## 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}