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Fourier transform is an linear operator: $$\mathscr{F}(\psi(x))=\hat \psi (k) =\frac{1}{\sqrt{2\pi}}\int \psi (x) e^{-ikx} dx$$ $$\mathscr{F}(\alpha \psi (x)) =\frac{1}{\sqrt{2\pi}}\int \alpha \psi (x) e^{-ikx} dx=\alpha\hat \psi (k)$$ Uncertainty principle is not any constraint since transform pair $\psi $ and $\hat \psi$ automatically obey Heisenberg's ...


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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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