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LATEX

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}

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