How does an outcome of photon measurement 1 affect Photon 2 in quantum entanglement? I know I am missing something and this question is probably very silly, but I would like to understand. Quoting an article:     If one photon is measured to be in a +1 state, the other must be in a   -1 state. Since the outcome of one photon affects the outcome of the other, the two are said to be entangled.......when you measure the state of one photon you immediately know the state of the other....If we’re light years apart, we each know the other’s outcome for entangled pairs of photons, but the outcome of each entangled pair is random (what with quantum uncertainty and all), and we can’t force our photon to have a particular outcome.   I just cannot see the "magic" here. Using a stupid analogy: There are two balls, black and white, wrapped in a piece of cloth. You take one and I take the other. Whenever and wherever I unwrap the one I took, I will immediately know which one you have.  What is so special about that in the world of particles, how does the outcome of the first affect the other?
Understanding quantum entanglement.. help me validate this analogy! I'm struggling to understand the concept of quantum entanglement. I've distilled my understanding into an analogy, and I need your help to validate it. Here it is:  Let's say I receive two envelopes. Both envelopes have this written on them:     Hello! Open this letter...   If the paper inside is red, then other envelope has blue paper.   If the paper inside is blue, then other envelope has red paper.   So, if I open one letter, and the paper inside is red, then I know the other letter contains the blue paper.  Is this equivalent to quantum entanglement? The fact that I will know the color of the other paper when I open (i.e. observe) one envelope?  In theory, if one of the letters "appeared" in a different galaxy, I would still know the color of the paper inside that letter, just by opening my letter, instantly, correct?  Before I open the letter, is the paper inside each envelope in superposition (??), i.e. 50/50, red or blue?  In this analogy, I am assuming that what's written on the letters is true. Is this a correct assumption in quantum physics? I think so yes?  In real quantum physics, would I be able to change the color of one paper, so that I change the color of the other one? No right?
How do I align an image to centre? How do I align an image to centre?  In a beamer presentation I have a 2-3 items on a slide followed by an image. The image is not wide enough to cover whole slide, so it leaves around 30% space from right. How do I adjust it to leave enough space on both sides?  \begin{frame} \frametitle{Outlook} \begin{itemize} \item apps \item apps \item apps \includegraphics[scale=0.3]{P2P}     \end{itemize} \end{frame} MyClass[] array; List&lt;MyClass&gt; list;   What are the scenarios when one is preferable over the other? And why? How can I avoid soft dull "spots" in my pictures? Recently, I've been doing a lot more photography, particularly with my 100-400mm telephoto and 16-35mm wide. I am skilled enough that I can capture the shot mostly as I want with just the camera gear, requiring minimal post processing. I'm a huge fan of ETTR now, so most of my shots are overexposed to support more shadow DR and improve the noise performance of my el-chepo 450D body.  I've noticed quite a bit lately during post processing in lightroom that my sky's or any other lighter areas of my shots tend have spots on them. They are soft and dull to one degree or another (seems to depend on the zoom factor), so probably something either on the sensor or the lens. I'm having a hard time figuring out exactly what they are, as I try to use my lens pen brush before any shoot, yet I still have them. I am having to make quite a few healing corrections post-process to get rid of them.  Is there an easy way to identify where the spots are on my lenses? Once identified, if a quick brush of the outer lens element with a lens pen (or soft camel hair brush) doesn't do the trick, does that mean the spots are on my sensor? Is it possible there is particulate inside the lens body (all of my gear is less than a year old outside of my 450D body, and its all Canon L-Series gear, so I would home that I don't already have detrimental particulates floating around inside my lenses messing up my shots. :|)  Here are a couple examples. The spots are harder to see on screen, and I first noticed them when I printed a few of my more recent shots. There are quite a few, and the shot the examples below were taken from, where the sky took up about 30% of the top edge of the photo, there were probably 10-12 different spots, some of them darker, many of them lighter. The darker ones are the ones that are very visible in prints, where as the softer ones are not as visible, but still apparent enough to ruin an expensive print. LaTeX template for typesetting a novel I'd like to use LaTeX to typeset a novel. Could you recommend me any template?  What I mean is a template (some .tex file) where I can include my chapterN.tex files (mainly plain text) and get a pretty good design with zero work :) How to interpret output from Linux 'top' command? What is a nicer, less immature saying?  I love using 'A Freudian slip', but that is only applicable in certain situations. How to integrate $\int\sqrt{x^2+1}dx$ Question  I calculated the classical heat capacity of a diatomic gas as $C_V = (9/2)Nk_B$, however the accepted value is $C_V = (7/2)Nk_B$.  I assumed the classical Hamiltonian of two identical atoms bound together as $$ H = \dfrac{1}{2m}( |\bar{p}_2|^2 + |\bar{p}_2|^2)+ \dfrac{\alpha}{2} |\bar{q}_1-\bar{q}_2|^2. $$ I calculated the partition function of $N$ particles as $$ Z = \left( \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} e^{-\beta H} ~d^3q_1~d^3p_1~d^3q_2~d^3p_2 \right)^N \propto V^N T^{(9/2)N}. $$ I calcuated the heat capacity as $$ C_V = \dfrac{\partial }{\partial T} \left( k_B T^2 \dfrac{\partial \ln(Z)}{\partial T} \right) = \dfrac{9}{2}k_BN. $$  Why does the classical argument fail?  Classical Derivation  The partition function is \begin{align} Z &amp;=&amp; \left( \frac{1}{h^6} \int \mathrm{e}^{- \beta H(\bar{q}_1,\bar{q}_2,\bar{p}_1,\bar{p}_2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1   ~d^{3}p_2 \right)^N \\&amp;=&amp; \left( \frac{1}{h^6} \int \mathrm{e}^{- \beta ((|\bar{p}_1|^2+|\bar{p}_2|^2)/(2m)+\alpha |\bar{q}_1-\bar{q}_2|^2/2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1   ~d^{3}p_2 \right)^N \end{align} A useful gaussian integral \begin{align} \int_{-\infty}^{\infty} e^{-\gamma (x-x_0)^2}dx = \sqrt{\dfrac{\pi}{\gamma}} \end{align} The partition function can be evaluated using separated integrals \begin{align} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_1|^2} ~d^{3}p_1 = \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_2|^2} ~d^{3}p_2 = \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \end{align} and \begin{align} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha |\bar{q}_1-\bar{q}_2|^2/2 } ~d^{3}q_1 ~d^{3}q_2 = \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 \iiint_{-\infty}^{\infty}  ~d^{3}q_1 = \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 V \end{align} The last set of integrals are improper integrals. One has to take the limit as the space approaches infinite containment. In that limit, integrating one set of variables $d^3q_2$ approaches the limit of a finite Gaussian term, while the other $d^3q_1$ approaches the diverging value of the total volume of the gas.  The partition function is \begin{align} Z &amp;=&amp; \left( h^{-6} \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 V \right)^N \\&amp;=&amp; \left( h^{-6} \left(k_B T \pi\right)^{9/2} \left( \dfrac{2}{\alpha} \right)^{3/2} V \right)^N \\&amp;=&amp; \left( h^{-6} \left(k_B \pi\right)^{9/2} \left( \dfrac{2}{\alpha} \right)^{3/2} \right)^N V^N T^{9N/2} \end{align} Shells have a useful feature where pressing the keyboard UP arrow replaces the contents of the current command line with the previous command that was entered. (And if you do it again, the command before that, and so on.)  I would like a similar feature, where pressing (some other button) replaces the contents of the current command line with the last line of output from the previous command (and if you do it again, the previous line of output , and so on.)  Is there any shell that provides a feature like this?  The motivation is that you often want to run a command, and then run another command with part of the previous commands's output as input. (No, I don't want to do anything involving piping the output of one command into the next. I'm looking for an interactive experience, where you run the first command, see its output, and then get a line of that output into your next command, without having to reach for the mouse to do a clumsy copy-and-paste.) Are Goodberries affected by the Life domain class features? Can not move polygons but only vertices of an object by bmesh! How does an attacker's reach interact with the Mounted Combatant feat's ability to force an attack against the mount to target the rider instead? Disable autostart on boot and restrict background execution What is the Physical Meaning of Commutation of Two Operators? I understand the mathematics of commutation relations and anti-commutation relations, but what does it physically mean for an observable (self-adjoint operator) to commute with another observable (self-adjoint operator) in quantum mechanics?   E.g. an operator $A$ with the Hamiltonian $H$? If $g$ is the generator of a group $G$, order $n$, when is $g^k$ a generator? Say a cyclic group of order $9$ is denoted by $G = \langle g \rangle$. I have a claim that  $\langle g^k \rangle = G $ whenever $k$ is a unit in $\mathbb{Z}_9$. I'm struggling to understand why this is true. Is this specific to order 9? In general, for a group of order $n$, do we require that $k$'s multiplicative inverse in $\mathbb{Z}_n$ be $n$ itself? Otherwise the generating group will be of order less than $n$, and thus cannot be equal to the group. How to make 'python' program command execute Python 3? The python program command executes Python 2. Python 3 can be executed using the python3 command.  How can Python 3 be executed using the python command? Does one work in or on an aeroplane? In an exam paper, there was a picture of an air stewardess in the aeroplane serving passengers. One of my pupils wrote the following:      The air stewardess works on an aeroplane.    Shouldn't it be the air stewardess works in an aeroplane? I have to figure out a way to count how many sequences are there s.t:   My alphabet is $1$ up to $49$. Each number that is chosen, is chosen only once. The sequence is $6$ digits long. The sequence is in increasing order.  e.g that is a valid sequence: $(1, 2, 40, 42, 43, 49)$  and this is not: $(1, 2, 40, 25, 25, 12)$   I know that for the first 3 rules the answer is: $\binom{49}{6}\cdot 6!$  How do I deal with the fourth? Finding out NSArray/NSMutableArray changes' indices I want to calculate $$ \int_{0}^{\infty} \frac{\arctan{2t}}{e^{2\pi t}+1} \, {\rm d}t $$ and am wondering if there at all is an analytical expression for the result. I haven't tried much so far, but expressing the denominator as the geometric series and then substituting $z=2\pi t (n+1)$. Equivalently I get $$ \sum_{n=0}^{\infty} \frac{{\rm si}\left(\pi(n+1)\right)}{2\pi (n+1)} $$ where $$ {\rm si}(x) = -\int_{x}^{\infty} \frac{\sin(t)}{t} \, {\rm d}t $$ is the shifted ${\rm Si}(x)$ function. The value is approximately $0.03618247146$, in fact I know from comparison with something else that it is exactly $\frac{\ln \pi -1}{4}$, but there should be some (direct) derivation, or?  Thanks for your input if you have any.  Edit: I tried to follow your instructions, but am now stuck. So I wrote $$ {\cal L}^{-1} \left(\frac{\psi(t)-\psi(1)}{t-1}\right)(s) =\sum_{n=0}^{\infty} \frac{e^{-sn}}{n+1} = e^s \int_s^\infty \frac{{\rm d}x}{e^{x}-1} $$ and taking the Laplace-transform \begin{align} \psi(t) &amp;= \psi(1) + (t-1)\int_0^\infty {\rm d}s \, e^{-s(t-1)} \int_s^\infty \frac{{\rm d}x}{e^{x}-1} \\ &amp;= \psi(1) + (t-1) \int_0^\infty {\rm d}x \int_{0}^{x} {\rm d}s \, \frac{e^{-s(t-1)}}{e^{x}-1} \\ &amp;= \psi(1) + \int_{0}^{\infty} {\rm d}x \, \frac{1-e^{-x(t-1)}}{e^x-1} \tag{1} \\ &amp;= -\frac{1}{t} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-tx}}{e^x-1} \tag{2} \\ &amp;= \dots \\ &amp;=-\sum_{k=0}^{n} \frac{1}{t+k} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-(t+n)x}}{e^x-1} \\ &amp;\stackrel{(\rm{2})}{=} -\frac{1}{2t} + \ln(t) - \int_0^\infty {\rm d}x \, \left\{ \frac{1}{2} - \frac{1}{x} + \frac{1}{e^x-1}\right\} e^{-tx} \tag{Binet 1} \end{align} where $\psi(1)=-\gamma$.  Now we integrate over $t$ from $1$ to $y$ yielding $$ \ln \Gamma(y) \stackrel{({\rm 1})}{=} \psi(1)\,(y-1) + \int_0^\infty {\rm d}x \, \frac{y-1+\frac{e^{-x(y-1)}-1}{x}}{e^x-1} $$ but what now? Substitute?