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The theory of probability used in QM is intrinsically different from the one commonly used for the following reason: The space of events is non-commutative (more properly non-Boolean) and this fact deeply affects the conditional probability theory. The probability that A happens if B happened is computed differently in classical probability theory and in ...

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This question gets to the heart of what makes quantum mechanical amplitudes different from classical probabilities. It is true that if you make measurements in the basis of states $\{\lvert 00 \rangle ,\lvert 11 \rangle\}$ then the two states have the same measurement statistics, and so cannot be distinguished. The interesting thing is that it is possible to ...

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This question was studied fairly recently by a team at Edinburgh University. Their paper is available here, though I'm not sure if you can get it without having to hand over some cash. The bottom line is that in principle the trajectory of a die can be calculated, but it is a chaotic system and that means tiny inaccuracies in the measured initial conditions ...

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If you look closely, you will find that the event "find the particle at $x$" always has zero probability of occuring: Since $\rho(x) = \lvert \psi(x) \rvert ^2$ is a probability density, it must be integrated over some subset of $\mathbb{R}$ to actually gain a probability. The probability to find the particles at one of the positions in a subset $S \subset ... 12 Another use of the SLE approach seems to be (I haven't read the papers below much beyond their abstracts) as a tool to probe for the presence of conformal invariance in various systems, when a direct (numerical or experimental) verification is difficult. In this approach, (i) one extracts suitable non self-crossing paths, (ii) one determines (empirically) ... 12 The total probability of all mutually excluding alternatives must always be 100%, so it is conserved. The conservation law in the spacetime tend to be "local", so just like for the charge conservation, we may derive the conservation of the probability in Schrödinger's equation from the local continuity equation $$\frac{\partial \rho}{\partial t} + \mathbf ... 11 The theory of quantum mechanics has been developed to explain observations, i.e. measurements. Without observations it is a floating mathematical construct. One of the postulates to connect the mathematics with reality is: To every observable there corresponds an operator which operating on the state function will give an eigenvalue. So the question ... 11 SLEs can be used in order to define a certain kind of QFT. You can check M. Douglas' talk, from page 28 forward: Foundations of Quantum Field Theory (PDF). There's also another great article, Conformal invariance and 2D statistical physics. You may also like SLE and the free field: Partition functions and couplings. Finally, there's an approach to try and ... 10 Probabilites are (squares of) probability amplitudes, which can be obtained as inner products of vectors on Hilbert space: \langle X|Y \rangle. Under a transformation U, the ket transforms as$$|Y\rangle \rightarrow |Y'\rangle = U |Y\rangle$$and the bra as$$\langle X| \rightarrow \langle X'| = \langle X| U^{\dagger}$$So if U is unitary, ... 10 Matty Hoban pointed me to a paper (PDF here) by Itamar Pitowsky from 1991 which looks the geometry of correlation polytopes and their symmetries. I haven't read the paper in full, but glancing through it, on page 400 (page 6 of the actual paper) under the statement of results the author seems to say that the cardinality of the symmetry group is n! 2^n ... 10 This relationship is called the Born rule and it's one of the postulates of quantum mechanics. There have been various attempts to derive it, but none of them having been terribly convincing so at the moment we have to assume it is true. Fortunately experiment supports this assumption. The rule was originally suggested by Max Born (hence the name) in 1926 ... 10 In position-space (that is, when your functions are functions of x), the function \int|\Psi|^2 gives the probability of finding the particle in a given range. The expectation value of x is where you'd expect to find the particle. It is often essentially the weighted average of all the positions where the probability density, |\Psi|^2, is the weighting ... 8 You might like this 110-page paper by me and Alex Arkhipov, which is all about a quantum bosonic analogue of Galton's board (we even use the same graphic you did -- see Section 1.1!). In particular, we gave strong evidence that such a board (with an arbitrary configuration of "pegs," and with multiple entry points for the "balls") is exponentially hard even ... 8 If you want the high school answer, then no, the numbering does not matter. If all faces are equally-likely, the probability is the same regardless of how you number the die, and similarly all derived quantities (such as variance or probability to be greater than 8) are the same because the underlying distribution is the same. If you want the real answer, ... 8 There are two that I know of in the context of state estimation. The first is for estimating the mean of P and is a Metropolis-Hasting MCMC algorithm here: Optimal, reliable estimation of quantum states. The second is also mainly for computing the mean (but can do other functions -- including the characteristic function of the region you are interested ... 8 With quantum mechanics, you have to ask your questions very, very carefully. Is it possible to have a Sun-sized star in your pocket? It depends on what you mean by "in". Do all of the atoms of the star need to be entirely in your pocket, or is it sufficient that some part of each atom's wave function be inside your pocket? It is possible to have any ... 7 If p=1 and d is non-empty, then clearly the random walk is not transient, because since you use the \ell^1 norm, there is a closed surface which always reflects the walker back into the enclosed finite region. If, however, p = 0 then the walk is transient, since simple random walks in 3 dimensions are transient. Now, if there is a p_c<1 its ... 7 Are you looking for a proof? If so, this link (which has some sign errors as pointed out in the comments) proves it as follows (without the sign errors): We start by differentiating the definition of the probability with respect to time only:$$ \frac{\partial P(x,t)}{\partial t} = \frac{\partial}{\partial t}\left (\psi^*(x,t) \psi(x,t)\right) = \left[ ... 7 The random potential is a model for small fluctuations in the local energy of an electron, when there are defects or impurities that raise the energy of an electron at certain spots by a little bit, and lower it at other spots. This is a good model, although it doesn't look like one at first, because the quantum mechanical solution is not sensitive to the ... 7 I think a reasonable first approximation can be made like this: choose an arbitrary orientation for the die, and figure out, if the die were released in that orientation with its lowermost point resting on a surface, which side would it fall on? That can be easily calculated; you just draw a line going straight down from the center of mass, and whichever ... 7 Before trying to understand quantum mechanics proper, I think it's helpful to try to understand the general idea of its statistics and probability. There are basically two kinds of mathematical systems that can yield a nontrivial formalism for probability. One is the kind we're familiar with from everyday life: each outcome has a probability, and those ... 7 The answer is negative for two distinct reasons. (1) In QM operator means linear operator and the map$\psi \mapsto |\psi(x)|^2$is not linear, evidently. (2) Wavefunctions are elements of$L^2(\mathbb R)$and these elements are defined up to zero-measure sets. I mean that, if$\psi(x) \neq \psi'(x)$for$x\in E$where$E$has zero measure, then ... 7 Standard deviation adds uncertainties to the measured value:$23.3\pm 0.4\,{\rm m}$. One can quickly look at the error (which has units of${\rm m}$in my case) and think, The value could be as low as$22.9\,{\rm m}$or as high as$23.7\,{\rm m}$without much thinking. Modifying this to being a percentage of the value would be confusing. Plus it would be ... 7 The ratio between avaible microstates is given by the following ratio: $$\frac{n_{B}}{n_{A}}=\frac{\int_{V/2}d^{3N}q d^{3N}p}{\int_{V}d^{3N}q d^{3N}p}=\frac{\int_{V/2}d^{3N}q}{\int_{V}d^{3N}q}=\frac{\int_{V/2}d^{3}q_1 \int_{V/2}d^3 q_{2}\dots }{\int_{V}d^{3}q_1 \int_{V}d^3 q_{2}\dots}=\left( V/2 \right)^N/V^N=\frac{1}{2^N}$$ where$N$is the number of ... 6 You will almost never encounter a calculation that is intended to account for every detail of a phenomenon with perfect accuracy. That isn't possible, and in fact many times adding more detail to a calculation only takes away from the insight it grants. Why make a complicated calculation when a simple one tells you everything you want to know? Gamow is ... 6 This is the exact reason why we do statistical hypothesis confidence testing. In essence, the confidence interval we get from this test is a quantitative measure of "how far we've converged". For example, consider an experiment to test whether a coin is imbalanced or not. Our null hypothesis is that it is not: in symbols, our hypothesis is${\rm Pr}(H) = ...

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Part of you problem is "Probability amplitude is the square root of the probability [...]" The amplitude is a complex number whose amplitude is the probability. That is $\psi^* \psi = P$ where the asterisk superscript means the complex conjugate.1 It may seem a little pedantic to make this distinction because so far the "complex phase" of the ...

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