\documentclass[11pt,fleqn]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% template that does not use Revtex4 %%% but allows special fonts %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% %%% Please use this template. %%% Edit it using e.g. Notepad %%% Ignore the header (do not change it) %%% Process the file in the Latex site %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Page setup \topmargin -1.5cm \oddsidemargin -0.04cm \evensidemargin -0.04cm \textwidth 16.59cm \textheight 24cm \setlength{\parindent}{0cm} \setlength{\parskip}{0cm} % Fonts \usepackage{latexsym} \usepackage{amsmath} \usepackage{amssymb} \usepackage{bm} % Math symbols I \newcommand{\sinc}{\mbox{sinc}} \newcommand{\const}{\mbox{const}} \newcommand{\trc}{\mbox{trace}} \newcommand{\intt}{\int\!\!\!\!\int } \newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ } \newcommand{\ar}{\mathsf r} \newcommand{\im}{\mbox{Im}} \newcommand{\re}{\mbox{Re}} % Math symbols II \newcommand{\eexp}{\mbox{e}^} \newcommand{\bra}{\left\langle} \newcommand{\ket}{\right\rangle} % Mass symbol \newcommand{\mass}{\mathsf{m}} \newcommand{\Mass}{\mathsf{M}} % More math commands \newcommand{\tbox}[1]{\mbox{\tiny #1}} \newcommand{\bmsf}[1]{\bm{\mathsf{#1}}} \newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}} \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} % Other commands \newcommand{\hide}[1]{} \newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}} \newcommand{\bitem}{$\bullet$ \ \ \ } \newcommand{\Cn}[1]{\begin{center} #1 \end{center}} \newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}} \newcommand{\Dn}{\vspace*{3mm}} % Figures \newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}} % heading \newcommand{\heading}[1]{\begin{center} \Large {#1} \end{center}} \newcommand{\auname}[1]{\begin{center} \bf Submitted by: #1 \end{center}} \begin{document} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \heading{Weak measurements} \auname{Adam Reichenthal} \Dn %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \Dn =\!=\!=\!=\!=\!=\!= \large {\bf [1] Pre and Post selection in strong measurements} \normalsize{ \Dn Lets take a state: \Dn \qquad $|\Psi \rangle= \displaystyle\sum_{n} a_n|n \rangle$. $\hfill{(1)}$ \Dn The probability for a "collapse" to the particular state $\langle n|$ is given by the projection formula: \Dn \qquad $Prob(n|\Psi)=|\langle n|\Psi\rangle|^2$ $\hfill{(2)}$ \Dn \Dn Now let us take an operator $\hat{A}$ that its result when measuring the state $|n \rangle$ is: \Dn \qquad $\hat{A}|n \rangle = a_n|n \rangle$ $\hfill{(3)}$ \Dn then the expectation value $\langle \hat{A} \rangle$ of $\hat{A}$ with respect to the state $|\Psi \rangle$ is defined by: \Dn \qquad $\displaystyle{\langle \hat{A} \rangle{^{strong}_{\Psi}} \equiv \frac{\textstyle\sum_{n}Prob(n|\Psi) \ a_n}{\textstyle\sum_{n}Prob(n|\Psi)}= \frac{ \langle \Psi|\hat{A}|\Psi \rangle}{\langle \Psi | \Psi \rangle}} \xrightarrow[states]{normalized} \langle\Psi|\hat{A}|\Psi\rangle$ $\hfill{(4)}$ \Dn \Dn Let us calculate what is the probability of (2) to finish at state $|\Phi\rangle$. We use the same method: \Dn \qquad $Prob(\Phi|n|\Psi) \equiv Prob(\Phi|n)Prob(n|\Psi)=|\langle\Phi|n\rangle|^2 \ |\langle n|\Psi\rangle|^2$ $\hfill{(5)}$ \Dn So we get that the expectation value now is: \Dn \qquad $\displaystyle{ \langle \hat{A}\rangle{^{strong}_{\langle\Phi|\Psi\rangle}} = \frac{\textstyle\sum_{n}|\langle\Phi|P^{(n)}|\Psi\rangle|^2 a_n}{\textstyle\sum_{n}|\langle\Phi|P^{(n)}|\Psi\rangle|^2} } $ $\hfill{(6)}$ \Dn Where: \Dn \qquad $P^{(n)} \equiv |\Psi_n\rangle\langle \Psi_n|$ \ is the "projection operator". $\hfill{(7)}$ \Dn \Dn } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \Dn =\!=\!=\!=\!=\!=\!= \large {\bf [2] The three boxes paradox} \normalsize{ \Dn Let us consider a particle that can be located in one of three orthogonal quantum boxes. We prepare the particle at time $t_{past}$ in an equally weighted superposition state: \Dn \qquad $|\Psi\rangle=\frac{1}{\sqrt{3}}(|1\rangle+|2\rangle+|3\rangle)$ $\hfill{(8)}$ \Dn We notice that every eigenstate has the same collapse probability. \Dn Next let us assume that at a later time $t_{future}$ the particle is found to be in the state: \Dn \qquad $|\Phi\rangle=\frac{1}{\sqrt{3}}(|1\rangle+|2\rangle-|3\rangle)$ $\hfill{(9)}$ \Dn If we preform a strong measurement of an observable $A$ which has degenerate eigenstates and we want to know the probability for the outcome $a_n$ eigenvalue, we can use: \Dn \qquad $\displaystyle{ Prob(A=a_n)=\frac{|\langle\Phi|P^{(n)}|\Psi\rangle|^2}{\displaystyle\sum_{i} |\langle \Phi|P^{(i)}|\Psi\rangle|^2} }$ $\hfill{(10)}$ \Dn \Dn This is called the ABL rule [1,2]. \Dn We'll calculate the probability of finding the particle in box 1: \Dn \qquad $\displaystyle{ P(P_1=1)=\frac{|\langle\Phi|1\rangle\langle1|\Psi\rangle|^2}{|\langle\Phi|1\rangle\langle1|\Psi\rangle|^2+|\langle\Phi|2\rangle\langle2 |\Psi\rangle+\langle\Phi|3\rangle\langle3|\Psi\rangle|^2}= \frac{|\frac{1}{3}|^2}{|\frac{1}{3}|^2+|0|^2}=1}$ $\hfill{(11)}$ \Dn So if we open box 1 we will definitely fine the particle there. The same goes for box 2. \Dn Lets check out box 3: \Dn \qquad $\displaystyle{P(P_3=1)=\frac{|\langle\Phi|3\rangle\langle3|\Psi\rangle|^2}{|\langle\Phi|3\rangle\langle3|\Psi\rangle|^2+|\langle\Phi|1\rangle\langle1|\Psi\rangle+\langle\Phi|2\rangle\langle2|\Psi\rangle|^2}= \frac{|\frac{1}{3}|^2}{|\frac{1}{3}|^2+|\frac{2}{3}|^2}=\frac{1}{5}}$ $\hfill{(12)}$ \Dn \Dn We can summarize this case as follows: in classical physics the particle has only one position therefore only opening one specific box will corresponds with the result finding the particle in the box with a probability of $\frac{1}{N}$. But a quantum particle acts differently since it does't matter which box of the N-1 (generalized case) boxes we open (always) we'll find the single particle there for sure. That is why we say that the particle is simultaneously in N-1 boxes. Finely we deduce that as more boxes there are the chance we'll find it in the last box decreases. \Dn A different way to emphasize this result is to let Alice prepare the pre selected state as (8) and let her measure the post selected state. If Bob sees the particle at 1 or 2 (he chooses a box) then if the post selected state is like (9) Alice knows that he saw it. If Bob doesn't see the particle for example he measured 1 then the particle at time t is at the state $\frac{1}{\sqrt{2}}(|2\rangle+|3\rangle)$ and that is what Alice will measure and because it's different from (9) she'll know he didn't see the particle for sure. \Dn } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \Dn =\!=\!=\!=\!=\!=\!= \large {\bf [3] Von Neumann with post selection and Weak measurements} \normalsize{ \Dn Von Neumann was first to proposed a theory modelling a measurement through interaction. Its principle is to obtain the value of a measured observable through the observable of measurement device such as a shift of a pointer variable during the interaction. \Dn In an ideal measurement the initial position of the pointer $q$ is well localized around the zero and therefore the conjugate momentum $\hat{p}$ has a very large uncertainty. In a weak measurement the initial state of the measuring device is such that $\hat{p}$ is localized around zero with small uncertainty so there is a large uncertainty in $q$ and therefore the measurement becomes inaccurate. So in contrast to the classic Von Neumann approach of measuring quantum states, weak measurement uses a measuring device whose pointer has a very large uncertainty when compared with the regular shift. \Dn In Ref [3] it has been suggested to define $\langle A\rangle{^{weak}_{\Psi \Phi}} = \frac{\langle\Phi|A|\Psi\rangle}{\langle\Phi|\Psi\rangle}$. We would like to interpret this result within the framework of the Von-Neumann measurement scheme which is used also in the study of weak continuous measurements [4-8]. In particular we would like to clarify the significance of the imaginary part. \Dn The coupling of the system to a Von-Nuemann pointer whose canonical coordinates are ${(\hat{x},\hat{q})}$ is described by the Hamiltonian: \Dn \qquad $H = -\lambda g(t) A \hat{x}$ $\hfill{(13)}$ \Dn where $g(t)$ is a short time normalized function, $\lambda$ is assumes to be very small and $A$ is the observable that is a measured Hermitian operator. \Dn In general $A$ is an observable that has a spectrum of values $\{a_i\}$ so that if $A=a$ this interaction shifts the pointer: \Dn \qquad ${q \mapsto q + \lambda a}$ $\hfill{(14)}$ \Dn Note that $x$ unlike $q$ is a constant of the motion. \Dn For a weak measurement we assume that $\lambda$ is small and obtain the following result: the evolution of the system with the pointer is described by: \Dn \qquad $U = \eexp{-i\int Hdt} = \eexp{-i\int -\lambda g(t) A \hat{x}dt}= \eexp{i\lambda A \hat{x}} \approx \ \ 1+i\lambda A \hat{x}$ $\hfill{(15)}$ \Dn The matrix elements of this evolution operator are: \Dn \qquad $\langle \Phi, x|U| \Psi, x_0 \rangle \ \ = \ \ U[x]_{\Phi,\Psi} \ \delta(x-x_0)$ $\hfill{(16)}$ \Dn where $U[x]$ is a system operator that depends on the constant parameter $x$. \Dn \Dn If the system is prepared in the state $P^{\Psi} = |\Psi\rangle\langle\Psi|$, and the pointer is prepared in state $\rho^{(0)}$, then after the interaction we get: \Dn \qquad $\mbox{state}= U P^{\Psi}\rho^{(0)} U^{\dag}$ $\hfill{(17)}$ \Dn The reduced state of the pointer after post selection is \Dn \qquad $\rho(x'',x') \ \ = \ \ \trc\Big[ \ P^{\Phi}Q^{x'',x'} \ U P^{\Psi}\rho^{(0)} U^{\dag} \ \Big]$ $\hfill{(18)}$ \Dn where $Q^{x'',x'}=|x' \rangle\langle x''|$. \Dn Note that the state is not normalized: the trace is the probability to find the system in state $\Phi$. \Dn The relation between $\rho$ and $\rho^{(0)}$ can be written as follows: \Dn \qquad $\rho(x'',x') \ \ = \ \ \tilde{K}(x'',x') \ \rho^{(0)}(x'',x')$ $\hfill{(19)}$ \Dn where: \Dn \qquad $\tilde{K}(x'',x') \ \ = \ \ \langle \Phi | U[x'] | \Psi \rangle \ \langle \Psi | U[x'']^{\dag} | \Phi \rangle$ $\hfill{(20)}$ \Dn \Dn Equivalently, we can transform to the Wigner representation by changing variables and then multiplication becomes a convolution: \Dn \qquad $\rho(X,r) = \tilde{K}(X,r) \ \rho^{(0)}(X,r)$ $\hfill{(21)}$ \Dn \qquad $\rho(X,q) = \int K(X,q-q') \ \rho^{(0)}(X,q') \ dq'$ \Dn where \Dn \qquad $ K(X,q-q') = \int \tilde{K}(X,r) \ \eexp{-i(q-q')r} dr$ $\hfill{(22)}$ \Dn \qquad $ \tilde{K}(X,r) = \Big\langle \Psi \Big| U\left[X-\frac{r}{2}\right]^{\dag} \Big| \Phi \Big\rangle \ \Big\langle \Phi \Big| U\left[X+\frac{r}{2}\right] \Big| \Psi \Big\rangle$ \Dn If we sum over $\Phi$ we get the standard result for weak measurement without post-selection. We receive: \Dn \qquad $\langle\Psi| (1-i\lambda (X-\frac{r}{2})A^{\dag})(1+i\lambda(X+\frac{r}{2}) A) |\Psi\rangle $ $\hfill{(23)}$ \Dn \qquad $=\langle\Psi| ( 1+i\lambda X(A-A^{\dag})+i\lambda\frac{r}{2}(A+A^{\dag})) |\Psi\rangle$ \Dn and because $A$ is a hermitian operator meaning $A^{\dag} = A$ and $\langle\Psi|\hat{A}|\Psi\rangle=\re\langle A\rangle$ we get that (23) is equal to: \Dn \qquad $\eexp{i\lambda r\re\langle A\rangle}$ $\hfill{(24)}$ \Dn So finally we can see that (22) becomes: \Dn \qquad $ K(X,q-q') = \int dr \eexp{-ir(q-q'-\lambda \re\langle A\rangle)}= 2\pi\delta(q-q'-\lambda \re\langle A \rangle)$ $\hfill{(25)}$ \Dn So indeed when $\lambda=1$ for strong measurement (14) is correct producing an eigenstate. \Dn \Dn Now let us go back and assume there is a post selected state: \Dn \qquad $\langle\Psi| (1-i\lambda (X-\frac{r}{2})A^{\dag}|\Phi\rangle \ \langle\Phi|(1+i\lambda (X+\frac{r}{2})A|\Psi\rangle$ $\hfill{(26)}$ \Dn \qquad $\displaystyle{ = \langle\Psi|\Phi\rangle \Big(1- i \lambda (X - \frac{r}{2})\frac{\langle\Psi|A^{\dag}|\Phi\rangle}{\langle\Psi|\Phi\rangle} \Big) \langle\Phi|\Psi\rangle \Big(1+i\lambda(X+\frac{r}{2})\frac{\langle\Phi|A|\Psi\rangle}{\langle\Psi|\Phi\rangle}\Big) }$ \Dn We define: \Dn \qquad $\displaystyle{ \frac{\langle\Psi|A^{\dag}|\Phi\rangle}{\langle\Psi|\Phi\rangle} \equiv A_1 \ , \ \frac{\langle\Phi|A|\Psi\rangle}{\langle\Phi|\Psi\rangle} \equiv A_2}$ $\hfill{(27)}$ \Dn So we get that (26) is: \Dn \qquad $1+i\lambda X(A_2-A_1)+i\lambda\frac{r}{2}(A_2+A_1)$ $\hfill{(28)}$ \Dn We notice that $A$ is a complex number: $A=\re\langle A\rangle+i\im\langle A\rangle$ and because $A$ is hermitian $A_1=A_2^{\dag}$ so (28) is: \Dn \qquad $1+i\lambda (X 2i\im\langle A_2\rangle + r\re\langle A_2\rangle) = \eexp{i\lambda(2Xi\im\langle A_2\rangle+r\re\langle A_2\rangle)}$ $\hfill{(29)}$ \Dn So (22) this time becomes: \Dn \qquad $K(X,q{-}q') = \eexp{-2\lambda \im\langle A_2 \rangle X} \ 2\pi\delta(q-q'-\lambda \re\langle A_2 \rangle)$ $\hfill{(30)}$ \Dn where $\langle A \rangle$ is define as the "weak value" by Yakir Aharonov. \Dn This means that if one starts with a minimal Gaussian of width $\sigma$, its center will be shifted as follows: \Dn \qquad $\mbox{q-shift} = \re[\langle A \rangle_{\Phi,\Psi}^{\tbox{weak}}](\lambda)$ $\hfill{(31)}$ \Dn \qquad $\mbox{x-shift} = \im[\langle A \rangle_{\Phi,\Psi}^{\tbox{weak}}](-2 \sigma^2)$ \Dn Now we can see once again that a weak measurement that doesn't have a post selected state is like a regular strong measurement: \Dn \qquad $\langle A \rangle{^{weak}_{\Psi}}= \displaystyle\sum_{\alpha} Prob(\alpha)A{^{weak}_{\Phi^{\alpha}|\Psi}}= \displaystyle\sum_{\alpha} |\langle\Phi^{\alpha}|\Psi\rangle|^2 \ \frac{\langle\Psi^{\alpha}|A|\Psi\rangle}{\langle\Phi^{\alpha}|\Psi\rangle} = \displaystyle\sum_{\alpha}\langle\Phi^{\alpha}|\Psi\rangle^* \langle\Phi^{\alpha}|A|\Psi\rangle = \langle\Psi|A|\Psi\rangle$ $\hfill{(32)}$ \Dn Which is the known result as we saw at (4). \Dn } \Dn %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \Dn =\!=\!=\!=\!=\!=\!= \large {\bf [4] Experiment preforming weak measurements} \normalsize{ \Dn $\bf{Stage 1}$ is preparing the pre selected state: lets take a beam of spin $\frac{1}{2}$ particles that moves in the $y$ direction. The particles' spins point in the $x-z$ plane in the direction $\hat{n}$ which makes an angle $\theta$ with the $z$ axis. \Dn $\bf{Stage 2}$ is to measure it weakly: the beam passes through a Stern-Gerlach device that measures the spin weakly in the z direction. It succeeds doing so because its gradient of the magnetic field is sufficiently small. Therefore the motion of the beam changes only slightly. \Dn $\bf{Stage 3}$ is having the post selected state: finally we pass the beam through another S-G device this time which has a normal gradient. Its job is to split the beam into two beams corresponding to the two values of $\sigma _x$. We take the particles with $\sigma_x=+1$ and the beam finally hits a screen. On the screen we obtain a wide spot whose displacement in the direction $\hat{z}$ is measured. This displacement will yield the weak value of $\sigma_z$. \Dn \Dn The initial spin state $|\Psi\rangle$ is: \Dn \qquad $|\Psi\rangle = cos(\frac{\theta}{2})|\uparrow\rangle+sin(\frac{\theta}{2})|\downarrow\rangle$ $\hfill{(33)}$ \Dn and the final state $|\Phi\rangle$ is: \Dn \qquad $|\Phi\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle+|\downarrow\rangle)$ $\hfill{(34)}$ \Dn So we get that: \Dn \qquad $ \langle\Phi|\Psi\rangle = \frac{1}{\sqrt{2}}(cos(\frac{\theta}{2})+sin(\frac{\theta}{2})) \ , \ \langle\Phi|\hat{\sigma}_z|\Psi\rangle=\frac{1}{\sqrt{2}}(cos(\frac{\theta}{2})-sin(\frac{\theta}{2})) $ $\hfill{(35)}$ \Dn We take the weak value of the spin component $\sigma_z$. This displacement will yield the weak value of the spin component $\sigma_z$: \Dn \qquad $\displaystyle{ \langle\hat{\sigma}_{z}\rangle{^{weak}_{\langle\Phi|\Psi\rangle}} \ = \frac{\langle \Phi|\sigma_z|\Psi\rangle}{\langle\Phi|\Psi\rangle} = \frac{cos(\theta)}{1+sin( \theta )}}$ $\hfill{(36)}$ \Dn Substituting this in (30) shows us the movement of the pointer. We notice that if $\theta \rightarrow -\pi$ we get very large values that exceeds the eigenvalue spectra of spin $\frac{1}{2}$. \Dn \Dn \Dn \Dn } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Appendix: Philosophical discussion on the TSVF} \Dn \Dn =\!=\!=\!=\!=\!=\!= \large {\bf [1] Description of a quantum system in the standard approach} \normalsize{ \Dn In quantum mechanics we denote a system at a given time $t_{now}$ by a quantum state: \Dn \qquad $|\Psi\rangle$ $\hfill{(1)}$ \Dn This state is defined by the result of a measurement that was performed at the past meaning $t_{measure}