Ian Jauslin
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\documentclass{ian-presentation}

\usepackage[hidelinks]{hyperref}
\usepackage{graphicx}
\usepackage{xcolor}
\usepackage{amsmath}
\usepackage{array}
\usepackage{ulem}


\definecolor{highlight}{HTML}{981414}
\long\def\high#1{{\color{highlight}#1}}

\begin{document}
\pagestyle{empty}
\hbox{}\vfil
\bf\Large
\hfil Bell's inequalities\par
\smallskip
\hfil \large and non-locality\par
\vfil
\large
\hfil Ian Jauslin
\rm\normalsize
\vfil
Youtube channel: {\tt \href{https://www.youtube.com/@ianjauslin9430}{ianjauslin}}
\hfill{\tt \href{http://ian.jauslin.org}{http://ian.jauslin.org}}
\eject

\setcounter{page}1
\pagestyle{plain}

\title{Introduction}
\begin{itemize}
  \item
  Bell's inequalities:
  \href{https://link.aps.org/pdf/10.1103/PhysicsPhysiqueFizika.1.195}{[Bell, 1964]},
  \href{https://doi.org/10.1103/PhysRevLett.23.880}{[Clauser, Horne, Shimony, Holt, 1969]},\par
  \href{https://cds.cern.ch/record/980036/files/197508125.pdf}{[Bell, 1975]}...

  \item
  Prove that quantum mechanics is a \high{non-local} theory.

  \item
  Actually, they are extremely general, and only assume some carefully chosen assumptions about the probabilistic nature of quantum theory.

  \item
  Often characterized as forbidding ``local hidden variable theories'', as we shall see, this is not exactly untrue, but is misleading.

  \item
  Nobel prize 2022: Aspect, Clauser, Zeilinger: experimental verification of the predictions of quantum mehcanics.

  \item
  Reference: \high{\href{http://www.scholarpedia.org/article/Bell\%27s_theorem}{{\it Scholarpedia} article by Goldstein, Norsen, Tausk, Zanghi.}}
\end{itemize}

\vfill
\eject

\hbox{}
\vfill
\hfil{\bf\Large Part I}\par\bigskip
\hfil{\bf\Large Bell, 1964}
\vfill
\eject

\title{Bell's theorem}
\begin{itemize}
  \item
  Step 1: the EPR argument: \href{https://doi.org/10.1103/PhysRev.47.777}{[Einstein, Podolsky, Rosen, 1935]}.
\end{itemize}
\vfill
\eject

\hfil
\includegraphics[trim=10 1in 2in 2in, clip, height=\textheight]{epr.pdf}
\eject

\title{EPR}
\begin{itemize}
  \item
  Two observers at distance $\Delta x$ measure the spin of electrons.

  \item
  Electron spin: can be measured in any \high{direction} in $\mathcal S^2$, returns \high{$+1$ or $-1$}.

  \item
  Before measurement, the electrons are in an \high{entangled state}: they are \high{anticorrelated} (if one returns $+1$, the other returns $-1$).

  \item
  In the usual approach to quantum mechanics, observables \high{do not have values before they are measured}.

  \item
  The observers perform their measurement at the same time.
  \high{If the world were local}, then, since the observers are spatially separated, one measurement cannot affect the other.

  \item
  But the outcomes are \high{perfectly anticorrelated}.
  Therefore, the outcomes had to be \high{determined before} the measurement was done (``hidden variables'').
\end{itemize}
\vfill
\eject

\title{Bell's theorem}
\begin{itemize}
  \item
  Step 1: the EPR argument: \href{https://doi.org/10.1103/PhysRev.47.777}{[Einstein, Podolsky, Rosen, 1935]}:
  $$
  \boxed{\mathrm{QM\ is\ local}\quad\Longrightarrow\quad \mathrm{observables\ have\ predetermined\ values}}
  $$

  \item
  Step 2: Bell's inequality.
\end{itemize}
\vfill
\eject

\hfil
\includegraphics[trim=10 1in 2in 2in, clip, height=\textheight]{epr.pdf}
\eject

\title{Bell's inequality (pigeonhole)}
\vskip-15pt
\begin{itemize}
  \item
  In the EPR setting, the observers each make three measurements (e.g. choosing three different directions of spin).
  The outcomes of the measurements are \high{random}.

  \item
  We \high{assume} that the outcomes of the measurements \high{exist} independently of the measurement (\high{predetermined values}).
  In other words, we can represent the outcome of measurements as random variables that exist \high{simultaneously}:
  $$
    Z_1^{(A)},Z_2^{(A)},Z_3^{(A)}\in\{-1,+1\}
    ,\quad
    Z_1^{(B)},Z_2^{(B)},Z_3^{(B)}\in\{-1,+1\}
  $$
  that are distributed according to a probability distribution $\mathbb P$.

  \item
  We assume \high{perfect anticorrelation}:
  $$
    \mathbb P(Z_i^{(A)}\neq Z_i^{(B)})=1
    .
  $$
\end{itemize}
\vfill
\eject

\title{Bell's inequality (pigeonhole)}
\begin{itemize}
  \item
  By the pigeonhole principle, at least two of the measurements must give the same answer:
  $$
    \mathbb P(Z_1^{(A)}= Z_2^{(A)})
    +
    \mathbb P(Z_1^{(A)}= Z_3^{(A)})
    +
    \mathbb P(Z_2^{(A)}= Z_3^{(A)})
    \geqslant 1
  $$

  \item
  Since $A$ and $B$ are anticorrelated: $\mathbb P(Z_i^{(A)}=Z_j^{(A)})=\mathbb P(Z_i^{(A)}\neq Z_j^{(B)})$, so
  $$\boxed{
    \mathbb P(Z_1^{(A)}\neq Z_2^{(B)})
    +
    \mathbb P(Z_1^{(A)}\neq Z_3^{(B)})
    +
    \mathbb P(Z_2^{(A)}\neq Z_3^{(B)})
    \geqslant 1
  }$$
\end{itemize}
\vfill
\eject

\title{Bell's theorem}
\begin{itemize}
  \item
  Step 1: the EPR argument: \href{https://doi.org/10.1103/PhysRev.47.777}{[Einstein, Podolsky, Rosen, 1935]}:
  $$
  \boxed{\mathrm{QM\ is\ local}\quad\Longrightarrow\quad \mathrm{spins\ have\ predetermined\ values}}
  $$

  \item
  Step 2: Bell's inequality:
  $$\boxed{
    \begin{array}{l}
      \mathrm{spins\ have\ predetermined\ values}
      \Longrightarrow\\
      \hskip30pt\Longrightarrow
      \mathbb P(Z_1^{(A)}\neq Z_2^{(B)})
      +
      \mathbb P(Z_1^{(A)}\neq Z_3^{(B)})
      +
      \mathbb P(Z_2^{(A)}\neq Z_3^{(B)})
      \geqslant 1
    \end{array}
  }$$

  \item
  Step 3: According to the laws of quantum mechanics,
\end{itemize}
\vfill
\eject

\title{Quantum mechanical prediction}
\begin{itemize}
  \item
  Suppose the three measurements are done at angles $\frac{2\pi}3$ from each other, then, for $i\neq j$,
  $$
    \mathbb P(Z_i^{(A)}\neq Z_j^{(B)})=\frac{1+\cos\frac{2\pi}3}2=\frac 14
    .
  $$

  \item
  Therefore,
  $$
    \mathbb P(Z_1^{(A)}\neq Z_2^{(B)})
    +
    \mathbb P(Z_1^{(A)}\neq Z_3^{(B)})
    +
    \mathbb P(Z_2^{(A)}\neq Z_3^{(B)})
    =\frac34<1
    .
  $$
\end{itemize}
\vfill
\eject

\title{Bell's theorem}
\begin{itemize}
  \item
  Step 1: the EPR argument: \href{https://doi.org/10.1103/PhysRev.47.777}{[Einstein, Podolsky, Rosen, 1935]}:
  $$
  \boxed{\mathrm{QM\ is\ local}\quad\Longrightarrow\quad \mathrm{spins\ have\ predetermined\ values}}
  $$

  \item
  Step 2: Bell's inequality:
  $$\boxed{
    \begin{array}{l}
      \mathrm{spins\ have\ predetermined\ values}
      \Longrightarrow\\
      \hskip30pt\Longrightarrow
      \mathbb P(Z_1^{(A)}\neq Z_2^{(B)})
      +
      \mathbb P(Z_1^{(A)}\neq Z_3^{(B)})
      +
      \mathbb P(Z_2^{(A)}\neq Z_3^{(B)})
      \geqslant 1
    \end{array}
  }$$

  \item
  Step 3: According to the laws of quantum mechanics,
  $$\boxed{
    \mathbb P(Z_1^{(A)}\neq Z_2^{(B)})
    +
    \mathbb P(Z_1^{(A)}\neq Z_3^{(B)})
    +
    \mathbb P(Z_2^{(A)}\neq Z_3^{(B)})
    \not\geqslant 1
  }$$
\end{itemize}
\vfill
\eject

\title{Bell's theorem}
\begin{itemize}
  \item
  Step 1: the EPR argument: \href{https://doi.org/10.1103/PhysRev.47.777}{[Einstein, Podolsky, Rosen, 1935]}:
  $$
  \boxed{\mathrm{QM\ is\ \high{not}\ local}\quad\Longleftarrow\quad \mathrm{spins\ \high{do\ not}\ have\ predetermined\ values}}
  $$

  \item
  Step 2: Bell's inequality:
  $$\boxed{
    \begin{array}{l}
      \mathrm{spins\ \high{do\ not}\ have\ predetermined\ values}
      \Longleftarrow\\
      \hskip30pt\Longleftarrow
      \mathbb P(Z_1^{(A)}\neq Z_2^{(B)})
      +
      \mathbb P(Z_1^{(A)}\neq Z_3^{(B)})
      +
      \mathbb P(Z_2^{(A)}\neq Z_3^{(B)})
      \high{\not\geqslant} 1
    \end{array}
  }$$

  \item
  Step 3: According to the laws of quantum mechanics,
  $$\boxed{
    \mathbb P(Z_1^{(A)}\neq Z_2^{(B)})
    +
    \mathbb P(Z_1^{(A)}\neq Z_3^{(B)})
    +
    \mathbb P(Z_2^{(A)}\neq Z_3^{(B)})
    \not\geqslant 1
  }$$
\end{itemize}
\vfill
\eject

\title{Reactions to Bell's theorem}
\begin{itemize}
  \item
  The violation of Bell's inequality shows that the values of the spin in different directions do not \high{simultaneously} exist.
  More generally, the values of \high{non-commuting operators} do not simultaneously exist.

  \item
  This has lead many to state that Bell's theorem shows there are no \high{hidden variables} in quantum mechanics.
  This is \high{not true}: it is only true that there non-commuting observables do not simultaneously have values (``no non-contextual hidden variables'').

  \item
  There \high{is} a theory of quantum mechanics with ``hidden variables'': \high{Bohmian}.

  \item
  Others have said that Bell's theorem shows there are no \high{local hidden variables} in quantum mechanics.
  While this is technically true, it misses the point: there is no \high{locality} in quantum mechanics.
\end{itemize}
\vfill
\eject

\title{More general statement}
\begin{itemize}
  \item
  The theorem discussed earlier requires \high{perfect anticorrelation} between spins in an entangled singlet.
  What if quantum mechanics is only very slightly wrong, and the anticorrelation is not exactly perfect?

  \item
  The proof of Bell's theorem goes through showing there are no pre-existing values, which has caused some confusion.
  Can we prove the theorem \high{without any reference to pre-existing values}?
\end{itemize}
\vfill
\eject


\hbox{}
\vfill
\hfil{\bf\Large Part II}\par\bigskip
\hfil{\bf\Large Bell, revisited}
\vfill
\eject

\hfil
\includegraphics[trim=10 1in 1in 2in, clip, height=\textheight]{setup.pdf}
\eject

\title{General setup}
\begin{itemize}
  \item
  Observers $A$ and $B$

  \item
  Each observer can set a tunable parameter on their measurement device: $\alpha,\beta$ (for instance, the direction of spin).

  \item
  Outcomes of a measurement: random variables $X_A,X_B\in[-1,1]$, distributed according to a \high{joint} probability distribution $\mathbb P_{\alpha,\beta}$ that \high{depends on $\alpha,\beta$} (the outcomes of measurements with different parameters $\alpha,\beta$ are not required to simultaneously exist).
\end{itemize}
\vfill
\eject

\title{Locality}
\begin{itemize}
  \item
  Naive definition:
  $$
    \mathbb P_{\alpha,\beta}(X_A,X_B)
    =\mathbb P_\alpha^{(A)}(X_A)
    \mathbb P_\beta^{(B)}(X_B)
  $$

  \item
  This does not allow for any correlation between $A$ and $B$ (in particular, it excludes the anticorrelation considered earlier).
\end{itemize}
\vfill
\eject

\title{General setup}
\begin{itemize}
  \item
  Observers $A$ and $B$

  \item
  Parameters $\alpha,\beta$.

  \item
  Outcomes of a measurement: random variables $X_A,X_B\in[-1,1]$, $\sim\mathbb P_{\alpha,\beta}$.

  \item
  Allow for an extra parameter $\lambda$, whose value is shared by both $A$ and $B$ and may affect the outcome of the measurements.
  For instance, $\lambda$ could be a shared state of the electrons (e.g. an anticorrelated entangled state).
  $\lambda$ is a random variable distributed according to $\mathbb Q$ (independent of $\alpha,\beta$).

  \item
  Locality:
  $$
    \mathbb P_{\alpha,\beta}(X_A,X_B|\lambda)
    =\mathbb P_\alpha^{(A)}(X_A|\lambda)
    \mathbb P_\beta^{(B)}(X_B|\lambda)
  $$
\end{itemize}
\vfill
\eject

\title{Bell's inequality}
\vfill
{\bf Theorem}: Under these assumptions, $\forall\alpha,\alpha',\beta,\beta'$,
$$
  |\mathbb E_{\alpha,\beta}(X_AX_B)
  -\mathbb E_{\alpha,\beta'}(X_AX_B)|
  +
  |\mathbb E_{\alpha',\beta}(X_AX_B)
  +\mathbb E_{\alpha',\beta'}(X_AX_B)|
  \leqslant 2
$$
where $\mathbb E_{\alpha,\beta}$ is the expectation value in the probability distribution $\mathbb P_{\alpha,\beta}$.
\vfill
\eject

\title{Proof of Bell's inequality}
$$
  \mathbb E_{\alpha,\beta}(X_AX_B)
  =\int d\mathbb Q(\lambda)\ \mathbb E_{\alpha,\beta}(X_AX_B|\lambda)
$$
by the locality condition,
$$
  \mathbb E_{\alpha,\beta}(X_AX_B|\lambda)
  =
  \mathbb E_{\alpha}^{(A)}(X_A|\lambda)
  \mathbb E_{\beta}^{(B)}(X_B|\lambda)
$$
so
$$
  |\mathbb E_{\alpha,\beta}(X_AX_B)
  \pm\mathbb E_{\alpha,\beta'}(X_AX_B)|
  \leqslant
  \int d\mathbb Q(\lambda)\ 
  \left|\mathbb E_\alpha^{(A)}(X_A|\lambda)\right|
  \left|\mathbb E_\beta^{(B)}(X_B|\lambda)\pm\mathbb E_{\beta'}^{(B)}(X_B|\lambda)\right|
  .
$$
Since $|X_A|\leqslant 1$, $|\mathbb E_\alpha^{(A)}(X_A|\lambda)|\leqslant 1$.
\vfill
\eject

\title{Proof of Bell's inequality}
If $x:=\mathbb E_\beta^{(B)}(X_B|\lambda)$ and $y:=\mathbb E_{\beta'}^{(B)}(X_B|\lambda)$, then
$$
  \begin{array}{>\displaystyle l}
    |\mathbb E_{\alpha,\beta}(X_AX_B)
    -\mathbb E_{\alpha,\beta'}(X_AX_B)|
    +
    |\mathbb E_{\alpha',\beta}(X_AX_B)
    +\mathbb E_{\alpha',\beta'}(X_AX_B)|
    \leqslant\\\hfill\leqslant
    \int d\mathbb Q(\lambda)\ 
    |x-y|+|x+y|
  \end{array}
$$
and since $|x|\leqslant 1$ and $|y|\leqslant 1$,
$$
  |x-y|+|x+y|\leqslant 2
  .
$$
\hfill$\square$
\vfill
\eject

\title{Quantum mechanical prediction}
\begin{itemize}
  \item
  $\alpha,\beta\in\mathcal S^2$: direction of spin:
  $$
    \mathbb E_{\alpha,\beta}(X_AX_B)=-\alpha\cdot\beta
    .
  $$

  \item
  $$
  \begin{array}{>\displaystyle l}
    |\mathbb E_{\alpha,\beta}(X_AX_B)
    -\mathbb E_{\alpha,\beta'}(X_AX_B)|
    +
    |\mathbb E_{\alpha',\beta}(X_AX_B)
    +\mathbb E_{\alpha',\beta'}(X_AX_B)|
    =\\[0.3cm]\hfill=
    |\alpha\cdot(\beta-\beta')|+|\alpha'\cdot(\beta+\beta')|
  \end{array}
  $$

  \item Choose $\beta'\cdot\beta=0$, $\alpha=(\beta-\beta')/\sqrt2$, $\alpha'=(\beta+\beta')/\sqrt2$:
  $$
    |\mathbb E_{\alpha,\beta}(X_AX_B)
    -\mathbb E_{\alpha,\beta'}(X_AX_B)|
    +
    |\mathbb E_{\alpha',\beta}(X_AX_B)
    +\mathbb E_{\alpha',\beta'}(X_AX_B)|
    =2\sqrt2>2
    .
  $$

  \item
  \high{Quantum mechanics violates Bell's inequality}.
\end{itemize}
\vfill
\eject

\title{General setup}
\begin{itemize}
  \item
  Observers $A$ and $B$

  \item
  Parameters $\alpha,\beta$.

  \item
  Outcomes of a measurement: random variables $X_A,X_B\in[-1,1]$, $\sim\mathbb P_{\alpha,\beta}$.

  \item
  Allow for an extra parameter $\lambda$, distributed according to $\mathbb Q$ (independent of $\alpha,\beta$).

  \item
  \high{\sout{Locality}}.
\end{itemize}
\vfill
\eject

\title{EPR}
{\bf Theorem}:
If $X_A,X_B\in\{-1,1\}$, if
$$
  \mathbb P_{\alpha,\alpha}(X_A\neq X_B)=1
$$
and $\mathbb P$ is \high{local}, then $\forall\lambda,\alpha$, there exists $Z_\alpha(\lambda)\in\{-1,1\}$ such that
$$
  \mathbb P_{\alpha,\beta}(X_A=Z_\alpha(\lambda)|\lambda)
  =\mathbb P_{\alpha,\beta}(X_B=-Z_\alpha(\lambda)|\lambda)
  =1
$$
and
$$
  \mathbb P_{\alpha}^{(A)}(X_A=Z_\alpha(\lambda)|\lambda)
  =\mathbb P_{\beta}^{(B)}(X_B=-Z_\alpha(\lambda)|\lambda)
  =1
  .
$$


\end{document}