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For the 1s orbital the highest probability density exists inside the nucleus. Welcome to the subatomic world! But there are many other orbitals where the density inside or near the nucleus is low. Look e.g. at http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html for details (note that a standard “hydrogen atom” mantra doesn’t signify much of ...


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Note that I didn't check your work leading up to the last integral. Assuming it is correct, first remove the $\cos(\omega_2-\omega_1)t$ term from the integral since it doesn't depend on $x$. You are left with evaluating $$ \int_{0}^{L/2} dx \ \sin^2\left(\frac{\pi x}{L}\right) \cos\left(\frac{\pi x}{L}\right) = \frac{L}{\pi} ...


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What you want is that $Pr(right)-Pr(left)>0$. But $$Pr(right)-Pr(left)=-2\times\frac{96}{41\pi}\int^{L/2}_{0}\sin^2(\frac{\pi x}{L})\cos(\frac{\pi x}{L})\cos(\omega_2-\omega_1)tdx$$ Realize, that $\sin^2(\frac{\pi x}{L})\cos(\frac{\pi x}{L})>0$, for $x\in[0,L/2]$, and so is the integral. Hence what you want is simply the smallest $t$ such that ...


1

What you can say is that you have a distribution of the results you are going to get (be it a discrete or continuous random variable), and when you calculate the average of a large sample, you are adding the random variables and multiplying by a constant. The addition of random variables translates into a convolution of the probability density functions, ...


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In general, we can say nothing about finite $n$, but most of the time, we can safely assume some "niceness" of the distributions in question. If, for example, we assume a finite Variance $\sigma^2$ (a quite common feature), we could use Chebyshev's inequality for a rough error estimation of the form $$P(|\bar{X_n} - µ| > \alpha) \leq ...


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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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The paper you point to describes the experimental test by Rowe et al., in Wineland’s group in 2001. This experiment was performed in an ion trap, where Alice and Bob are separated by 3 µm, and therefore it is impossible in practice to close the locality loophole. As they say in the methods section, even though all known interactions would cause ...


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As mentioned in a comment, the observable universe is finite. However, it is reasonable to think that the total universe may be infinite. Although we don't know this for certain, there is no experimental evidence that definitively contradicts that (eg. the curvature of space is very close to zero, within the precision of our measurements). In fact, Brian ...


2

QM probabilities do obey $\sigma$-additivity, however your sample space and probability measure change depending on what measurements you are making. To use the classic example, consider a double slit experiment in which a particle can pass through two slits, $A$ and $B$, and hits a screen at point $x$. If we don't measure which slit the particle went ...


3

In general if you have m out of n atoms on average of a certain composition, then if you take a sample of size N, you can approximate the distribution of atoms with a binomial distribution with mean $$mean = N \frac{m}{n}$$ and standard deviation $$std = \sqrt{N\left(\frac{m}{n}\right)\left(1-\frac{m}{n}\right)}$$ This is on the assumption that the atoms ...


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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 ...


1

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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