# The $I_{3322}$ Inequality

I am trying to understand the $$I_{3322}$$ inequality which is an another example of Bell inequalities and which is different from the famous CHSH inequality. I haven't got hold of any standard reference for that and I'm trying to glean through the original papers. However I am getting confused because of many different formulations of the same inequality, and hence I came here to understand why they all represent the same thing.

My first reference is a paper by Sliwa. The original paper is here, and an arxiv version is here. The author expresses the inequality in terms of expectations - $$-E(A_1B_1)-E(A_1B_2)-E(A_1B_3)-E(A_2B_1)-E(A_2B_2)+E(A_2B_3)-E(A_3B_1)+E(A_3B_2)-E(A_1)-E(A_2)-E(B_1)-E(B_2) \leq 4.$$

The next reference I referred to is by Pal and Vertesi (available here and on the arxiv) where the inequality is also expressed in terms of expectations - $$E(A_1B_1)+E(A_1B_2)-E(A_1B_3)+E(A_2B_1)+E(A_2B_2)+E(A_2B_3)-E(A_3B_1)+E(A_3B_2)-E(A_2)-E(B_1)-2E(B_2)\leq 0.$$

My third reference is a paper by Collins and Gisin (available here and on the arxiv). They express in terms of probabilities - $$p_{11}+p_{12}+p_{13}+p_{21}+p_{22}-p_{23}+p_{31}-p_{32}-p_1^A-2p_1^B-p_2^B\leq 0.$$

The last reference is the survey paper by Brunner et al (available here and on arxiv). They also express in terms of probabilities - $$p_{11}+p_{12}+p_{13}+p_{21}+p_{22}-p_{23}+p_{31}-p_{32}-p_1^A-p_1^B\leq 1.$$

Now if one glances over them, one finds that each one of them is different from others by few terms. I feel that there may be some simplification to derive one from the others but I don't see how. I would kindly urge the physics community to shed some light on this issue.

Also in the last reference by Brunner et al, they say that the CHSH inequality can also be viewed as a game. Is there a similar kind of game from which the $$I_{3322}$$ inequality can be derived.

P.S.- I am not a physicist, but a mathematician, so it would be be helpful if minimal physics is used.

Thank you.

• I don't have an answer but I am interested. Forgive me if I am totally off subject but I don't quite get some of the examples you give. I thought there where nine combinations that Alice and Bob (A) and (B) could select from a three position slit. For example 11,12,13, 21, 22, 23, 31, 32, 33. Where I get confused is when it is derived that the chances of correlation are YYY, YYX, YXX, YXY, XYY, XXY, and XXX. Where (Y)=yes and (X)=no. – Bill Alsept Feb 4 '16 at 7:04
• I am not sure I understand your comment properly. Perhaps an expert in the field will help us out. Sorry! – Nirakar Neo Feb 11 '16 at 1:12

Bell inequalities are equivalent if you can map one to another by

1. Relabelling the parties (in this case A and B)
2. Relabelling the inputs (in this case 1,2,3)
3. Relabelling the outputs (in this case -1 and 1 for Śliwa version, 0 and 1 for the Pál and Vértesi version, and undefined for the other two versions)
4. Scaling by a nonzero constant.

These equivalences are explored here, for example. The last two conditions are not relevant for this question, as the notations used here, correlators and Collins-Gisin, do not represent them.

We start by showing that the Śliwa version is equivalent to the Collins and Gisin's version. For that, we pass the latter from Collins-Gisin notation to correlator notation via the formulas $$4p_{ij} = E(A_iB_j) + E(A_i) + E(B_j) +1$$ $$2p_i^A = E(A_i)+1$$ $$2p_j^B = E(B_j)+1$$ Scaling the inequality by 4, for convenience, we end up with $$E(A_1B_1)+E(A_1B_2)+E(A_1B_3)+E(A_2B_1)+E(A_2B_2)-E(A_2B_3)+E(A_3B_1)-E(A_3B_2)+E(A_1)+E(A_2)-E(B_1)-E(B_2) \leq 4,$$ where we see that the asymmetry in the coefficients became an asymmetry in the signs. We can take care of that by relabelling Alice's outputs, mapping -1 to +1 and vice-versa. This changes the expectation values as $$E(A_iB_j) \mapsto -E(A_iB_j)$$ $$E(A_i) \mapsto -E(A_i)$$ $$E(B_j) \mapsto E(B_j)$$ And the inequality becomes $$-E(A_1B_1)-E(A_1B_2)-E(A_1B_3)-E(A_2B_1)-E(A_2B_2)+E(A_2B_3)-E(A_3B_1)+E(A_3B_2)-E(A_1)-E(A_2)-E(B_1)-E(B_2) \leq 4,$$ which is Śliwa's version.

Now, we can obtain Brunner et al.'s version either from Śliwa's version by changing back to Collins-Gisin notation, or directly from Collins and Gisin's version by relabelling Alice's output. It's more interesting to do the latter.

The idea of the Collins-Gisin notation is to show the probabilities only of $$n-1$$ outcomes out of $$n$$; the other probability is implicit from the normalisation condition. In this case $$n=2$$, so all the probabilities refer to obtaining the outcome $$0$$, and the probabilities of outcome $$1$$ are implicit (here the labels are arbitrary). Thus after relabelling Alice's outcome we have to rewrite the probabilities as functions of the outcome $$0$$. This is done via the identities $$p(01) = p^A(0)-p(00)$$ $$p(10) = p^B(0)-p(00)$$ $$p(11) = 1+p(00)-p^A(0)-p^B(0)$$ $$p^A(1) = 1-p^A(0)$$ $$p^B(1) = 1-p^B(0)$$ So if we relabel Alice's outcome from 0 to 1, this changes the probabilities in the inequality as $$p_{ij} \mapsto p_j^B-p_{ij}$$ $$p^A_{i} \mapsto 1-p^A_i$$ $$p^B_j \mapsto p^B_j$$ Applying this transformation to the Collins and Gisin's version, we obtain $$-p_{11}-p_{12}-p_{13}-p_{21}-p_{22}+p_{23}-p_{31}+p_{32}+p_1^A+p_1^B\leq 1.$$ Which is not equal to Brunner et al.'s version. They made a sign mistake.

This is a partial answer, that I will be updating as I get more insight.

First, in Pal and Vertesi's article they use correlator notation, but the inequality and bounds they give to it only hold if the operators $A_i$ and $B_j$ are actually projectors, so what they really want to mean is $E(A_iB_j)=p_{ij}$ and $E(P_i)=p^P_i$. This can be checked not only when computing the bounds they give for $I_{3322}$, but is also the case for the CHSH inequality they give in Eqs. (1) and (2).

Then, the inequality is the same as that given by Collins and Gisin when changing the labels of the measurements $1\leftrightarrow 2$ in both parties.

Second, if one transforms Sliwa's version to probability notation (given that the measurements are binary, we can substitute $E(P_i) = 2p^P_i - 1$ and $E(A_iB_j)=4p_{ij}-2(p^A_i+p^B_j)+1$) one obtains the form that appears in Brunner's article.

So, we have grouped the four expressions in two families. Now what is left is to connect them both.

• Actually, you don't obtain the form that appears in Brunner's article. See my answer. – Mateus Araújo Apr 18 at 13:03