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The times that the Wigner function is positive does not mean it should be interpreted as a probability distribution. (What events would it be the probability distribution of? Certainly not a fuzzy joint measurement of position and momentum, e.g, with coherent states; that's the Husimi Q function.) Is there a simple system (hopefully a simple harmonic ...

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$$<\hat{A}> = \int \psi^*(x)\hat{A}\psi(x) dx$$ now $\hat{A}\psi(x)=a(x)\psi(x)$ so, $$<\hat{A}>=\int a(x)|\psi(x)|^2dx$$ Let $|\psi(x)|^2dx = d \mu$ now $\int|\psi(x)|^2dx=\int d\mu = 1$ $$<\hat{A}> = \int a(\mu)d\mu$$

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Let us use bra-ket notation. Suppose that the operator $\hat A$ has discrete and bounded eigenvalues $\mathcal A =\{A_1, A_2, \ldots\}$ with eigenkets $|A_1\rangle, A_2\rangle,\ldots$. The eigenkets form a complete orthogonal set since $\hat A$ is symmetric. Then any ket $|\psi\rangle$ can be expanded, $$|\psi\rangle = \sum \psi_n |A_n\rangle.$$ Since the ...

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Observables are associated with linear operators, not measureable functions, so how can we talk about the expectation of a linear operator? The "expectation of a linear operator" is a term from quantum theory. It is defined by the integral $$\int \psi^*(x)\hat{A}\psi(x) dx$$ or similar. The meaning of this number is not necessarily the same as ...

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There are two answers to this. One answer simply points out that the probability of the jth outcome specified by the Born rule $p_j = tr(\rho\hat{P}_j)$, where $\hat{P}_j$ is the projector onto the jth outcome, satisfy the axioms of probability: http://mathworld.wolfram.com/ProbabilityAxioms.html. Another answer is that the Born rule can be explained ...

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Since you want a bit of mathematical rigor: A quantum state is a self-adjoint positive trace class operator on a Hilbert space with trace 1. This is called density matrix $\rho$. In its simplest form, given $\psi\in \mathscr{H}$, $\rho$ is the orthogonal projector on the subspace spanned by $\psi$. Let $E_\rho(\cdot):D_\rho\subset\mathcal{A}(\mathscr{H})\to ... 0 I do not think probability distributions are preserved by the Hamiltonian flow...consider a probability distribution that is a$\delta$-function on the phase space at initial time (you have just one point with probability one), so it is a particle with fixed coordinate and momentum. If you evolve in time by the Hamiltonian flow, you will find yourself at the ... 1 For$N$particles that can occupy$M$different states$\left(M\geq N\right)$, it seems to me that the answer is $$M\cdot M-1 \cdot \dots \cdot M-(N-1)=M!/(M-N)!\$$ if one imposes that each must occupy a different state. That is since, to the first particle,$M$states are available. For the second,$M-1$. We continue thus until we run out of particles, ... 3 All the definitions you've posted are correct, and they aren't in conflict with each other, although they are a bit imprecise. I'll try to explain in more detail what these concepts are. Probability I'll assume we don't need to attempt to define what probability is. :) I'll just note that it's formalized in mathematics under the aegis of measure theory. ... 3 To understand the difference between probability and probability density consider the difference between mass and density. Density is the mass per unit volume, so to find the mass you multiply the density by the volume: $$mass = density \times volume$$ In some cases the density will be a function of position and we have to write it as a function of the ... -1 It has been many years since I have used this information (so please forgive my inaccuracies, but I wanted to get you PART of the way there with an answer (since no one has responded yet). The probability density is the integral (area under the curve) of the probability. I think the reason for the funky discrepancy between the two definitions is because.. ... 3 The Projection Operator$\delta_{\varepsilon\, a}$is$1$for some particular$|a\rangle$and is$0\$ on any orthogonal state. The definition of the expectation in classical probability theory for this operator is given by \left\langle \delta_{\varepsilon\, a}\right\rangle = 1\Pr(|x\rangle = |a\rangle) + 0\Pr(|x\rangle \perp |a\rangle) = ...

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Fermi's golden rule originates from first-order perturbation theory, which means that it can be derived from Schrödinger's equation under the condition that the perturbing hamiltonian be weak enough that there isn't a significant depletion of the initial state (for a proof see e.g. this link). As such, it only holds when the transition probabilities are on ...

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Well, there's three different types of answers I can think of, with regards to what you asked. One case we can have is that there's some systems that are so complicated, that if you change their initial states only slightly, the system changes drastically. This is called chaos. James Gleick's book is a good one for laypeople. Any graduate level mechanics ...

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So I was wondering when physics takes this approach to this subject matter is it saying that the movement of an atom is actually random or that the details of explaining may be to cumbersome and be approximated with great accuracy just by using these distributions? Depends on the physicist. Some like to think there is fundamental randomness in nature, ...

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The quantum world is really random. There are no local hidden variables.

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