\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}