implausible) that Nature does give us for free such black boxes with an infinite amount of qubits which are infinitely entangled. The vaccum state of QFT is, after all, entangled, even though it is a small amount of entanglement that is very hard to use for a Bell inequality violation.

    What I don’t find conceivable is that we could actually implement the unitaries necessary to manipulate these infinitely complex systems. As you point out, that would probably require infinite energy, and some other infinite resources as well.

  48. Boaz Barak Says:

    Scott #40: thank you – I didn’t realize that the energy would blow up and that’s a very significant point.

    The one point that might still make the result potentially useful is that it’s not merely a separation but a reduction. That is, rather than merely refuting the Connes embedding conjecture by a single counterexample, they actually show a general recipe to give a protocol for every problem in RE. That might give some hope that there is more than one set of “black boxes with knobs” that could be used in their result. In any case, the situation is of course very different from Bell since I’m not a physicist, but my understanding is that there aren’t many natural theories with these black boxes lying around.

    On a different note, can this result be used to give quantum 2 prover “proofs of work” or “delegating computation” where the verifier complexity is an absolute constant which is completely independent of the complexity of the computation but only depends on the error probability? That is, we can have an absolute constant time verifier that can be used to certify that you spent T units of computation for an arbitrarily large T, or where you can certify to it the result of computing F(x) for F in TIME(T)? I guess this is the question if whether in the case that the Turing machine does halt in time T, is the prover strategy computable in time polynomial (or even any function of) T?

  49. Tobias Fritz Says:

    Boaz #39: I agree with that very nice summary, but my point still stands: actual Bell experiments are not a good tool to tell us anything new about physical reality, regardless of which theory of physics is correct.

    The reason that Bell tests are not interesting as actual experiments is because we already know with close to certainty what the result will be. And why is that? It’s because, as in your step 3, we already need to have a good theoretical model of our experiment in order to design it [1]. Now suppose that we conduct a Bell test based on such a model, either a vanilla one or a fancy one of the kind that could distinguish between the commuting-operator paradigm and the tensor product paradigm. Then again we will already know the result before performing the experiment, simply because we’ve modelled it. And if our model is not completely correct, then there has to be a discrepancy between model and reality somewhere in the big table of probabilities that describes the Bell experiment. In other words, our model fails to be correct already for one particular choice of setting. So in order to witness this discrepancy, it’s enough to do the experiment with that choice of setting only, so that performing the whole Bell test is moot. Even worse, in all likelihood we would have noticed the failure of our model already much earlier, namely while working with the individual parts that make up our Bell experiment apparatus.

    This is all notwithstanding crazy conspiracy theories in which the laws of nature change as soon as we decide to conduct a Bell-type experiment. But anyone who entertains such theories also needs to explain why they nevertheless find it plausible to assume that the choices of setting are random and not correlated with the entangled state, since without this assumption a Bell test is once again useless. Thus this amounts to believing in a conspiracy and in the absence of a conspiracy at the same time.

    [1] Of course this applies to any experiment in physics. However, one characteristic of an interesting experiment is that it probes the limits of applicability of our model, like a particle accelerator probing higher energies than those at which we have confidence in our model. Perhaps you would argue that an experiment like we’re discussing here would probe the limits of complexity. This may or may not be the case, but I think that my point above applies regardless.

    ———

    I’m a bit worried that this discussion may be too much of a rabbit hole deflecting from the amazing result that has been achieved. So I’ll leave it at this for now, perhaps unless Scott encourages further discussion.

  50. fred Says:

    Scott #16

    “The whole point of the new result is that, as long as the two QCs share an unlimited amount of entanglement, they can convince the verifier”

    If two provers/QCs are sharing an “infinite amount of entanglement”, aren’t they effectively one prover/QC?

    How do you define the boundaries of a QC that’s sharing entanglement with another QC?
    We can trivially connect two classical computers to make a bigger machine (Moore’s law is one result of this), but given two QCs, it’s difficult to join them to create a bigger QC, but “sharing entanglement” seems to be one way to do it, gradually, no?

  51. mr_squiggle Says:

    re: (Not the “Aaronson-Ambainis Conjecture,” a different conjecture.)

    Have you considered calling this other thing the “Ambainis-Aaronson Conjecture”, or is that something else again?

    (I’m sorry, I love reading your blog, Scott, although I confess most of it goes over my head.)

  52. Scott Says:

    fred #50:

      If two provers/QCs are sharing an “infinite amount of entanglement”, aren’t they effectively one prover/QC?

    No, they’re not—for one thing, because they still can’t exchange messages with each other.

    The correlations that you can produce using quantum entanglement—yes, even an infinite amount of entanglement—are only a tiny subset of all the correlations that are a-priori possible. (And far from being some throwaway remark, the preceding sentence is at the core of a large fraction of the advances in quantum protocols and quantum foundations that we’ve seen in the last 15 years.)

  53. Greg Kuperberg Says:

    William #32 – Do you expect that there are groups that are not hyperlinear?

  54. Mateus Araújo Says:

    Fritz #22: Thanks for posting the link to arXiv:0812.1517, that was a delightful read. The key quote from that paper is “… the only known models in which even the distal split property does not hold are physically pathological models, such as models with noncompact global gauge group and models of free particles such that the number of species of particles grows rapidly with mass.”

    “Distal split property” is the property that algebras on space-like separated regions that are at least a distance d apart from eachother have a tensor-product structure.

    Therefore, the problem is worse than I thought in #47 and #24; there I’m just complaining about essentially infinitely dimensional systems. Correlations in C_qc must come from algebras that fail this basic locality condition, and there isn’t even a non-pathological model where it fails! This is like somebody violating the Tsirelson bound because your spacetime has CTCs. Well, duh, but violating the Tsirelson bound is the least of your problems.

    Until someone comes up with a physical QFT that fails the (distal) split property we should assume that the set of physically possible correlations is C_qa, not C_qc.

  55. Mateus Araújo Says:

    Fritz #49: I’m sorry, I’m afraid your argument ignores the actual history of Bell experiments. The thinking at the time the earlier experiments were done (as recounted by Aspect and Clauser in several conferences) was that entanglement might as well be just an idealisation, and that in reality it might collapse when we tried to spatially separate two entangled particles.

    How could you test that without measuring spatially separated observables? And how could you tell which basis the entangled state had collapsed to, without making measurements in different bases? It’s not as if there was any good model of this “entanglement collapses with distance” idea. At the very least one would need to measure an entanglement witness to be sure no separable state could have given rise to those results.

    Couple that with the hope that quantum mechanics could be replaced with a more fundamental deterministic theory (which was still taken seriously at the time), and you have a very strong justification for performing a Bell test.

  56. Thomas Vidick Says:

    Boaz #48, regarding “on a different note”: Yes, it is the case that if the TM M halts in T steps, a perfect winning strategy is computable in time poly(T).
    This said, if you do want the verifier to accept for runtimes at least T, and not below, then you need the verifier to be able to compute T. So I think its size will not be constant; it will be related to the size of a TM that computes T (i.e. the Kolmogorov complexity of T).
    Note our result doesn’t say anything about “runtime” of the provers. It would say that they need to have at least ~T EPR pairs, and make some measurement on all of them.

  57. William Slofstra Says:

    Another fun result to throw into the ring when thinking about speculative laws of nature is Coudron and Vidick’s result that if the provers’ algebras are only approximately commuting, rather than exactly commuting (more precisely, the commutator of Alice’s and Bob’s observables is bounded by some constant delta which decreases with the instance size), then the resulting complexity class MIP*_delta has a deterministic time upper bound which is doubly exponential in delta. So if we make delta small enough that it’s beyond what we could plausible observe, there would still be a separation between MIP*_delta and MIP*.

    Mateus @ 43: The point is that for the RE-hard family of games given in the paper, we are promised that the tensor-product value is either equal to 1, at most 1/2. So if we had a sequence converging to the tensor-product value from above, where (like the NPA hierarchy) the terms of the sequence are computable, and we compute the terms of this sequence while simultaneously computing the lower bounds given by the brute-force method, then eventually we will find either an upper bound below 1, or a lower bound > 1/2, giving us an algorithm for the halting problem. So no such computable sequence can exist.

    However, since we don’t know yet whether MIP^co = coRE, your question is still a good one. You just have to turn it around and make it about lower bounds on the commuting-operator value.

  58. Tobias Fritz Says:

    Mateus #55: I can’t resist one further quick reply. You clearly know the history better than I do, but you’re attacking a straw man; I never claimed that we should not measure spatially separated observables. My claim is that doing that within the framewok of a Bell experiment is not very interesting. Indeed, we’ll get more comprehensive and relevant results if we do a complete tomography of the putative entangled state, or even better we just measure all observables that are experimentally accessible (regardless of whether they’re themselves separable or entangled) and check whether these are compatible with our model. Of course we can then compute the value of a Bell inequality from that, but that’s just a mere byproduct of having probed the limits of our model. And it’s more informative at that, because the model may still be wrong even if the Bell inequality value conforms with our expectations.

  59. Sniffnoy Says:

    Mateus #46:

    Ah, thanks, that clarifies things! I didn’t realize that C_qa was also the closure of C_q. So we just have C_c (or the empty set?) generating MIP = NEXP; C_q, C_qs, or C_qa all generating MIP* = RE; C_qs generating MIP^co, speculated to equal coRE; C_ns generating MIP_ns = EXP; and all possible correlations generating IP = EXP. Yay!

  60. Mateus Araújo Says:

    Slofstra #57: I don’t see the argument. The lowerbounding algorithm might stay below 1/2 arbitrarily long, and the upperbounding algorithm might stay at 1 arbitrarily long. What difference does it make having both? We still cannot have any upper bound on the running time of this “algorithm”.

  61. matt Says:

    Scott #43: you write “And so, for example, if you hand the verifier an n-bit program that runs for T steps before halting, then just by running a poly(n)-time protocol, the verifier can check that the provers must have shared at least T entangled qubits. (Or even a literally infinite number of qubits, if we’re willing to go outside the world of tensor-product Hilbert spaces.) Insane.” Suppose that the result in the paper had only been that “for any finite number T (i.e as large as BB(n) ), the verifier can check that the provers shared at least T entangled qubits”, what would have been the consequences? Would that still disprove Connes embedding? Presumably it would still imply uncomputability of the value of the game?

  62. Scott Says:

    matt #61: Interesting question! I know the Connes Embedding Conjecture implied that MIP* has a computable upper bound. So once you can solve the halting problem in MIP*, you’ve refuted Connes Embedding, full stop. On the other hand, there might be some work that needs to be done to get from “verifying that the provers share at least BB(n) entangled qubits” to “making them solve the halting problem for you.”

  63. Vitor Says:

    Gentzen #44 and Scott #45:

    It would take solving EXP-complete problems (at least!) to convince me that a prover is an actual divine being. Otherwise, my priors would massively favor the “mundane” explanation involving a secret proof of P = PSPACE by the NSA or whoever.

    Furthermore, there’s a technicality with this MIP* protocol that makes me think it can’t prove godhood. Let’s assume 2 beings claim they’re omniscient. You ask them to run the protocol on a hard problem, and they successfully do. This shows that they either have arbitrarily much computation power (and entanglement) available, or that they read the content of each other’s minds and can thus cheat on the protocol. Let’s say that you’re able to prevent any mundane ways of cheating, by only using your finite resources.

    In conclusion, they can convince you that they have at least some weird nonlocal mindreading available, but not actual omniscience. Note that I’m taking as a given that the MIP* protocol breaks down completely if its assumptions are violated in any way (I hope that’s reasonable, please correct me if I’m wrong).

    Followup question: is it possible to build a protocol that degrades gracefully when the communication assumptions between the provers only hold approximately?

  64. Scott Says:

    Vitor #63: Yes, concluding the presence of the divine from empirical observations is a notoriously tricky business. 😀

    If some magic box seemed quickly to solve arbitrary instances of NP-complete problems before my eyes—to say nothing of EXP-complete problems or the halting problem!—then in practice, I’d probably want to start out with most of my probability mass on all sorts of mundane explanations, like ordinary stage magic (e.g., swapping out the instance I chose for a different, easier instance without my noticing), or even a drug-fueled hallucination.

  65. William Slofstra Says:

    Greg @53: I’d also like to know the answer to that question. As far as I can tell, the current paper doesn’t immediately provide an example of a non-hyperlinear group. The existence of such a group is equivalent to the existence of a linear-system nonlocal game with commuting operator value 1 and tensor-product value less than 1. So a natural thing to try would be to adapt the techniques of this paper so that all the games involved are linear system games. While some parts of the construction can be done with linear system games, other parts (for instance, the conditionally linear probability distributions) seem more difficult, at least at the moment. Maybe someone will figure out a workaround for these parts?

    On the other hand, I think it’s pretty likely that the construction in this paper could be done with binary constraint system games. This would give a binary constraint system which is satisfiable in a II1 factor, but not approximately satisfiable in finite-dimensions. This raises the question of which families of binary constraint systems would admit an instance with this type of satisfiability gap. Atserias, Kolaitis, and Severini have a paper about operator satisfiability problems for binary constraint systems, where they observe that LIN, the class of linear constraint systems, reduces to almost all other families of binary constraint systems. Showing that LIN has such an instance should answer this question for all other families. So this gives another reason to look for a non-hyperlinear group, beyond natural curiosity about group algebras.

  66. Halting Is Poly-Time Quantum Provable | Gödel's Lost Letter and P=NP Says:

    […] developments have been covered by Scott Aaronson here, Lance here, Boaz Barak here, and in a personal way by Vidick here. The new paper weights in at 165 […]

  67. . Says:

    So does it mean there is nothing beyond quantum models?

  68. Scott Says:

    . #67: No. It means that there’s one particular quantum complexity class that captures RE, but most quantum complexity classes fall far short of RE, and in any case RE is not everything.

  69. domotorp Says:

    Scott #17: I meant something more interesting, for example, suppose that the two Provers share an infinitely long sequence of random bits that are hidden from the Verifier. Would this give PSPACE, NEXP or RE?

  70. domotorp Says:

    OK, I’ve found that this latter version would be the same as NEXP.

  71. William Slofstra Says:

    Mateus @60: It doesn’t matter how long it takes; since the two series converge to the same value, this procedure will eventually halt for every game, implying that the promise problem is decidable. (This is exactly why Connes being true would have implied that MIP* is decidable.)

    matt @61 and Scott @62: Yes, if Connes was true then the procedure I’ve been talking about with Mateus (finding lower bounds from below using the brute-force method, and upper bounds from above using the NPA hierarchy) would halt on every game, not just games satisfying the promise that the value is equal to 1 or less than 1/2. Since the set of games is computable, this would imply that the running time of this algorithm is some computable function. So to falsify Connes, it would be enough to show that all decidable languages are contained in MIP*. (You can also make the same argument with the dimension of the Hilbert spaces required in place of running time.)

    An interesting complexity class is the refinement of MIP* in which the proof systems have to be complete in the tensor product model, but sound in the commuting-operator model (so for yes instances, the provers have to be able to win with probability greater than the completeness threshold c in the tensor-product model, and not be able to win with probability greater than the soundness threshold s in the commuting-operator model). This class contains only decidable languages. Similarly to the idea that MIP^co could equal coRE, it seems plausible to conjecture that this new class contains all decidable languages. (I think it would be pretty neat if this is true.)

  72. Mateus Araújo Says:

    Fritz #58: You are still being anachronistic. The first Bell test was performed by Freedman and Clauser in 1972. As far as I know quantum state tomography was introduced in 1989 by Vogel and Risken.

    You have to take into account that at the time quantum theory was far from being well-developed and trusted; in fact a lot of our trust in quantum theory comes from the dramatic certification of entanglement in the Bell experiments. Of course doing state tomography, as you suggest, would give us much more information about what is wrong with the theory (if that were the case), but to get to this point first people needed to determine whether it was worth it developing the theory so much.

    Even if we already had quantum state tomography back then, I still think performing a Bell test explicitly would have been a good idea, for pedagogical reasons. Without it, experts would know that local hidden-variable models were dead, but with the experiment the message would percolate to the wider community.

  73. Mateus Araújo Says:

    Slofstra #71: I see. If one assumes that it does not halt, it would mean that for all n,d we would have that tensorNPA(n) > Tarksi(d) + \epsilon, contradicting the assumption that both sequences converge to the same value, so it must halt.

    Shit, there’s really no hope for the Tsirelson bound then. If converging sequences of computable numbers are not possible, what’s the best we can do about upperbounding it without violating undecidability?

  74. David E Speyer Says:

    Are there any expositions out there of the Connes conjecture suitable for reading by a naive algebraist who doesn’t know what an operator algebra is? From this thread, I gather that involves some sort of inequalities analogous to the CHSH inequality which are satisfied by commuting measurements on a finite dimensional Hilbert space, but not an infinite dimensional one.

    That seems amazing to think of, but maybe it isn’t more surprising than the fact that the CHSH inequality exists in the first place. Is there any hope of writing down an actual inequality of this sort?

  75. Scott Says:

    David Speyer #74: Try Thomas Vidick’s recent survey for the AMS Notices. Come to think of it, I’ll also add a link from the main post.

  76. Henry Yuen Says:

    David Speyer #74: In Section 12.4 the paper does provide a construction of an explicit nonlocal game whose commuting operator value is 1, while the tensor product value (i.e. what is approximable by finite dimensional strategies) is at most 1/2. This nonlocal game can be written as an inequality in the same way the CHSH game corresponds to an inequality.

    However, while “explicit”, the nonlocal game encodes a Turing machine that runs a hierarchy of semidefinite programs, so the description of such an inequality is somewhat cumbersome! There is probably a much simpler nonlocal game that separates commuting operator model from tensor product model…

  77. David E Speyer Says:

    Thanks! That is very helpful. I remember starting that article in the Notices and stopping half way through but, with the motivation of the new result, I couldn’t put it down.

    I still find it incredibly counterintuitive. I’ll see if I can think of an actual argument for why I feel that way.

  78. gentzen Says:

    That survey for the AMS Notices is really nice. The explanations for MIP = NEXP and the PCP theorem were very intuitive and helpful for me. I only read the first 6 pages for far, but I am confident that the remaining pages will be intuitive and helpful as well.

    I wonder a bit where the references to “godlike powers” and “an actual divine being” in response to my comment come from. It is true that I did read Henry Yuen’s story on the visit of Alice and Bob, and that his comparison of Alice and Bob to “computational gods” made sense to me. It probably triggered the association with Joel David Hamkins’ question. But when answering his question, one must admit that even godlike powers are not enough.

    I still try to come to terms with MIP* = RE. Here is my next attempt: If the given computation halts, and the two provers have done that computation and therefore know that it halts, then they can convince a polynomial classical verifier that they have done that computation, and that it did halt. (And if that computation had a meaningful results in addition to merely halting, then they also can tell the verifier that result and convince him that it is the actual result.)

    It’s still “sort of unbelievable”, but maybe not as unbelievable as the Banach-Tarski paradox.

  79. HL Says:

    This reminds me of Marcus-Spielman-Srivastava solving the Kadison-Singer problem by working on a seemingly unrelated problem. These kind of moments are a true inspiration to people pursuing new branches far from the “trunk” of mathematics. Without connection to the Connes embedding problem, mathematical establishment would probably ignore this result no matter how important it is within the field.

    By the way, Gilles Pisier has been writing a book on Connes problem, so there might be connections to various other classical questions in functional analysis: https://www.math.tamu.edu/~pisier/TPCOS.pdf

  80. David E Speyer Says:

    So, if Loki and Star Trek’s Q wanted to convince me as to the value of the middle digit of Ackermann(100), and I could lock them in two boxes which prevented them from communicating with each other, and they had infinitely many entangled particles in their boxes, they could succeed. Do I have this right?

  81. Scott Says:

    David #80: Yes, you have it right. And they wouldn’t even need infinitely many entangled qubits for that purpose, but just a huge (comparable to Ackermann(100)) finite number. With the proviso—a big one—that they would also need to be able to apply the requisite unitary transformations to those qubits. And that (of course) your belief about the middle digit would remain conditional on your beliefs in no-signaling and in the validity of quantum mechanics.

  82. David E Speyer Says:

    I’m trying to think through why this seems so crazy. One could imagine studying:

    (Classical) Provers who have planned a classical, perhaps probabilistic, strategy in advance, but have no entangled particles.

    (FT) Entanglement of particles modeled as the tensor product of finite dimensional vector spaces

    (FC) Entanglement of particles modeled as commuting operator on finite dimensional vector spaces

    (IT) Entanglement of particles modeled as the tensor product of infinite dimensional vector spaces

    (IC) Entanglement of particles modeled as commuting operators on infinite dimensional vector spaces

    Bell’s theorem is that Classical < FT. Since I don’t have rocks in my head, it bothers me. It is nonobvious but apparently not deep that FT=FC=IT. And the new result is that IT < IC.

    This bothers me in two ways. One is an algebraic intuition about bimodules that seems like I should be able to show IT=IC, but that’s probably too technical to be worth hashing out in a comment thread.

    But the other is philosophical — it seems crazy that provers should be able to demonstrate they command literally infinite resources as opposed to a ridiculous amount of resources. Whatever Hilbert space they are working with presumably has a countable basis. How could they ever convince me they didn’t just access enough basis elements to answer the question I wanted to ask?

  83. Scott Says:

    David #82:

      It is nonobvious but apparently not deep that FT=FC=IT.

    Not quite—what’s true is that anything in IT is expressible as a limit of things in FT=FC.

      it seems crazy that provers should be able to demonstrate they command literally infinite resources as opposed to a ridiculous amount of resources. Whatever Hilbert space they are working with presumably has a countable basis. How could they ever convince me they didn’t just access enough basis elements to answer the question I wanted to ask?

    I’m with you that this is crazy. But certainly a key part of the answer is that, while the Hilbert space relevant to the commuting-operator model does have a countable basis, it’s not decomposable into a tensor product of “Alice’s part” and “Bob’s part.” So our intuitions about what it even means for Alice and Bob to act on the Hilbert space in a “spatially separated manner” are liable to mislead us, if those intuitions are ported over from the finite-dimensional world, where commutativity simply means having a tensor product decomposition. Or at least that’s how I’ve bracketed off my own confusion about this. 🙂

  84. RandomOracle Says:

    Does the proof of MIP*=RE (or even the previous one of NEEXP ⊆ MIP*) relativize? I suspect not, but is there some easy way to see this?

  85. Rand Says:

    When Anand visited QuICS a few months ago (to talk about his big NEEXP < MIP* result), he suggested to me that this might be true and my mind was boggled.

    The bogglement persists.

  86. Joshua B Zelinsky Says:

    This may be a silly question: Given that we’re talking about infinite amounts of entanglement being used in the MIP* = RE result, has anyone thought carefully about how dependent this result is on one’s axioms? The techniques used seem to all live in ZFC, but I’d be particularly interested if the result depends on the axiom of choice, especially in regards to the behavior of the relevant Hilbert space.

  87. Scott Says:

    RandomOracle #84: That’s a superb question—I don’t know why I didn’t think to ask it myself.

    The issue is that we have two heuristics locked in direct conflict. On the one hand, nothing about interactive proofs relativizes, and MIP*=RE is about interactive proofs. On the other hand, everything in computability theory relativizes, and MIP*=RE is (ultimately) about computability theory.

    In this case, I can prove that (as you guessed) the first heuristic has to win. That is, there exists an oracle relative to which not even coNP, let alone RE, is contained in MIP*. To see this, let’s consider the coNP problem of deciding whether f:{0,1}n→{0,1} is the all-zeroes function. Suppose there’s something that the two entangled provers can do to cause the BQP verifier to accept with high probability. Now change f(x) from 0 to 1 at a single randomly-chosen x∈{0,1}n, and let the provers run exactly the same strategy as before. By the BBBV Theorem (i.e., the optimality of Grover’s algorithm), the verifier won’t “notice” the change, and will still accept with high probability. Notice that the amount of entanglement between the provers, and even the number of provers, were completely irrelevant here.

    Of course, this still leaves the challenge of articulating which step(s) in the proof of MIP*=RE fail to relativize. The intuitive difficulty is that, if we’re talking about Turing machines that run for arbitrary lengths of time, and which therefore (it seems) can only be simulated in a step-by-step manner anyway, what difference could it possibly make if those Turing machines also query an oracle?

    The answer, I think, is that despite that intuition, nonrelativizing stuff happens (and needs to happen) in the “compression lemma” that’s at the core of the new result. Indeed, the Time Hierarchy Theorem relativizes, while the compression lemma says informally that in the land of MIP*, the Time Hierarchy Theorem is as nothing … since at no cost to yourself, you can always just force the provers to go up another exponential in how many entangled qubits they use.

    And indeed, the compression lemma is exactly where the authors bring in PCP and low-degree polynomial machinery—well-known sources of non-relativizing goodness.

    If I’m right, then MIP*=RE is the first fully convincing example I’ve ever seen of a non-relativizing computability theorem.

    (Update: In response to my inquiry, Thomas Vidick was kind enough to confirm the above understanding—adding that their proof uses NEXP⊆MIP* as an ingredient, which already fails to relativize—although that’s almost certainly not the only non-relativizing part of the proof.)

  88. Sniffnoy Says:

    Josh #86: I mean, we’re only dealing here with separable Hilbert spaces, right? So such issues shouldn’t be relevant…

  89. Boaz Barak Says:

    Scott #87: By your argument above, also the prior result that MIP* contains NEEXP doesn’t relativize, right?

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  91. Scott Says:

    Boaz #89: Right.

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  93. Simon Apers Says:

    What was the common belief regarding Connes’ embedding conjecture? Ken Regan’s blog post states that it was “apparently widely believed”, but I’ve also heard otherwise.

  94. Amazing: Zhengfeng Ji, Anand Natarajan, Thomas Vidick, John Wright, and Henry Yuen proved that MIP* = RE and thus disproved Connes 1976 Embedding Conjecture, and provided a negative answer to Tsirelson’s problem. | Combinatorics and more Says:

    […] Optimized: MIP*=RE; Other posts on blogs: GLL (Ken Regan); WIT (Boaz Barak); CC (Lance Fortnow); WN; Posts on an […]

  95. . Says:

    Does it show two entangled provers can convince a verifier in polynomial time that a mathematical proof of a statement in Sigma^0_1 exists without revealing the proof?

  96. . Says:

    Also does it show MIP* is not closed under complement?

  97. Scott Says:

    .: Yes and yes. Those are both immediate implications of being equal to RE.

  98. Tobias Fritz Says:

    David #82, Scott #83: Perhaps it’s instructive to have an explicit example of two commuting algebras of operators on an infinite-dimensional Hilbert space which cannot be written as a (direct sum or direct integral of) tensor product of representations of the individual algebras. The following is based on Example B.13 here, which is due to Takesaki. I will use the fact that the spectrum of an entangled operator acting on a tensor product Hilbert space is independent of the particular representations of the two operator algebras under consideration (as long as the representations are faithful).

    As a prelude, let G be a finitely generated non-amenable group. For example, a non-abelian free group of finite rank will do just fine. Then consider the random walk on the Cayley graph of G with respect to a given finite set of generators. A result of Kesten says that the spectrum of this random walk does not contain 1. This is intuitive: the Cayley graph branches out too much to have almost invariant distributions.

    Now consider G acting on itself both on the left and on the right. This defines a unitary representation of G x G on ℓ²(G), or equivalently two commuting representations of the reduced group C*-algebra C*_r(G). Now let S be a symmetric generating set in G, and consider the entangled operator given by the Laplace operator

    L = |S|¯¹ Σ_g g ⊗ g¯¹,

    where the sum is over generators g ∈ S. This hermitian operator has the basis vector δ₁ ∈ ℓ²(G) as an eigenvector of eigenvalue one, where 1 ∈ G is the neutral element.

    Now let us consider L in any tensor product representation, and show that the spectrum of L is bounded away from one. By my statement in the first paragraph above, it is enough to consider the tensor product of the two representation where G acts on itself from the left with the one where it acts on itself from the right. Using the isomorphism g ↦ g¯¹ to turn the second representation into a left action of G on itself as well, we essentially get the Laplace operator associated to the action of G on G x G by left multiplication. Since this decomposes into disjoint orbits given by G acting on merely one copy of itself, the above statement about random walks on the Cayley graph applies, showing that the spectrum of L is bounded away from one.

    ———

    We could also try to use the above algebras of commuting operators to construct a counterexample to Tsirelson’s problem. However, it’s far from obvious how to go about that; for example, the Kirchberg factorization property implies that this does not work as soon as G is residually finite, a property that is satisfied e.g. by free groups. This illustrates the difficulty of the problem.

  99. asdf Says:

    This is also an interesting bypass of Gödel’s speed-up theorem.

  100. Amazing: Zhengfeng Ji, Anand Natarajan, Thomas Vidick, John Wright, and Henry Yuen proved that MIP* = RE and thus disproved Connes 1976 Embedding Conjecture, and provided a negative answer to Tsirelson’s problem. | Combinatorics and more Says:

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  101. Daniel Says:

    This sounds pretty cool. However, I would like to know if it has any significant implications to the foundational questions?

  102. Rel Says:

    (2) There is a two-prover game, analogous to the Bell/CHSH game, for which Alice and Bob can do markedly better with a literally infinite amount of entanglement than they can with any finite amount of entanglement.

    (4) There are types of correlations between Alice and Bob that can be produced using infinite entanglement, but that can’t even be approximated using any finite amount of entanglement.

    It seems like (4) follows from (2) and wasn’t (2) was already known in 1711.10676 (Sloftra and Vidick). Or is there a deeper connection that I did not catch?

  103. Scott Says:

    Daniel #101: Not, like, for Many-Worlds vs Copenhagen or anything like that. But it has huge implications for the foundational question of how we could ever know, in principle, if we were living in an infinite-dimensional Hilbert space—see the discussion above for more.

  104. Scott Says:

    Rel #102: The key was my adjective “markedly.” 🙂 It was already known, from Slofstra’s 2014 breakthrough, that there are correlations that you can achieve with infinite entanglement, but can’t perfectly achieve with any finite amount of it. The new result shows, much more strongly, that there are correlations that you can achieve with infinite entanglement (in the commuting operators model), but can’t even approximate with any finite amount.

  105. Sebastian Oberhoff Says:

    Having thought about this result for a while now I believe there are also a number of epistemological issues (never thought I’d use that term) that I think ought to be addressed more directly.

    When I first heard about this result the image I had in my mind was something like the following: One fine day two strangers, Alice and Bob, introduce themselves to us, claiming that through extensive meditation they’ve achieved full enlightenment and are now capable of instantaneously determining the truth of any statement whatsoever. “Cool”, we respond and lead them to separate rooms in each of which there’s a little apparatus which upon pressing a button spits out half of an EPR-pair taken from an unlimited supply. (The other half would, upon request, appear in the other room after the same number of button presses.)

    Now, if Alice and Bob did indeed have the powers that they claimed, they ought to be able to reveal utterly amazing things to us. For instance, assuming that the Riemann Hypothesis is provable in ZFC, they could prove that this is so, even if the proof was too large to write down in the observable universe, simply by (quickly!) demonstrating that the program which searches for the proof halts.

    Is it really that simple though? Recall that the protocol crucially hinges on our belief that Alice and Bob can’t communicate. And the justification for that was presumably that we have an axiomatic belief in the truth of quantum mechanics. However, Alice and Bob have just given us very strong evidence that quantum mechanics doesn’t apply to them. If it did, they would be bound by the Church-Turing thesis, which they apparently aren’t. On the other hand, if we’re simply going to take their word for it that they’re not cheating with telepathy, then we might as well also take their word for it that the Riemann Hypothesis is provable and be done.

    And it gets even stranger. As Scott already alluded to, Alice and Bob need a LOT of entanglement for this protocol. In fact the number of qubits that their protocol consumes as a function of time must provably grow faster than ANY computable function (though, as I understand, it’s not infinite for any particular input as Scott seems to think). Alice and Bob have to really mash that button.

    There’s of course nothing wrong with this at the mathematical level. But I still feel like somewhere along the line here the concept of “prover” has been silently promoted from being merely omniscient to now also being effectively omnipotent. Whereas a quantum computer built by mortals has to laboriously dictate “Hadamard gate here please, next CNOT on these qubits, …” Alice and Bob instead have an infinite bandwidth channel between their minds and the environment.

    This then raises one more scary question: Given that Alice and Bob have just showcased the ability to operate on unimaginably many qubits in an instant, how can we still rule out the possibility that they just unitarily operated on us and that all our memories of taking part in this protocol are actually planted? It seems like our encounter with Alice and Bob, far from giving us new knowledge, actually undermined the whole concept.

    PS: Just to be clear, this is in no way a criticism of the brilliant mathematics in the paper. It’s just something that I think people should be clear about when discussing the result using the “prover” and “verifier” antropomorphisms.

  106. Lauri Love Says:

    Riddle me this…

    what is the kolmogorov complexity of a transcript between a polytime verifier and two MIP* provers establishing a case of undecidable problem?

    what kind of compression is this? something extra (‘magic’) happens between the provers, infinitely so in the CEC-refuting case, and potentially unbounded entanglement resources are consumed, but the veracity is still contained quite finitely within the transcript, the magic non-local coordination being merely catalytic…

    nevertheless the transcript proves the having-happenedness of a infinite-dimensional passage to commutativity. so what would this very mundane and humdrum record of transcendence look like, and how could it not be simulated by enumeration?

    amongst all strings up to some length, what proportion of them are interpretable as MIP* proof transcripts of the inclusion or exclusion of a case of an undecidable problem in a language, necessitating an infinite dimensional Hilbert space to compute?

    (apologies for anything phrased incorrectly. i am lay, lax and lackadaisical)

    -lauri

  107. David E Speyer Says:

    Tobias Fritz #98 thanks! I had already came up with what I think is an example of a group representation which can’t be written as a direct sum of finite dimensional representations: Use the same L^2(G) construction with the Higman group, which has no nontrivial finite dimensional representations at all. I hadn’t realized that L^2 of the free group would have issues, but your explanation makes a lot of sense.

    I tried to imagine a corresponding communication game and came up with “Alice and Bob each receive a word in group G. They must return the same bit if their words have the same product and different bits otherwise.” Given an infinite entangled state encoding L^2(G), I think they can succeed with probability 1 on the same product and 1/2 on different products.

    But it seemed to me that they could cheat without using entanglement at all: Use classical computation to replace the word they are given with a word in a standard form, so that they both have the same word if and only if they have the same product. Then compute the same 1 bit hash of that word and return the hash.

    But maybe I am somewhere in the right neighborhood anyway.

  108. fred Says:

    Probably a dumb question, but when we say “a literally infinite amount of entanglement”, what sort of infinity are we talking about?

    The idea of a computation relying on infinite resources can only work in a universe that’s itself infinite, so only a fraction of the universe resources are used?
    But then, an infinite amount of computational resources means that the computational time is unbounded (the more bits, the more time it takes to “reach” them), so I’m not clear at all how any meaningful conclusion about computational complexity can even be derived from this.

  109. fred Says:

    Scott #104

    “The new result shows, much more strongly, that there are correlations that you can achieve with infinite entanglement (in the commuting operators model), but can’t even approximate with any finite amount.”

    Anything involving infinities can be very counter-intuitive.
    E.g. the Riemann rearrangement theorem (series converging to an arbitrary number), or in the topology of fractals (volume of a Menger sponge, related to the Cantor set).

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  111. Timothy Chow Says:

    Scott, does MIP* = RE algebrize?

  112. Scott Says:

    fred #108:

      Probably a dumb question, but when we say “a literally infinite amount of entanglement”, what sort of infinity are we talking about?

    There are two different notions in play here. For MIP*, one can actually assume without loss of generality that the provers share only a finite amount of entanglement—since any success probability that they could achieve with infinite entanglement, is simply a limit of success probabilities that they can achieve with larger and larger finite amounts.

    But then there’s this other complexity class, MIP*co (the co standing for “commuting operator”), where you do allow the provers to share an arbitrary infinite-dimensional Hilbert space, subject only to the constraint that their operations commute with each other (which is a way to formalize the requirement that “the provers can’t communicate”).

    Now, a huge implication of the new result is to separate MIP* from MIP*co—showing that there are cases where two commuting provers can achieve a success probability, in getting the verifier to accept, that’s strictly greater than their success probabilities with larger and larger finite amounts of entanglement.

    Thus, if you like, “the two different kinds of infinity” in play here—the limit kind and the commuting-operator kind—are different. This is what solves Tsirelson’s problem as well as Connes’ embedding conjecture.

    Neither MIP* nor MIP*co is “physically realistic,” but both of them are perfectly well-defined mathematically. In both cases, we have a polynomial-time verifier, and then take the supremum over the whole infinity of possible prover strategies that satisfy the appropriate constraints (which are different for the two classes). It’s a well-defined supremum.

  113. Scott Says:

    Timothy #111:

      Scott, does MIP* = RE algebrize?

    That’s a superb question! I don’t know.

    The closest related question for which we know the answer is, “does MIP=NEXP algebrize?” There the answer is “it’s complicated and depends on your exact definition of algebrization—but for a suitable definition, yes, it does” (for details, see my and Avi’s paper, as well as the followup papers by Impagliazzo-Kabanets-Kolokolova and Aydindioglu-Bach). Let me conjecture that the situation for MIP*=RE might be similar. Does anyone else have insight to add?

  114. . Says:

    So two Gods can convince you of a proof that you cannot know? Divine intervention? Theological mathematics as a formal branch of respectable study?

  115. . Says:

    Does it show things beyond realizability (such as presence of supernatural) exists in a concrete sense?

  116. Mateus Araújo Says:

    I’ve been thinking about the intuition that allowing the provers to have stronger and stronger correlations should make the power of the complexity class weaker and weaker, as it becomes progressively easier for the provers to cheat. It is very persuasive. We have that \(C_c \subset C_q \subset C_{ns} \subset C_s \) so we would expect that the corresponding complexity classes would obey \( \text{MIP} \supset \text{MIP}^* \supset \text{MIP}^{ns} \supset \text{IP} \), and that does hold apart from MIP*, that blows everything up. I’d wager, though, that this intuition is actually correct (and would be delighted if a computer scientist with too much time in their hands would bother to prove), if the correlations are all we are using.

    You see, unlike the other complexity classes, MIP* doesn’t just use the quantum mechanical correlations, but also assumes that the provers are bound by the laws of quantum mechanics. This is what makes it much harder for them to cheat. For instance, we know that nonlocal games have a wonderful property called self-testing, which means that if (e.g.) they achieve the value \((2+\sqrt{2})/4\) in the CHSH game, the provers must have made measurements on the X and Z basis in a 2-qubit maximally entangled state (modulo some uninteresting isometry). And indeed, self-testing is an essential ingredient of the proof by Zhengfeng et al. that MIP* = RE, and also in the proof by Natarajan and Wright before them that MIP* is more powerful than MIP.

  117. Scott Says:

    . #114-115:

      So two Gods can convince you of a proof that you cannot know?

    Two entangled Gods, yes.

      Divine intervention?

    No, divine computation followed by ordinary polynomial-time intervention.

      Does it show things beyond realizability (such as presence of supernatural) exists in a concrete sense?

    No.

      Theological mathematics as a formal branch of respectable study?

    Yes.

    Sorry, but further questions at this level will go unanswered.

  118. fred Says:

    . #114

    “So two Gods can convince you of a proof that you cannot know? Divine intervention? Theological mathematics as a formal branch of respectable study?”

    But isn’t it true of any “zero knowledge” proofs?
    https://en.wikipedia.org/wiki/Zero-knowledge_proof

    “one party (the prover) can prove to another party (the verifier) that they know a value x, without conveying any information apart from the fact that they know the value x. “

  119. fred Says:

    A lot of implicit assumptions are being thrown around, which aren’t all that obvious to me.
    Not as to whether it’s valid to assume infinite resources, but more whether it’s ever valid to sweep causality under the rug.
    And the idea of a perfect randomness generator is also questionable (without invoking the magic of QM).

    Shouldn’t we consider that the actors (provers, verifiers) are part of the same closed system, which itself is setup by an external experimenter. For example the system is running as a digital simulation on a computer (a sort of game of life), the provers and verifiers being patterns of bits in the memory, following some evolution rules.
    Then “pure” randomness can be supplied at any time from the outside of the simulation, in a way that’s uncorrelated with the state of the simulation (that’s not strictly possible since the simulation + external system is itself a deterministic system where everything is correlated… hence invoking pure randomness is really god-like magic).
    In this light it’s clear that provers and verifiers have no real power to make any choice (choice is always an illusion), it all depends on the initial conditions setup by the external experimenter (who himself has no power of choice).
    Maybe randomness can be achieved (from the external experimenter point of view) by implementing the whole simulation as a many world simulation, so whenever the verfier relies on a random number, the experimenter just splits the simulation, each fork seeing a specific dice roll value.

    Then we say the provers have infinite computing resources.
    That makes it clear that the simulation system itself is no longer finite.
    With or without infinite resources, the provers can’t themselves simulate the verifier (a prover can’t simulate itself without triggering an infinite recusion).
    So provers can never have “god-like” powers (at least not matching the power of the external experimenter).

    So answering any sort of question about what the provers/verifiers can or can’t do is really asking about the connection between integers describing the state of the simulation and its evolution – for any sort of initial condition the experimenter can setup and for all the possible values of dice rolls that can be injected, will the system ever end up in a certain state (the verifier is convinced).

  120. David E Speyer Says:

    @fred As I understand it, classical zero knowledge proofs exist for problems in PSPACE, meaning problems which could be solved using a polynomial amount of memory, but an arbitrary amount of time. Indeed, if there were a zero knowledge proof protocol for a statement S then there would also be a PSPACE procedure for verifying the truth of S, namely, “check all possible responses the prover could make to my queries, and see if any of them lead to acceptance.” (I think this isn’t quite right, since proof protocols are only required to work with high probability, not certainty. But I think this is the right idea.)

    That’s why I chose computing the middle digit of Ackermann(N) as my example — I don’t see how to do it without computing and simultaneously storing all the digits of Ackermann(N-1), which will require more than polynomial space.

  121. . Says:

    @DavidESpeyer and @Fred and @Scott “No, divine computation followed by ordinary polynomial-time intervention.”. Who else can do infinite computation other than God(s)? Who else can convince a polynomial time verifier of a proof infinite computation? This is convincingly borderline however proper theology. A theology that supports ‘morethanmono’theism.

  122. Sniffnoy Says:

    Surprised nobody’s yet linked to where Scott previously discussed complexity (and in particular interactive proof) results in terms of gods…

  123. Sebastian Oberhoff Says:

    Scott #112: This sounds wrong to me. First, if the provers only share a finite amount of entanglement, then it ought to be possible for the verifier to give them a program which iterates through all possible things the provers might do and then diagonalizes against them, contradicting the equality with RE.
    Second, how does showing that MIP* is really powerful separate it from MIP*co?

  124. Scott Says:

    Sebastian #123: It’s not wrong. I said a finite amount of entanglement. I didn’t say that the amount was less than BB(n). 🙂

    As for your second question, it was known that MIP*⊆RE and MIP*co⊆coRE. So once it was shown that MIP*=RE … need I say more?

  125. . Says:

    Ahhh there is a catch. The amount of entanglement is a function of n? We can dismantle this proof probably.

  126. Scott Says:

    . #125: Obviously the amount of entanglement depends on n. It did so even in the earlier, less shocking MIP* protocols. What the new result shows is that, when you approximately optimize over prover strategies, the amount of entanglement can’t even be upper-bounded by any computable function of n (!!), but only by something like the Busy Beaver function.

    You can’t “dismantle” a mathematical construction by pointing to something that’s infeasible about it from an engineering standpoint, certainly if no one ever claimed the latter.

    You are hereby banned from this thread—not for ignorance and not for arrogance, but precisely for the combination of the two.

  127. Bogdan Says:

    Consider 2 (or more if needed) classical all-powerful provers. Let S be the set of all possible ways we can restrict their communication. So, S contains two extremes (no communication and unrestricted communication) and everything in between. For s in S, let MIP(s) be the class of languages for which there is a corresponding protocol. Let MIP! = max_(s \in S) MIP(s) (or may be better definition MIP! = union_(s \in S) MIP(s)). The definition of MIP! is completely classical and involves nothing quantum. Yet if “the provers were limited to communicating with each other in all and only the ways that perfectly simulated quantum entanglement” (comment 17), then MIP! contains MIP*, and, by the Theorem, MIP! contains RE. Can MIP! be even larger?

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  129. Scott Says:

    Bogdan #127: “All possible ways to restrict the provers’ communication” is not a well-defined concept, any more than “the set of all sets.” You’d need to spell out more precisely what you mean.

    Having said that, if we considered only ways to restrict the provers’ communication where there was still a computable enumeration of all their possible strategies, then RE would still be an upper bound on your class. And since your class would contain MIP*=RE, we’d then have MIP!=RE as well.

  130. fred Says:

    David #120

    Oh, I see, thanks!

    . #121

    But what’s the meaning of “convince” (or “choose”) in a system that’s evolves deterministically (with or without a bit of pure randomness thrown in)?
    We’re only asking whether a digital computer (running the prover and verifier algorithms and states, as a bunch of bits in memory) can start in one given state and reach another state. There’s no place for free will and gods in there (but as I was saying, imo relying on infinites and pure randomness is ill-defined in the context of a digital machine).

  131. fred Says:

    “relying on infinites and pure randomness is ill-defined in the context of a digital machine”

    What I mean by ill-defined is that as soon as we’re saying that a digital machine has access to an unbounded amount of resources (like memory) and we’re assuming there’s an actual effective way to access (read/write) that unlimited pool of resources (not taking an infinite amount of time), like “let me load the entirety of the number PI into my infinite memory”, we can always rearrange the resources (bits and gates) to realize an infinite grid of local ALUs with dedicated finite memory (could even be infinite local memory since any infinity can be sliced into more infinities) that could then solve any NP-hard problem in poly time by just trying all solutions in “parallel”.

  132. Scott Says:

    fred #130-131: Your idea that “relying on infinites and pure randomness is ill-defined in the context of a digital machine” will come as news to the generations of theoretical computer scientists who have worked with perfectly well-defined models of computation involving randomness, infinities, or both.

    The fact that the provers could build multiplexers to solve NP-complete problems in polynomial time is irrelevant. Of course they can solve NP-complete problems in polynomial time—they’re the f-ing provers! The only constraints are on their correlations (they share unlimited entanglement but can’t communicate), and on the verifier’s running time.

    Do you agree that for any language L, there’s a mathematically well-defined question about whether MIP* contains L or not? If you do, then MIP* is a well-defined class, and its equality with RE is a well-defined question. Indeed, it’s even better than that: because all MIP* strategies are limits of finite strategies, MIP*=RE can be phrased as an arithmetical question! Which makes manifest, even before the new paper, that the answer couldn’t possibly have depended on the Axiom of Choice or anything like that.

    I regret that I’m traveling and exhausted and sick, and will not engage a single more “yes, but…” about the well-definedness of MIP*. If you don’t want to accept what someone who knows is unequivocally telling you, then my advice is to read the MIP* papers (starting with the early ones) until it makes sense. Good luck.

  133. Sebastian Oberhoff Says:

    Scott #124: In #112 you characterized MIP* as MIP but with f(n) ebits of entanglement for some arbitrary, yet finite, function f. On the other hand, MIP*co permits a literal infinite amount to be used on a single input. So it would seem that anything MIP* can do, MIP*co can do as well. But then we’d have RE = MIP* ⊆ MIP*co ⊆ coRE. Where’s the catch here?

  134. Scott Says:

    Sebastian #133: A priori, adding more entanglement could make an MIP class stronger or weaker. On the one hand, it gives the provers more power in the honest case; on the other hand, it also gives them more power to cheat. It takes a lot of work to see how these two opposing considerations will interact. In 2012, Ito and Vidick won a FOCS Best Paper Award just for showing that MIP* contains MIP=NEXP.

    So your proposed containment of MIP* in MIP*co just doesn’t follow—and in fact, we know from the new result that it’s false. Yes, it’s counterintuitive that going from finite to infinite (commuting-operator) entanglement takes away the ability to verify that programs halt—though it might turn out to give back, as compensation, the even more impressive ability to verify that programs run forever!

    While the containment of MIP* in RE is immediate, the containment of MIP*co in coRE is far from obvious (at least to me). But it follows from a paper by Navascues, Pironio, and Acin on semidefinite programming hierarchies.

  135. matt Says:

    Is there any hope for an interactive proof system which can prove “the halting problem for Turing machines with an oracle for the halting problem”?

  136. Scott Says:

    matt #135: Interesting question! Clearly, one would need an interactive proof system for which there’s no computable enumeration of the provers’ strategies (which would give an upper bound of RE), and also no computable enumeration of possible obstructions to prover strategies (which would give an upper bound of coRE). So if it exists, it’s presumably a pretty weird-looking creature!

  137. SDP Says:

    Scott, what is a good introductory paper (or papers) on the connection between semidefinite programming hierarchies and computational complexity? I have a background in optimizing SDPs, but barely an undergrad background in complexity. Thanks.

  138. asdf Says:

    Matt #135, I don’t see how that is possible, since you send two queries A,B (both finite strings) and get back two responses X,Y (also finite strings, bounded in length by a polynomial) and a mapping from finite strings to finite strings of bounded size must be computable. That isn’t so bad for the RE case since “yes” RE instances are also computable (i.e. the MIP* system might be able to convince you that Goldbach’s conjecture is false, but not that it is true). But you’re asking for something that can convince you that the Twin Prime Conjecture is false (TPC is a so-called Pi-0-2 sentence and its negation is Sigma-0-2). There can’t(?) be such a computable map in general.

    Maybe there can be an IP system that can convince a verifier with a halting oracle (aka Pi-0-1 oracle) of the falsity of a Pi-0-2 sentence, and so on upward. That is probably easier though I don’t know how to approach it.

    Scott or anyone: any idea how the operator algebra community is reacting to thise? Has the paper been submitted to a journal? The preprint page doesn’t mention anything about that. It does seem analogous to the situation with the Kadison-Singer problem, where it took a while for the “mainstream” to catch up.

  139. Interaction + Entanglement = Efficient Proofs of Halting | Says:

    […] more popular introductions to our result, see the blog posts of Scott Aaronson, Dick Lipton, and Gil Kalai and reporting by Davide Castelvecchi for Nature and Emily Conover for […]

  140. asdf Says:

    Wat?! Ah ok. It looked from the above pingback that Thomas Vidick was blogging on a weird Bitcoin site under the handle “Plato”, but it turns out “Plato” is some kind of aggregator that pulled in Thomas’s post from quantumfrontiers.com. It was confusing enough that I decided to remark on it to save others some possible questioning.

  141. Alan Says:

    Hi Scott,

    I actually came following a Google search of you after watching your 6.045 lectures on Youtube. Just wanted to say they helped quite a bit in my own studies as a 4th yr CS student at the University of Toronto.

    So thank you for that and for allowing plebs like me insight into Davos etc thru your blogs!

  142. gentzen Says:

    gentzen #78: Finished studying the AMS survey and the paper intro:

    That survey for the AMS Notices is really nice. The explanations for MIP = NEXP and the PCP theorem were very intuitive and helpful for me. I only read the first 6 pages for far, but I am confident that the remaining pages will be intuitive and helpful as well.

    The remaining 4 pages were helpful as well, but harder going. Nerd sniping like:

    … there exist nine 4×4 Hermitian matrices X_1 ,…,X_9 that each square to identity and such that moreover the three matrices in any row (resp., column) (i) commute and (ii) multiply to +Id (resp., −Id). (The solution is not too hard to find; as a hint, it suffices to consider matrices with coefficients in {0,±1,±j}.)

    slowed me down significantly, and triggered me to look into papers from the References section. And also triggered me to try my luck again with the paper intro. (Actually an interesting strategy for a survey paper, to try to trigger the reader to dig deeper into the references. I don’t remember anything similar from other survey papers, or AMS notices.)

    The first page of that intro is hard going, but then the intro suddenly becomes very readable, intuitive and helpful starting with section “1.1 Interactive proof systems”. It covers some of the same material as the AMS survey, but in a different complementary way. Especially the following explanation of equation (5) in the paragraph on “Nonlocal games” helped me understanding the role of 𝜓 in the quantum versions of the game:

    This definition captures the intuitive notion that a classical strategy for the players is specified by (i) a distribution ν on Λ that represents some probabilistic information shared by the players that is independent of the verifier’s questions, and (ii) two functions A^λ , B^λ that represent each players’ “local strategy” for answering given their shared randomness λ and question x or y.

    Before I had struggled with the hint: “By taking direct sums of PVMs and scaled vectors it is not hard to see that both sets are convex.” Now I got it that for two elements from the set, you could take the direct sum to get a larger Hibert space where both elements are represented by the same operators, but different states normalized 𝜓_1, 𝜓_2 orthogonal to each other (and even orthogonal to the images of the other state by the operators). Then the state sqrt(p_1)𝜓_1 + sqrt(p_2)𝜓_2 for p_1 + p_2 = 1 will represent the convex combination (with weights p_1 and p_2) of the two elements.

    Why do I write such details? Any reader of the AMS survey or the paper intro will probably either have understood that hint directly, or could it have easily worked out for himself, or maybe did not really cared too much about such details anyway. My point for going into such details is more to make it clear what those texts being very intuitive and helpful can mean for concrete example, like getting an intuition of the role of different parts definitions, or helping to fill in concrete mathematical details (on my own).

    And why do I write another comment in this old thread at all? Good point. I actually finished the reading already some time ago, but was unsure whether I should write that comment (and hence did not write it). Partly it was Yoni’s comment #166 under the Davos post which encouraged me, partly it the “aggregator feed” from Thomas Vidick’s own post from 30. January (+asdf’s clarification with a link to the original post). And I thought I could also share some independent information in this comment: The paper now has 231 scites on SciRate, giving it a solid lead over Ronald de Wolf’s Quantum Computing: Lecture Notes with 149 scites and Ewin Tang’s A quantum-inspired classical algorithm for recommendation systems with 131 scites.

  143. fred Says:

    Hi Scott,

    just to make sure I’m not losing my mind here…

    When it comes to NP hard problems, we do not know of any efficient way to find out whether a solution (or more than one) exists vs there are no solution at all, right?

    So the method would say “no solution exists” or “at least one solution exists”, but without giving back any actual solution.

  144. Scott Says:

    fred #143: I don’t understand what you’re asking. In MIP*, there are two all-powerful (albeit untrustworthy) provers. Either one of them could just hand the verifier a satisfying assignment for an NP-complete problem instance, assuming one exists.

    In addition, though, if you have an oracle for telling you whether an NP-complete problem instance is satisfiable or not, a standard exercise shows that you can also use that oracle to find satisfying assignments in polynomial time, whenever they exist. (Hint: use binary search.)

  145. A quantum strategy could verify the solutions to unsolvable problems — in theory – WEBSITE OF SCIENCE RESEARCH Says:

    […] paper generated a lot of excitement from scientists on social media and on blogs. Whenever new research gets scientists talking, it’s a good sign that it’s worth looking into. […]

  146. AQFT Says:

    Just regarding the split property in QFT. I don’t think it means that the total algebra of two regions is the tensor product of the individual local algebras. That’s generally not possible due to the Reeh-Schlieder theorem which means in QFT there are no local pure states.

    Rather the split property implies the two separate properties: Operational Independence and Commensurability. Together these ensure that despite the ubiquitous entanglement present in QFT two separate regions are operationally local. Operational Independence means the states preparable for one region and the operations that can be performed there are not dependent on the second region. Commensurability is just the usual commutativity for spacelike regions, i.e. choice of observable in region A doesn’t disturb statistics in region B.

  147. Polymath Says:

    I’m trying to imagine how giving the benefit of every conceivable doubt this could be seen to threaten Church’s Thesis, but I don’t think it can. Am I getting this right?

    Informally, two entangled gods who can’t communicate classically could convince you whether any given TM halts or not; but only because you are assuming that they can’t communicate classically. How do you become convinced that they can’t communicate classically? Unless you are doing the thing in one step, sending two superbly organized queries off in opposite directions in such a way that the answers come back soon enough that you can be sure there was no communication because you are already sure of how far away each of the gods really is from you and from the other one and know they haven’t moved closer to you in the meantime, I don’t see how you could know they weren’t communicating classically. Does the protocol account for this?

  148. Infinity in Nature? | More Quantum Says:

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  149. Mathematicians Are Studying Planet-Sized Quantum Computers With God-Like Powers – Shirley Coyle Says:

    […] prominent computer scientist and professor at the University of Texas at Austin, Scott Aaronson, called the discovery one of the biggest theoretical “surprises so far in this […]

  150. fred Says:

    Thanks Scott, you answered my question (finding solutions just from oracle yes/no answer)!

  151. How a quantum technique highlights math’s mysterious link to physics - Chemist.gr Says:

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  152. From Science News: “How a quantum technique highlights math’s mysterious link to physics” | sciencesprings Says:

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  153. Shtetl-Optimized » Blog Archive » Freeman Dyson and Boris Tsirelson Says:

    […] The spectacular answer—no—was only announced one month ago, as a corollary of the MIP*=RE breakthrough, something that Tsirelson happily lived to see although I don’t know what his reaction was. […]

  154. Justin Says:

    With regard to the above result I couldn’t help noticing a possible ‘spooky’ connection to a 2018 paper by Xi Dong, Eva Silverstone and Gonzalo Torroba on De Sitter Holography and Entanglement Entropy. To cut a long story short they have shown that you can stitch together two Ads/CFT’s and get a de sitter space. Whether an actual connection exists I don’t know but when I read it recently the first thing that popped into my mind was MIP*=RE and felt it was worth communicating. Is it worth investigating? Here’s the paper if anyone’s interested https://arxiv.org/abs/1804.08623

  155. Shtetl-Optimized » Blog Archive » Paperz Says:

    […] Hamoon Mousavi, Seyed Nezhadi, and Henry Yuen have now taken the MIP*=RE breakthrough even a tiny step further, by showing that “zero-gap MIP*” (that is, quantum multi-prover interactive proofs with an arbitrarily small gap between the completeness and soundness probabilities) takes you even beyond the halting problem (i.e., beyond Recursively Enumerable or RE), and up to the second level of the arithmetical hierarchy (i.e., to the halting problem for Turing machines with oracles for the original halting problem). This answers a question that someone asked in the comments section of this blog. […]

  156. Pablo Says:

    Hi Scott, perhaps you’ve already been asked this. Can you point to what would you suggest to read in order to learn the main techniques that are used to learn complexity analysis (for the lack of a better name to this kind of results)?
    I mean, for anyone with decent knowledge of QC, like anyone who took your Quantum Information Science course.

  157. Scott Says:

    Pablo #156: Understanding MIP*=RE would be a long journey even for “professional quantum complexity theorists,” let alone for those just finishing an undergrad course. But if you wanted to start on that journey: I’d begin with the classical MIP=NEXP Theorem (Babai, Fortnow, and Lund, 1991), as well as the original work that connected Bell inequality violation with multiprover interactive proofs (Cleve, Hoyer, Toner, and Watrous, 2004). Then the NEXP⊆MIP* theorem of Ito and Vidick (2012), then NEEXP in MIP* (Natarajan and Wright, last year), and the related compression theorems interpolating between that result and Slofstra’s, and then you could be ready to start on MIP*=RE.

  158. Pablo Says:

    Scott #157: Thanks for your answer! Anyway I was asking the more general question of what techniques are key for the study of ‘quantum complexity’, in which MIP*=RE is just a case.
    Thanks!

  159. Graced With Knowledge, Mathematicians Seek to Understand – Book Booster Says:

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  160. Graced With Knowledge, Mathematicians Seek to Understand | Unhinged Group Says:

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  161. Graced With Knowledge, Mathematicians Seek to Understand - Abstractions on Nautilus - NSO News Says:

    […] of 2020, a team of computer scientists posted a sweeping proof that has been hailed as one of the top results in its field this century. Yet the proof went far beyond computer science. Through a long chain of […]

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