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diff --git a/Jauslin_gmathphys_2023.tex b/Jauslin_gmathphys_2023.tex new file mode 100644 index 0000000..55e7fb5 --- /dev/null +++ b/Jauslin_gmathphys_2023.tex @@ -0,0 +1,552 @@ +\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} |