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Any increasing sequence $(\Lambda_n)_{n\geq 1}$ of finite subsets of $\mathbb{Z}^d$, $d\geq 2$, such that $\bigcup_{n\geq 1} \Lambda_n =\mathbb{Z}^d$ will do. All sequences $(\mu_{\Lambda_n}^+)_{n\geq 1}$ of finite-volume Gibbs measures in $\Lambda_n$ with $+$-boundary condition converge to the same infinite-volume Gibbs measure $\mu^+$, under which there is ...


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This is simply a consequence of basic properties of probabilities. Let us take an event $A$ made of independent events $A_i$ (so that $p(A) = \sum\limits_i p(A_i)$), and an event $B$ made of independent events $B_j$ (so that $p(B) = \sum\limits_j p(B_j)$) Then you have : $p(A/B) = \dfrac {\sum\limits_{i,j} p(A_i/B_j) \,p(B_j)}{\sum\limits_{j}\,p(B_j)}$


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For the incoming state, you don't know which spin state the particle is in, so you should average over the possible states. But you can measure the spin of the outgoing state, so to get the total cross section you should add up the cross sections for each spin. More formally, an unpolarized incoming particle should be described as a density matrix, $$ ...


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Imagine an experiment where you are sending one particle of spin 1/2 into a "black box" and waiting for the result. You know you are sending in one particle at a time, but you do not know the spin and presume it's going to be fifty fifty - so you say in fifty percent of cases you will have the spin "up" (or better "right" for relativistic particles) and add ...


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For me, the thing that really gives Shannon's definition legs is Shannon's Noiseless Coding Theorem. It proves the following remarkable and important fact: Let an information source send a message comprising statistically independent symbols and suppose that these symbols belong to an $N$ letter alphabet, and that the propability of transmission of the ...


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There are several possible answers to this. One is to look at Shannon's definition of the entropy, $$ H = -\sum_i p_i \log p_i, $$ and note that it has the form of an expectation: $H$ is the expected value of $-\log p_i$, so it makes sense to give a name to that latter quantity. This is nice if you understand the value of the entropy. In Shannon's paper ('A ...


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Classical physics is developed based on this idea that understanding of a phenomenon will help in predicting the future and the past of a system. It actually works on the prejudice that quantities which are measurable can be measured with utmost accuracy by developing better methodology for experiments. So, the based on classical physics i.e.ignoring the ...


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Yes. Yes, it is true. More generally, for any state $\lvert \psi \rangle$, the probability to find it in the state $\lvert\phi\rangle$ is $\lvert \langle \phi \vert \psi \rangle \rvert^2$, and so, since $\lvert \mathrm{e}^{\mathrm{i}\phi_n}\rvert^2 = 1$, phases do not influence probability in this case.


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Certainly not. I was already known for a long time that certain microscopic phenomena were best described by probabilistic theories, the prime example being radioactivity. Even on the classical level, statistical mechanics (canonical example: Brownian motion) had prepared some physicists to relax their classical conceptions of reality. However, it was of ...


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There is a statistical test for exactly this kind of problem--it is called Pearson's Chi-Squared test. It is able to determine if an observed distribution of results, such as the results of rolling a die, matches a theoretical distribution, such as one where every number occurs equally as often. The test requires at least 5 expected occurrences for each ...



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