*In order to meet the experimental results with entangled photons, it is defined*:
$$\begin{array}{}{\displaystyle B(\delta ,\lambda )=-A(\delta ,\lambda ).}\end{array}$$(7)

*Purpose: accounting for the entanglement of the singlet state*.

This explains property #3: The mechanism of entanglement by a parameter and property #4: The locality of the polarization measurements with entangled photons.

The rules for the distribution of the incoming photons onto the two output directions of the polarizer are also valid for the partner photon on side 2. With equation (7) it follows:

For incoming photon 1 with polarization *ϕ*_{1} = 0^{∘} having incoming photon 2 with polarization *ϕ*_{2} = 90^{∘} and polarizer P2 setting *β* = *α* + *π*/2 we obtain from equation (4)
$$\begin{array}{}B(\delta ,\lambda )=-1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-1<\lambda <\mathrm{cos}(2\delta ),\end{array}$$(8)

*meaning photon 2 is in p-state* *α*+*π*/2 and from equation (5)
$$\begin{array}{}B(\delta ,\lambda )=+1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(2\delta )<\lambda <1,\end{array}$$(9)

*meaning photon 2 is in p-state* *α*, where *δ* = *β* – *π*/2 = *α*.

For incoming photon 1 with polarization *ϕ*_{1} = 90^{∘} having incoming photon 2 with polarization *ϕ*_{2} = 0^{∘} and polarizer P2 setting *β* = *α* we obtain from equation (2)
$$\begin{array}{}B(\delta ,\lambda )=+1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-1<\lambda <\mathrm{cos}(2\delta ),\end{array}$$(10)

*meaning photon 2 is in p-state α* and from equation (3)
$$\begin{array}{}B(\delta ,\lambda )=-1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(2\delta )<\lambda <1,\end{array}$$(11)

*meaning photon 2 is in p-state **α* + *π*/2, where *δ* = *β*−0^{∘} = *α*.

With equations (2) and (5) it follows A(*δ*, *λ*) = 1 for each value of *δ* and for –1 < *λ* < 1 although from different incoming photons. From equations (8) and (11) we get *B*(*δ*, *λ*) = –1 for –1 < *λ* < 1, again from different incoming photons. Photons 1 with *A*(*δ*, *λ*) = +1 are part of an ensemble, selected by polarizer P1, with the polarization *α* and the partner photons 2 with *B*(*δ*, *λ*) = –1 are part of an ensemble with the polarization *α*+*π*/2 coupled with photon 1 by equation (7). This follows from model assumption M3.

With photon 1 in state *ϕ*_{1}= *α* and photon 2 in state *ϕ*_{2} = *α* +*π*/2 we can rotate the coordinate system by an
angle *α* and get ${\varphi}_{1}^{\prime}={\varphi}_{1}-\alpha =\alpha -\alpha ={0}^{\circ}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}and\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}{\varphi}_{2}^{\prime}={\varphi}_{2}-\alpha =\alpha +\pi /2-\alpha ={90}^{\circ}$ and *β*′ = *β* – *α*.

This explains Property #5: Rotational invariance of the polarization measurements with entangled photons.

For incoming photon 1 with polarization ${\varphi}_{1}^{\prime}={0}^{\circ}$ having incoming photon 2 with polarization ${\varphi}_{2}^{\prime}={90}^{\circ}$ and polarizer P2 setting *β*′ we obtain from equation (8)

$$\begin{array}{}B(\delta ,\lambda )=-1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-1<\lambda <\mathrm{cos}(2\delta ),\end{array}$$(12)

*meaning photon 2 is in p-state β*′ and from equation (9)
$$\begin{array}{}B(\delta ,\lambda )=+1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(2\delta )<\lambda <1,\end{array}$$(13)

*meaning photon 2 is in p-state β*′ –*π*/2, with *δ* = *β*′ – *π*/2.

For incoming photon 1 with polarization ${\varphi}_{1}^{\prime}={90}^{\circ}$ having incoming photon 2 with polarization ${\varphi}_{2}^{\prime}={0}^{\circ}$ and polarizer P2 setting *β*′–*π*/2 we obtain from equation (10)

$$\begin{array}{}B(\delta ,\lambda )=+1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-1<\lambda <\mathrm{cos}(2\delta ),\end{array}$$(14)

*meaning photon 2 is in p-state β′–**π*/2 and from equation (11)
$$\begin{array}{}B(\delta ,\lambda )=-1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(2\delta )<\lambda <1,\end{array}$$(15)

*meaning photon 2 is in p-state β*′ with *δ* = *β*′ – *π*/2 – 0^{∘}.

The expectation value of a common measurement with polarizers P1 and P2 is
$$\begin{array}{}{\displaystyle E(\alpha ,\beta )=1/2\underset{-1}{\overset{1}{\int}}A\left(\alpha ,\lambda \right)\ast B(\beta ,\lambda )d\lambda ,}\end{array}$$(16)

where *α* and *β* are the polarizer settings at side 1 and side 2 respectively and
$$\begin{array}{}A\left(\alpha ,\lambda \right)=A\left(\delta \left(\alpha \right),\lambda \right)\end{array}$$

where *δ*(*α*) = *α* – *α* = 0 and
$$\begin{array}{}B\left(\beta ,\lambda \right)=\phantom{\rule{thinmathspace}{0ex}}B(\delta (\beta ),\lambda )\end{array}$$

where *δ*(*β*)= *β* – *α* – *π*/2.

With *A*(0,*λ*) = 1 and *B*(*β*,*λ*) *from equations* (12) and (13) where *δ* = *β*′ – *π*/2 = *β* – *α* – *π*/2 it follows from equation (16)
$$\begin{array}{}E(\alpha ,\beta )\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}& {\displaystyle =-1/2\underset{-1}{\overset{\mathrm{cos}(2\delta )}{\int}}d\lambda +1/2\underset{\mathrm{cos}(2\delta )}{\overset{1}{\int}}d\lambda}\\ & =-{\mathrm{cos}}^{2}(\delta )+{\mathrm{sin}}^{2}(\delta )=-{\mathrm{sin}}^{2}(\beta -\alpha )+{\mathrm{cos}}^{2}(\beta -\alpha ).\end{array}$$(17)

The same result is obtained using *A*(*α*,*λ*) = –1 and *B*(*β*,*λ*) equations (14) and (15).

The probability P_{α,β} to measure a photon 2 polarization at *β* after photon 1 was measured at *α* can be obtained from
$$\begin{array}{}E(\alpha ,\beta )=+1\ast (1-{P}_{\alpha ,\beta})-1\ast {P}_{\alpha ,\beta}\end{array}$$(18)

yielding
$$\begin{array}{}{P}_{\alpha ,\beta}=1/2(1-E(\alpha ,\beta ))={\mathrm{sin}}^{2}(\beta -\alpha ).\end{array}$$(19)

This explains property #6: The QM predictions of polarization measurements with entangled photons.

As the expectation value *E*(*α*,*β*) from equation (17) does exactly match the predictions of quantum physics it also violates Bells [4] inequality. This explains property #7: The model violates Bells inequality as quantum physics does.

What is the reason for this? With *δ* = *β* – *α* – *π*/2 the value of *B*(*β*,*λ*) in equation (16) depends on the setting angle *α* of the polarizer at side 1 whereas in Bells model the measured value at side 2 does not depend on the setting of polarizer at side 1. This difference is the reason that the model presented here violates Bells inequality whereas Bells model does not. Bell thought, “the result B for particle 2 should not depend on the setting … for particle 1”. But we have seen with model assumptions M1-M6 that a local model is possible without Bells restriction cited above. Bell had cited Einstein [12] “But one supposition we should in my opinion, absolutely hold fast: the real factual situation of the system S_{2} is independent of what is done with the system S_{1}, which is spatially separated from the former.” This supposition is fulfilled with the model above. With model assumption M3 the polarization of the incoming photon at side 2 depends on the setting of the polarizer at side 1 although there is no nonlocal action. As only those photons 2 are considered which match with a measured photon 1 it is not unexpected to see the result at side 2 depending on the polarizer setting at side 1. The rule that determines which polarizer exit a photon will take is the same for both sides. Dependencies between the photons on either side originate from the shared parameter *λ* and not from a nonlocal influence of photon 1 upon photon 2.

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