The theorem of unified locality

Abstract

In this study, a theorem about the vectorization of the entangled-photons trajectories is presented, and through it, an effect equivalent to the unification of the individual localities of the entangled particles is evidenced, which will be confirmed in two scenarios: a theoretical demonstration, and four simple experiments carried out on an optical table. In this way, the existence of this possibility, in terms of entanglement, will be scientifically established when explaining the instantaneous synchronization of non-local outcomes as a result of local measurements from the vectorization of the entangled-photons trajectories without resorting to local hidden variables, or faster-than-light arguments. Finally, this explanation will be completely contained within the Theory of Special Relativity, eliminating entanglement as a showdown scenario between the two main pillars of Physics: Special Relativity, and Quantum Mechanics.

Article highlights

This study provides a new and revealing approach to the non-locality of entanglement, giving a justification for its instantaneity without resorting to faster-than-light arguments, and thus expanding the space of possibilities of quantum communication for the creation of new and more performing protocols.

1 Introduction

During the last 100 years, there has been a great debate about entanglement [1,2,3,4], in particular, due to the instantaneous synchronization of non-local outcomes from local measurements of entangled particles. The debate has focused on the fact that an instantaneous and non-local phenomenon only seems to be possible, prima facie, by resorting to faster-than-light (FTL) arguments [5,6,7], which contradicts the original arguments of the Theory of Special Relativity [8]. The debates between Albert Einstein and Neils Bohr on the matter are famous. On one hand, Einstein argued that such a phenomenon undermined the very foundations of the Theory of Special Relativity [8] since according to it, nothing can travel faster than light; on the other hand, he argued that it was absurd to think that reality took place exclusively from the observation and that it should take place independently of any measurement [9]. Specifically, Einstein considered the random nature of Quantum Mechanics [10,11,12], that is, its purely probabilistic basis before the observation, as inadmissible [9]. These arguments were refuted by Bohr, who thought in the opposite way to Einstein, that is, in the opinion of Bohr, the reality before the observation of the entanglement is purely probabilistic, becoming deterministic since the measurement, and only from that moment. Moreover, while Bohr did not delve into superluminal arguments to explain the behavior of entanglement from the observation of entangled particles; Albert Einstein, Boris Podolski, and Nathan Rosen [13] (EPR) considered that a theory that left in the hands of probability the behavior of a non-local phenomenon violates the first commandment of the Theory of Special Relativity [8] (i.e., nothing travels faster than light): at best, it could never be considered a complete theory [13]. In the second half of the twentieth century, the first efforts to elucidate this problem began to take shape, and so in 1964 [14], John Bell presented a proposal for an experimental test in the form of a theorem based on inequality, in such a way that if that inequality is not fulfilled, it automatically means that the entanglement cannot be explained by local hidden variables (LHV), and therefore it is a purely non-local phenomenon, thus contradicting EPR. In 1969 [15], John Clauser, Michael Horne, Abner Shimony, and Richard Holt presented an improvement to Bell's theorem based on the same criterion, that is, if the inequality they proposed does not hold, then both the locality and the LHV are completely excluded from the explanation of the entanglement. In 1982 [16, 17], Alan Aspect carried out the first two experimental verifications of Bell’s Theorem, although a great part of the scientific community questioned his experiments, considering them vitiated by loopholes. Just in 2015 [18], Ronald Hanson carried out the first experiment related to Bell’s Theorem, where he proclaimed the absence of loopholes. Notwithstanding any heated debate, the current consensus is that entanglement instantaneously synchronizes nonlocal outcomes from local measurements, i.e., it is a non-local phenomenon, which does not allow voluntarily sending information using these attributes, since the result of a quantum measurement is random, and it is precisely randomness which saves the Theory of Special Relativity [8]; since information cannot be sent voluntarily between two entangled particles based on local measurements of them. This argument is at the heart of the current debate about the very nature of entanglement for two reasons:

• 1. Randomness saves Special Relativity [8] because man cannot voluntarily send information to two distant points by making use of the instantaneous attributes of entanglement without resorting to a violation of the speed of light as nature’s maximum speed limit. Instead, if nature does this without man’s control, then isn’t there such a violation?

• 2. If Special Relativity [8] needs randomness to save it, then this does not speak very well of this theory, since it would depend on a kind of Russian roulette.

Regarding the first argument, Physics must explain an instantaneous transmission between two distant points, regardless of whether or not a man can obtain a benefit from this phenomenon, and as far as possible through concomitant actions with the pre-established theories that make it up, such as Special Relativity [8], which, and already concerning the second argument, should not need anything to save it, as it should happen with one of the most successful theories of Physics.

There have been innumerable efforts to try to explain how entanglement [1,2,3,4] works. Einstein himself resorted to an explanation of this phenomenon based on local hidden variables (LHV) [9], while Bohm did so in the 1950s based on non-local hidden variables (NLHV) [19]. Subsequently, other efforts were directed at bolder approaches such as superdeterminism [20], the multiverse [21], and even time retrieval [22]. Instead, in this study, we will base ourselves on a new approach, the vectorization of the entangled-photons trajectories, or space vectorization.

The outline of the paper is as follows: In Sect. 2, the space vectorization concept is introduced via a structure of Hypothesis, Thesis, and two Demonstrations, where the first one is theoretical, and the second one consists of four experimental verifications on an optical table. Section 3 presents a comparative discussion of the results obtained at the demonstrations. Finally, Sect. 4 deals with the general conclusions of this study.

2 The theorem of unified locality

From here on, the rest of the paper will be organized as a theorem, through which it will be shown that the individual localities of two entangled particles merge into a single and integrating locality that encompasses both entangled particles, i.e., arises the concept of unified locality. In this way, we will be able to overcome the pre-existing notion of non-locality [14,15,16,17,18], commonly associated with entangled particles, which leads to a head-on collision between Special Relativity [8, 23, 24], and Quantum Mechanics [10,11,12].

2.1 Entangled-photons trajectories vectorization (hypothesis)

Figure 1a represents an ultra-violet (UV) laser beam incident on two beta barium borate (BBO) Type-I crystals (together, one next to the other) rotated by 90º to produce Type-II down-conversion, i.e., a Bell state of type [1]:

$$\left| {\beta_{00} } \right\rangle = {{\left( {\left| {00} \right\rangle + \left| {11} \right\rangle } \right)} \mathord{\left/ {\vphantom {{\left( {\left| {00} \right\rangle + \left| {11} \right\rangle } \right)} {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}$$

(1)

Fig. 1

Similarities between a fragment of an optical table, specifically the one related to the resulting beams (signal and idler) of the double Type-I down-conversion (0º, and 90º) process, and the light-cone from the Special Relativity theory, where: a is the optical table seen from above, in particular, the process known as down-conversion whereby both beams of entangled photons are generated (signal, and idler), b is the light-cone with its isotemporal hyperplanes represented by circles, c represents the vectorization of the entangled-photons trajectories involved in the optical table, all in spatial units, and d is a slice of the light-cone parallel to the drawing plane in which the isotemporal hyperplanes can be seen as horizontal dot lines, where the similarity with the graph (c) can be appreciated, although unlike the (c) case the units are not homologated due to the fact that the vertical axis represents time (blue) while the other two axes (green) represent distances, with c being a simple relationship between distance and time

At time t₀, two beams start from point C, the left beam (signal) goes to point A, while the right beam (idler) goes to point B. Both points (A and B) are hit by both beams at the same time at instant t_m, at which point the quantum measurement [25] is performed. Photons from both beams travel at the speed of light c, so the angle between beam $\overline{AC}$ and the vertical blue line $\overline{DC}$ is the same as that between beam $\overline{BC}$ and the aforementioned vertical. Thus, the Euclidean distance $\left| {\overrightarrow {d}_{AB} } \right|$ is made up of two identical halves, i.e., $\left| {\overrightarrow {d}_{AD} } \right| = {{\left| {\overrightarrow {d}_{AB} } \right|} \mathord{\left/ {\vphantom {{\left| {\overrightarrow {d}_{AB} } \right|} 2}} \right. \kern-0pt} 2}$ and $\left| {\overrightarrow {d}_{BD} } \right| = {{\left| {\overrightarrow {d}_{BA} } \right|} \mathord{\left/ {\vphantom {{\left| {\overrightarrow {d}_{BA} } \right|} 2}} \right. \kern-0pt} 2}$, that is, $\left| {\overrightarrow {d}_{AD} } \right| = \left| {\overrightarrow {d}_{BD} } \right|$, where $\left| {\overrightarrow {d}_{AB} } \right| = \left| {\overrightarrow {d}_{BA} } \right|$, and | • | the modulus of “•”.

Based on the aforementioned geometric relationships, we are in a position to formulate the hypothesis of the theorem.

Hypothesis

Every beam corresponding to the path of an entangled photon is vectorized.

When we say that the paths of the entangled photons are vectorized, we mean that they have a direction and magnitude, which can be measured in meters or feet. Then, returning to Fig. 1, we can observe the extraordinary similarity between the down-conversion process of Fig. 1a, which takes place on an optical table, and represents the most conspicuous setting for experiments with entangled photons, with its corresponding counterpart, i.e., the light-cone of Fig. 1b, resulting from Special Relativity [8], where the trajectories of the photons are also represented in red, and which maintains a complete correspondence with experiments like the one in Fig. 1(a). However, the similarity between Figs. 1a and b is only apparent, as not all physical units involved are the same. For example, in Fig. 1a, the diagonal lines (red) (photon trajectories), the horizontal line (black), and the vertical line (blue) represent spatial dimensions. However, in Fig. 1b, the horizontal line (black) is an Euclidean distance, but the vertical line (blue) represents time, while the oblique lines (red) represent the relationship between space and time, which in this case is the speed of light, i.e., c = d/t. Figure 1b is completed as follows: (a) the horizontal circles represent isotemporal planes, in which different instances of the experiment take place, e.g., the horizontal line (black) between points A and B is the diameter of the circle corresponding to the instant t_m, (b) the upper part of the figure is the future light cone, (c) the lower part constitutes the past light cone, and (d) the coordinate center where the two spatial axes (green) intersect with the vertical timeline (blue) for instant t₀ is a point in the plane known as the hypersurface of the present generated from the two spatial axes (green). In the four plots of Fig. 1, the experiment starts at t₀ (point C) and ends at t_m (point D), i.e., with the quantum measurement. Both inside the upper cone (future light cone) and the lower one (past light cone) in Fig. 1b a subluminal process takes place. On the other hand, on the cones, any process is luminous, while outside the cones, all processes are superluminal, i.e., FTL [5,6,7]. Figure 1c corresponds exclusively to the upper part of Fig. 1a, that is, from point C (t₀) to point D (t_m), where the angle γ has to do with the starting angle of the photons from the BBO according to their wavelength (in this case, 810 nm). This angle occurs because both sides of the triangle, i.e., the photon's path (red) and its vertical projection (blue) are spatial dimensions. On the other hand, we do not see that angle in Fig. 1d, corresponding to a cut of the future light cone according to the space–time plane of Fig. 1b from t₀ to t_m, since all the sides of both triangles have different physical units. In Figs. 1a and c, the trajectories of the entangled photons (red) in both beams (signal and idler) have direction, with a magnitude and an angle, as well as a specific speed c = d/t. Therefore, these trajectories are vectorized. Consequently, we conjecture that the vectorization of the trajectories of the entangled photons can occur due to a phenomenon of persistence or memory of the state of each photon during each instant of its trajectory, and where the set of all these states are chained in space–time constituting an apparent rigid structure (where, we must understand by rigid, a linear ray that only changes direction if the direction of the BBO changes) which is vectorized with direction and module. In other words, the compilation of the states, at each instant of their trajectories, of all the photons emitted by the BBO, gives rise to a vectorial structure with module and angle (lattice). In consequence, the vectorization of the trajectories of the entangled photons (red) implies the vectorization of the quantum channel. At this point in the analysis, it does not matter if the Geometry of the trajectory of the entangled photons is not linear on the optical table, as it happens when we use optical fibers, which can have remarkably irregular paths. This is purely symbolic since what really counts is that both paths (those of both beams: signal and idler) are identical in length, so there will always exist an equivalent linear path like the one in Fig. 1a for an irregular trajectory, as long as both beams have the same length (see Appendix A), even considering the fact that in an optical fiber the light travels at a speed of only 2/3 of the one which takes place in a vacuum. In this context, where the paths of the entangled photons are vectorized for all instants t₀ ≤ t < t_m, we say that the entanglement is also vectorized until the photons arrive at points A and B, where the quantum measurement is made at instant t_m, and thus the collapse of the wave function occurs, all traces of vectorization disappearing, as well as the entanglement. In this way, during t₀ ≤ t < t_m, the trajectories of the entangled photons (red diagonal vectors) can be decomposed into their horizontal (black) and vertical (blue) components, facilitating the development of the following section.

2.2 Null equivalent channel and unified locality (thesis)

Next, we will address the development of the thesis from two points of view to conclude this section with a rigorous definition of it.

Geometry point of view: Based on Fig. 1c, $\forall t \in \left[ {t_{0} ,t_{m} } \right)$, in the triangle on the left, we have,

$$\overrightarrow {d}_{c} = \overrightarrow {d}_{t} + \overrightarrow {d}_{AD}$$

(2)

while in the triangle on the right, we have,

$$\overrightarrow {d}_{c} = \overrightarrow {d}_{t} + \overrightarrow {d}_{BD}$$

(3)

In the previous subsection, we saw that $\left| {\overrightarrow {d}_{AD} } \right| = \left| {\overrightarrow {d}_{BD} } \right|$, and since we are dealing with vector magnitudes in opposite directions, it results:

$$\overrightarrow {d}_{AD} + \overrightarrow {d}_{BD} = 0 = \overrightarrow {d}_{effective}$$

(4)

We call effective distance ($\overrightarrow {d}_{effective}$) to the physical magnitude of Eq. (4) resulting from the sum of both equal and opposite vectors, that we can see it is null regardless of the Euclidean distance between points A and B ($\overrightarrow {d}_{AB}$) in every moment. Consequently, the effective time (t_effective) that it takes for Bob to learn of a local measurement made by Alice is:

$$t_{effective} = \frac{{\left| {\overrightarrow {d}_{effective} } \right|}}{c} = 0$$

(5)

The result of Eq. (5) implies that Bob is instantly notified of the outcome of the measurement made by Alice, while the entanglement is in effect. Then, $\forall t/t \ge t_{m}$, the wave function collapses, the entanglement disappears, and therefore,

$$\overrightarrow {d}_{effective} = \overrightarrow {d}_{AB}$$

(6)

that is, the effective distance between A and B is simply the Euclidean distance between points A and B. Figure 2 shows the effective distance between both photons in blue dot lines, in such a way that, $\forall t \in \left[ {t_{0} ,t_{m} } \right)$ both photons are entangled, so that the $\overrightarrow {d}_{effective} = 0$, while for the instant t = t_m, the photons are mutually independent with a $\overrightarrow {d}_{effective} = \overrightarrow {d}_{AB}$, which can be observed both in the optical table of Fig. 2a, and in a cross-section parallel to the drawing plane of the light cone of Fig. 2b.

Fig. 2

Graphic representation of the effective distance (d_effective) between both photons in blue dot lines, while the entanglement is in force (t₀ ≤ t < t_m), and since the collapse of the wave function (t ≥ t_m), both for, a) the optical table, as well as for b) the light cone (cut parallel to the plane of the drawing). During the period t₀ ≤ t < t_m, d_effective = 0, while from the moment of measurement (t_m), d_effective = d_AB, that is, it is equal to the Euclidean distance between both points A and B

Figure 3a shows entangled photons with a $\overrightarrow {d}_{effective} = 0$ on the interval [t₀, t_m), such that points A and B are effectively coincident, where the qubit q[0] leaving the BBO is distributed to Alice and q[1] to Bob, however, it seems that they had never been separated, since an equivalent qubit arises as a result of the fusion of both qubits which we will call q[0–1]. It is as if both were mutually local. On the other hand, when applying a rotatable polarizer P(θ), which turns out to be a kind of weak measurement, the wave function collapses, both photons become independent, and points A and B recover their respective original locations (from the $\overrightarrow {d}_{effective}$ point of view, since in reality points A and B never stopped being in their original positions). In the latter case, both photons are non-local since the $\overrightarrow {d}_{effective} = \overrightarrow {d}_{AB}$, that is, equal to the Euclidean distance between points A and B. Figure 3b shows the $\overrightarrow {d}_{effective}$ as a function of time, with entanglement and hence locality unified before t_m, and not locality and independence of both photons from t_m. Moreover, Fig. 3a implies that if the locality is unified before t_m, then there should be no difference between applying a polarizer to only one beam, let’s say Alice, or applying a polarizer to each beam at the same time. This is a direct consequence that we will demonstrate in the next two sections, however, in experiments like the one in Fig. 3a, the one that really separates the photons is the “rotatable polarizer” and not the BBO, since the separation (distribution) from the latter is effectively apparent.

Fig. 3

Analysis on the optical table of the relationship among the d_effective, the locality, and the degree of correlation between both photons: a during the period t₀ ≤ t < t_m, both photons are entangled, with a d_effective = 0 between them, it is as if both qubits q[0], and q[1], were merged into a single equivalent qubit q[0–1], with a unified locality, while since the instant t_m, the action of a rotatable polarizer causes the collapse of the wave function, then, d_effective = d_AB, the photons become independent and each one recovers its own locality, and b representation of the d_effective as a function of time

Therefore, given that both beams (signal and idler) corresponding to the trajectories of two entangled particles are vectorized during the interval t₀ ≤ t < t_m, then the effective distance between both particles vanishes, since if we are talking about two entangled photons where one goes to point A and the other to point B, both photons being instantly synchronized in mirror paths, then the effective (equivalent) distance between A and B is zero. This is equivalent to a contraction of the $\overrightarrow {d}_{effective}$, or what is the same, a unification of the locality. As a direct consequence of this, any notification of a local measurement carried out on an entangled photon is instantly received by its counterpart, as Eq. (5) shows us. With the intervention of a rotatable polarizer P(θ) in any of the two beams (or in both at the same time) the unification of the locality disappears and both photons become independent as well as their localities. In this context, no FTL arguments [5,6,7] are needed for the aforementioned instant notification to take place.

Entropy point of view: Fig. 4 shows an entropic interpretation of the locality as a function of the degree of correlation between both photons. Therefore, given two subsystems (A, and B) that interact with each other in some way, their density matrices treated individually are,

$$\rho_{A} = \rho_{B} = \frac{1}{2}\left( {\left| 0 \right\rangle \left\langle 0 \right| + \left| 1 \right\rangle \left\langle 1 \right|} \right) = \frac{1}{2}I = \frac{1}{2}\left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]$$

(7)

while their von Neumann entropies [1] are,

$$S_{A} = S_{B} = - tr\left[ {\rho_{A} \;log\left( {\rho_{A} } \right)} \right] = - tr\left[ {\rho_{B} \;log\left( {\rho_{B} } \right)} \right] = - tr\left[ {\frac{1}{2}\left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]log\left( {\frac{1}{2}\left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]} \right)} \right] = 1$$

(8)

where tr(•) is the trace of the square matrix (•), and log(•) is logarithm base 2 of (•). Then, the entropy of the composed system $S_{A \cup B}$, or entropic distance δ_AB between them is,

$$\delta_{AB} = S_{A \cup B} = - tr\left[ {\rho_{A \cup B} \;log\left( {\rho_{A \cup B} } \right)} \right]$$

(9)

Fig. 4

Entropic distance δ_AB = S_A∪B, for: (a) t₀ ≤ t < t_m, in the presence of entanglement the locality is unified, then δ_AB = S_A∪B = 0, while for (b) t ≥ t_m, the entropic distance becomes maximum, i.e., δ_AB = S_A∪B = 2, with an entropic configuration compatible with that of two totally independent photons

The entopic distance δ_AB depends on the degree of correlation between both subsystems. In this analysis, we will only consider two cases: completely independent, or maximally entangled, as can be seen in Fig. 4. Besides, in the classical and quantum worlds, the correlations between the subsystems are those established by the additional information. In the case of composite quantum systems, mutual information $S_{A \cap B}$ is introduced to quantify that additional information, allowing us to obtain the degree of correlation between both subsystems [2, 26], i.e., the entropy of the composite system $S_{A \cap B}$ indicates that the uncertainty of a state $\rho_{A \cup B}$ is less than the two subsystems $S_{A}$ and $S_{B}$ added together. Therefore, the entropic distance, δ_AB in terms of the mutual information $S_{A \cap B}$, will be,

$$\delta_{AB} = S_{A \cup B} = S_{A} + S_{B} - S_{A \cap B} \ge 0$$

(10)

Now, if the mutual information $S_{A \cap B}$ defines the entropic distance, then, for completely independent photons, we have,

$$S_{A \cap B} = 0 \Rightarrow \delta_{AB} = 2$$

(11)

This result is perfectly logical, because if subsystems A and B are completely independent, then they do not share any information, for example, outcomes resulting from local measurements. This leads to a maximum entropic distance δ_AB of 2. While for maximally entangled photons,

$$S_{A \cap B} = 2 \Rightarrow \delta_{AB} = 0$$

(12)

This is the most important case since the mutual information or information exchanged is all the information available, and accordingly, the entropic distance between A (sender) and B (receiver) becomes zero. In other words, from the point of view of entropy, A and B are stuck, that is, in the same place, as if there were no sender and receiver or distance between them, i.e., the roles die and in reality, no one transmits or receives information, giving an appearance of total locality, because why should Alice transmit to Bob what he already has?

Then, during the interval t₀ ≤ t < t_m, Fig. 4a shows that both photons are entangled, and as a consequence of this the mutual information or $S_{A \cap B}$ is maximum, therefore the total entropy of both photons, or $S_{A \cup B}$, and the one that we call entropic distance or δ_AB is null, which is concomitant with the null effective distance of Figs. 2, and 3. On the other hand, in Fig. 4b we can see that since the moment t_m, the wave function collapses, then both photons do not have mutual information, i.e., $S_{A \cap B} = 0$, and therefore the total entropy of both photons $S_{A \cup B}$ or entropic distance δ_AB becomes maximum, which indicates a total independence between both photons. In this last case, the absence of mutual information is represented by the non-existence of an intersection between both sets, which is equivalent to saying that both locations are independent.

In short, two entangled photons have zero effective and entropic distances, therefore they both share the same locality regardless of the Euclidean distance that separates them. This allows us to define the thesis.

Thesis: Two entangled photons share the same locality, therefore the action of a given polarizer (at an arbitrary angle θ) on one or both beams at the same time should yield the same results.

2.3 Theoretical demonstration

Figure 5a shows us the complete setup needed for this demonstration. Next, we will analyze the different instances of a state $\left| {\Omega \left( t \right)} \right\rangle$, which crosses Fig. 5a, from left to right, according to its timeline. Therefore:

Fig. 5

Rotatable polarizers and strong measurements on a pair of entangled particles of the $\left| {\beta_{00} } \right\rangle$-type: (a) complete configuration with a control signal “c”, which selects the intervention of Bob’s rotatable polarizer, while θ establishes the angle of the polarizer, (b) c = 0, i.e., Bob without a rotatable polarizer, and (c) c = 1, Bob with a rotatable polarizer

At t₀:

$$\left| {\Omega \left( {t_{0} } \right)} \right\rangle = \left| {\beta_{00} } \right\rangle$$

(13)

At t₁:

$$\left| {\Omega \left( {t_{1} } \right)} \right\rangle = \left( {P\left( \theta \right)_{A} \otimes P\left( \theta \right)_{B}^{c} } \right)\left| {\beta_{00} } \right\rangle$$

(14)

where “$\otimes$” is Kronecker’s product [1], and

$$P\left( \theta \right) = R_{y} \left( \theta \right)P\left( {0^{ \circ } } \right)R_{y} \left( \theta \right)^{T}$$

(15)

with

$$R_{y} \left( \theta \right) = \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} & { - \sin \left( \theta \right)} \\ {\sin \left( \theta \right)} & {\cos \left( \theta \right)} \\ \end{array} } \right]$$

(16)

$$P\left( {0^{ \circ } } \right) = \left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 0 \\ \end{array} } \right]$$

(17)

and (•)^T means transpose of (•). So if c = 0, then, Bob's rotatable polarizer is deactivated, i.e., it disappears from the path of Bob's beam. So that

$$P\left( \theta \right)_{B}^{0} = I_{2 \times 2} = \left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]$$

(18)

Instead, if c = 1, then, Bob's rotatable polarizer is activated, i.e., it intersects the path of Bob's ray, then

$$P\left( \theta \right)_{B}^{1} = P\left( \theta \right)$$

(19)

Now, replacing Eqs.(16) and (17) into Eq. (15), we have

$$\begin{gathered} P\left( \theta \right) = R_{y} \left( \theta \right)P\left( {0^{ \circ } } \right)R_{y} \left( \theta \right)^{T} \hfill \\ \quad \quad \;{\kern 1pt} = \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} & { - \sin \left( \theta \right)} \\ {\sin \left( \theta \right)} & {\cos \left( \theta \right)} \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} & { - \sin \left( \theta \right)} \\ {\sin \left( \theta \right)} & {\cos \left( \theta \right)} \\ \end{array} } \right]^{T} \hfill \\ \quad \quad \;{\kern 1pt} = \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} & { - \sin \left( \theta \right)} \\ {\sin \left( \theta \right)} & {\cos \left( \theta \right)} \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} & {\sin \left( \theta \right)} \\ { - \sin \left( \theta \right)} & {\cos \left( \theta \right)} \\ \end{array} } \right] \hfill \\ \quad \quad \;{\kern 1pt} = \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} & { - \sin \left( \theta \right)} \\ {\sin \left( \theta \right)} & {\cos \left( \theta \right)} \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} & {\sin \left( \theta \right)} \\ 0 & 0 \\ \end{array} } \right] \hfill \\ \quad \quad \;{\kern 1pt} = \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)^{2} } & {\cos \left( \theta \right)\sin \left( \theta \right)} \\ {\sin \left( \theta \right)\cos \left( \theta \right)} & {\sin \left( \theta \right)^{2} } \\ \end{array} } \right]. \hfill \\ \end{gathered}$$

(20)

Therefore, for the case of Fig. 5b with c = 0, at t₁, and considering Eq. (20), we have

$$\begin{gathered} \left( {P\left( \theta \right) \otimes I_{2 \times 2} } \right)\left| {\beta_{00} } \right\rangle = \left( {\left( {R_{y} \left( \theta \right)P\left( {0^{ \circ } } \right)R_{y} \left( \theta \right)^{T} } \right) \otimes I_{2 \times 2} } \right)\left| {\beta_{00} } \right\rangle \hfill \\ = \left( {\left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)^{2} } & {\cos \left( \theta \right)\sin \left( \theta \right)} \\ {\sin \left( \theta \right)\cos \left( \theta \right)} & {\sin \left( \theta \right)^{2} } \\ \end{array} } \right] \otimes \left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}} \\ 0 \\ 0 \\ {{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}} \\ \end{array} } \right] \hfill \\ = \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)^{2} } & 0 & {\cos \left( \theta \right)\sin \left( \theta \right)} & 0 \\ 0 & {\cos \left( \theta \right)^{2} } & 0 & {\cos \left( \theta \right)\sin \left( \theta \right)} \\ {\sin \left( \theta \right)\cos \left( \theta \right)} & 0 & {\sin \left( \theta \right)^{2} } & 0 \\ 0 & {\sin \left( \theta \right)\cos \left( \theta \right)} & 0 & {\sin \left( \theta \right)^{2} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}} \\ 0 \\ 0 \\ {{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}} \\ \end{array} } \right] \hfill \\ = \left[ {\begin{array}{*{20}c} {{{\cos \left( \theta \right)^{2} } \mathord{\left/ {\vphantom {{\cos \left( \theta \right)^{2} } {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}} \\ {{{\cos \left( \theta \right)\sin \left( \theta \right)} \mathord{\left/ {\vphantom {{\cos \left( \theta \right)\sin \left( \theta \right)} {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}} \\ {{{\sin \left( \theta \right)\cos \left( \theta \right)} \mathord{\left/ {\vphantom {{\sin \left( \theta \right)\cos \left( \theta \right)} {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}} \\ {{{\sin \left( \theta \right)^{2} } \mathord{\left/ {\vphantom {{\sin \left( \theta \right)^{2} } {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}} \\ \end{array} } \right] = \frac{{\left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} \\ {\sin \left( \theta \right)} \\ \end{array} } \right] \otimes \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} \\ {\sin \left( \theta \right)} \\ \end{array} } \right]}}{\sqrt 2 } = \frac{{\left| {\Psi \left( \theta \right)} \right\rangle \otimes \left| {\Psi \left( \theta \right)} \right\rangle }}{\sqrt 2 }, \hfill \\ \end{gathered}$$

(21)

where

$$\left| {\Psi \left( \theta \right)} \right\rangle = \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} \\ {\sin \left( \theta \right)} \\ \end{array} } \right]$$

(22)

Now, for the case of Fig. 5c with c = 1, at t₁, and considering again Eq. (20), we have

$$\begin{gathered} \left( {P\left( \theta \right) \otimes P\left( \theta \right)} \right)\left| {\beta _{{00}} } \right\rangle = \left( {\left( {R_{y} \left( \theta \right)P\left( {0} \right)R_{y} \left( \theta \right)^{T} } \right) \otimes \left( {R_{y} \left( \theta \right)P\left( {0} \right)R_{y} \left( \theta \right)^{T} } \right)} \right)\left| {\beta _{{00}} } \right\rangle \hfill \\ = \left( {\left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)^{2} } & {\cos \left( \theta \right)\sin \left( \theta \right)} \\ {\sin \left( \theta \right)\cos \left( \theta \right)} & {\sin \left( \theta \right)^{2} } \\ \end{array} } \right] \otimes \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)^{2} } & {\cos \left( \theta \right)\sin \left( \theta \right)} \\ {\sin \left( \theta \right)\cos \left( \theta \right)} & {\sin \left( \theta \right)^{2} } \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ 0 \\ 0 \\ {{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ \end{array} } \right] \hfill \\ = \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)^{4} } & {\cos \left( \theta \right)^{3} \sin \left( \theta \right)} & {\cos \left( \theta \right)^{3} \sin \left( \theta \right)} & {\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } \\ {\cos \left( \theta \right)^{3} \sin \left( \theta \right)} & {\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } & {\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } & {\cos \left( \theta \right)\sin \left( \theta \right)^{3} } \\ {\cos \left( \theta \right)^{3} \sin \left( \theta \right)} & {\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } & {\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } & {\cos \left( \theta \right)\sin \left( \theta \right)^{3} } \\ {\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } & {\cos \left( \theta \right)\sin \left( \theta \right)^{3} } & {\cos \left( \theta \right)\sin \left( \theta \right)^{3} } & {\sin \left( \theta \right)^{4} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ 0 \\ 0 \\ {{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ \end{array} } \right] \hfill \\ = \left[ {\begin{array}{*{20}c} {{{\cos \left( \theta \right)^{4} } \mathord{\left/ {\vphantom {{\cos \left( \theta \right)^{4} } {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }} + {{\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } \mathord{\left/ {\vphantom {{\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ {{{\cos \left( \theta \right)^{3} \sin \left( \theta \right)} \mathord{\left/ {\vphantom {{\cos \left( \theta \right)^{3} \sin \left( \theta \right)} {\sqrt 2 + }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 + }}{{\cos \left( \theta \right)\sin \left( \theta \right)^{3} } \mathord{\left/ {\vphantom {{\cos \left( \theta \right)\sin \left( \theta \right)^{3} } {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ {{{\cos \left( \theta \right)^{3} \sin \left( \theta \right)} \mathord{\left/ {\vphantom {{\cos \left( \theta \right)^{3} \sin \left( \theta \right)} {\sqrt 2 + }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 + }}{{\cos \left( \theta \right)\sin \left( \theta \right)^{3} } \mathord{\left/ {\vphantom {{\cos \left( \theta \right)\sin \left( \theta \right)^{3} } {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ {{{{{\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } \mathord{\left/ {\vphantom {{\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }} + \sin \left( \theta \right)^{4} } \mathord{\left/ {\vphantom {{{{\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } \mathord{\left/ {\vphantom {{\cos \left( \theta \right)^{2} \sin \left( \theta \right)^{2} } {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }} + \sin \left( \theta \right)^{4} } {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ \end{array} } \right] \hfill \\ = \left[ {\begin{array}{*{20}c} {{{\cos \left( \theta \right)^{2} } \mathord{\left/ {\vphantom {{\cos \left( \theta \right)^{2} } {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ {{{\cos \left( \theta \right)\sin \left( \theta \right)} \mathord{\left/ {\vphantom {{\cos \left( \theta \right)\sin \left( \theta \right)} {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ {{{\sin \left( \theta \right)\cos \left( \theta \right)} \mathord{\left/ {\vphantom {{\sin \left( \theta \right)\cos \left( \theta \right)} {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ {{{\sin \left( \theta \right)^{2} } \mathord{\left/ {\vphantom {{\sin \left( \theta \right)^{2} } {\sqrt 2 }}} \right. \kern-\nulldelimiterspace} {\sqrt 2 }}} \\ \end{array} } \right] = \frac{{\left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} \\ {\sin \left( \theta \right)} \\ \end{array} } \right] \otimes \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} \\ {\sin \left( \theta \right)} \\ \end{array} } \right]}}{{\sqrt 2 }} = \frac{{\left| {\Psi \left( \theta \right)} \right\rangle \otimes \left| {\Psi \left( \theta \right)} \right\rangle }}{{\sqrt 2 }}, \hfill \\ \end{gathered}$$

(23)

since $\cos \left( \theta \right)^{2} + \sin \left( \theta \right)^{2} = 1.$

The equality between Eqs. (21) and (23) proves the theorem. This means that when the wave function collapses both in the configuration of Fig. 5b and in that of Fig. 5c, they do so in the same way. It is as if it were the same qubit.

The only way that the polarizer P(θ) applied on the qubit q[0] of Alice's beam affects the result on the qubit q[1] on Bob's beam is for q[0] and q[1] to be mutually local. It is as if it were a unified qubit q[0–1], which is concomitant with the unification of the individual localities into a single one. For this reason, applying the polarizer to one beam or both at the same time gives the same result. As we can see, the previous demonstration is totally independent of the value of the angle θ of the polarizer. However, in the following experimental demonstration on the optical table, we are forced to select some notable examples of angles such as 0º, 90º, 45º, and 135º, hoping, even with experimental errors involved, to access results similar to those of the theoretical proof.

2.4 Experimental demonstration on the optical table

Next, we proceed to experimentally demonstrate the thesis implementing the protocol of Fig. 5a on the optical table of Fig. 6. For this, we use an input laser beam or pump laser of 405 nm, with a range of power between 20 and 50 mW, which is used as the source of power. A gallium-nitride (GaN) diode laser is used for two reasons: (1) it has greater stability and temperature control, and (2) its short wavelength allows us to work with efficient detectors of 810 nm. Then, the blue diode laser beam (405 nm, 50 mW) passes through a narrow bandpass filter or quartz plate of 405 nm. Subsequently, the laser beam passes through a zero-order half-wave plate (HWP) with a phase of 22.5º, that is, which represents a Hadamard matrix [1]. This last step allows obtaining photons with a state of polarization of the diagonal type,

$$\left| + \right\rangle = {{\left( {\left| 0 \right\rangle + \left| 1 \right\rangle } \right)} \mathord{\left/ {\vphantom {{\left( {\left| 0 \right\rangle + \left| 1 \right\rangle } \right)} {\sqrt 2 }}} \right. \kern-0pt} {\sqrt 2 }}$$

(24)

Fig. 6

Implementation of the protocol of Fig. 5a on an optical table, where it is possible to observe that one of the brown cables coming out of the polarizers-controllers module (blue) acts on a motor that makes its rotatable polarizer appear and disappear from the path of Bob's beam, by making it move on some rails, regardless of the original function of both brown cables consisting of transmitting the signal corresponding to the θ angle

Then, the laser beam is directed to a solid block (5 × 5 × 3 mm³) of beta barium borate (BBO) to produce a Bell state of the type $\left| {\beta_{00} } \right\rangle$ like that of Eq. (1). For this, we need two BBO Type-I crystals together, one next to the other and where the second is rotated 90º respect to the first, to produce two Type-I down-conversion. The state $\left| + \right\rangle$ of Eq. (24) entering the first crystal, which generates a pair $\left| {HH} \right\rangle$, while the second crystal generates a pair $\left| {VV} \right\rangle$, in such a way that together they generate the state $\left| {\beta_{00} } \right\rangle$.

When using a 405 nm laser pump, the double Type-I down-conversion produces a 6º cone at the output of the second crystal, that is, 3º for the beam known as signal (810 nm) and 3º opposite for the beam known as idler (810 nm), with a phase matching angle for Type-I down-conversion of approximately 29º. In Fig. 6, the angle of both beams (signal and idler) with respect to an imaginary horizontal line was exaggerated to better appreciate the layout, and the beam path dimensions are not to scale. Moreover, several adjustment elements have not been incorporated into Fig. 6 so as not to complicate it.

Both in Alice’s beam and in Bob’s, we use rotatable polarizers of 810 nm with controllable angles through an individual waveplate controller connected to a laptop. The 810 nm calcite film rotatable polarizers are good throughout the visible spectrum and have high extinction ratios. On Bob’s side, his controller allows moving a cart that moves on rails and on which his rotatable polarizer is mounted to control its entry into the scene through a control signal c. In this way, when the control signal c = 0, corresponding to Fig. 5(b), only Alice will have a rotatable polarizer in the path of her beam, while if the control signal is c = 1, corresponding to Fig. 5c, both beams will hit their respective rotatable polarizers. Thus, it is possible to reproduce the configurations of Figs. 5b and c just by giving the control signal c the appropriate value. Two dual-wavelength 405/810 nm polarizing beam-splitter (PBS) of 0.5″ × 0.5″ × 0.5″ are used. When working at 810 nm, the avalanche photo-diodes (APD) have an efficiency of 60%, and we have worked with acquisition times ranging from 50 ms to 1 s, with and without an 810 nm filter before the APDs. A four-channel time-tagger device is used. Both wave plate controllers only control the angles of both rotatable polarizers, while the signal c is generated by the laptop through an Arduino circuit [27], which is not represented in Fig. 6 in order not to complicate the configuration of it.

This experiment can be reproduced by replacing the P(θ) rotatable polarizers with two possible alternatives:

$$P\left( \theta \right) = HWP\left( {\theta /2} \right) \times P\left( {0^{{\text{o}}} } \right) \times HWP\left( {\theta /2} \right) = EOM\left( \theta \right) \times P\left( {0^{{\text{o}}} } \right) \times EOM\left( \theta \right)$$

(25)

where HWP means zero-order half-wave plates of 808 nm, EOM is an electro-optical modulator, and P(0º) is an 810 nm calcite film polarizer fixed at horizontal polarization, i.e., 0º degree.

To prove the thesis, we will resort to four critical angles, i.e., 0º, 90º, 45º, and 135º, which will be replaced in Eq. (25), where θ = 0º corresponds to the horizontal polarization (H), 90º to the vertical (V), 45º to the diagonal (D), and 135º to the antidiagonal (A).

At this point, we must define the performance of the outcomes obtained with respect to the APDs, both for Alice and Bob in relation to the HV base of both PBS of Fig. 6. With these performances, we build the fidelities with which we will evaluate whether or not the thesis is verified experimentally.

Therefore, Alice’s performance with respect to the horizontal output of her PBS results from the photon counting carried out by their respective APDs (1, and 2) is then:

$$\eta_{H}^{A} = \frac{{n_{1} }}{{n_{1} + n_{2} }}$$

(26)

while for the case of the vertical output of her PBS, we have,

$$\eta_{V}^{A} = \frac{{n_{2} }}{{n_{1} + n_{2} }}$$

(27)

where n₁ represents the result of the photon count in the APD of Alice's exit #1, and n₂ corresponds to the photon count in the APD of Alice's exit #2. With similar criteria, on Bob's side result the following performances,

$$\eta_{H}^{B} = \frac{{n_{3} }}{{n_{3} + n_{4} }}$$

(28)

and

$$\eta_{V}^{B} = \frac{{n_{4} }}{{n_{3} + n_{4} }}$$

(29)

Being $\eta^{T} = \left\{ {\eta_{H}^{T} ,\eta_{V}^{T} } \right\}$ the expected theoretical performance in each case, the fidelities are:

$$F_{H}^{A} = 1 - \frac{{\left| {\eta_{H}^{T} - \eta_{H}^{A} } \right|}}{{max\left\{ {\eta_{H}^{T} ,\eta_{V}^{T} } \right\}}}$$

(30)

$$F_{V}^{A} = 1 - \frac{{\left| {\eta_{V}^{T} - \eta_{V}^{A} } \right|}}{{max\left\{ {\eta_{H}^{T} ,\eta_{V}^{T} } \right\}}}$$

(31)

$$F_{H}^{B} = 1 - \frac{{\left| {\eta_{H}^{T} - \eta_{H}^{B} } \right|}}{{max\left\{ {\eta_{H}^{T} ,\eta_{V}^{T} } \right\}}}$$

(32)

and

$$F_{V}^{B} = 1 - \frac{{\left| {\eta_{V}^{T} - \eta_{V}^{B} } \right|}}{{max\left\{ {\eta_{H}^{T} ,\eta_{V}^{T} } \right\}}}$$

(33)

Except for experimental error, the results regardless of the value of the control signal c should be similar and thus verify the thesis experimentally. Therefore, based on Fig. 7, which represents the experimental reproduction performances of the theoretical values for both Alice and Bob with the rotatable polarizer only on Alice's side for θ = 0º, Fig. 7a, and for θ = 90º, Fig. 7c, and with rotatable polarizers on both sides for θ = 0º, Fig. 7b, and θ = 90º, Fig. 7d, we proceed to calculate the respective fidelities.

Fig. 7

2D bars with the performances of the experiments corresponding to Fig. 5 for θ = 0º, and 90º: a c = 0, and θ = 0º, b c = 1, and θ = 0º, c c = 0, and θ = 90º, and d c = 1, and θ = 90º

For c = 0, and θ = 0º, Fig. 7a, yields:

$$F_{H}^{A} = 1 - \frac{{\left| {1 - 0.9651} \right|}}{{max\left\{ {0,1} \right\}}} = 1 - \frac{{\left| {1 - 0.9651} \right|}}{1} = 0.9651$$

(34)

and

$$F_{H}^{B} = 1 - \frac{{\left| {1 - 0.8109} \right|}}{{max\left\{ {0,1} \right\}}} = 1 - \frac{{\left| {1 - 0.8109} \right|}}{1} = 0.8109$$

(35)

For c = 1, and θ = 0º, Fig. 7b:

$$F_{H}^{A} = 1 - \frac{{\left| {1 - 0.9940} \right|}}{{max\left\{ {0,1} \right\}}} = 1 - \frac{{\left| {1 - 0.9940} \right|}}{1} = 0.9940$$

(36)

and

$$F_{H}^{B} = 1 - \frac{{\left| {1 - 0.9939} \right|}}{{max\left\{ {0,1} \right\}}} = 1 - \frac{{\left| {1 - 0.9939} \right|}}{1} = 0.9939$$

(37)

For c = 0, and θ = 90º, Fig. 7c:

$$F_{V}^{A} = 1 - \frac{{\left| {1 - 0.9766} \right|}}{{max\left\{ {1,0} \right\}}} = 1 - \frac{{\left| {1 - 0.9766} \right|}}{1} = 0.9766$$

(38)

and

$$F_{V}^{B} = 1 - \frac{{\left| {1 - 0.8148} \right|}}{{max\left\{ {1,0} \right\}}} = 1 - \frac{{\left| {1 - 0.8148} \right|}}{1} = 0.8148$$

(39)

For c = 1, and θ = 90º, Fig. 7d:

$$F_{V}^{A} = 1 - \frac{{\left| {1 - 0.9956} \right|}}{{max\left\{ {1,0} \right\}}} = 1 - \frac{{\left| {1 - 0.9956} \right|}}{1} = 0.9956$$

(40)

and

$$F_{V}^{B} = 1 - \frac{{\left| {1 - 0.9938} \right|}}{{max\left\{ {1,0} \right\}}} = 1 - \frac{{\left| {1 - 0.9938} \right|}}{1} = 0.9938$$

(41)

Figure 8 shows the performances of Fig. 7 via a 3D bars representation, where in all cases it is evident which theoretical value is being induced.

Fig. 8

3D bars with the performances of the experiments corresponding to Fig. 5 for θ = 0º, and 90º: a c = 0, and θ = 0º, b c = 1, and θ = 0º, c c = 0, and θ = 90º, and d c = 1, and θ = 90º

Figures 7 and 8 show that θ = 0º gives rise to $\left| 0 \right\rangle$ in both outputs, while θ = 90º gives rise to $\left| 1 \right\rangle$ in both outputs, regardless of whether we use rotatable polarizers on only Alice's side or on both sides at the same time. The first case turns out to be the most interesting, in such a way that with θ = 0º we have a state $\left| 0 \right\rangle$ in both outputs, while if θ = 90º, we have a $\left| 1 \right\rangle$ in both outputs, which can be interpreted as an instantaneous induction of binary information from Alice to Bob, without the intervention of a classical disambiguation channel, as happens in the quantum teleportation protocol [28], and without resorting to FTL arguments [17, 18], but rather the unification of the locality by the action of the vectorization of the trajectories of the entangled photons.

Next, we will evaluate the fidelities for the cases of inducing diagonal and antidiagonal states.

Therefore, based on Fig. 9, which represents the experimental reproduction performances of the theoretical values for both Alice and Bob with the rotatable polarizer only on Alice's side for θ = 45º, Fig. 9a, and for θ = 135º, Fig. 9c, and with rotatable polarizers on both sides for θ = 45º, Fig. 9b, and θ = 135º, Fig. 9d, we proceed to calculate the respective fidelities.

Fig. 9

2D bars with the performances of the experiments corresponding to Fig. 5 for θ = 45º, and 135º: a c = 0, and θ = 45º, b c = 1, and θ = 45º, c c = 0, and θ = 135º, and d c = 1, and θ = 135º

For c = 0, and θ = 45º, corresponding to Fig. 9a:

$$F_{H}^{A} = 1 - \frac{{\left| {0.5 - 0.5274} \right|}}{{max\left\{ {0.5,0.5} \right\}}} = 1 - \frac{{\left| {0.5 - 0.5274} \right|}}{0.5} = 0.9726$$

(42)

and

$$F_{H}^{B} = 1 - \frac{{\left| {0.5 - 0.4860} \right|}}{{max\left\{ {0.5,0.5} \right\}}} = 1 - \frac{{\left| {0.5 - 0.4860} \right|}}{0.5} = 0.9720$$

(43)

For c = 1, and θ = 45º, Fig. 9b:

$$F_{H}^{A} = 1 - \frac{{\left| {0.5 - 0.4864} \right|}}{{max\left\{ {0.5,0.5} \right\}}} = 1 - \frac{{\left| {0.5 - 0.4864} \right|}}{0.5} = 0.9728$$

(44)

and

$$F_{H}^{B} = 1 - \frac{{\left| {0.5 - 0.5} \right|}}{{max\left\{ {0.5,0.5} \right\}}} = 1 - \frac{{\left| {0.5 - 0.5} \right|}}{0.5} = 1$$

(45)

For c = 0, and θ = 135º, Fig. 9c:

$$F_{V}^{A} = 1 - \frac{{\left| {0.5 - 0.5323} \right|}}{{max\left\{ {0.5,0.5} \right\}}} = 1 - \frac{{\left| {0.5 - 0.5323} \right|}}{0.5} = 0.9354$$

(46)

and

$$F_{V}^{B} = 1 - \frac{{\left| {0.5 - 0.4945} \right|}}{{max\left\{ {0.5,0.5} \right\}}} = 1 - \frac{{\left| {0.5 - 0.4945} \right|}}{0.5} = 0.9890$$

(47)

For c = 1, and θ = 135º, Fig. 9d:

$$F_{V}^{A} = 1 - \frac{{\left| {0.5 - 0.5126} \right|}}{{max\left\{ {0.5,0.5} \right\}}} = 1 - \frac{{\left| {0.5 - 0.5126} \right|}}{0.5} = 0.9748$$

(48)

and

$$F_{V}^{B} = 1 - \frac{{\left| {0.5 - 0.5255} \right|}}{{max\left\{ {0.5,0.5} \right\}}} = 1 - \frac{{\left| {0.5 - 0.5255} \right|}}{0.5} = 0.9490$$

(49)

Figure 10 shows the performances of Fig. 9 via a 3D bars representation, where in all cases it is evident which theoretical value is being induced.

Fig. 10

3D bars with the performances of the experiments corresponding to Fig. 5 for θ = 45º, and 135º: a c = 0, and θ = 45º, b c = 1, and θ = 45º, c c = 0, and θ = 135º, and d c = 1, and θ = 135º

As for the case of the induction of states {H, V}, and taking into account the experimental error, similar results have been obtained regardless of the value of the control signal c, which proves the thesis experimentally. This is because in all cases fidelities exceeding 80% have been obtained when a state is induced both on Alice's side and on both sides at the same time.

3 Discussion

In the four examples of Fig. 7, we can see that given an alphabet with which to code messages to be transmitted between Alice and Bob, {H, V}, and considering that in all the examples the fidelities are never less than 80%, there is total discrimination when discerning whether the state that has been tried to be induced is a $\left| 0 \right\rangle$ or a $\left| 1 \right\rangle$ regardless of the value of the control signal c.

As we can see from the experiments carried out in Fig. 6, when Alice wishes to transmit computational basis states (CBS) to Bob, that is to say $\left\{ {\left| 0 \right\rangle ,\left| 1 \right\rangle } \right\}$, he will receive states of the type: $\left\{ {\left( {0.83 \pm 0.02} \right)\left| 0 \right\rangle + \left( {0.17 \mp 0.02} \right)\left| 1 \right\rangle ,\left( {0.17 \pm 0.02} \right)\left| 0 \right\rangle + \left( {0.83 \mp 0.02} \right)\left| 1 \right\rangle } \right\}$, respectively. This can be resolved in two ways:

• 1. Bob can perform post-processing whereby:

when he gets a $\left( {0.83 \pm 0.02} \right)\left| 0 \right\rangle + \left( {0.17 \mp 0.02} \right)\left| 1 \right\rangle \to {\text{post - processing}} \to \left| 0 \right\rangle$, and when he gets a $\left( {0.17 \pm 0.02} \right)\left| 0 \right\rangle + \left( {0.83 \mp 0.02} \right)\left| 1 \right\rangle \to {\text{post - processing}} \to \left| 1 \right\rangle$, or

• 2. Alice can apply an intensity amplifier in order to recover the 3 dB drop suffered as a result of applying the rotatable polarizer. See Eq. (21). In this last case, the action of the intensity amplifier frees Bob from having to apply post-processing, since instead of receiving the states mentioned above, he will receive:$\left( {0.96 \pm 0.02} \right)\left| 0 \right\rangle + \left( {0.04 \mp 0.02} \right)\left| 1 \right\rangle$, when Alice transmits a $\left| 0 \right\rangle$, and $\left( {0.04 \pm 0.02} \right)\left| 0 \right\rangle + \left( {0.96 \mp 0.02} \right)\left| 1 \right\rangle$, when Alice transmits a $\left| 1 \right\rangle$.In both cases, the fidelity goes from $\left( {83 \pm 2} \right)\%$ to $\left( {96 \pm 2} \right)\%$.

Another important aspect highlighted in this study is the concept of vectorization of the trajectories from which the annulment of the equivalent distance between entangled particles is directly derived, and therefore of the effective time, Eqs. (4), and (5), respectively. Moreover, as a consequence of the vectorization of the trajectories of Fig. 1 (i.e., the conjecture on which all this study revolves), which involves the distances between the entangled particles and the midpoint that separates them, the effective distance between them is canceled because those distances are of equal magnitude and in the opposite direction. See Eq. (4) and Fig. 1. Then, replacing this effective distance in Eq. (5), a null time results for the notification of a local measurement carried out, for example, in A to its counterpart in B. Therefore, Eq. (5) is a direct consequence of the cancellation of the effective distance. Besides, the theoretical demonstration of Sect. 2.3, also reflects this instantaneity, since if it is the same to put the polarizer at an angle θ only in A or in both (A and B) at the same time. This automatically tells us about the cancellation of the effective distance between both entangled particles, or what is the same in the unification of the locality. This completely coincides with what we know about entanglement, that is, its main attribute, instantaneity regardless of the distance that separates the entangled particles and that is confirmed in the experiments carried out in Sect. 2.4.

On the other hand, when we talk about null entropic distance, this is not a statement about the observers or the particles but about the channel and therefore the unification of the locality. There are many possible definitions of entropic distance; however, in this study, we use an extension of the Shannon-based information distance defined in Aslmarand-Roklin-Rajski [29,30,31]. See Eq. (10). In such a way that when the entropy of a system formed by two subsystems (A and B) is zero, i.e., $S_{A \cup B} = \delta_{AB} = 0$, this automatically implies that both subsystems (A and B) are maximally entangled and therefore share the same information. In the context of this study, this means that there is no entropic distance between the qubits q[0] and q[1]. See Fig. 3. Therefore, zero entropic distance means zero uncertainty about what is transmitted because the channel is minimized to zero length since both entangled qubits are equivalent to a single qubit.

Figure 3 shows the instances that both particles go through in time and their relationship with the level of correlation between them, in such a way that between t₀ and t_m, both particles are entangled and their locality is unified (according to trajectories vectorization), but from t_m both particles are completely independent and therefore non-local.

Finally, in Appendix B the relationship between the formalism presented in this study and Bell's.

Theorem [14] is fully developed, in which case we can observe that the unification of locality is concomitant with non-locality, to the point that the equality of the Eqs. (21) and (23) is in a way a proof of non-locality without inequalities, which was confirmed experimentally with the experiment in Fig. 6.

4 Conclusion

Through two demonstrations, one purely theoretical and the other experimental, it has been verified that states can be instantly induced from a transmitter to a receiver using an EPR channel without resorting to a classical disambiguation channel, as in the case of the quantum teleportation protocol [28], and without resorting to FTL arguments [17, 18], i.e., without deepening the rift between Quantum Mechanics [10,11,12] and Special Relativity [8]. Specifically, the theorem of unified locality is completely contained within the Theory of Special Relativity [8], elimina-ting entanglement as a showdown scenario between two of the main pillars of Physics: Special Relativity [8], and Quantum Mechanics [10,11,12]. This is because the individual locations of the entangled photons become entangled by uniting into a single equivalent locality, as shown in Fig. 3(a). This curious phenomenon is due to the vectorization of the trajectories of the entangled photons, which gives rise to the cancellation of the effective distance between the emitter and receiver during the entanglement period. Instead, the effective distance becomes equal to the Euclidean distance between the emitter and the receiver once the wave function has collapsed due to the quantum measurement [25], at which point both photons become completely independent, that is, non-local. Finally, the theoretical and experimental verification of the unified locality theorem will have a decisive impact on the understanding of the intrinsic mechanisms of entanglement [1, 4] for the creation of new quantum communication [32,33,34,35] and cryptography [35, 36] protocols with a strong commitment to the future quantum Internet [37,38,39,40,41,42,43,44,45,46,47,48,49,50].

Appendices

Appendix A

Fig. A1 see Fig. 11

Fig. 11

Validity of the analysis of Sect. 2 for irregular trajectories (although of equal length) of the entangled photons

Based on Fig. 11, suppose that immediately after the down-conversion process, that takes place in the BBO, two fiber optic adapters are applied in such a way that the entangled photons never move through free space, but always do through fiber optic cables. Then, two black-colored fiber optic cables ($\overline{A^{\prime}C}$ and $\overline{B^{\prime}C}$), with exaggerated zigzagging in their trajectories and of equal length, transport the entangled photons, from point C to A' and from the same point C to B', in such a way that if we linearly extend both fiber optic cables reaching the respective points A and B through linear trajectories equivalent in length of red color, we will see that the distance traveled by both entangled photons is the same, i.e., $\left| {\overrightarrow {d}_{AC} } \right| = \left| {\overrightarrow {d}_{BC} } \right|$. In other words, the black zigzagging path $\overline{A^{\prime}C}$ is of equal Euclidean length to the red linear equivalent path $\overline{AC}$, and the black zigzagging path $\overline{B^{\prime}C}$ is of the same Euclidean length as the red linear equivalent path $\overline{BC}$. However, the linear red paths between points A' and C, and B' and C have different lengths, i.e., $\left| {\overrightarrow {d}_{A^{\prime}C} } \right| \ne \left| {\overrightarrow {d}_{B^{\prime}C} } \right|$. If we now draw a blue bisector from point C that cuts segment $\overline{AB}$ in the middle, i.e., at point D, this generates two equal angles γ, as in Fig. 1c, such that $\left| {\overrightarrow {d}_{AD} } \right| = \left| {\overrightarrow {d}_{BD} } \right|$. Therefore, for zigzagging fiber optic paths, the analysis in Sect. 2 is still valid, the same as for the case of mixed configurations, i.e., free space with optical fiber.

Appendix B

Next, we will describe the relationship between Bell's Theorem [14] and the formalism proposed in this study in order to explore whether both are compatible.

Then, based on Eq. (22),

$$\left| {\psi \left( \theta \right)} \right\rangle = \cos \left( \theta \right)\left| 0 \right\rangle + \sin \left( \theta \right)\left| 1 \right\rangle = \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} \\ {\sin \left( \theta \right)} \\ \end{array} } \right]$$

(50)

and its partner in quadrature, that is,

$$\left| {\psi \left( {\theta + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)} \right\rangle = \cos \left( {\theta + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 0 \right\rangle + \sin \left( {\theta + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 1 \right\rangle = \left[ {\begin{array}{*{20}c} {\cos \left( {\theta + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)} \\ {\sin \left( {\theta + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - \sin \left( \theta \right)} \\ {\cos \left( \theta \right)} \\ \end{array} } \right]$$

(51)

we can compose the Bell state $\left| {\beta_{00} } \right\rangle$ of Eq. (1) in terms of the formalism used throughout this study, i.e. the qubit model of Eqs. (50) and (51),

$$\begin{gathered} \left| {\beta_{00} } \right\rangle = \frac{{\left| {\psi \left( \theta \right)} \right\rangle \left| {\psi \left( \theta \right)} \right\rangle + \left| {\psi \left( {\theta + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)} \right\rangle \left| {\psi \left( {\theta + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)} \right\rangle }}{\sqrt 2 } \hfill \\ \quad \quad {\kern 1pt} = \frac{{\left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} \\ {\sin \left( \theta \right)} \\ \end{array} } \right] \otimes \left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)} \\ {\sin \left( \theta \right)} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} { - \sin \left( \theta \right)} \\ {\cos \left( \theta \right)} \\ \end{array} } \right] \otimes \left[ {\begin{array}{*{20}c} { - \sin \left( \theta \right)} \\ {\cos \left( \theta \right)} \\ \end{array} } \right]}}{\sqrt 2 } \hfill \\ \quad \quad {\kern 1pt} = \frac{1}{\sqrt 2 }\left[ {\begin{array}{*{20}c} {\cos \left( \theta \right)^{2} + \sin \left( \theta \right)^{2} } \\ {\cos \left( \theta \right)\sin \left( \theta \right) - \sin \left( \theta \right)\cos \left( \theta \right)} \\ {\sin \left( \theta \right)\cos \left( \theta \right) - \cos \left( \theta \right)\sin \left( \theta \right)} \\ {\sin \left( \theta \right)^{2} + \cos \left( \theta \right)^{2} } \\ \end{array} } \right] = \frac{1}{\sqrt 2 }\left[ {\begin{array}{*{20}c} 1 \\ 0 \\ 0 \\ 1 \\ \end{array} } \right] = \frac{{\left| {00} \right\rangle + \left| {11} \right\rangle }}{\sqrt 2 }. \hfill \\ \end{gathered}$$

(52)

Once the aforementioned correspondence between Bell's state and the proposed formalism has been established, we will proceed to establish the link between the formalism and Bell's Theorem. For this, we will resort to the simplest proof of Bell's inequality in the literature proposed by Preskill [51], following Mermin's suggestions [52], and conveniently exposed by Lorenzo Maccone [53]; that is, under the conditions that three arbitrary two-valued properties A, B, C satisfy counterfactual definiteness and locality, and that P(X, X) = 1 for X = A, B, C (i.e. the two objects have same properties), the following inequality among correlations holds,

$$P\left( {A,B} \right) + P\left( {A,C} \right) + P\left( {B,C} \right) \ge 1$$

(53)

namely, a Bell inequality, where the individual probabilities are:

$$P\left( {A,A} \right)\, = \,P\left( {B,B} \right)\, = \,P\left( {C,C} \right)\, = \,{1}$$

(54)

In such a way that if any experiment linked to entanglement yields the sum of the three combined probabilities of Eq. (53) a value less than one, then it would be violating Bell's inequality, which would automatically indicate the non-locality of the entanglement.

Therefore, and in order to prove inequality, we choose the same examples from Maccone [53] but slightly adapted to the formalism of this study. Then, taking into account Eqs. (50) to (52), we have the following triad of two-value properties $A(\theta = 0^{ \circ } )$, $B\left( {{{\theta = \pi } \mathord{\left/ {\vphantom {{\theta = \pi } 3}} \right. \kern-0pt} 3}} \right)$, and $C\left( {{{\theta = - \pi } \mathord{\left/ {\vphantom {{\theta = - \pi } 3}} \right. \kern-0pt} 3}} \right)$:

$$A(\theta = 0^{ \circ } ):$$

$$\left| {a_{0} } \right\rangle \equiv \cos \left( {0^{ \circ } } \right)\left| 0 \right\rangle + \sin \left( {0^{ \circ } } \right)\left| 1 \right\rangle = \left| 0 \right\rangle$$

(55)

$$\begin{gathered} \left| {a_{1} } \right\rangle \equiv \cos \left( {0^{ \circ } + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 0 \right\rangle + \sin \left( {0^{ \circ } + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 1 \right\rangle \hfill \\ \quad \;\;{\kern 1pt} = \cos \left( {{\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 0 \right\rangle + \sin \left( {{\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 1 \right\rangle = \left| 1 \right\rangle , \hfill \\ \end{gathered}$$

(56)

$$B(\theta = {\pi \mathord{\left/ {\vphantom {\pi 3}} \right. \kern-0pt} 3}):$$

$$\left| {b_{0} } \right\rangle \equiv \cos \left( {{\pi \mathord{\left/ {\vphantom {\pi 3}} \right. \kern-0pt} 3}} \right)\left| 0 \right\rangle + \sin \left( {{\pi \mathord{\left/ {\vphantom {\pi 3}} \right. \kern-0pt} 3}} \right)\left| 1 \right\rangle = {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| 0 \right\rangle + {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| 1 \right\rangle$$

(57)

$$\begin{gathered} \left| {b_{1} } \right\rangle \equiv \cos \left( {{\pi \mathord{\left/ {\vphantom {\pi 3}} \right. \kern-0pt} 3} + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 0 \right\rangle + \sin \left( {{\pi \mathord{\left/ {\vphantom {\pi 3}} \right. \kern-0pt} 3} + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 1 \right\rangle \hfill \\ \quad \;\;{\kern 1pt} = \cos \left( {{{5\pi } \mathord{\left/ {\vphantom {{5\pi } 6}} \right. \kern-0pt} 6}} \right)\left| 0 \right\rangle + \sin \left( {{{5\pi } \mathord{\left/ {\vphantom {{5\pi } 6}} \right. \kern-0pt} 6}} \right)\left| 1 \right\rangle = - {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| 0 \right\rangle + {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| 1 \right\rangle , \hfill \\ \end{gathered}$$

(58)

$$C(\theta = {{ - \pi } \mathord{\left/ {\vphantom {{ - \pi } 3}} \right. \kern-0pt} 3}):$$

$$\left| {c_{0} } \right\rangle \equiv \cos \left( {{{ - \pi } \mathord{\left/ {\vphantom {{ - \pi } 3}} \right. \kern-0pt} 3}} \right)\left| 0 \right\rangle + \sin \left( {{{ - \pi } \mathord{\left/ {\vphantom {{ - \pi } 3}} \right. \kern-0pt} 3}} \right)\left| 1 \right\rangle = {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| 0 \right\rangle - {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| 1 \right\rangle$$

(59)

$$\begin{gathered} \left| {c_{1} } \right\rangle \equiv \cos \left( {{{ - \pi } \mathord{\left/ {\vphantom {{ - \pi } 3}} \right. \kern-0pt} 3} + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 0 \right\rangle + \sin \left( {{{ - \pi } \mathord{\left/ {\vphantom {{ - \pi } 3}} \right. \kern-0pt} 3} + {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0pt} 2}} \right)\left| 1 \right\rangle \hfill \\ \quad \;\;{\kern 1pt} = \cos \left( {{\pi \mathord{\left/ {\vphantom {\pi 6}} \right. \kern-0pt} 6}} \right)\left| 0 \right\rangle + \sin \left( {{\pi \mathord{\left/ {\vphantom {\pi 6}} \right. \kern-0pt} 6}} \right)\left| 1 \right\rangle = {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| 0 \right\rangle + {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| 1 \right\rangle , \hfill \\ \end{gathered}$$

(60)

Then, according to the first line of Eq. (52), we have,

$$\left| {\beta_{00} } \right\rangle = \frac{{\left| {a_{0} } \right\rangle \left| {a_{0} } \right\rangle + \left| {a_{1} } \right\rangle \left| {a_{1} } \right\rangle }}{\sqrt 2 } = \frac{{\left| {b_{0} } \right\rangle \left| {b_{0} } \right\rangle + \left| {b_{1} } \right\rangle \left| {b_{1} } \right\rangle }}{\sqrt 2 } = \frac{{\left| {c_{0} } \right\rangle \left| {c_{0} } \right\rangle + \left| {c_{1} } \right\rangle \left| {c_{1} } \right\rangle }}{\sqrt 2 }$$

(61)

We are now in a position to calculate the inequality of Eq. (53), by putting $\left| {a_{0} } \right\rangle$ and $\left| {a_{1} } \right\rangle$ in terms of $\left| {b_{0} } \right\rangle$ and $\left| {b_{1} } \right\rangle$ for P(A, B), then $\left| {a_{0} } \right\rangle$ and $\left| {a_{1} } \right\rangle$ in terms of $\left| {c_{0} } \right\rangle$ and $\left| {c_{1} } \right\rangle$ for P(A, C), and $\left| {a_{0} } \right\rangle$ and $\left| {a_{1} } \right\rangle$ in terms of $\left| {b_{0} } \right\rangle$, $\left| {b_{1} } \right\rangle$, $\left| {c_{0} } \right\rangle$ and $\left| {c_{1} } \right\rangle$, simultaneously for P(B, C).

Then, if.

$$\left| {a_{0} } \right\rangle = {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {b_{0} } \right\rangle - {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {b_{1} } \right\rangle ,\,\,{\text{and}}$$

(62)

$$\left| {a_{1} } \right\rangle = {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {b_{0} } \right\rangle + {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {b_{1} } \right\rangle$$

(63)

we can replace both into the first term of Eq. (61),

$$\begin{gathered} \left| {\beta_{00} } \right\rangle_{{P\left( {A,B} \right)}} = \frac{{\left| {a_{0} } \right\rangle \left( {{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {b_{0} } \right\rangle - {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {b_{1} } \right\rangle } \right) + \left| {a_{1} } \right\rangle \left( {{{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {b_{0} } \right\rangle + {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {b_{1} } \right\rangle } \right)}}{\sqrt 2 } \hfill \\ \quad \quad \quad \;\;\;{\kern 1pt} = \frac{{\left| {a_{0} } \right\rangle \left| {b_{0} } \right\rangle - \sqrt 3 \left| {a_{0} } \right\rangle \left| {b_{1} } \right\rangle + \sqrt 3 \left| {a_{1} } \right\rangle \left| {b_{0} } \right\rangle + \left| {a_{1} } \right\rangle \left| {b_{1} } \right\rangle }}{2\sqrt 2 }. \hfill \\ \end{gathered}$$

(64)

Now, if

$$\left| {a_{0} } \right\rangle = {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {c_{0} } \right\rangle + {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {c_{1} } \right\rangle$$

(65)

and

$$\left| {a_{1} } \right\rangle = {{ - \sqrt 3 } \mathord{\left/ {\vphantom {{ - \sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {c_{0} } \right\rangle + {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {c_{1} } \right\rangle$$

(66)

we can replace both into the first term of Eq. (61),

$$\begin{gathered} \left| {\beta_{00} } \right\rangle_{{P\left( {A,C} \right)}} = \frac{{\left| {a_{0} } \right\rangle \left( {{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {c_{0} } \right\rangle + {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {c_{1} } \right\rangle } \right) + \left| {a_{1} } \right\rangle \left( {{{ - \sqrt 3 } \mathord{\left/ {\vphantom {{ - \sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {c_{0} } \right\rangle + {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {c_{1} } \right\rangle } \right)}}{\sqrt 2 } \hfill \\ \quad \quad \quad \;\;\;{\kern 1pt} = \frac{{\left| {a_{0} } \right\rangle \left| {c_{0} } \right\rangle + \sqrt 3 \left| {a_{0} } \right\rangle \left| {c_{1} } \right\rangle - \sqrt 3 \left| {a_{1} } \right\rangle \left| {c_{0} } \right\rangle + \left| {a_{1} } \right\rangle \left| {c_{1} } \right\rangle }}{2\sqrt 2 }. \hfill \\ \end{gathered}$$

(67)

Finally, replacing Eqs. (62), (63), (65), and (66) into the first term of Eq. (61),

$$\begin{gathered} \left| {\beta_{00} } \right\rangle_{{P\left( {B,C} \right)}} = \frac{{\left( {{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {b_{0} } \right\rangle - {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {b_{1} } \right\rangle } \right)\left( {{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {c_{0} } \right\rangle + {{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {c_{1} } \right\rangle } \right) + \left( {{{\sqrt 3 } \mathord{\left/ {\vphantom {{\sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {b_{0} } \right\rangle + {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {b_{1} } \right\rangle } \right)\left( {{{ - \sqrt 3 } \mathord{\left/ {\vphantom {{ - \sqrt 3 } 2}} \right. \kern-0pt} 2}\left| {c_{0} } \right\rangle + {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}\left| {c_{1} } \right\rangle } \right)}}{\sqrt 2 } \hfill \\ \quad \quad \quad \;\;\;{\kern 1pt} = \frac{{\left| {b_{0} } \right\rangle \left| {c_{0} } \right\rangle - \sqrt 3 \left| {b_{1} } \right\rangle \left| {c_{0} } \right\rangle + \sqrt 3 \left| {b_{0} } \right\rangle \left| {c_{1} } \right\rangle - 3\left| {b_{1} } \right\rangle \left| {c_{1} } \right\rangle - 3\left| {b_{0} } \right\rangle \left| {c_{0} } \right\rangle - \sqrt 3 \left| {b_{1} } \right\rangle \left| {c_{0} } \right\rangle + \sqrt 3 \left| {b_{0} } \right\rangle \left| {c_{1} } \right\rangle + \left| {b_{1} } \right\rangle \left| {c_{1} } \right\rangle }}{4\sqrt 2 } \hfill \\ \quad \quad \quad \;\;\;{\kern 1pt} = \frac{{ - 2\left| {b_{0} } \right\rangle \left| {c_{0} } \right\rangle - 2\sqrt 3 \left| {b_{1} } \right\rangle \left| {c_{0} } \right\rangle + 2\sqrt 3 \left| {b_{0} } \right\rangle \left| {c_{1} } \right\rangle - 2\left| {b_{1} } \right\rangle \left| {c_{1} } \right\rangle }}{4\sqrt 2 } \hfill \\ \quad \quad \quad \;\;\;{\kern 1pt} = \frac{{ - \left| {b_{0} } \right\rangle \left| {c_{0} } \right\rangle - \sqrt 3 \left| {b_{1} } \right\rangle \left| {c_{0} } \right\rangle + \sqrt 3 \left| {b_{0} } \right\rangle \left| {c_{1} } \right\rangle - \left| {b_{1} } \right\rangle \left| {c_{1} } \right\rangle }}{\sqrt 2 }. \hfill \\ \end{gathered}$$

(68)

For Eq. (64), the probability of obtaining zero for both properties is the square modulus of the coefficient of $\left| {a_{0} } \right\rangle \left| {b_{0} } \right\rangle$, namely $\left( {\tfrac{1}{2\sqrt 2 }} \right)^{2} = \tfrac{1}{8}$, while the probability of obtaining one for both is the square modulus of the coefficient of $\left| {a_{1} } \right\rangle \left| {b_{1} } \right\rangle$, again 1/8. Hence, P(A, B) = 1/8 + 1/8 = 1/4. Analogously, we find that P(A, C) = 1/4 and that P(B, C) = 1/4. Finally, the three probabilities are replaced in Eq. (53),

$$P\left( {A,B} \right) + P\left( {A,C} \right) + P\left( {B,C} \right) = \frac{3}{4} < 1$$

(69)

That is, the Bell ‘s inequality is violated through the formlism presented in this study, which is completely concomitant with the Bell's Theorem [14].