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First of all, if there are $p^N$ possible states, this means that the particles and the compartments are distinguishable. For undistinguishable particles (and distinguisable compartments, which is probably reasonable physically speaking) the number of possible states, or configurations, has to be divided. For instance the configuration $(n_1,n_2,\dots,n_p)$ ...


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The probability for one compartment being empty is actually the probability that at least one compartment is empty. Let's call it $\Pi^{(1)}$, $$ \Pi^{(1)} = \frac{(p-1)^N}{p^N} $$ When you write the probability for any of the compartments to be empty as $$ \Pi = \sum_{j=1}^{p}{\Pi^{(1)}} = p\frac{(p-1)^N}{p^N} $$ you are overcounting. Longer argument: ...


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From here, how do we say that probability distribution function is constant as we flow in the phase-space? More accurately, value of probability distribution function $f_t(p,q)$ at representative point $p^*(t),q^*(t)$ moving along any Hamiltonian trajectory in phase space is constant in time. The function itself generally changes in time. This value is ...


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To talk about $\langle \psi_i \vert U(t) \vert \psi_i \rangle$ as the "expectation value of the time evolution operator" is probably the least insightful way to talk about this quantity. Since $U(t) = \exp(-\mathrm{i}Ht)$ for time-independent Hamiltonians, if you want to look at expectation values, you could as well look at that of $H$ directly. Note that, ...


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Your statement on a pseudo-probability interpretation is slightly off, in that you most certainly did not write the Wigner quasi-probability distribution in your edit. Taking your two bases to be $x$ and $p$ for specificity, what you wrote, $ρ(x,p)$, up to a phase: $e^{(ixp/ħ)}$, is the "standard ordering prescription" for quasi-distribution functions, ...


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The probability of N atoms left is \begin{equation} P(N) = \left( \begin{array}{c} 100 \\ N \end{array} \right) 0.5^{100} \approx \frac{1}{\sqrt{50\pi}}\exp\left(-\frac{(N-50)^2}{50}\right) \end{equation} The expected value will be 50 which is also the most probable value. The standard deviation is 5, so we expect the number of atoms left to be 50 ± 5.


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It's not "multiplied by $r^2$ to get the probility density". The issue is that the volume element in spherical coordinates is $$ \mathrm{d}V = r^2\sin(\theta)\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi$$ and since the probability to find a particle in a subspace $X\subset \mathbb{R}^3$ is $$ P(X) = \int_X\lvert \psi(r)\rvert^2\mathrm{d}V$$ by definition of a ...


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If quantum mechanics is probabilistic, there is no reason for a particle to be in one place and not the other, but particles do make up their minds... but how? This topic is obscured by common confusions about what quantum mechanics entails. There are several different explanations of what is happening in reality, if anything, to bring about the ...


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In fact, for Quantum Mechanics Nature is intrinsically probabilistic: there is no law describing with probability $1$ a single event. In accordance with Copenhagen interpretation of the quantum formalism this must not be considered as a limitation to our knowledge, but it is simply the manner as things exist in Nature, it is an ontological claim. In this ...


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Quantum mechanics is uncertain, but it's not probabilistic. There are some very important differences between the two terms. In quantum mechanics nature limits our knowledge about which particular outcome a future measurement will have, but the possible outcomes of given initial conditions and their expectation values are perfectly deterministic (and ...



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