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In chsh we can explain why quantum violate the inequality but it tells nothing about nonlocality : The operator S=AB-AB'+A'B'+A'B The measurement process in copenhagen is used : the wavefunction collapses in an eigenstate of A remeasuring A will give the same result noted a The same for A' noted result a' Then a strange thing happens for the B side the ...

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Consider a molecule of oxygen in a balloon. You know that at nonzero temperature all those molecules are bouncing around in all directions. Of course, the mass of air doesn't have any net motion in any direction. Indeed, the average velocity of each molecule is zero: $$\langle v \rangle = 0 \, .$$ Of course, the average energy of any particular molecule is ...

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No. Let's look at a spin measurement in detail. You can choose any direction $\hat n = n_x\hat x+n_y\hat y+n_z\hat z$ and take any spin, $\alpha |+\rangle+\beta |-\rangle,$ where we define $|\pm\rangle$ by $$\left(n_x\sigma_x+n_y\sigma_y+n_z\sigma_z\right) |\pm\rangle=\pm|\pm\rangle).$$ Then you can send a beam of particles into a small region with an ...

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One way to facilitate this discussion is to think of what's classically forbidden but quantumly permissible. My favorite so far is a game that I call Betrayal. Let me explain that in this answer. Betrayal: Game Rules The players are a cooperative three-person team, they will either all win or they will all lose. They will be put through some number $N \gg ... 1 I still don't quite understand the reasoning behind the conclusion that entangled particles somehow can communicate their state to each other instantaneously, even though they are separated by a substantial distance This isn't correct, they occupy a joint state. From what I gather [...] upon observation of one of the particles, it immediately (and ... 0 I've been taught that when you measure the value of an observable for some state of a system described by$|\psi\rangle$then the state of the sistem "collapses" to the eigenvector associated with the eigenvalue measured. ... Following this logic, when you measure the position of a particle described by$|\psi\rangle$and you get a value$x_0$... 0 You can use the vector, just expressing it at a "lower level" is difficult When you say$\langle x| x_0\rangle = \delta(x - x_0)$you are already succeeding, even if you're a little confused about what the "square root of a$\delta$-function" looks like in terms of a real, honest-to-goodness wavefunction. So for example once you have the above result you ... 0 How do you describe a state with a density matrix after measuring position? If you already have a density matrix, you use the Heisenberg picture. In the Heisenberg picture the state never changes. Instead the operators corresponding to observables change in time. So if you measure position at time$t_1$, then density matrix and state does not change, but ... 0 There exists a valid derivation the Born rule, but here you start from a weaker assumption, it's not the sort of an ab initio derivation from only the other postulates as the MWI advocates would like to have. This argument works as follows. We start by asking how we know that the Born rule is in fact valid. The answer must involve doing experiments, ... 2 I don't think there is any great meaning to this statement - I think the author(s) are indicating how measurement is treated in QM formalism, rather than any practical issues introduced by measuring such systems. Because this is dangerously close to the topic of the uncertainty principle / observer effect, it's worth stating that quantum uncertainty is not ... 1 First of all I think I should clarify the confusion of the terminology "uncertainty in$x$". What we mean by uncertainty is simply the standard deviation of the observable defined by: $$\Delta x= \sqrt{ \left< x^2 \right> - \left< x \right>^2}$$ where$<A>$denotes the expectation value of the operator$A$is some state$\psi\$. You should ...

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