A quantum state \(\left| \psi \right\rangle \in \mathcal {H} = \mathcal {H}_1 \otimes \mathcal {H}_2\), where \(\mathcal {H}_1\) and \(\mathcal {H}_2\) are Hilbert spaces and \(\mathcal {H}\) is their tensor product space, is said to be entangled (with respect to \(\mathcal {H}_1\) and \(\mathcal {H}_2\)) if there do not exist substates \(\left| \psi _1\right\rangle \in \mathcal {H}_1\) and \(\left| \psi _2\right\rangle \in \mathcal {H}_2\) such that \(\left| \psi \right\rangle = \left| \psi _1\right\rangle \otimes \left| \psi _2\right\rangle \). A state which is not entangled is said to be separable.
This is a mathematical definition concerning properties of vector spaces which, taken at face value, suggests that there are plenty of classical systems that are entangled. For example, the acoustic field of a general, coupled mode solution to sound propagation in the ocean is, in this sense, an entangled state (though perhaps non-separable would be a more accurate description). The term classical entanglement has been suggested to describe certain (classical) coherent optical states that exhibit properties similar to that of entangled quantum systems [32–34], though the lack of actual statistical outcomes makes this association somewhat dubious. Classical entanglement, as it is construed in these works, refers only to nonseparability. In the context of quantum mechanics, and in keeping with its original use by E. Schrödinger, entanglement connotes statistical correlations [35].
In what follows, we will consider statistical manifestations of entanglement arising from the proposed detector-based model. Given an entangled (i.e., non-separable) design state \(\left| \psi \right\rangle \) and corresponding random vector \(\varvec{a}\), we would like to know whether the latter is entangled, and in what sense. Rather than appeal to vector space properties, we look instead to an equivalence of statistical predictions. In particular, we will examine the statistics of “detector clicks” for various observables.
A bit of notation may help. Let \(\left| \uparrow \right\rangle = [1, 0]^\mathsf {T}\) and \(\left| \downarrow \right\rangle = [0, 1]^\mathsf {T}\) denote the eigenvectors of \(Z\), with eigenvalues \(+1\) and \(-1\), respectively. These may be interpreted as orthogonal polarizations of a single photon. The states \(\left| \uparrow \right\rangle \) and \(\left| \downarrow \right\rangle \) may also be viewed as the qubit states \(\left| 0\right\rangle \) and \(\left| 1\right\rangle \), respectively, in the computational basis [36]. If \(\mathcal {H}_1 = \mathcal {H}_2 = \mathrm {span}\{\left| \uparrow \right\rangle , \left| \downarrow \right\rangle \}\), then \(\mathcal {H} = \mathcal {H}_1 \otimes \mathcal {H}_2\) is spanned by the following (separable) orthonormal basis states
$$\begin{aligned} \left| 1\right\rangle&= \left| \uparrow \right\rangle \otimes \left| \uparrow \right\rangle = \left| \uparrow \uparrow \right\rangle , \quad \left| 2\right\rangle = \left| \uparrow \right\rangle \otimes \left| \downarrow \right\rangle = \left| \uparrow \downarrow \right\rangle \\ \left| 3\right\rangle&= \left| \downarrow \right\rangle \otimes \left| \uparrow \right\rangle = \left| \downarrow \uparrow \right\rangle , \quad \left| 4\right\rangle = \left| \downarrow \right\rangle \otimes \left| \downarrow \right\rangle = \left| \downarrow \downarrow \right\rangle \end{aligned}$$
This four-dimensional (i.e., two-qubit) space is the one we will investigate.
Bell States
The Bell states are a set of four maximally entangled states that also form an orthonormal basis for our four-dimensional Hilbert space, \(\mathcal {H}\). They are given by
$$\begin{aligned} \left| \phi _1\right\rangle&= \frac{1}{\sqrt{2}} \left[ \left| \uparrow \uparrow \right\rangle + \left| \downarrow \downarrow \right\rangle \right] , \quad \left| \phi _2\right\rangle = \frac{1}{\sqrt{2}} \left[ \left| \uparrow \downarrow \right\rangle + \left| \downarrow \uparrow \right\rangle \right] \\ \left| \phi _3\right\rangle&= \frac{1}{\sqrt{2}} \left[ \left| \uparrow \uparrow \right\rangle - \left| \downarrow \downarrow \right\rangle \right] , \quad \left| \phi _4\right\rangle = \frac{1}{\sqrt{2}} \left[ \left| \uparrow \downarrow \right\rangle - \left| \downarrow \uparrow \right\rangle \right] \end{aligned}$$
Suppose \(\left| \psi \right\rangle = \left| \phi _2\right\rangle \), so the component vector in the standard basis is \(\varvec{\alpha } = [0,1,1,0]^\mathsf {T}/\sqrt{2}\). Now consider the 16 Hermitian matrices composed of pair-wise tensor products of the four Pauli matrices \(I\), \(X\), \(Y\), and \(Z\); i.e., \(I\otimes I, I\otimes X, \ldots , Z\otimes Z\). It is readily verified that the Hilbert-Schmidt inner product between any two different pairs is zero, as this property holds for the Pauli matrices themselves. Between themselves, they take on the value \(4\). Hence these operators, when suitably normalized, form a tomographically complete quorum set, which may be used to deduce the quantum state from their expectation values using QST.
Going further, one finds that the application of each member of the quorum on \(\varvec{\alpha }\) results in a vector that, like \(\varvec{\alpha }\), has components that are either zero or of equal magnitude. With suitable choices of \(s\), \(\gamma \), and \(\varvec{{w}}\), then, Theorem 1 or 3 holds, depending upon the choice of distribution for \(\varvec{{w}}\), and the distribution of detected outcomes follows the Born rule. Consequently, the expectation values of the random variables corresponding to each observable match the quantum predictions exactly (in the case of Theorem 3) or asymptotically (in the case of Theorem 1). Tomographically, then, the random vector \(\varvec{a} = s\varvec{\alpha } + \varvec{{w}}\) is equivalent to the entangled state \(\left| \psi \right\rangle \). It is straightforward to show that a similar result holds for the other three Bell states.
For example, let \(s = (\sqrt{2}-1)\sigma \), \(\gamma = \sigma > 0\), and \(\varvec{{w}} = \sigma \varvec{z}/\left\| \varvec{z}\right\| \). Now suppose we measure \(Z\) on both the first (left) particle and second (right) particle.. For the first particle, we will obtain either a projection onto the subspace \(\mathrm {Span}\{\left| \uparrow \uparrow \right\rangle , \left| \uparrow \downarrow \right\rangle \}\), if the result is \(+1\), or \(\mathrm {Span}\{\left| \downarrow \uparrow \right\rangle , \left| \downarrow \downarrow \right\rangle \}\), if the result is \(-1\). On the other hand, for the second particle we will obtain a projection onto either the subspace \(\mathrm {Span}\{\left| \uparrow \uparrow \right\rangle , \left| \downarrow \uparrow \right\rangle \}\), if one measures \(+1\), or \(\mathrm {Span}\{\left| \uparrow \downarrow \right\rangle , \left| \downarrow \downarrow \right\rangle \}\), if one measures \(-1\). Of course, it is also possible that neither measurement yields a detection, but let us suppose that they both do. Since the random amplitudes of outcomes \(\left| \uparrow \uparrow \right\rangle \) and \(\left| \downarrow \downarrow \right\rangle \) are \(|a_1| \le \sigma \) and \(|a_4| \le \sigma \), respectively, these outcomes will never occur. Furthermore, since \(\mathrm {E}[|a_2|] = \mathrm {E}[|a_3|]\), the outcomes \(\left| \uparrow \downarrow \right\rangle \) and \(\left| \downarrow \uparrow \right\rangle \) will be equally likely. Since, by Theorem 3, no more than one detection is possible, we conclude that, if we obtain \(\left| \uparrow \right\rangle \) (i.e., \(+1\)) for particle 1, then we must obtain \(\left| \downarrow \right\rangle \) (i.e., \(-1\)) for particle 2, and vice versa. The detected outcomes are thus perfectly anti-correlated, as one might expect for an entangled pair.
Of course, this doesn’t work for every observable. Consider the Hermitian matrix \(I\otimes B_+\), which is diagonalized by the unitary matrix \(I\otimes W_+\). For \(\left| \psi \right\rangle = \left| \phi _2\right\rangle \), the quantum probabilities \(|\left\langle n\right| (I\otimes W_+)\left| \psi \right\rangle |^2\), for \(n = 1, \ldots , 4\), are approximately \(0.0732, \, 0.4268, \, 0.4268, \, 0.0732\). The conditional detection probabilities are found numerically to be approximately \(0.0364, \, 0.4608, \, 0.4641, \, 0.0388\), with an uncertainty of about \(0.004\). Though similar, these values are only an approximation to the quantum predictions.
Violations of Bell’s Inequality
Another hallmark of entanglement is the possibility of violating Bell’s inequality. More precisely, the CHSH inequality is given by [37]
$$\begin{aligned} S_\mathrm{B} = \bigl | \mathrm {E}[AB] + \mathrm {E}[AB'] \bigr | + \bigl | \mathrm {E}[A'B] - \mathrm {E}[A'B'] \bigr | \le 2 \; , \end{aligned}$$
(24)
where \(A\), \(A'\), \(B\), \(B'\) are random variables bounded by unity and \(\mathrm {E}[\,\cdot \,]\) is the expectation with respect to some probability measure \(P\). It is important to note that, in order for this inequality to hold, the same probability measure \(P\) is used for all four expectation values. Thus,
$$\begin{aligned} \mathrm {E}[AB]&= \int A(\varvec{a}) B(\varvec{a}) dP(\varvec{a}) \end{aligned}$$
(25a)
$$\begin{aligned} \mathrm {E}[AB']&= \int A(\varvec{a}) B'(\varvec{a}) dP(\varvec{a}) \end{aligned}$$
(25b)
$$\begin{aligned} \mathrm {E}[A'B]&= \int A'(\varvec{a}) B(\varvec{a}) dP(\varvec{a}) \end{aligned}$$
(25c)
$$\begin{aligned} \mathrm {E}[A'B']&= \int A'(\varvec{a}) B'(\varvec{a}) dP(\varvec{a}) \end{aligned}$$
(25d)
The analogous expressions for quantum mechanics replace \(\mathrm {E}[AB]\), say, with \(\langle AB\rangle = \left\langle \psi \right| AB\left| \psi \right\rangle \) for a fixed quantum state \(\left| \psi \right\rangle \). In particular, if \(\left| \psi \right\rangle \) is the Bell state \(\left| \phi _2\right\rangle \), and the four observables are
$$\begin{aligned} A = Z\otimes I, \quad B&= -I\otimes (X+Z)/\sqrt{2} \\ A' = X\otimes I, \quad B'&= +I\otimes (X-Z)/\sqrt{2} \end{aligned}$$
then
$$\begin{aligned} S_\mathrm{Q} = \left| \left\langle AB\right\rangle + \left\langle AB'\right\rangle \right| + \left| \left\langle A'B\right\rangle - \left\langle A'B'\right\rangle \right| = 2\sqrt{2} \; . \end{aligned}$$
(26)
How can we reconcile this result with the CHSH inequality? As we have noted, the theorem applies to expectations that are with respect to the same probability measure. If the probability measures differ for each pair of observables, then the inequality need no longer hold. Now, in the model proposed here, expectations are with respect to conditional probability distributions, conditioned, that is, on single detections. Let these be denoted \(\mathrm {E}_1[AB]\), \(\mathrm {E}_2[AB']\), \(\mathrm {E}_3[A'B]\), and \(\mathrm {E}_4[A'B']\). This means that Bell’s theorem does not apply and violations of the inequality are possible. It remains to ask whether
$$\begin{aligned} S_\mathrm{D} = \bigl | \mathrm {E}_1[AB] + \mathrm {E}_2[AB'] \bigr | + \bigl | \mathrm {E}_3[A'B] - \mathrm {E}_4[A'B'] \bigr | \end{aligned}$$
(27)
can ever be greater than 2.
A numerical study was performed to investigate this possibility. As before, a random realization of \(M = 2^{20}\) independent noise vectors of the form \(\varvec{{w}} = \sigma \varvec{z}/\left\| \varvec{z}\right\| \), with \(\sigma = 1\), was drawn. Using \(\varvec{\alpha } = [0, \, 1, \, 1, \, 0]^\mathrm {T}\), a set of \(M\) complex-valued vectors of the form \(\varvec{a} = s\varvec{\alpha } + \varvec{{w}}\), with \(s = (\sqrt{2}-1)\sigma \), was created. A detection threshold of \(\gamma = \sigma \) was used.
To this set the Hermitian conjugates of the unitary matrices \(U_1 = I\otimes W_+\), \(U_2 = I\otimes W_-\), \(U_3 = H\otimes W_+\), and \(U_4 = H\otimes W_-\) were applied separately to \(\varvec{a}\) for the observables \(AB\), \(AB'\), \(A'B\), and \(A'B'\), respectively. For each of the four measurements, the diagonalized matrix of eigenvalues is
$$\begin{aligned} U_1^\dagger (AB) U_1&= \mathrm {diag}([-1, \, +1, \, +1, \, -1]) \end{aligned}$$
(28a)
$$\begin{aligned} U_2^\dagger (AB') U_2&= \mathrm {diag}([+1, \, -1, \, -1, \, +1]) \end{aligned}$$
(28b)
$$\begin{aligned} U_3^\dagger (A'B) U_3&= \mathrm {diag}([-1, \, +1, \, +1, \, -1]) \end{aligned}$$
(28c)
$$\begin{aligned} U_4^\dagger (A'B') U_4&= \mathrm {diag}([+1, \, -1, \, -1, \, +1]) \end{aligned}$$
(28d)
Of the \(M\) realizations of \(\varvec{a}\), typically only about 5 % resulted in a detection, though it was a different 5 % for each of the four observables. (This should not be confused with the detector efficiency, which is a measure of coincidence rates and no meaning in this context.) Let \(I_1\) denote the set of values of \(\varvec{a}\), a subset of all \(M\) realizations, that resulted in a detection for the observable \(AB\). Define \(I_2\), \(I_3\), and \(I_4\) similarly for \(AB'\), \(A'B\), and \(A'B'\), respectively. Furthermore, let \(I_{ij}\) denote the subset of \(I_i\) for which the \(j^\mathrm {th}\) component exceeded the threshold. Note that \(I_{i1}\), \(I_{i2}\), \(I_{i3}\), and \(I_{i4}\) are mutually exclusive, and their union is \(I_i\). Finally, let \(n_{ij}\) denote the cardinality of \(I_{ij}\) and \(n_i\) the cardinality of \(I_i\).
The results of the numerical simulation are summarized in Table 1. From these results we deduce mean values of \(\mathrm {E}_1[AB] = 0.8497\), \(\mathrm {E}_2[AB'] = 0.8440\), \(\mathrm {E}_3[A'B] = -0.8486\), and \(\mathrm {E}_4[A'B'] = 0.8481\), each with an uncertainty of about \(0.004\), computed as \(1/\sqrt{n_i}\). Combining these results, we find
$$\begin{aligned} S_\mathrm{D} = 3.39 \pm 0.016 \; , \end{aligned}$$
(29)
where the uncertainties of the four correlations were added to arrive at the final uncertainty in \(S_\mathrm{D}\). This is clearly greater than \(2\) and, in fact, greater than the Tsirelson bound of \(2\sqrt{2} = 2.8284\), which is an upper bound on quantum violations of the CHSH inequality.
Table 1 Results of a numerical simulation to test violations of Bell’s inequality using \(\varvec{{w}} = \sigma \varvec{z} / \left\| \varvec{z}\right\| \)
A similar numerical study was performed using \(\varvec{{w}} = \sigma \varvec{z}\), with \(\sigma = 1\), \(s = \sigma \), and \(\gamma = 3\sigma \). The results of this study are summarized in Table 2, from which we deduce the following mean values: \(\mathrm {E}_1[AB] = 0.6403\), \(\mathrm {E}_2[AB'] = 0.6484\), \(\mathrm {E}_3[A'B] = -0.6695\), and \(\mathrm {E}_4[A'B'] = 0.6733\), each with an uncertainty of about \(0.02\). Combining these results, we find
$$\begin{aligned} S_\mathrm{D} = 2.63 \pm 0.08 \; . \end{aligned}$$
(30)
Although the correlations are not as strong, we do find again that Bell’s inequality is violated.
Table 2 Results of a numerical simulation to test violations of Bell’s inequality using \(\varvec{{w}} = \sigma \varvec{z}\). About 0.25 % of the \(M = 2^{20}\) realizations resulted in a single detection
These results demonstrate that a (classical) deterministic model can violate Bell’s inequality. Such violations are made possible by the fact that the model is contextual, and this contextuality is, in turn, a consequence of our conditioning on single-detection events. In some cases, these violations can be larger than those predicted by quantum mechanics. This is so despite the fact that the Born rule is not perfectly reproduced for all observables concerned.
Local Realism
The notion that quantum mechanics is at odds with local realism first arose in the context of the Einstein–Podolsky–Rosen (EPR) paradox [38] and later by Bohm in terms of discrete states [39]. This paradox was recast by Bell [4] into an inequality that, he concluded, no local realistic theory could violate. A variation of this inequality was first tested by Clauser[40] and, in a later landmark experiment, by Aspect [41], with results in agreement with quantum predictions. This is generally regarded as conclusive evidence that quantum mechanics, and hence nature, is fundamentally nonlocal, despite the fact that it has been known for some time that violations of Bell’s inequality are possible under the so-called detection loophole [42–44]. The analysis of the previous section reconfirms this through a specific example, but it did not address local realism, as the two observables were effectively measured as one. Although recent experiments, with detector efficiencies over 70 %, claim to have closed the detection loophole for photons, these results do not address violations of Bell’s inequality but, rather, a little-known inequality due to Eberhard used to test local realism while accounting for detector inefficiencies [45–47].
Let us, then, consider an alternate scheme whereby the two observables are measured separately and independently. This corresponds to the usual sense of local realism in the context of Bell’s inequality, namely, that the choice of \(A\) or \(A'\) and its outcome are not influenced by (nor do they influence) the choice of \(B\) or \(B'\) and its outcome. As is common in such discussions, we will describe this situation in terms of two familiar actors.
Suppose Alice and Bob, who live in different cities, both receive a letter upon which is written a particular realization, say, \(s\varvec{\alpha }+\varvec{{w}} = [-0.165+0.2046i, \, 0.8316+0.6696i, \, 0.5690-0.2230i, \, 0.2321-0.1111i]^\mathsf {T}\), of the Bell state \(\varvec{\alpha } = [0, 1, 1, 0]^\mathsf {T}/\sqrt{2}\) with \(s = (\sqrt{2}-1)\sigma \), \(\varvec{{w}} = \sigma \varvec{z}/\left\| \varvec{z}\right\| \), and \(\sigma = 1\). At a previously agreed upon time they each select a measurement to perform on it. Alice chooses either \(A = Z\otimes I\) or \(A' = X\otimes I\), while Bob chooses either \(B = -I\otimes (X+Z)/\sqrt{2}\) or \(B' = I\otimes (X-Z)/\sqrt{2}\). The choice and outcome are written down but not shared until later.
These measurements are performed, not by a device, but by simple pencil-and-paper calculations whereby Alice multiplies \(s\varvec{\alpha }+\varvec{{w}}\) by either \((I\otimes I)^\dagger \), to measure \(A\), or \((H\otimes I)^\dagger \), to measure \(A'\), thereby obtaining the result \(\varvec{a}\) or \(\varvec{a}'\), respectively. Using a detection threshold \(\gamma = \sigma \), she determines the result of the measurement for, say, \(A\) by noting if either \(|a_1|^2 + |a_2|^2 > \gamma ^2\), in which case she writes “\(+1\)”, or \(|a_3|^2 + |a_4|^2 > \gamma ^2\), in which case she writes “\(-1\)”. This is so because
$$\begin{aligned} (I\otimes I)^\dagger A (I\otimes I) = \mathrm {diag}([+1, +1, -1, -1]) \; . \end{aligned}$$
(31)
If neither is the case, or if both are true, she simply writes “NaN” (Not a Number). The same procedure is followed when measuring \(A'\), since
$$\begin{aligned} (H\otimes I)^\dagger A' (H\otimes I) = \mathrm {diag}([+1, +1, -1, -1]) \; . \end{aligned}$$
(32)
Bob proceeds in much the same manner as Alice, multiplying \(s\varvec{\alpha }+\varvec{{w}}\) by either \((I\otimes W_+)^\dagger \), to measure \(B\), or \((I\otimes W_-)^\dagger \), to measure \(B'\) and denoting the respective results \(\varvec{b}\) and \(\varvec{b}'\). To determine the outcome of measuring, say, \(B\), Bob examines whether \(|b_2|^2 + |b_4|^2 > \gamma ^2\), in which case he writes “\(+1\)”, or \(|b_1|^2 + |b_3|^2 > \gamma ^2\), in which case he writes “\(-1\)”. This is so because
$$\begin{aligned} (I\otimes W_+)^\dagger B (I\otimes W_+) = \mathrm {diag}([-1, +1, -1, +1]) \; . \end{aligned}$$
(33)
To measure \(B'\), Bob examines whether \(|b'_1|^2 + |b'_3|^2 > \gamma ^2\), in which case he writes “\(+1\)”, or \(|b'_2|^2 + |b'_4|^2 > \gamma ^2\), in which case he writes “\(-1\)”, since
$$\begin{aligned} (I\otimes W_-)^\dagger B' (I\otimes W_-) = \mathrm {diag}([+1, -1, +1, -1]) \; . \end{aligned}$$
(34)
Like Alice, if there are no threshold crossings, or two threshold crossings, he simply writes “NaN” for the outcome.
Suppose Alice chooses to measure \(A\). In this case, \(\varvec{a} = s\varvec{\alpha }+\varvec{{w}}\) and she finds that \(|a_1|^2 + |a_2|^2 = 1.209 > \gamma ^2\) while \(|a_3|^2 + |a_4|^2 = 0.4397 \le \gamma ^2\), so she writes down “\(+1\).” Suppose further that Bob independently chooses to measure \(B\) and thereby computes \(\varvec{b} = [0.1658+0.4453i, \, 0.8314+0.5403i, \, 0.6145-0.2485i, \, -0.0033-0.0173i]^\mathsf {T}\). Since \(|b_2|^2 + |b_4|^2 = 0.9836 \le \gamma ^2\) and \(|b_1|^2 + |b_3|^2 = 0.6651 \le \gamma ^2\), he records “NaN” — no valid outcome was obtained. Had Bob chosen to measure \(B'\) instead, he would have found \(\varvec{b}' = [0.7052 + 0.6969i, \, 0.4707 + 0.0672i, \, 0.4322 - 0.1880i, \, -0.4369 + 0.1635i]^\mathsf {T}\). In that case, since \(|b'_2|^2 + |b'_4|^2 = 0.4436 <= \gamma ^2\) and \(|b'_1|^2 + |b'_3|^2 = 1.2051 > \gamma ^2\), he would have recorded “\(+1\)” instead of “NaN” for the outcome.
Now suppose Alice and Bob play this game many, many times. For each instance of \(s\varvec{\alpha }+\varvec{w}\) they record which measurement they performed and whether they obtained “\(+1\),” “\(-1\),” or “NaN” as an outcome. When the game is over, they compare notes. All items on the list in which either Alice or Bob recorded “NaN” are struck out. Next, the results are grouped into four categories, corresponding to the four measurement combinations. Finally, the correlation is computed for each group.
A numerical study was performed along these lines, with \(M = 2^{20}\) realizations of \(s\varvec{\alpha }+\varvec{{w}}\) for each of the four measurement choices. In each case, about 30 % of the \(M\) realizations resulted in a single detection for either Alice or Bob, and about 10 % of the \(M\) realizations resulted in a single coincidence detection. This corresponded to a detector efficiency of about \(\eta = 0.33\) (the ratio of coincident to single detections), which is comparable to that of a good quantum optics experiment.
Table 3 summarizes the results of this study. For example, the number of times Alice obtained \(\uparrow (+1)\) and Bob obtained \(\downarrow (-1)\) when she measured \(A\) and he measured \(B\) was 12069. The total number of coincidences for this pair of observables was \(118251\), resulting in a mean correlation of \(\mathrm {E}_1[AB] = 0.5833\). Computing these four mean correlations allows us to compute \(S_\mathrm{D}\), which was found to be
$$\begin{aligned} S_\mathrm{D} = 2.3405 \pm 0.004 \; . \end{aligned}$$
(35)
This result is, of course, larger than 2 and, so, violates Bell’s inequality. As before, this was made possible by the fact that not all measurements resulted in a single coincidence detection for both Alice and Bob. This example, however, shows that local measurements made within a fully deterministic (i.e., classical) model can still violate Bell’s inequality.
Table 3 Results of a numerical simulation to test violations of Bell’s inequality under the local measurements
Interestingly, the violation is not as large as was found in Sect. 6.2. It seems, therefore that separated measurements have weaker correlations than joint measurements. Furthermore, if one uses \(\varvec{{w}} = \varvec{z}\) instead of \(\varvec{{w}} = \varvec{z}/\Vert \varvec{z}\Vert \), as was done above, no violation is observed. Introducing correlations in the initial noise term therefore seems to have the effect of strengthening the correlations in the measured outcomes. This suggests that a different choice of noise distribution could lead to a higher efficiency and an even larger violation.