Bell inequalities for $S_4$ group: classical and quantum bounds for two-orbit case

Abstract

Within Güney–Hillery approach, a number of examples of classical and quantum bounds on the sum of probabilities resulting from two orbits of $S_4$ is considered. It is shown that the violation of Bell’s inequalities is rather rare and gentle.

Appendix

1.1 The explicit form of standard representation of $S_4$

Below we present the explicit form of standard representation of $S_4$ in an unitary basis. To this end, it is sufficient to know the matrices representing transpositions. They read [13]:

$$\begin{aligned} {\begin{matrix} &{} D\left( 12\right) =\left[ \begin{array}{ccc} 1 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 1 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad -1 \end{array}\right] ,\qquad \qquad \quad D\left( 13\right) =\left[ \begin{array}{ccc} 1 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad -\frac{1}{2} &{}\quad -\frac{\sqrt{3}}{2}\\ 0 &{}\quad -\frac{\sqrt{3}}{2} &{}\quad \frac{1}{2} \end{array}\right] \\ &{} D\left( 14\right) =\left[ \begin{array}{ccc} -\frac{1}{3} &{}\quad -\frac{\sqrt{2}}{3} &{}\quad -\frac{\sqrt{6}}{3}\\ -\frac{\sqrt{2}}{3} &{}\quad \frac{5}{6} &{} \quad -\frac{\sqrt{3}}{6}\\ -\frac{\sqrt{6}}{3} &{}\quad -\frac{\sqrt{3}}{6}&{}\quad \frac{1}{2} \end{array}\right] ,\qquad D\left( 23\right) =\left[ \begin{array}{ccc} 1 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad -\frac{1}{2} &{}\quad \frac{\sqrt{3}}{2}\\ 0 &{}\quad \frac{\sqrt{3}}{2} &{}\quad \frac{1}{2} \end{array}\right] \\ &{} D\left( 24\right) =\left[ \begin{array}{ccc} -\frac{1}{3} &{}\quad -\frac{\sqrt{2}}{3} &{}\quad \frac{\sqrt{6}}{3}\\ -\frac{\sqrt{2}}{3} &{}\quad \frac{5}{6} &{}\quad \frac{\sqrt{3}}{6}\\ \frac{\sqrt{6}}{3} &{}\quad \frac{\sqrt{3}}{6}&{}\quad \frac{1}{2} \end{array}\right] ,\qquad \quad D\left( 34\right) =\left[ \begin{array}{ccc} -\frac{1}{3} &{}\quad \frac{\sqrt{8}}{3} &{}\quad 0\\ \frac{\sqrt{8}}{3} &{}\quad \frac{1}{3} &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 1 \end{array}\right] .\end{matrix}} \end{aligned}$$

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1.2 The specification of orbits

We consider the orbits of $S_4$ in the threedimensional space carrying the standard representation. To this end, we use the general scheme outlined in Sect. 2. The cyclic subgroup $H=\left\{ e,g,g^2\right\}$ is generated by $g=(2314)$; therefore, $m=3$ and $k=8$. In order to simplify notation, we put below $D(\tilde{g})\equiv \tilde{g}$ for any $\tilde{g}\in S_4$. The initial vector v is given by Eq. (23); using the explicit form of the representation one finds that $g^lv$, $l=0,1,2$, are mutually orthogonal. The elements of the orbit under consideration are of the form

$$\begin{aligned} v_{\alpha l}\equiv g_\alpha g^l v, \quad \alpha =1,\ldots , 8,\quad l=0,1,2; \end{aligned}$$

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It remains to select the elements $g_\alpha$ representing the left cosets from $S_4/H$. The particular choice adopted here is described in Table 1.

Table 1 The coset representatives

1.3 The eigenvalues of $X(\tilde{g},v)$

The eigenvalues of the operators $X(\tilde{g},v)$ (cf. Eq. 11) are given by Eq. (13). To find their actual values one has only to know the projections $(v_A\otimes v_B)_s$ of the product vectors on the subspaces carrying irreducible representations entering the decomposition (12). To this end, one should compute the matrix of the relevant Clebsh-Gordan coefficients; this is quite straightforward and the final result reads [13]

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The rows of C correspond to the consecutive basic vectors of block-diagonal basis while the columns to the product vectors. More explicitly, the relevant projections of $\mathbf {v}\otimes \mathbf {v}'$ onto the irreducible subspaces read [13]:

$$\begin{aligned} {\begin{matrix} &{}D_0:\quad \frac{1}{\sqrt{3}}\left( \mathbf {v}\cdot \mathbf {v}'\right) \\ &{} D_2:\quad \frac{1}{\sqrt{3}}\left( v_1v_3'+v_3v_1'\right) -\frac{1}{\sqrt{6}}\left( v_2v_3'+v_3v_2'\right) \\ &{} \qquad \quad \frac{1}{\sqrt{3}}\left( v_1v_2'+v_2v_1'\right) +\frac{1}{\sqrt{6}}\left( v_2v_2'-v_3v_3'\right) \\ &{}\widetilde{D}:\quad \frac{1}{\sqrt{2}}\left( \mathbf {v}\times \mathbf {v}'\right) \\ &{} D: \quad \sqrt{\frac{2}{3}}v_1v_1'-\frac{1}{\sqrt{6}}\left( v_2v_2'+v_3v_3'\right) \\ &{} \qquad \quad \frac{1}{\sqrt{3}}\left( v_2v_2'-v_3v_3'\right) -\frac{1}{\sqrt{6}}\left( v_1v_2'+v_2v_1'\right) \\ &{} \qquad \quad -\frac{1}{\sqrt{3}}\left( v_2v_3'+v_3v_2'\right) -\frac{1}{\sqrt{6}}\left( v_1v_3'+v_3v_1'\right) . \end{matrix}} \end{aligned}$$

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The eigenvalues of $X(\tilde{g},v)$, for v given by Eq. (23), are presented in Table 2. For convenience, all elements $\tilde{g}\in S_4$ are grouped into equivalence classes. In the following two columns the elements of $S_4$ are written out together with their $(\alpha ,l)$ indices; then the eigenvalues are presented which correspond to the successive irreducible components entering the Clebsch–Gordan decomposition of $D\otimes D$; finally, in the last column the maximal eigenvalues are given.

In order to obtain the classical estimate on the sum (16) we may use the algorithm described in Ref. [13]. We use the property that the joint probability maximizing S can be chosen in the form described by Eq. (17). Therefore, for any joint configuration $(\underline{a},\underline{b})=(a_1, \ldots ,a_8;b_1,\ldots ,b_8)$ (cf. Eq. 15) we have to determine the number of times it appears in the configurations by vectors entering both orbits. To this end, we start with some element $(\alpha ,l)$ viewed as an Alice state. We select from the first orbit the corresponding Bob’s element $(\alpha _{\tilde{g}_1},l_{\tilde{g}_1})$; then we look for the element of the second orbit containing $(\alpha _{\tilde{g}_1},l_{\tilde{g}_1})$ as a second factor. Its first factor serves then for the search of an appropriate Alice’s element of the first orbit and the procedure is repeated. As a result, we obtain a closed cycle (see below). Then we select any element which does not belong to the cycle and repeat all steps. We arrive at the disjoint set of cycles of the same length and all 48 elements of both orbits belong to some cycle. Having this decomposition at hand we have to select the maximal set of cycles such that the corresponding set of vertices has the following property: for any $\alpha$, $\alpha '$ it contains at most one vertex $v_{\alpha l}\otimes v_{\alpha ' l'}$. The total number of edges gives the classical bound on the sum $S(\tilde{g}_1,\tilde{g}_2)$.

In grouptheoretical language, we start with the first orbit and select some element $g_0v\otimes g_0\tilde{g}_1v$; then we look for the element of the second orbit containing $g_0\tilde{g}_1v$ as its second factor. It reads $g_0\tilde{g}_1\tilde{g}_2^{-1}v\otimes g_0\tilde{g}_1\tilde{g}_2^{-1}\tilde{g}_2v$. Coming back to the first orbit, we find the vector with the same first factor of tensor product, $g_0\tilde{g}_1\tilde{g}_2^{-1}v\otimes g_0\tilde{g}_1\tilde{g}_2^{-1}\tilde{g}_1v$ and repeat the procedure. In this way, we arrive at the sequence of vectors of the form

$$\begin{aligned} g_0(\tilde{g}_1\tilde{g}_2^{-1})^{k}v\otimes g_0(\tilde{g}_1\tilde{g}_2^{-1})^{k}\tilde{g}_1v, \quad k=0,1,\ldots \end{aligned}$$

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from the first orbit and

$$\begin{aligned} g_0(\tilde{g}_1\tilde{g}_2^{-1})^{k+1}v\otimes g_0(\tilde{g}_1\tilde{g}_2^{-1})^{k+1}\tilde{g}_2v, \quad k=0,1,\ldots \end{aligned}$$

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from the second one. These cycle doses if $(\tilde{g}_1\tilde{g}_2^{-1})^k = e$; therefore, for the length of the cycle (the number of vertices or edges) equals twice the order of $\tilde{g}_1\tilde{g}_2^{-1}$.

Below we give the examples of this algorithm for $B=8,\,12,\,14$ and 16.

$B=8$: the pair of orbits containing the vectors $(v\otimes v_{42},v\otimes v_{70})$

$B=12$: the pair of orbits containing the vectors $(v\otimes v_{40},v\otimes v_{72})$

Table 2 The eigenvalues of $X(\tilde{g},v)$ for all elements $\tilde{g}\in S_4$

$B=14$: the pair of orbits containing the vectors $(v\otimes v_{20},v\otimes v_{32})$

$B=16$: the pair of orbits containing the vectors $(v\otimes v_{62},v\otimes v_{82})$