Observing the violation of Bell inequalities, be it in their original formulation, or in the nowadays more commonly used CHSH formulation, involves computing averages of specific experimentally measurable quantities. In the CHSH formulation, these are for example the averages of the products of the experimental outcomes.

Are there scenarios in which Bell nonlocality can be observed without such averages, that is, in a single-shot scenario? By this I mean a scenario in which a single measurement outcome (rather than a collection of outcomes used to compute averages, as is done in CHSH formulation) is enough to rule out a local hidden variable explanation.

I seem to recall having seen this kind of thing in some paper, possibly in a scheme involving three or more parties (or maybe it was with systems with three or more outcomes?). I can't, however, find the reference right now, so I'm not sure whether I'm remembering this correctly or not.


4 Answers 4


Are there scenarios in which Bell nonlocality can be observed without such averages, that is, in a single-shot scenario?

No, you have to collect statistics. Any single result you see could have been due to classical players picking completely at random and getting lucky. Making the chance of classical luck arbitrarily close to zero requires repetition (or arbitrarily large games).

For a single-shot experiment to work, you need a game with an outcome that has a Bayes factor of infinity between a quantum strategy and all classical strategies. This means every classical strategy must have a 0% success rate while some quantum strategy has a non-zero success rate. That's not possible, because any move a quantum strategy makes can also be made by accident by a random strategy.

You can have game where entangled strategies have a 100% success rate, such as the mermin-peres game, but the Bayes factor of that game is not infinity because the classical strategy has a success rate of 8/9 > 0.

  • $\begingroup$ This is not correct. You can't demand a Bayes factor of infinity to have a violation, otherwise no amount of statistics will be enough to convince you. You can only ever demand the Bayes factor to be above some threshold, but for any given threshold it is possible to attain it in a single-shot experiment. See my answer. $\endgroup$ Commented Jul 1, 2020 at 12:28
  • $\begingroup$ @MateusAraújo The question was asking for a Bayes factor of infinity and this answer is explaining why you can't have it. Your comment agrees with my answer, so I'm not sure why you're saying it's not correct. Maybe I just don't see an interesting distinction between running a simple game multiple times and running a big complicated game once (I do mention that case in the answer). $\endgroup$ Commented Jul 1, 2020 at 19:48
  • $\begingroup$ The question wasn't asking for a Bayes factor of infinity. On the contrary, the OP implied that they accept the usual statistical violation as valid, which means that they are fine with a finite Bayes factor. I'm saying that your answer is incorrect because you claim that a Bayes factor of infinity is necessary, and this is not true. Neither in the single-shot scenario, nor in the usual multi-round scenario. $\endgroup$ Commented Jul 1, 2020 at 21:11
  • $\begingroup$ There is a world of difference between running a simple game multiple times and a big complicated game once; the latter case is much harder to analyse, and the particular case of parallel repetition is of crucial importance for complexity theory; it is a whole subfield of research. $\endgroup$ Commented Jul 1, 2020 at 21:14

I believe you're thinking of the all-versus-nothing proofs based on GHZ states. You start with a state such as $$ |\Psi\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle) $$ and you select at random one of the four measurements to implement: $X\otimes X\otimes X$, $X\otimes Y\otimes Y$, $Y\otimes X\otimes Y$ or $Y\otimes Y\otimes X$. Assuming every one of these can be implemented perfectly every time, you must always get the answer -1 (just calculate $\langle\Psi|M|\Psi\rangle$ where $M$ is one of the measurement operators) for any of the measurements with two $Y$s, and +1 for the $XXX$ measurement.

However, if the outcomes are associated with a local hidden variable model, this outcome is impossible. Let $x_i\in\{-1,1\}$ and $y_i\in\{-1,1\}$ be the hidden variables associated with the outcomes of measuring $X$ or $Y$ respectively on qubit $i$.

Now let us consider what would happen if we made each of the four measurements, and took the product of the outcomes. The quantum model says we get answer $-1$. The local hidden variable model, however, says that answer must be $$ (x_1x_2x_3)(x_1y_2y_3)(y_1x_2y_3)(y_1y_2x_3)=x_1^2x_2^2x_3^2y_1^2y_2^2y_3^2=1. $$ A local hidden variable model is incompatible with always getting the predicted set of quantum answers.

So, the idea is that whatever measurement you make, you know there is a fixed answer that you must get if the system is behaving properly quantumly (and assuming no errors whatsoever in the system). If you ever get a disagreement, the system cannot be behaving perfectly quantumly. So, that gives you the non-probabilstic answer you mention.

Of course, a more reliable way of running this is to repeat the experiment many times. If you ever get a false answer, that's a proof of non-quantumness. However, the conclusion of quantumness (i.e. always getting the expected answers) still has an associated probability because after $n$ runs, there's a probability of $1/4^n$ that the local hidden variable version gives all the expected quantum answer.

  • $\begingroup$ so if I understand correctly, two measurements can be enough to rule the quantum model, but I don't understand how you rule out the LHVs. Your argument about multiplying the outcomes only seems to work for a deterministic LHV, otherwise you are not ensured to get the same, say, $x_1$, on different measurement runs, no? $\endgroup$
    – glS
    Commented Jan 15, 2020 at 10:59
  • $\begingroup$ @glS No, it does the opposite. The quantum model gives specific answers. If you don't get those specific answers, you have ruled out quantum. All the LHV argument does is justify that LHVs are compatible with not always getting the right answer (and bound the probability of right answer as 3/4). Yes, I only analysed deterministic LHVs but if no deterministic LHV always gives the right answer, no convex combination always gives the right answer either. $\endgroup$
    – DaftWullie
    Commented Jan 15, 2020 at 11:06
  • $\begingroup$ Also, the LHV argument is more subtle than I think you're making it. The assumption is that variables such as $x_i$ are fixed before the measurement choice is made, and the answer to the measurement corresponds to those variables. So it's a property of a single experiment if you could perform all 4 measurements in one go (which of course we can't). In that single experiment, could all 4 possible LHV outcomes have been the same as all 4 possible quantum answer? No. Will we catch them? Only with probability 3/4. That's why we repeat many times. $\endgroup$
    – DaftWullie
    Commented Jan 15, 2020 at 11:09
  • $\begingroup$ yes, I meant to say "rule out the quantum model", sorry. I understand that if no deterministic LHV is compatible with the measurements, no LHV can be, but does this apply here? You are defining a function $f$ of the exp outcomes (which multiplies them all together), and argue that QM always gives $f=-1$, fine. Then argue that for any deterministic LHVs with $\lambda$, $f_\lambda=1$, ok. Then if you take mixtures of possible $\lambda$s you also get $f=1$. What I don't see is how this covers the case in which $\lambda$ changes in between the four experimental runs needed to compute $f$ each time $\endgroup$
    – glS
    Commented Jan 15, 2020 at 11:54
  • $\begingroup$ because if $\lambda$ is allowed to change between different experimental runs, then how is $f_\lambda$ well-defined? We could easily find that the product of the outcomes is not $1$: say e.g. that when measuring $YYX$ the LHV has changed so that now we find value of $y_1,y_2,x_3$ different than the ones observed in the previous three measurement settings. $\endgroup$
    – glS
    Commented Jan 15, 2020 at 11:57

Yes, there are. I just wrote a paper about it, actually.

You need to define carefully what you mean by obtaining a Bell violation or ruling out local hidden variables. You can't demand to have a result which is impossible to explain with local hidden variables: the local bound of a Bell inequality is inherently probabilistic, so it is possible to obtain any statistic you observe merely by chance. What you can do is demand it to be unlikely, with $p$-value smaller than some threshold, or Bayes factor above some threshold. For example, in one of the loophole-free Bell tests they reported that the $p$-value of their data under the local hidden variables hypothesis was below $10^{-30}$.

In a single-shot scenario, the $p$-value of winning a single round of a nonlocal game is simple its local bound, the maximal probability of winning it with local hidden variables. If you can make the local bound smaller than this threshold, while keeping the quantum probability of victory close to 1, then you'll have a single-shot violation.

It turns out that it is possible to construct (families of) nonlocal games such that the local bound becomes arbitrarily close to 0, while the Tsirelson bound becomes arbitrarily close to 1, so yes, a single-shot violation is possible for any threshold you want.

The simplest way of constructing such a game is by doing parallel repetition of a pseudo-telepathy game, that is, playing $n$ times in parallel a nonlocal game with Tsirelson bound 1. The local bound goes down to zero exponentially in $n$ (this is highly nontrivial to prove), while the Tsirelson bound stays equal to 1, so there you have it.

This construction is physically meaningless though, because the local bound here is the probability of winning all parallel instances simultaneously, which you'll never be able to do in reality, so the fact that it is exponentially unlikely with local hidden variables doesn't help you. There are more sophisticated constructions that solve this problem: instead of considering the probability of winning all instances, you can also consider the probability of winning a fraction of instances higher than what you'd expect from the local bound: it turns out that this still goes down to zero exponentially with $n$ under the local hidden variables hypothesis, as proven by Rao's concentration bound, and goes up to one exponentially with $n$ under quantum mechanics. Now this is robust to experimental errors, so it is the way to do it in reality.

You might find unsatisfactory the use of parallel repetition; well there also exists a nonlocal game, the Khot-Vishnoi game, such that the local bound is arbitrarily close to zero and the Tsirelson bound arbitrarily close to one, and it is not based on parallel repetition. This game is very hard to implement experimentally, though.


Original paper: Going beyond Bell's theorem

Papers by Mermin:

We can give a non-probabilistic proof that local hidden variables theories are incompatible with Quantum Mechanics.

The proof does not explain how experimentally decide which theory is correct; it just shows more clearly than Bell or CHSH that Quantum Mechanics and local hidden variables theories are incompatible.


Suppose we have 3-qubit GHZ state $$|GHZ\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle)$$ For visibility sake think of a qubit as of a spin $1/2$ particle.

We can measure spin in $X$ and $Y$ directions; $X$ direction means Hadamard basis $\{|+\rangle, |-\rangle\}$, $Y$ direction means $\{\frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle), \frac{1}{\sqrt{2}}(|0\rangle-i|1\rangle)\}$ basis.

First, Quantum Mechanics predicts that if we measure 2 qubits in $Y$ direction and 1 qubit in $X$ direction, each qubit's measurement outcome is $50\% / 50\%$ random, but there is strong correlation: the outcomes of $Y$ measurements determine the outcome of $X$ measurement; namely, if $Y$ measurements are identical, $X$ measurement is $|-\rangle$, otherwise $|+\rangle$. Following Einstein's "elements of reality" concept, it means that each qubit has deterministic "instructions" how to behave if measured in $X$ and $Y$ direction, and the set of legal instructions for all 3 qubits in GHZ experiment is limited.

Following the 1st cited Mermin paper, the legal instructions are (first line $X$ measurement, second line $Y$ measurement):

 ---  -++  +-+  ++-  -++  ---  ++-  +-+
 ---  -++  +-+  ++-  +--  +++  --+  -+-

It now turns out that if we apply legal instructions approach to measuring all 3 qubits in $X$ direction, the only possible outcomes are $|---\rangle$, $|-++\rangle$, $|+-+\rangle$ and $|++-\rangle$.

On the other hand, quantum mechanics predicts that the only possible outcomes are $|+++\rangle$, $|+--\rangle$, $|-+-\rangle$ and $|--+\rangle$, because

$$|GHZ\rangle = \frac12(|+++\rangle+|+--\rangle+|-+-\rangle+|--+\rangle)$$ and we have no overlap with legal instructions (hidden variables) approach.


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