What is the no-signaling set and how can it be related to other types of correlations?

The paper Bell nonlocality by Brunner et. al includes a striking diagram on page 7:

This is fascinating to me because it suggests a framework of categorizing correlations that encompasses classical, quantum, "no-signaling", and maybe full signaling. What is the theoretical framework underlying this diagram, the relations of the sets, and the limits of the polytope? Especially, what is the no-signaling set? Is it accurate to say it's a set of correlations that are super-quantum but sub-signaling which have not been found to exist in physical reality?

• – glS
Commented Nov 23, 2021 at 6:44

I'll describe the case of two party correlations but this can straightforwardly extended to more parties.

Let's give a box to Alice and a box to Bob. Alice and Bob can interact with their boxes by providing them classical inputs $$x \in \mathcal{X}$$ and $$y \in \mathcal{Y}$$ respectively. They can also receive outputs from their boxes, $$a \in \mathcal{A}$$ and $$b \in \mathcal{B}$$ respectively. The inputs and outputs all come from finite sets. Overall, without describing exactly how the boxes produce their outputs given their inputs, we can describe this setting by the correlations that the boxes produce. In particular we can look at the conditional distribution $$p(a,b|x,y).$$ We can view this distribution as a vector in $$\mathbb{R}^{\mathcal{|A||B||X||Y|}}$$. For example, in the simplest nontrivial case all the inputs and outputs are binary and the distribution is viewed as a vector in $$\mathbb{R}^{16}$$.

Now comes the fun bit. Suppose you know something about how the boxes produce their inputs given their outputs, what are the possible probability vectors that said devices can produce? In the following we assume that the devices are spacelike separated during the experiment, this imposes no-signalling constraints which I'll detail below.

Local correlations (classical) $$\mathcal{L}$$

The local set assumes that the devices act in the following manner. Alice and Bob's devices share a copy of some "hidden" random variable $$\Lambda$$ which they can use to correlate their devices. When given an input, Alice's device samples $$\Lambda$$ and uses the value of her input $$x$$ in order to choose an output perhaps probabilistically, i.e., Alice's device produces output $$a$$ with probability $$p(a|x,\lambda)$$. Similarly Bob's device will produce output $$b$$ when the input was $$y$$ with some probability $$p(b|y,\lambda)$$. Overall their distribution must decompose as $$p(a,b|x,y) = \sum_{\lambda} p(\lambda) p(a|x,\lambda) p(b|y,\lambda).$$ Any distribution that can be written as above is called a local distribution. In the diagram it is denoted as the set $$\mathcal{L}$$. The correlations that can be obtained in this manner form a convex polytope (called the local polytope) where the extreme points of the polytope are given by the local deterministic distributions, i.e., $$\Lambda$$ is trivial and $$p(a|x,\lambda), p(b|y,\lambda) \in \{0,1\}$$ (Alice and Bob's device produce an output deterministically given an input). As the local set is a convex polytope we can also describe it as an intersection of a collection of halfspaces, the inequalities defining these halfspaces are known as Bell-inequalities.

Quantum correlations $$\mathcal{Q}$$

Suppose now we can we can build our devices from quantum systems. The most general strategy we can do is to share some entangled state between the two devices and perform local measurements on it. More formally a distribution $$p(a,b|x,y)$$ is called quantum if there exists a bipartite Hilbert space $$H_{AB}$$, a state $$|\psi\rangle \in H_{AB}$$ and POVMs $$\{M_{a|x}\}_{a\in \mathcal{A}}, \{N_{b|y}\}_{b \in \mathcal{B}}$$ for each $$x \in \mathcal{X}$$ and $$y \in \mathcal{Y}$$ such that $$p(a,b|x,y) = \mathrm{tr}[(M_{a|x} \otimes N_{b|y})|\psi\rangle\langle\psi|].$$ We denote the set of distributions that can be written like this by $$\mathcal{Q}$$. Bell's theorem essentially states that $$\mathcal{L}$$ is a strict subset of $$\mathcal{Q}$$.

No-signalling correlations $$\mathcal{NS}$$

We made a physical assumption that the device's were spacelike separated. We do this to impose that the inputs of one party do not affect the local distribution of the other party. That is, Alice cannot choose her input in order to send information to Bob faster than the speed of light and vice versa. Formally, these are described by the no-signalling conditions $$$$\label{eq:ns} \sum_a p(a,b|x,y) = \sum_a p(a,b|x',y) \qquad \forall b,x,x',y$$$$ $$$$\label{eq:ns2} \sum_b p(a,b|x,y) = \sum_b p(a,b|x,y') \qquad \forall b,x,y,y' .$$$$ Again but verbosely, the inputs of Alice are Bob do not change the marginal distribution observed by the other party -- they cannot transmit information. It's a good exercise to check that local and quantum distributions satisfy the no-signalling constraints.

Now suppose you don't care about how your boxes produce their outputs given their inputs except that you want them to not signal. I.e., you allow any distribution $$p(a,b|x,y)$$ that satisfies the above conditions. This is precisely the no-signalling set of correlations depicted in the diagram. As to how the authors made the diagram, I expect that they took some simple choice of inputs and outputs and then took a 2 dimensional slice of the correlation sets and this is what they are displaying. In general we have $$\mathcal{L} \subset \mathcal{Q} \subset \mathcal{NS}$$.

Signalling and other sets

Well if you want to drop the no-signalling constraints then you're pretty much at any valid probability distribution. The no-signalling constraints are motivated by relativity. If you add other constraints (like dimension constraints in the quantum setting) you can generate new correlation sets and explore their properties. For instance if we assume the quantum state shared between the devices is a two-qubit state then the set of achievable correlations is no longer a convex set!

• Why do you say if the shared quantum state between devices is a two-qubit state the set of correlations is not convex? From the beginning you were talking about distributing a two-qubit system between to parties, weren't you? (As one qubit is only for one device)
– Pegi
Commented Nov 25, 2021 at 11:54
• @Pegi said nothing about the dimensions of the quantum systems until the very end. I only said that in the quantum setting we distribute some entangled state between the two parties. The dimensions of the local Hilbert spaces can be arbitrary. In particular this is need to ensure that the set of quantum correlations is convex. Commented Nov 25, 2021 at 12:06
• Oh ok. But I thought based on the paper, in the case that we have two parties there the correlation set is convex.
– Pegi
Commented Nov 25, 2021 at 12:11
• Yes, the set is convex. The set only becomes nonconvex when you impose a constraint on the dimension. The paper does not constrain the dimension in the diagram as far as I am aware. the number of parties is independent of the dimension of the local Hilbert spaces. Commented Nov 25, 2021 at 13:34
• Thank you. How do you prove it gets noconvex? I guess when for example the dimension of the local Hilbert spaces of the parties are different, we can not have a linear combination and it become nonconvex?
– Pegi
Commented Nov 26, 2021 at 10:37

The other answer already covered most of the bases. I'd just add an explicit example of a no-signalling, non-quantum distribution, because I think it's useful to have some in mind when discussing these things. Consider a two-two behaviour, that is, a conditional probability distribution $$p(ab|xy)$$ with $$a,b,x,y\in\{0,1\}$$, such that $$p(a,a|x,y)=1/2$$ for $$xy=0$$, and $$p(a,1-a|1,1)=1/2$$. More explicitly, we can represent this as the "matrix": $$\begin{array}{c|cccc} & 00 & 01 & 10 & 11 \\\hline 00 & 1/2 & 1/2 & 1/2 & 0 \\ 01 & 0 & 0 &0 & 1/2 \\ 10 & 0 & 0 &0 & 1/2 \\ 11 & 1/2 & 1/2 & 1/2 & 0 \\ \end{array},$$ where the columns represent the possible values of $$x,y$$, and the rows those of $$a,b$$.

To check that this is no-signalling, you need to ensure that summing over $$a$$ gives the same distribution regardless of $$x$$, and same for $$b$$ and $$y$$. In terms of the matrix, this means to check that summing first and third, and second and fourth rows, makes first and third and second and fourth columns identical. In this case, all columns become identical, so the condition is trivially verified. In fact, we get $$\sum_{a'} p(a'b|xy)=\sum_{b'} p(ab'|xy)=1/2,$$ for all $$a,b,x,y$$.

One of course then needs to show that this behaviour is non-local, and also non-quantum.

• CHSH comes out $4$ if we redefine $a,b\in\{-1, +1\}$, so this is indeed non-quantum by Tsirelson's bound. Commented Nov 23, 2021 at 6:31