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## New answers tagged probability

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There is no derivation of the probabilistic nature of the wavefunction: it is an interpretation (postulate), the only one that makes the theory consistent. From Sakurai, Modern Quantum Mechanics: SchrÃ¶dinger published his famous wave equation in February 1926 in the famous paper Quantisierung als Eigenwertproblem (Quantization as an Eigenvalue Problem),...

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This is just a quick stab, and it might show my ignorance more than anything else. Since you are working with a two, level spin system i'm actually giessting $m,n=\pm\frac{1}{2}$ . You can then explicitly write your density matrix as $$\rho\left(t\right)=\begin{pmatrix}\rho_{\frac{1}{2},\frac{1}{2}} & \rho_{-\frac{1}{2},\frac{1}{2}}e^{-i\omega t}e^{-\... 5 Whilst it is certainly true that Quantum Probability Theory (QPT) is an entirely different framework from Classical (Kolmogorovian) Probability Theory (CPT) (specifically because the event structure is non-Boolean and the random-variable structure is non-commutative), we can still identify enough formal similarity to borrow the classical terminology. In ... 0 Maybe you can be interested in another interpretation of Hermitian matrices. In a recent paper we have proposed to see them as gambles on a quantum experiment. We have then enforced rational behaviour in the way a subject accepts/rejects these gambles by introducing few simple rules. These rules yield, in the classical case, the Bayesian theory of ... 1 A quantum system can be described by a set of evolving quantum mechanical observables. This is not the same as describing a system in terms of a stochastic quantity described by a single number chosen at random. A quantum system really does have multiple values of any unsharp observable, see https://arxiv.org/abs/quant-ph/0104033. Those different versions ... 7 I'm going to try to explain why and how density operators in quantum mechanics correspond to random variables in classical probability theory, something none of the other answers have even tried to do. Let's work in a two-dimensional quantum space. We'll use standard physics bra-ket notation. A quantum state is a column vector in this space, and we'll ... 6 I believe it is misguided to think that classical probability makes sense any more than quantum mechanics, with its "peculiar" probability calculations, makes sense. I'm going to be slightly mischievous here and make a friendly attack your first paragraph: does really make sense? Of course it makes perfect sense as a measure-theoretic definition, but how ... 11 Quantum mechanics is indeed a probability theory, but it is a non-commutative probability theory. So it is not just a matter of having signed/complex measures, but really of having a non-commutative probabilistic framework. Quantum mechanics was developed, historically, before non-commutative probability theories and I think that people in probability ... 9 You could certainly model any one quantum observable as a random variable. The problem comes in when you have multiple observables, which you might attempt to model as classical random variables with some joint distribution. From this joint distribution, you can compute various probabilities (like \textrm{Prob}(Y\neq X), for example), according to the ... 3 I suggest you read about the Binomial distribution. In general, you are correct that if p=1/2, the distribution and its properties are symmetric under exchange of successes and failures (or forward and backwards in your example). If the number of trials is even, the mean, mode (most probable single outcome) and median coincide at zero in your example. If ... 8 You are mixing here two kinds of most likely. First, you are correct, that the same amount of forward and backward steps will be the most probable of all possible outcomes. Second, if you add probabilities of all outcomes with unequal result, you will see that getting some unequal result will be much more likely. The outcomes of coin tossing are described ... 1 The problem is identical to tossing a coin four times and looking for 2 heads and 2 tails. There are 16 possible outcomes. Four heads has a probability of \frac {1}{16} Four tails has a probability of \frac {1}{16} Three heads and one tail in any order has a probability of \frac {4}{16} Three tails and one head in any order has a probability of \frac ... 2 I think the most interesting approach in this direction is Caticha's entropic dynamics, for example in his "Entropic Dynamics, Time and Quantum Theory", arxiv:1005.2357. "Quantum mechanics is derived as an application of the method of maximum entropy. No appeal is made to any underlying classical action principle whether deterministic or stochastic. ... Both ... 0 Bayesian probability is based on the classical logic of plausible reasoning, as described by Jaynes, Probability Theory: The Logic of Science. One can and should try to find a Bayesian interpretation of the wave function. Here I can recommend Caticha's "entropic dynamics" as one possible approach. But I don't think "quantum logic" will be helpful here. ... 1 This refers to the Feynman rule that crossing fermionic lines produce relative minus signs between amplitudes. That is, if you have some process that is mediated by two different Feynman graphs and one graph is obtained from the other through an odd number exchanges of fermionic endpoints, you must subtract instead of add the amplitudes corresponding to ... 1 Like gonenc pointed out your assumption that normalizing your wave function does not imply continuity. And yes you'll probably won't need the normalization factor in your further calculations. The reason for you doing this could be consistency with the Interpretation of the wave function squared as a probability amplitude:$$ P = \int|\psi(r,t)|^2 dr 

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ANSWER: Mutually-exclusive events cannot exist before measurement, in the probabilistic formulation of quantum mechanics (Copenhagen interpretation-CIQM), because, maximally, CIQM is required to violate local realism and, minimally, it might break the principle of locality. And after measurement, the problem you mentioned does not exist because it is ...

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To answer your question we have to keep in mind, that holes are quasi particles. They are a mathematical formalism. They are introduced as empty states in the valence band. From a physical point of view it makes sense to construct these particles, as they really have the properties of real charge carriers. The hole energy is at its minimum at the top of the ...

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