Consider the standard 2-2 Bell scenario, with two parties each one choosing between two measurement settings, with each measurement setting leading to one of two possible measurement outcomes. Consider the space of possible corresponding behaviours, that is, conditional probability distributions $(p(ab|xy))_{a,b,x,y\in\{0,1\}}$.

As also discussed in What is the no-signaling set and how can it be related to other types of correlations? and What are the vertices of the no-signalling set $\mathcal{NS}$?, it is standard in this context to distinguish between different classes of behaviours. I'll mention in particular the set $\mathcal L$ of local behaviours, the set $\mathcal Q$ of quantum behaviours, and the set $\mathcal{NS}$ of no-signalling behaviours.

A standard representation of the relations between these sets, given in the review by Brunner et al., is

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One notable feature of this is that while $\mathcal L$ and $\mathcal{NS}$ are polytopes, that is, the convex hull of a finite number of points, $\mathcal Q$ is represented as having a boundary that is non-flat in some sections.

I wouldn't call this feature surprising, but is there an easy way to see why this is the case?

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    $\begingroup$ The quantum set is not a polytope, you might find this paper interesting: Geometry of the set of quantum correlations. $\endgroup$
    – Rammus
    Nov 26, 2021 at 9:34
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    $\begingroup$ $\mathcal{NS}$ is defined as the intersection of the set of all probability distribution with the hyperplanes that represent the no-signalling conditions. An intersection of a polytope with a hyperplane is a polytope. $\endgroup$ Nov 27, 2021 at 13:53
  • $\begingroup$ Why $\mathcal L$ is a convex polytope? I mean, yes, we can find hyperplanes to divide the convex set of unrestricted probability distribution(PD), and one side of those found hyperplanes is the $\mathcal L$. But I can't exclude this case: the shape of $\mathcal L$ is nearly a polytope but one face of it is a curve. $\endgroup$
    – narip
    Feb 5, 2022 at 1:52

1 Answer 1


So, the first comment is that we know that the quantum body itself is not a polytope because we may find cases in which the extremal points form a continuum of points instead of just a finite set of points. In this recent paper here they have even plotted sections of this body, and you can see the boundaries. In particular, the ones I liked the most are the elliptopes. The results there also show that this body cannot be a polytope. This is rigorously proved there in full detail, and I here outline a particular argument that might be more 'easy' as the question asks for.

One can see that for the restricted case, the quantum correlations satisfy the equation $1-x^2-y^2-z^2+2xyz \geq 0$ as in the above cited paper. The procedure is the following:

Let $c=(\langle A_1B_1\rangle,\langle A_1B_2\rangle,\langle A_2B_1\rangle,\langle A_2B_2\rangle) = (1,x,y,z)$. Now, since we have dichotomic measurements we can model $\langle A_iB_j \rangle = 2p(A_i=B_j)-1$. The set $p(A_i=B_j) = \text{Tr}(P^{A_i}_0 P^{B_j}_0)$ if we have set of outcomes $\{0,1\}$. All information in the correlations is then encoded in these traces.

Let's then make $\langle A_1B_1\rangle = 1$. We may then fix $P_0^{A_1} = P_0^{B_1} = \vert 0 \rangle \langle 0 \vert$. Instead of writing the projector I can equivalently write the vectors $\vert A_2^0 \rangle, \vert B_2^0 \rangle$ for the remaining states $P_0^{A_2} = \vert A_2^0 \rangle \langle A_2^0 \vert$ and same for $P_0^{B_2}$. I want then to optimize this section of the quantum body. In general, correlations can be described if we choose the vectors,

$$\vert A_2^0 \rangle = \cos(\beta)\vert 0 \rangle + \sin(\beta)\vert 1 \rangle$$

$$\vert B_2^0 \rangle = \cos(\gamma)\vert 0 \rangle + \sin(\gamma)\cos(\alpha)e^{i\theta}\vert 1 \rangle + \sin(\gamma)\sin(\alpha)\vert 2 \rangle$$

In this description, CHSH becomes $$1 \geq c_{12}+c_{21}-c_{22} = \langle A_1B_2\rangle + \langle A_2B_1\rangle-\langle A_2B_2\rangle (A_1=B_1) \\= 2\text{Tr}(P^{A_1}_0 P^{B_2}_0)-1 + 2\text{Tr}(P^{A_2}_0 P^{B_1}_0)-1-(2\text{Tr}(P^{A_2}_0 P^{B_2}_0)-1) \\\implies 2\text{Tr}(P^{A_1}_0 P^{B_2}_0)-1 + 2\text{Tr}(P^{A_2}_0 P^{B_1}_0)-1-(2\text{Tr}(P^{A_2}_0 P^{B_2}_0)-1) \leq 1 \\\implies 2\text{Tr}(P^{A_1}_0 P^{B_2}_0)-1 + 2\text{Tr}(P^{A_2}_0 P^{B_1}_0)-2\text{Tr}(P^{A_2}_0 P^{B_2}_0)\leq 1 \\\implies 2\text{Tr}(P^{A_1}_0 P^{B_2}_0) + 2\text{Tr}(P^{A_2}_0 P^{B_1}_0)-2\text{Tr}(P^{A_2}_0 P^{B_2}_0)\leq 2 \\\implies \text{Tr}(P^{A_1}_0 P^{B_2}_0) + \text{Tr}(P^{A_2}_0 P^{B_1}_0)-\text{Tr}(P^{A_2}_0 P^{B_2}_0)\leq \frac{1}{2}2 \\\implies \text{Tr}(P^{A_1}_0 P^{B_2}_0) + \text{Tr}(P^{A_2}_0 P^{B_1}_0) - \text{Tr}(P^{A_2}_0 P^{B_2}_0) \leq 1 \implies f \leq 1$$ but this implies that the maximum quantum set can be obtained from maximizing the CHSH term $f$ for this fact of the convex body. Meaning we may maximize the quantity:

$$f(\alpha,\beta,\gamma,\theta) = cos^2(\gamma)+\cos^2(\beta)-\cos^2(\gamma)\cos^2(\beta)+\sin^2(\gamma)\sin^2(\beta)\sin^2(\alpha)-2\sin(\gamma)\cos(\gamma)\sin(\beta)\cos(\beta)\sin(\alpha)\cos(\theta)$$

This can be done analytically (see the appendix of this paper here, published here) and the resulting equation is an equation that is not a polytope but corresponds to a continuous boundary curve, which is the same as the elliptope curve, but now for the traces instead of for the correlations (but this is a dual maximization representation for this particular case).

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    $\begingroup$ thanks, these answers are very much appreciated! To make sure I understand, what you are saying here is that, in the space of correlations between the considered observables (or equivalently, in the space of conditional probability distributions, i.e. behaviours), the boundary of the quantum set is a quadratic surface. I don't quite understand the way you are showing this though. The states you are using are not qubits, I take it? $|B^0_2\rangle$ has at least dim 3? Maybe I don't quite understand the setting $\endgroup$
    – glS
    Apr 1, 2022 at 16:04
  • $\begingroup$ They are not qubits indeed, they are any states. But since I took the particular case of 3 different operators I can map them into a qubit subspace associated with their span. My argument is that for specific cuts you can actually see the quantum set and find the boundary analytically as they have done in those papers. $\endgroup$
    – R.W
    Apr 1, 2022 at 16:12
  • $\begingroup$ This is the simplest argument I know for the fact that the quantum body cannot always be a polytope: because for the CHSH it is not and we can numerically or analytically see this. Now, this is not a proof nor a good understanding of the physical reason of why this is so, but I fear that physical principles explaining the quantum boundary are still to be found. $\endgroup$
    – R.W
    Apr 1, 2022 at 16:16
  • $\begingroup$ @glS Thanks for the questions, hopefuly I am being able to help a little. I am also interested to know if a simple and more elegant argument exists. $\endgroup$
    – R.W
    Apr 1, 2022 at 16:24

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