$$
\begin{equation}
\begin{aligned}
\mathbf{P} \{|y + \mathbf{v}_{y\pm}^A(\mathbf{v}_0, \theta_0^A)t| \le \epsilon\} &= 1, \\
\mathbf{P} \{|y + \mathbf{v}_{y\pm}^A(\mathbf{v}_0, \theta_0^A)t| > \epsilon\} &= 0, \\
\mathbf{P} \{|y - \mathbf{v}_{y\pm}^B(\mathbf{v}_0, \theta_0^B)t| \le \epsilon\} &= 1, \\
\mathbf{P} \{|y - \mathbf{v}_{y\pm}^B(\mathbf{v}_0, \theta_0^B)t| > \epsilon\} &= 0,
\end{aligned}
\tag{4.6.4}
\end{equation}
\[
\epsilon \ll 1,
\]
$$

where

$$
\begin{align}
\mathbf{v}_{y\pm}^A(\mathbf{v}_0, \theta_0^A) &= \eta_{\theta_0^A}^{\pm} \mathbf{v}_0, \mathbf{v}_{y\pm}^B(\mathbf{v}_0, \theta_0^B) = \eta_{\theta_0^B}^{\pm} \mathbf{v}_0, \nonumber \\
\eta_{\theta_0^A}^{+} &= \cos^2 \frac{\theta_0^A}{2}, \quad \eta_{\theta_0^A}^{-} = \sin^2 \frac{\theta_0^A}{2}, \tag{4.6.5} \\
\eta_{\theta_0^B}^{+} &= \cos^2 \frac{\theta_0^B}{2}, \quad \eta_{\theta_0^B}^{-} = \sin^2 \frac{\theta_0^B}{2}. \nonumber
\end{align}
$$

The wave function Ψ(**r**<sub>A</sub>, **r**<sub>B</sub>, *t*) of the two identical particles **A** and **B**, electrically neutral and with magnetic moments µ<sub>0</sub>, subject to magnetic fields E<sub>A</sub> and E<sub>B</sub>, admits on the basis of |±<sub>A</sub>⟩ and |±<sub>B</sub>⟩ four components Ψ<sup>a,b</sup>(**r**<sub>A</sub>, **r**<sub>B</sub>, *t*) and satisfies the two-body Pauli equation

$$
i\hbar \frac{\partial \Psi^{a,b}(t)}{\partial t} = \left( -\frac{\hbar^2}{2m} \Delta_A - \frac{\hbar^2}{2m} \Delta_B \right) \Psi^{a,b}(t) + \mu B_j^{\mathrm{E}_A}(\sigma_j)_c^a \Psi^{c,b}(t) + \mu B_j^{\mathrm{E}_B}(\sigma_j)_d^b \Psi^{a,d}(t) \quad (4.6.6)
$$

with the initial conditions:

$$
\Psi^{a,b}(0, \mathbf{r}_A, \mathbf{r}_B) = \Psi_0^{a,b}(\mathbf{r}_A, \mathbf{r}_B), \quad (4.6.7)
$$

where Ψ<sup>a,b</sup><sub>0</sub>(**r**<sub>A</sub>, **r**<sub>B</sub>) corresponds to the singlet state (4.6.1). To obtain an explicit solution of the EPR-B experiment, we take the numerical values of the Stern-Gerlach experiment, see subsection III.4

Below we explain the EPR-B experiment by using nonlocal two-body Pauli
equation

$$
\begin{equation}
\begin{split}
& \int d\mathbf{r}_A d\mathbf{r}_B \int dt \left[ -i\hbar \frac{\partial \Psi^{\#a,b}(t,t',\mathbf{r}_A,\mathbf{r}_B)}{\partial t} + \left( -\frac{\hbar^2}{2m}\Delta_A - \frac{\hbar^2}{2m}\Delta_B \right) \Psi^{\#a,b}(t,t',\mathbf{r}_A,\mathbf{r}_B) \right. \\
& \qquad \left. + \mu B_j^{\mathrm{E}_A}(\sigma_j)_c^a \Psi^{\#c,b}(t,t',\mathbf{r}_A,\mathbf{r}_B) + \mu B_j^{\mathrm{E}_B}(\sigma_j)_d^b \Psi^{\#a,d}(t,t',\mathbf{r}_A,\mathbf{r}_B) \right] = O(\hbar^a),
\end{split}
\tag{4.6.8}
\end{equation}
\[
d\mathbf{r}_A = dx_A dy_A z_A, \quad d\mathbf{r}_B = dx_B dy_B z_B
\]
$$

with a boundary condition

$$
\int d\mathbf{r}_A d\mathbf{r}_{BZ_A}(t_1) |\Psi^{\#}(t_1, t', \mathbf{r}_A, \mathbf{r}_B)|^2 = - \int d\mathbf{r}_A d\mathbf{r}_{BZ_B}(t_2) |\Psi^{\#}(t_2, t', \mathbf{r}_A, \mathbf{r}_B)|^2. \quad (4.6.9)
$$

One of the difficulties of the canonical interpretation of the EPR-B experiment is the
existence of two simultaneous measurements. By doing these measurements one
after the other, the interpretation of the experiment will be facilitated. That is the
purpose of the two-step version of the experiment EPR-B studied below.

A. First step EPR-B: Spin measurement of A

Consider that at time *t*₀ the particle **A** arrives at the entrance of electromagnet **E**<sub>A</sub>.
**Remark 4.6.5.** We assume that a particle **A** collapses in a magnetic field **E**<sub>**A**</sub> at some instant *t'* into two particles **A**<sub>+</sub> and **A**<sub>-</sub>, i.e. the spinor Ψ(z,y,t) collapses in a magnetic field **E**<sub>**A**</sub> at some instant *t'* into two spinors Ψ<sub>+</sub>(z,y,t,t',δ) and Ψ<sub>-</sub>(z,y,t,t',δ) given by Eq.(3.4.9.a)-Eq.(3.4.9.b), see Assumption 3.4.1.

**Remark 4.6.6.** The particles **A**<sub>+</sub> and **A**<sub>-</sub> stays within the magnetic field for a time Δ*t'* ≤ Δ*t* = Δ*t*/V<sub>0</sub>.

Thus after exit of the magnetic field E<sub>A</sub>, at time t<sub>1</sub> = t<sub>0</sub> + Δt + t, the wave functions Ψ<sub>+</sub>(z,y,t<sub>0</sub> + Δt + t, δ) and Ψ<sub>-</sub>(z,y,t<sub>0</sub> + Δt + t, δ) becomes

$$
\Psi_{+}(\mathbf{r}_{A_{+}}, \mathbf{r}_{B_{-}}, t_{0} + \Delta t + t) = f(\mathbf{r}_{B_{-}}) \times (f^{+}(\mathbf{r}_{A_{+}}, t)|_{+\mathbf{A}} \rangle \otimes |-\mathbf{B}\rangle) \quad (4.6.10.a)
$$

and

$$
\Psi_{-}(\mathbf{r}_{A_{-}}, \mathbf{r}_{B_{+}}, t_{0} + \Delta t + t) = f(\mathbf{r}_{B_{+}}) \times (f^{-}(\mathbf{r}_{A_{-}}, t)|_{-\mathbf{A}} \rangle \otimes |+\mathbf{B}\rangle) \quad (4.6.10.b)
$$

respectively, with

$$
\begin{align}
f^+(\mathbf{r}, t) &= \cos \frac{\theta_0}{2} f(x, z - z_\Delta - ut) \exp \left[ i \left( \frac{muz}{\hbar} + \varphi^+(t) \right) \right] \tag{4.6.11} \\
f^-(\mathbf{r}, t) &= \sin \frac{\theta_0}{2} f(x, z + z_\Delta + ut) \exp \left[ i \left( -\frac{muz}{\hbar} + \varphi^-(t) \right) \right] \nonumber
\end{align}
$$

where z<sub>Δ</sub> and u are given by

$$
z_{\Delta} = \frac{\mu_B B_0' (\Delta t)^2}{2m} = 10^{-5} m, \quad u = \frac{\mu_B B_0' (\Delta t)}{m} = 1 \text{m/s.} \qquad (4.6.12)
$$

**Remark 4.6.7.** We deduce that: the beam of particle **A** is divided into two **A**<sub>+</sub> and **A**<sub>-</sub> , and the beam of particle **B** is divided into two **B**<sub>+</sub> and **B**<sub>-</sub> .

**Remark 4.6.8.** Our first conclusion is: the position of **B**<sub>+</sub> and **B**<sub>-</sub> does not depend on the spin measurement of **A**<sub>+</sub> and **A**<sub>-</sub>, only the spins are involved. We conclude from equation (4.6.10) that the spins of **A**<sub>+</sub> and **B**<sub>-</sub> (**A**<sub>-</sub> and **B**<sub>+</sub>) remain opposite throughout the experiment. These are the two properties used in the relaxed causal interpretation.

**Remark 4.6.9.** By "relaxed locality principle" and decoherence it follows that the interection between **A**<sub>+</sub>, **A**<sub>-</sub> **B**<sub>+</sub>, and **B**<sub>-</sub> is absens, we assume the existance of wave functions

$$
\Psi_0^{A_+}(\mathbf{r}_{A_+}, \theta_0^{A_+}, \varphi_0^{A_+}), \Psi_0^{A_-}(\mathbf{r}_{A_-}, \theta_0^{A_-}, \varphi_0^{A_-}), \Psi_0^{B_+}(\mathbf{r}_B, \theta_0^{B_+}, \varphi_0^{B_+}), \Psi_0^{B_-}(\mathbf{r}_B, \theta_0^{B_-}, \varphi_0^{B_-}). \quad (4.6.13)
$$

B. Second step EPR-B: Spin measurement of B

The second step is a continuation of the first and corresponds to the EPR-B experiment broken down into two steps. On a pairs of particles **A**<sub>+</sub>, **B**<sub>-</sub> and **A**<sub>-</sub>, **B**<sub>+</sub> in a singlet state, first we made a Stern and Gerlach measurement on the **A**<sub>+</sub>and **A**<sub>-</sub> atom at instant *t*<sub>1</sub> between *t*<sub>0</sub> and *t*<sub>0</sub> + Δ*t* + *t*<sub>D</sub> :

$$
t_0 < t_1 < t_0 + \Delta t + t_D. \tag{4.6.14}
$$

Secondly, we make a Stern and Gerlach measurement on the **B**<sub>+</sub> and **B**<sub>-</sub> atom with
an electromagnet $E_B$ forming an angle $\delta$ with $E_A$ at instant $t_2$ between $t_0 + \Delta t + t_D$ and $t_0 + 2(\Delta t + t_D)$:

$$t_0 + \Delta t + t_D < t_2 \le t_0 + 2(\Delta t + t_D) \qquad (4.6.15)$$

At the exit of magnetic field $E_A$, at time $t_0 + \Delta t + t_D$, the pair of particles wave functions is given by Eq.(4.6.10.a) and Eq.(4.6.10.b) respectively.Immediately after the measurements of $A_+$ and $A_-$, still at time $t_0 + \Delta t + t_D$, the wave functions of $B_+$ and $B_-$ depends on the measurements $\pm$ of $A$ respectively such that:

$$\Psi_{B_{-/+}A}(\mathbf{r}_{B_{-}}, t_0 + \Delta t + t_1) = f(\mathbf{r}_{B_{-}})|_{-B} \rangle, \qquad (4.6.16.a)$$

and

$$\Psi_{B_{+/-}A}(\mathbf{r}_{B_{+}}, t_0 + \Delta t + t_1) = f(\mathbf{r}_{B_{+}})|_{+B} \rangle. \qquad (4.6.16.b)$$

Then, the measurement of $B_+$ and $B_-$ at time $t_2 > t_0 + 2(\Delta t + t_D)$ yields, in this two-step version of the EPR-B experiment, the same results for spatial quantization and correlations of spins as in the EPR-B experiment.

## Resolution of the EPR-B experiment in de Broglie-Bohm interpretation by the "relaxed locality principle"

We assume, at the creation of the two entangled particles $A$ and $B$, that each of the two particles $A$ and $B$ has an initial wave function with opposite spins:

$$\Psi_0^A(\mathbf{r}_A, \theta_0^A, \varphi_0^A) = f(\mathbf{r}_A) \left( \cos \frac{\theta_0^A}{2} |+A\rangle + \sin \frac{\theta_0^A}{2} e^{i\varphi_0^A} |-A\rangle \right) \qquad (4.6.17)$$

and

$$\begin{align}
\Psi_0^B(\mathbf{r}_B, \theta_0^B, \varphi_0^B) &= f(\mathbf{r}_B) \left( \cos \frac{\theta_0^B}{2} |+B\rangle + \sin \frac{\theta_0^B}{2} e^{i\varphi_0^B} |-B\rangle \right) = \nonumber \\
&\quad f(\mathbf{r}_B) \left[ \cos \left( \frac{\pi}{2} - \frac{\theta_0^A}{2} \right) |+B\rangle + \sin \left( \frac{\pi}{2} - \frac{\theta_0^A}{2} \right) e^{i(\varphi_0^A - \pi)} |-B\rangle \right] = \tag{4.6.18} \\
\Psi_0^B(\mathbf{r}_B, \theta_0^B, \varphi_0^B) &= f(\mathbf{r}_B) \left( \sin \frac{\theta_0^A}{2} |+B\rangle + \cos \frac{\theta_0^A}{2} e^{i\varphi_0^A} |-B\rangle \right) \nonumber
\end{align}$$

with $\theta_0^B = \pi - \theta_0^A$ and $\varphi_0^B = \varphi_0^A - \pi$, see Remark 4.6.4. The two particles $A$ and $B$ are statistically prepared as in the Stern and Gerlach experiment. Then the Pauli principle tells us that the two-body wave function must be antisymmetric; after calculation we find the same singlet state (4.6.1):

$$\Psi_0(\mathbf{r}_A, \theta^A, \varphi^A, \mathbf{r}_B, \theta^B, \varphi^B) = -e^{i\varphi^A} f(\mathbf{r}_A) f(\mathbf{r}_B) \times (|+A\rangle \otimes |-B\rangle - |-A\rangle \otimes |+B\rangle). \qquad (4.6.19)$$

Thus, we can consider that the singlet wave function is the wave function of a family of two fermions $A$ and $B$ with opposite spins: the direction of initial spin $A$ and $B$ exists, but is not known. It is a local hidden variable which is therefore necessary to add in the initial conditions of the model.

Here, we assume that at the initial time we know the spin of each particle (given by each initial wave function) and the initial position of each particle.

### Step 1: spin measurement of $A$ in de Broglie-Bohm interpretation

In the equation (4.6.19) particle $A$ can be considered independent of $B$. We can therefore give it the wave function
$$ \Psi^A(\mathbf{r}_A, t_0 + \Delta t + t) = \cos \frac{\theta_0^A}{2} f^+(\mathbf{r}_A, t)|_{+A} \rangle + \sin \frac{\theta_0^A}{2} e^{i\phi_0^A} f^-(\mathbf{r}_A, t)|_{-A} \rangle \quad (4.6.20) $$

which is the wave function of a free particle in a Stern Gerlach apparatus and whose initial spin is given by $(\theta_0^A, \phi_0^A)$. For an initial polarization $(\theta_0^A, \phi_0^A)$ and an initial position $z_0^A$, we obtain, in the de Broglie-Bohm interpretation [63] of the Stern and Gerlach experiment, an evolution of the position $z_{A\pm}(t)$ and of the spin orientation of $A_\pm$, $\theta^{A\pm}(z_{A\pm}(t), t)$, see [65].

The case of particles $B_\pm$ is different. $B_\pm$ follows a rectilinear trajectories with $y_{B\pm}(t) = \mathbf{v}_y^\pm(\mathbf{v}_0, \theta_0)t$, $z_{B\pm}(t) = z_0^B$ and $x_{B\pm}(t) = x_0^B$. By contrast, the orientation of its spin moves with the orientation of the spin of $A_\pm$:

$$ \theta^{B\mp}(t) = \pi - \theta^{A\pm}(z_{A\pm}(t), t) \quad (4.6.21) $$

and

$$ \varphi^{B\mp}(t) = \varphi^{A\pm}(z_{A\pm}(t), t) - \pi. \quad (4.6.22) $$

**Remark 4.6.10.** Let $A_\pm(t, \mathbf{r}_{A\pm}(t), \theta^{A\pm}(z_{A\pm}(t), t), \varphi^{A\pm}(z_{A\pm}(t), t))$ denote events such that: "at instant $t$ particle $A_\pm$ obtain the position coordinates $\mathbf{r}_{A\pm}(t) = \{x_{A\pm}(t), y_{A\pm}(t), z_{A\pm}(t)\}$ and spin orientation $\theta^{A\pm}(t) = \theta^{A\pm}(z_{A\pm}(t), t)$ and $\varphi^{A\pm}(t) = \varphi^{A\pm}(z_{A\pm}(t), t)$. Let $B_\mp(t, \mathbf{r}_{B\mp}(t), \theta^{B\mp}(z_{B\mp}(t), t))$ denote events such that: "at instant $t$ particle $B_\mp$ obtain the position coordinates $\mathbf{r}_{B\mp}(t) = \{x_{B\mp}(t), y_{B\mp}(t), z_{B\mp}(t)\}$ and spin orientation $\theta^{B\mp} = \theta^{B\mp}(z_{B\mp}(t), t)$ and $\varphi^{B\mp}(z_{B\mp}(t), t)$. Then in accordance with the relaxed principle of locality (see subsection IV.1) we assume that

$$ \{A_{\pm}(t_1, \mathbf{r}_{A_{\pm}}(t), \theta^{A_{\pm}}, \varphi^{A_{\pm}}(t)), B_{\mp}(t, \mathbf{r}_{B_{\mp}}(t), \theta^{B_{\mp}}(t)), \varphi^{B_{\mp}}(t)\}_{\text{s.l.s.}} \in [\mathcal{F}_{M_4}^{\#}, \{(t_1, \mathbf{r}_1), (t_2, \mathbf{r}_2)\}]_{\text{s.l.s.}} \rightleftharpoons \quad (4.6.23) $$

see subsection IV.1, Definition 4.1.2. We can then associate the wave functions:

$$ \Psi^{B_+}(\mathbf{r}_{B_+}, t_0 + \Delta t + t) = f(\mathbf{r}_{B_+}) \cos \frac{\theta^{B_+}(t)}{2} |_{+B} \rangle \quad (4.6.24) $$

and

$$ \Psi^{B-}(\mathbf{r}_{B-}, t_0 + \Delta t + t) = f(\mathbf{r}_{B-}) \sin \frac{\theta^{B-}(t)}{2} e^{i\varphi^{B-}(t)} |_{-B-} \rangle \quad (4.6.25) $$

This wave functions is specific, because it depends upon initial conditions of $A$ (position and spin). The orientation of spin of the particles $B_\pm$ is driven by the particles $A_\mp$ respectively through the singlet wave functions.

## Step 2: Spin measurement of $B_\mp$ in de Broglie-Bohm interpretation

### (I) The prediction of the result of the spin measurement of $B_\mp$ under assumption of canonical postulate of locality

At the time $t_0 + \Delta t + t_D$, immediately after the measurement of $A$,
$\theta^{B_{\mp}}(t_0 + \Delta t + t_D) = \pi$ or 0 in accordance with the value of $\theta^{A_{\pm}}(z_{A_{\pm}}(t), t)$ and the wave functions of $B_{\mp}$ is given by Eq.(4.6.16.a) and Eq.(4.6.16.b) respectively. The frame $(Ox'yz')$ corresponds to the frame $(Oxyz)$ after a rotation of an angle $\delta$ around the $y$-axis (see Fig.4.5.1). $\theta^{B_{\pm}}$ corresponds to the $B_{\mp}$-spin angle with the $z$-axis, and $\theta'^{B_{\pm}}$ to the $B$-spin angle with the $z'$-axis, then $\theta'^{B_{\pm}}(t_0 + \Delta t + t_D) = \pi + \delta$ or $\delta$. In this second step, we are exactly in the case of a particle in a simple Stern and Gerlach experiment (with magnet $E_B$) with a specific initial polarization equal to $\pi + \delta$ or $\delta$ and not random like in step 1. Then, the measurement of $B$, at time $t_0 + 2(\Delta t + t_D)$), gives, in this interpretation of the two-step version of the EPR-B experiment, the same results as in the EPR-B experiment above, III.4.1. Thus we obtain EPR-B paradox again in de Broglie-Bohm interpretation.

**Remark 4.6.11.** Note that derivation EPR-B paradox in de Broglie-Bohm interpretation completely based on canonical postulate of locality

## Step 2: Spin measurement of $B_{\mp}$ in de Broglie-Bohm interpretation

### (II) The prediction of the result of the spin measurement of $B_{\mp}$ under assumption of postulate of nonlocality

We assume now a weak or strong postulate of nonlocality, see subsection I.9. At the time $t_1 = t_0 + \Delta t + t_D$, immediately after the spin measurement of $A_{\pm}$, $\theta^{B_{\mp}}(t_0 + \Delta t + t_D) = \pi$ or 0 in accordance with the value of $\theta^{A_{\pm}}(z_{A_{\pm}}(t), t)$ and the wave functions of $B_{\mp}$ is given by Eq.(4.6.16.a) and Eq.(4.6.16.b) respectively.

**Remark 4.6.12.** In accordance with postulate of nonlocality it follows:

(i) Whenever a measurement of the spin of a particle $A_+$ is performed at instant $t_1$ and

particle $A_+$ is found in the state $|\uparrow\rangle_z$, i.e., a state $|\psi_{t_1}\rangle_{A_+}$ collapses at instant $t_1$ to the state

$|\uparrow\rangle_{z,A_+}$ with respect of the Heisenberg spin uncertainty relations (1.9.5), then a state $|\psi_{t_1}\rangle_{B_-}$

immediately collapses at instant $t_1$ to the state $|\downarrow\rangle_{z,B_-}$ with respect of the Heisenberg spin

uncertainty relations (1.9.5), and this is true independent of the distance in Minkovski

spacetime that separates the particles, e.g.,

$$|\psi_{t_1}\rangle_{A_+} \xrightarrow{\text{col lapse}} |\uparrow\rangle_{z,A_+} \Rightarrow |\psi_{t_1}\rangle_{B_-} \xrightarrow{\text{col lapse}} |\downarrow\rangle_{z,B_-} \quad (4.6.26)$$

In accordance with Heisenberg spin uncertainty relations (1.9.5) spin of a particle $B_-$ obtain

an uncertainty along direction $Oz'$ (see Fig.4.5.1) and therefore EPR-B paradox disappears.

(ii) Whenever a measurement of the spin of a particle $A$ is performed at instant
$t_1$ and

particle $A_-$ is found in the state $|↓⟩_z$, i.e., a state $|\psi_{t_1}\rangle_{A_−}$ collapses at instant $t_1$ to the state

$|↓⟩_{z,A_−}$ with respect of the Heisenberg spin uncertainty relations (1.9.5), then a state $|\psi_{t_1}\rangle_{B_+}$

immediately collapses at instant $t_1$ to the state $|↑⟩_{z,B_+}$ with respect of the Heisenberg spin uncertainty relations (1.9.5), and this is true independent of the distance in Minkovski spacetime that separates the particles, e.g.,

$$|\psi_{t_1}\rangle_{A_−} \xrightarrow{\text{col lapse}} |↓\rangle_{z,A_−} \Rightarrow |\psi_{t_1}\rangle_{B_+} \xrightarrow{\text{col lapse}} |↑\rangle_{z,B_+} \quad (4.6.27)$$

In accordance with Heisenberg spin uncertainty relations (1.9.5) spin of a particle $B_−$ obtain

an uncertainty along direction $Oz'$ (see Fig.4.5.1) and therefore EPR-B paradox disappears.

## Physical explanation of non-local influences using the relaxed principle of locality

From the wave function of two entangled particles, we find spins, trajectories and also a wave function for each of the two particles. In this interpretation, the quantum particle has a local position like a classical particle, but it has also a non-local behavior through the wave function. So, it is the wave function that creates the non classical properties. We can keep a view of a local realist world for the particle, but we should add a non-local vision through the wave function. As we saw in step 1, the non-local influences in the EPR-B experiment only concern the spin orientation, not the motion of the particles themselves. Indeed only spins are entangled in the wave function but not positions and motions like in the initial EPR experiment. This is a key point in the search for a physical explanation of non-local influences.

The simplest explanation of this non-local influence given above by using the relaxed principle of locality (see subsection IV.1)

**Conclusion:** A new quantum mechanical formalism based on the probability representation of quantum states is proposed. This paper in particular deals with the special case of the measurement problem, known as Schrödinger's cat paradox. We pointed out that Schrödinger's cat demands to reconcile Born's rule. Using new quantum mechanical formalism we find the collapsed state of the Schrödinger's cat always shows definite and predictable outcomes even if cat also consists of a superposition

$$|\text{cat}\rangle = c_1 |\text{live cat}\rangle + c_2 |\text{death cat}\rangle \\ |c_1|^2 + |c_2|^2 = 1.$$
Using new quantum mechanical formalism the EPRB-paradox is considered successfully. We find that the EPRB-paradox can be resolved by nonprincipal and convenient relaxing of the Einstein's locality principle.

**Appendix. A.**

The time-dependent Schrödinger equation governs the time evolution of a quantum mechanical system is:

$$i\hbar \frac{\partial \Psi(\mathbf{x}, t)}{\partial t} = \hat{\mathbf{H}} \Psi(\mathbf{x}, t). \qquad (A.1)$$

The average, or expectation, value $\langle x_i \rangle$ of an observable $x_i$ corresponding to a quantum mechanical operator $\hat{x}_i$ is given by:

$$\langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) = \frac{\int_{\mathbb{R}^d} x_i |\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)|^2 d^d\mathbf{x}}{\int_{\mathbb{R}^d} |\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)|^2 d^d\mathbf{x}}. \qquad (A.2)$$
$$i = 1, \dots, d.$$

**Remark A.1.** We assume now that: the solution $\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)$ of the time-dependent Schrödinger equation (A.1) has a good approximation by a delta function such that

$$|\Psi(\mathbf{x}, t, \mathbf{x}_0, t_0; \hbar)|^2 \approx \prod_{i=1}^{d} \delta(x_i - x_i(t, \mathbf{x}_0, t_0)), \qquad (A.3)$$
$$x_i(t, \mathbf{x}_0, t_0) = x_{i,0},$$
$$i = 1, \dots, d.$$

**Remark A.2.** Note that under conditions given by Eq.(A.3) QM-system which governed by Schrödinger equation Eq.(A.1) completely evolve quasiclassically i.e. estimating the position $\{x_i(t, \mathbf{x}_0, t_0; \hbar)\}_{i=1}^d$ at each instant $t$ with final error $\delta$ gives $|\langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) - x_i(t, \mathbf{x}_0, t_0)| \le \delta, i = 1, \dots, d$ with a probability
$$ \mathbf{P}\{|\langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) - x_i(t, \mathbf{x}_0, t_0)| \le \delta\} \approx 1. $$

Thus from Eq.(A.2) and Eq.(A.3) we obtain

$$ \langle x_i \rangle(t, \mathbf{x}_0, t_0; \hbar) \approx $$

$$ \begin{aligned} & \approx \frac{\int_{\mathbb{R}^d} x_i \prod_{i=1}^{d=1} \delta(x_i - x_i(t, \mathbf{x}_0, t_0)) d^d x}{\int_{\mathbb{R}^d} \prod_{i=1}^{d=1} \delta(x_i - x_i(t, \mathbf{x}_0, t_0)) d^d x} = x_i(t, \mathbf{x}_0, t_0). \\ & i = 1, \dots, d. \end{aligned} \quad (A.4) $$

Thus under condition given by Eq.(A.3) one obtain

$$ \begin{aligned} \langle x_{i,t} \rangle(t, \mathbf{x}_0, t_0; \hbar) &\approx x_i(t, \mathbf{x}_0, t_0), \\ i &= 1, \dots, d. \end{aligned} \quad (A.5) $$

**Remark A.3.** Let $\Psi_i(\mathbf{x}, t, \mathbf{x}_0, t_0), i = 1, 2$ be the solutions of the time-dependent Schrödinger equation (A.1). We assume now that $\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)$ is a linear superposition such that

$$ \begin{aligned} \Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= c_1 \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) + c_2 \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0). \\ |c_1|^2 + |c_2|^2 &= 1. \end{aligned} \quad (A.6) $$

Then we obtain

$$ \begin{aligned} |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 &= (\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0) \Phi^*(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)) = \\ &= ([c_1 \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) + c_2 \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)]) \times \\ &\quad \times ([c_1^* \Psi_1^*(\mathbf{x}, t, \mathbf{x}_0, t_0) + c_2^* \Psi_2^*(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)]) = \\ &= |c_1|^2 (|\Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0)|^2) + c_1^* c_2 (\Psi_1^*(\mathbf{x}, t, \mathbf{x}_0) \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)) + \\ &\quad |c_2|^2 (|\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2) + c_1 c_2^* (\Psi_1(\mathbf{x}, t, \mathbf{x}_0) \Psi_2^*(\mathbf{x}, t, \mathbf{y}_0, t_0)). \end{aligned} \quad (A.7) $$
**Definition A.1.** Let $\langle \mathbf{x} \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0)$ be a vector-function

$$ \langle \mathbf{x} \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) : [0, T] \times \mathbb{R}^d \times \mathbb{R}^d \times [0, T] \to \mathbb{R}^d $$

$$ \langle \mathbf{x} \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) = \{ \langle x_1 \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0), \dots, \langle x_d \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) \}, \quad (A.8) $$

where

$$ \begin{aligned} \langle x_i \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \int_{\mathbb{R}^d} x_i |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 d^d x = \\ &= |c_1|^2 \int_{\mathbb{R}^d} x_i |\Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0)|^2 d^d x + \\ &\quad + |c_2|^2 \int_{\mathbb{R}^d} x_i |\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2 d^d x + \\ &\quad + c_1^* c_2 \int_{\mathbb{R}^d} x_i \Psi_1^*(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2^*(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d x + \\ &\quad + c_1 c_2^* \int_{\mathbb{R}^d} x_i \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d x. \end{aligned} \quad (A.9) $$

**Definition A.2.** Let $\Delta(t, \mathbf{x}_0, \mathbf{y}_0, t_0)$ be a vector-function

$$ \Delta(t, \mathbf{x}_0, \mathbf{y}_0, t_0) : [0, T] \times \mathbb{R}^d \times \mathbb{R}^d \to \mathbb{R}^d $$

$$ (\Delta(t, \mathbf{x}_0, \mathbf{y}_0, t_0)) = \{ \delta_1(t, \mathbf{x}_0, \mathbf{y}_0, t_0), \dots, \delta_d(t, \mathbf{x}_0, \mathbf{y}_0, t_0) \}, \quad (A.10) $$

where
$$
\begin{align}
\delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) = \delta[x_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0)] = \nonumber \\
&= c_1^* c_2 \int_{\mathbb{R}^d} x_i \Psi_1^*(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d x + \tag{A.11}
\end{align}
$$

$$
+c_1 c_2^* \int_{\mathbb{R}^d} x_i \Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0) \Psi_2^*(\mathbf{x}, t, \mathbf{y}_0, t_0) d^d x.
$$

Substituting Eqs.(A.11) into Eqs.(A.9) gives

$$
\begin{align}
\langle x_i \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \int_{\mathbb{R}^d} x_i |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 d^d x = \nonumber \\
&= |c_1|^2 \int_{\mathbb{R}^d} x_i |\Psi_1(\mathbf{x}, t, \mathbf{x}_0, t_0)|^2 d^d x + \tag{A.12} \\
&\quad + |c_2|^2 \int_{\mathbb{R}^d} x_i |\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2 d^d x + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) = \nonumber \\
&= |c_1|^2 \langle x_i \rangle(t, \mathbf{x}_0, t_0) + |c_2|^2 \langle x_i \rangle(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0). \nonumber
\end{align}
$$

$$
+ |c_2|^2 \int_{\mathbb{R}^d} x_i |\Psi_2(\mathbf{x}, t, \mathbf{y}_0, t_0)|^2 d^d x + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) = \\
= |c_1|^2 \langle x_i \rangle(t, \mathbf{x}_0, t_0) + |c_2|^2 \langle x_i \rangle(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0).
$$

Substitution equations (A.5) into equations (A.12) gives

$$
\begin{align}
\langle x_i \rangle(t, \mathbf{x}_0, \mathbf{y}_0, t_0) &= \int_{\mathbb{R}^d} x_i |\Phi(\mathbf{x}, t, \mathbf{x}_0, \mathbf{y}_0, t_0)|^2 d^d x = \nonumber \\
&= |c_1|^2 \langle x_i \rangle(t, \mathbf{x}_0, t_0) + |c_2|^2 \langle x_i \rangle(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0) \tag{A.13} \\
&\approx |c_1|^2 x_i(t, \mathbf{x}_0, t_0) + |c_2|^2 x_i(t, \mathbf{y}_0, t_0) + \delta_i(t, \mathbf{x}_0, \mathbf{y}_0, t_0). \nonumber
\end{align}
$$
**Appendix. B.**

The Schrödinger equation (2.1) in region I = {x|x < 0} has the following form

$$ \hbar^2 \frac{\partial^2 \Psi_I(x)}{\partial x^2} + 2mE\Psi_I(x) = 0. \qquad (B.1) $$

From Schrödinger equation (B.1) follows

$$ \hbar^2 \int_{-\infty}^{0} \frac{\partial^2 \Psi_I(x)}{\partial x^2} dx + 2mE \int_{-\infty}^{0} \Psi_I(x) dx = 0. \qquad (B.2) $$

Let $\Psi_I^{\#}(x)$ be a function

$$ \Psi_I^{\#}(x) = \phi(x)\Psi_I(x), \qquad (B.3) $$

where

$$ \phi(x) = (\pi r_c^2)^{-1/4} \exp\left(-\frac{x^2}{2r_c^2}\right) \qquad (B.4) $$

see Eq.(2.9). Note that

$$ \frac{\partial^2[\phi(x)\Psi_I(x)]}{\partial x^2} = \frac{\partial}{\partial x}\left[\Psi_I(x)\frac{\partial\phi(x)}{\partial x} + \phi(x)\frac{\partial\Psi_I(x)}{\partial x}\right] = 2\frac{\partial\Psi_I(x)}{\partial x}\frac{\partial\phi(x)}{\partial x} + \Psi_I(x)\frac{\partial^2\phi(x)}{\partial x^2} + \phi(x)\frac{\partial^2\Psi_I(x)}{\partial x^2}. \qquad (B.5) $$

Therefore substitution (B.2) into LHS of the Schrödinger equation (B.1) gives
$$
\begin{align*}
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \Psi_I^{\#}(x)}{\partial x^2} dx + 2mE \int_{-\infty}^{0} \Psi_I^{\#}(x) dx = \\
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \phi(x) \Psi_I(x)}{\partial x^2} dx + 2Em \int_{-\infty}^{0} \phi(x) \Psi_I(x) dx = \tag{B.6} \\
& 2h^2 \int_{-\infty}^{0} \frac{\partial \Psi_I(x)}{\partial x} \frac{\partial \phi(x)}{\partial x} dx + h^2 \int_{-\infty}^{0} \Psi_I(x) \frac{\partial^2 \phi(x)}{\partial x^2} dx + \\
& \qquad + \int_{-\infty}^{0} \phi(x) \left\{ h^2 \frac{\partial^2 \Psi_I(x)}{\partial x^2} + 2Em \int_{-\infty}^{0} \Psi_I(x) \right\} dx.
\end{align*}
$$

Note that

$$
\int_{-\infty}^{0} \phi(x) \left\{ h^2 \frac{\partial^2 \Psi_I(x)}{\partial x^2} + 2Em \int_{-\infty}^{0} \Psi_I(x) \right\} dx = 0. \quad (B.7)
$$

Therefore from Eq.(B.6) and Eq.(2.3)-Eq.(2.4) one obtains

$$
\begin{align*}
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \Psi_I^{\#}(x)}{\partial x^2} dx + 2mE \int_{-\infty}^{0} \Psi_I^{\#}(x) dx = \\
& h^2 \int_{-\infty}^{0} \frac{\partial^2 \phi(x) \Psi_I(x)}{\partial x^2} dx + 2Em \int_{-\infty}^{0} \phi(x) \Psi_I(x) dx = \tag{B.8} \\
& = 2h^2 \int_{-\infty}^{\infty} \frac{\partial \Psi_I(x)}{\partial x} \frac{\partial \phi(x)}{\partial x} dx + h^2 \int_{-\infty}^{\infty} \Psi_I(x) \frac{\partial^2 \phi(x)}{\partial x^2} dx.
\end{align*}
$$

From Eq.(B.6) one obtains
$$
\begin{align*}
\frac{\partial \phi(x)}{\partial x} = (\pi r_c^2)^{-1/4} \frac{\partial}{\partial x} \exp\left[-\frac{x^2}{2r_c^2}\right] &= -(\pi r_c^2)^{-1/4} r_c^{-2} x \exp\left[-\frac{x^2}{2r_c^2}\right], \\
\frac{\partial^2 \phi(x)}{\partial x^2} &= -(\pi r_c^2)^{-1/4} r_c^{-2} \exp\left[-\frac{x^2}{2r_c^2}\right] + \tag{B.9} \\
&\quad + (\pi r_c^2)^{-1/4} r_c^{-4} x^2 \exp\left[-\frac{x^2}{2r_c^2}\right].
\end{align*}
$$

From Eq.(B.9) and Eq.(2.3)-Eq.(2.4) one obtains

$$
\begin{align*}
& \hbar^2 \int_{-\infty}^{0} \frac{\partial \Psi_I(x)}{\partial x} \frac{\partial \phi(x)}{\partial x} dx = \\
& - \frac{\hbar^2}{(\pi r_c^2)^{1/4} r_c^2} \int_{-\infty}^{0} \frac{\partial \exp(ikx)}{\partial x} x \exp\left[-\frac{x^2}{2r_c^2}\right] dx = \tag{B.10} \\
& - \frac{2\pi \sqrt{2mE} \hbar}{(\pi r_c^2)^{1/4} r_c^2} \int_{-\infty}^{0} x \exp\left(i \frac{2\pi \sqrt{2mE}}{\hbar} x\right) \exp\left[-\frac{x^2}{2r_c^2}\right] dx, \\
& k = \frac{2\pi}{\hbar} \sqrt{2mE}.
\end{align*}
$$

and

$$
\begin{align}
\hbar^2 \int_{-\infty}^{0} \Psi_I(x) \frac{\partial^2 \phi(x)}{\partial x^2} dx &= - \frac{\hbar^2}{(\pi r_c^2)^{3/4} r_c^2} \int_{-\infty}^{0} \exp(ikx) \exp\left[-\frac{x^2}{2r_c^2}\right] dx + \tag{B.11} \\
&\quad + \frac{\hbar^2}{(\pi r_c^2)^{1/4} r_c^2} \int_{-\infty}^{0} x^2 \exp(ikx) \exp\left[-\frac{x^2}{2r_c^2}\right] dx. \nonumber
\end{align}
$$
Appendix E. Calculating the spinor evolution in the Stern-Gerlach experiment

In the magnetic field $B = (B_x, 0, B_z)$, the Pauli equation gives coupled Schrödinger equations for each spinor component

$$i\hbar \frac{\partial \psi_{\pm}}{\partial t}(x,y,z,t) = -\frac{\hbar^2}{2m} \nabla^2 \psi_{\pm}(x,y,z,t) \pm \mu_B (B_0 - B'_0 z) \psi_{\pm}(x,y,z,t) \mp i \mu_B B'_0 x \psi_{\mp}(x,y,z,t). \quad (E.1)$$

Under transformation

$$\psi_{\pm}(x,z,t) = \exp\left(\pm \frac{i\mu_B B_0 t}{\hbar}\right) \bar{\psi}_{\pm}(x,y,z,t) \quad (E.2)$$

equation (E.1) becomes

$$i\hbar \frac{\partial \bar{\psi}_{\pm}}{\partial t}(x,y,z,t) = -\frac{\hbar^2}{2m} \nabla^2 \bar{\psi}_{\pm}(x,y,z,t) \mp \mu_B B'_0 z \bar{\psi}_{\pm}(x,y,z,t) \\ \mp i \mu_B B'_0 x \bar{\psi}_{\mp}(x,y,z,t) \exp\left(\pm i \frac{2\mu_B B_0 t}{\hbar}\right) \quad (E.3)$$

The coupling term oscillates rapidly with the Larmor frequency

$\omega_L = \frac{2\mu_B B_0}{\hbar} = 1,4 \times 10^{11} s^{-1}$. Since $|B_0| \gg |B'_0 z|$ and $|B_0| \gg |B'_0 x|$, the period of oscillation is short compared to the motion of the wave function. Averaging over a period that is long compared to the oscillation period, the coupling term vanishes, which entails

$$i\hbar \frac{\partial \bar{\psi}_{\pm}}{\partial t}(x,y,z,t) = -\frac{\hbar^2}{2m} \nabla^2 \bar{\psi}_{\pm}(x,y,z,t) \mp \mu_B B'_0 z \bar{\psi}_{\pm}(x,y,z,t). \quad (E.4)$$

The initial wave function $\bar{\psi}_{\pm}^0(x,y,z) = \psi_{\pm}^0(x,y,z) = \psi_x^0(x)\psi_y^0(y)\psi_{\pm}^0(z)$ with

$$\begin{align*}
\psi_x^0(x) &= \begin{cases} (2\pi\sigma_0^2)^{-\frac{1}{4}} e^{-\frac{x^2}{4\sigma_0^2}} & \text{if } |x| \le \delta \\ 0 & \text{if } |x| > \delta \end{cases} \\
\psi_y^0(y) &= \begin{cases} (2\pi\sigma_0^2)^{-\frac{1}{4}} e^{-\frac{y^2}{4\sigma_0^2}} & \text{if } |x| \le \delta \\ 0 & \text{if } |x| > \delta \end{cases} \\
\psi_+^0(z) &= \begin{cases} (2\pi\sigma_0^2)^{-\frac{1}{4}} e^{-\frac{z^2}{4\sigma_0^2}} \cos\left(\frac{\theta_0}{2} e^{i\frac{\varphi_0}{2}}\right) & \text{if } |z| \le \delta \\ 0 & \text{if } |z| > \delta \end{cases} \\
\psi_-^0(z) &= \begin{cases} (2\pi\sigma_0^2)^{-\frac{1}{2}} e^{-\frac{z^2}{4\sigma_0^2}} i \sin\left(\frac{\theta_0}{2} e^{-i\frac{\varphi_0}{2}}\right) & \text{if } |z| \le \delta \\ 0 & \text{if } |z| > \delta \end{cases}
\end{align*} \quad (E.5)$$

allows a separation of variables $x, y$ and $z$. Then we can compute explicitly the preceding equations for $t \in [0, \Delta t]$. We obtain: $\bar{\psi}_{\pm}(x,y,z,t) = \bar{\psi}_x(x,t)\bar{\psi}_y(y,t)\bar{\psi}_{\pm}(z,t)$ with
$$ \bar{\psi}_x(x,t) = (2\pi\sigma_t^2)^{-\frac{1}{4}} e^{-\frac{x^2}{4\sigma_t^2}} \exp\left\{\frac{i}{\hbar}\left[-\frac{\hbar}{2}\tan^{-1}\left(\frac{\hbar t}{2m\sigma_0^2}\right) + \frac{x^2\hbar^2t^2}{8m\sigma_0^2\sigma_t^2}\right]\right\}, $$

$$ \bar{\psi}_+(z,t) = \psi_K(z,t) \cos \frac{\theta_0}{2} e^{i \frac{\varphi_0}{2}} \quad \text{and} \quad K = -\mu_B B'_0, \qquad (E.6) $$

$$ \bar{\psi}_-(z,t) = \psi_K(z,t) i \sin \frac{\theta_0}{2} e^{-i \frac{\varphi_0}{2}} \quad \text{and} \quad K = +\mu_B B'_0, $$

$$ \sigma_t^2 = \sigma_0^2 + \left( \frac{\hbar t}{2m\sigma_0} \right)^2 $$

and

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