LATEX
تناظرات الدورانات و لورنتز و المتقطعة و بوانكريه قريبا على اليوتوب ان شاء الله
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EPR-Bohm-Bell
Bell's theorem2| مبرهنة بال 2: مسلمات اينشتاين-بودولسكى-روزن و متراجحة بال
Bell's theorem
EPR-Bohm Experiment
A pion decays at rest: $\pi^0\longrightarrow e^-+e^+$.
The pair electron-positron is maximally entangled in the singlet state (linear superposition):
\begin{eqnarray}|e^-e^+\rangle&=&\frac{1}{\sqrt{2}}(|+\rangle|-\rangle-|-\rangle|+\rangle)\nonumber\\&=&\frac{1}{\sqrt{2}}(|+\rangle_x|-\rangle_x-|-\rangle_x|+\rangle_x).\end{eqnarray}
Let us suppose that the measurement of the spin $S_z$ by Alice leads to the state:
\[|+\rangle|-\rangle=|+\rangle\frac{1}{\sqrt{2}}(|+\rangle_x-|-\rangle_x).\]
The measurement of the spin $S_z$ by Bob is unecesseary and Bob can instead measure the spin $S_x$. Bob can then determined the two spins $S_z$ and $S_x$ at the same time which is forbidden by Heisenberg's uncertainty principle since the operators $S_z$ and $S_x$ are incompatible.
This is the paradox and quantum mechanics is incomplete.
Classical Realism
$\lambda$: hidden variable.
Probability density $\rho(\lambda)$ : $\rho(\lambda)>0$ , $\int \rho(\lambda)d\lambda=1$.
Local Causality (Free Will)
Measurement Alice $\vec{a}$.
Measurement Bob $\vec{b}$.
$\vec{a}$, $\vec{b}$ are freely and independently chosen.
Alice: $ S_a=\pm 1$....$f(\vec{a},\lambda)=\pm 1$.
Bob: $S_b=\pm 1$...$g(\vec{b},\lambda)=\pm 1$.
If $\vec{b}=\vec{a}$....$g(\vec{a},\lambda)=-f(\vec{a},\lambda)$.
Expected value of product of Alice and Bob measurements:
\begin{eqnarray}P(a,b)&=& \int f(\vec{a},\lambda)g(\vec{b},\lambda) \rho(\lambda)d\lambda\nonumber\\&=&-\int f(\vec{a},\lambda)f(\vec{b},\lambda) \rho(\lambda)d\lambda\nonumber\\\end{eqnarray}
\begin{eqnarray} P(a,b)-P(a,c)&=&-\int \bigg[f(\vec{a},\lambda)f(\vec{b},\lambda)-f(\vec{a},\lambda)f(\vec{c},\lambda)\bigg] \rho(\lambda)d\lambda\nonumber\\&=&-\int f(\vec{a},\lambda)f(\vec{b},\lambda)\bigg[1-f(\vec{b},\lambda)f(\vec{c},\lambda)\bigg] \rho(\lambda)d\lambda\nonumber\\\end{eqnarray}
\[-1\leq f(\vec{a},\lambda)f(\vec{b},\lambda)\leq 1\Rightarrow 0\leq 1-f(\vec{a},\lambda)f(\vec{b},\lambda)\leq 2\]
Bell's inequality (hidden variables):
\begin{eqnarray} |P(a,b)-P(a,c)|&\leq &\int |f(\vec{a},\lambda)f(\vec{b},\lambda)|\bigg[1-f(\vec{b},\lambda)f(\vec{c},\lambda)\bigg] \rho(\lambda)d\lambda\nonumber\\&\leq &\int \bigg[1-f(\vec{b},\lambda)f(\vec{c},\lambda)\bigg] \rho(\lambda)d\lambda\nonumber\\&\leq & 1+P(b,c).\end{eqnarray}
Quantum Mechanics:
\begin{eqnarray}P(a,b)=-\vec{a}.\vec{b}.\end{eqnarray}