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I'd like to perhaps a slightly different viewpoint to your question and maybe turn it around a little. Probability is hard. Very hard. Defining the foundations of probability and statistics so that they are altogether sound and rigorous is actually a work in progress. It definitely is not complete. On the other hand Quantum mechanics is easy. Very easy! I'm ...

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Let me try to give you a kitchen-table explanation. I can't help you with statistics vis-a-vis quantum mechanics, but probability is very basic. The underlying "real stuff" in quantum mechanics are numbers that, when squared, produce probabilities of seeing things. Typically, these numbers are complex, but they don't always have to be. These numbers are ...

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I'm going to explain roughly what the Born Rule, following Stan Liou's comment. One of the Postulates of Quantum Mechanics relates a mathematical quantity, the wave function (or state $\psi$ of a Hilbert space, $\mathcal{H}$) to a measurable entity, the probability of a given event to happen. The idea goes like this: if you want to measure a quantity ...

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Does this mean, that the probability of detecting the particle it the SAME everywhere? No, it does not. This is quite a common mistake, stemming from the idea that the Green function $\mathcal{M}$ can be used in the role of the $\psi$ function of free particle with the Born interpretation of $|\psi|^2$ as probability density. But that is not possible, ...

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This is wholly analogous to the evanescent optical field that arises in the classically (i.e. computed by raytracing) forbidden region beyond a totally internally reflecting interface between two optical mediums. I analyse this situation in my answer here and there is also a great plot of the situation in Ruslan's answer here. Let's think of a 1D barrier ...

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If the particles are all the same (we don't need to invoke indistinguishability at that stage I think), then there is no loss of information since the permutation of two particles in a reduced distribution function will ask exactly the same question as before the permutation probability-wise. Note that usually, reduced distribution functions are defined ...

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I think this assertion would be more correct if it was something like: There are more 1 cubic-meters of matter units in a googolplex-meter-wide universe than there are possible quantum states of 1 cubic-meters of matter units, so at least one of those quantum states of 1 cubic-meters of matter occurs in more than one 1 cubic-meters of matter somewhere. In ...

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I'm just going to quote Wikipedia here: For the case of two colliding bodies in two dimensions, the overall velocity of each body must be split into two perpendicular velocities: one tangent to the common normal surfaces of the colliding bodies at the point of contact, the other along the line of collision. Since the collision only imparts force along ...

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If I have a velocity which has some component $v_x$ in the $x$-direction, then is there any reason for you to assume you know anything anything about the component of my velocity which might be in a perpendicular direction, $v_y$? No. So you can see that it is reasonable to assume that, if you know my $v_x$, my $v_y$ is still unconstrained, i.e. you have ...

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Spin eigenvectors are the same for the electron and the positron. The transition amplitude between the singlet state $s$, and for instance, a state up for electron and down for positron may be written (up to a complex unit phase) : $A = ((up_1)^\dagger \otimes (down_2)^\dagger) s \\=((\chi_+(\theta_1, \phi_1))^\dagger \otimes \chi_-(\theta_2, ... 0 The confusion arises because there are two kinds of classical limits, depending of the system under study. Let's start with fermions, which distribution is$n_F(\epsilon)=\frac{1}{e^{(\epsilon-\mu)/T}+1}$. The first classical limit (corresponding to the case mentioned in the question) is$T\gg \epsilon-\mu\$. This corresponds to the case where the ...

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The introductory paragraph you quote with horror says temperature ''high enough'' to avoid quantum effects. (It did not say anything like ''arbitrarily large''.) If the temperature is too low, things like Bose--Einstein condensation can occur, which invalidate Maxwell--Boltzmann statistics. The temperature should be high enough so that it is unlikely to ...

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The most probable position would be such as where the global maximum of the distribution is located. This is different to the expectation value of a distribution, but it happens that for a Gaussian function the mean and the most probable value are the same.

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