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The post measurement state is computed in the usual way: If you measure an observable with spectral decomposition $X = \sum_x x \ P_x$, $P$ being the projectors, the post measurement state if the outcome is $x$ is simply $\rho_x = \ P_x \ \rho \ P_x $ (up to normalization) To derive this in a neat way from the original postulate about pure states, you ...


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Intuitively, if the potential energy is a function only of the position, if you measure the position precisely, you can just calculate the potential energy using that precise measurement. More formally, if $V(\hat x)$ is any function of $\hat x$, the position operator $$[V(\hat x), \hat x] = 0$$ which really is just that any operator commutes with itself.


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This is an excellent set of questions. The basic thing to realize here is that a wave function described by $Ae^{i(kx - \omega t)}$, where here $\omega \equiv E/\hbar = \hbar k^2 \ 2m$, extends with equal weight through all of the entire universe. These waves are called "plane waves". Because they are of infinite extent these wave functions are Not ...


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The particle is described by a wave that is spread out in position and momentum space. It is not possible to produce a wavepacket that does not respect the HUP. You can make such wavefunctions undergo interference and when it's undergoing interference the square amplitudes don't respect the calculus of probability, for an example see ...


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Classical mechanics assumes that the state of a particle can be described, with arbitrary precision, by its position and momentum. That is in classical mechanics the state of the particle is, mathematically speaking, a phase vector. (And, the physical laws governing the particle's motion can be thought of as a trajectory in this vector space.) But quantum ...


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From my understanding of your questions, you are confusing the "scientific method" and the "uncertainty principle. The scientific method says that "given the same starting conditions, within a controlled environment, etc., the "results" should be the same (ei. repeatable within some degree of accuracy). The uncertainty principle "deals" with an entirely ...


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As you said if two operators commute they share eigenvectors. Physically this means that you can have a definite value for both. For example in the hydrogen atom the Hamiltonian $H$, which is the energy, and $J^2$, the magnitude of angular momentum, commute. A hydrogen atom can be in a state of definite energy and definite angular momentum. However, the ...


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Different trajectory each time due to probabilistic nature inherent in quantum mechanics and the uncertainty principle. The uncertainty principle is not "removable", it is not a constraint based on practical limitations in an experiment, but inherent to quantum mechanics calculations. There was an interesting thought experiment, given absolute control over ...


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The problem with this question is that, even if you perfectly controlled the conditions at the macroscopic level -- including somehow releasing the leaf in exactly the same way every time (nearly impossible), and using leaves in identical starting state (the same leaf may have lost some water by the time you repeat the experiment, and suppressing all ...


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No. The Uncertainity Principle states the following: The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time. $$\Delta x \Delta p \geq ...


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You are misunderstanding the Uncertainty Principle. The Uncertainty Principle says that a particle cannot simultaneously have a definite momentum and a definite position. This is not due to our incomplete knowledge of parameters. This is a fundamental law of the universe and arises from the fact that the momentum and position operators do not commute in ...


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In the case of a leaf's trajectory, yes, it'd be same in both cases if the environment is exactly the same. And you can generalize this to any macroscopic event. As for the formation of Earth simulating the universe since its beginning, it's complex because it's not a macroscopic event. In the quantum world, everything is probability driven. In a heap of ...


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At least to me, it is unclear what it means to be "measured by the environment". As far as decoherence is concerned the situation is however quite clear. Already the original "einselection" framework of Zurek is applicable to bipartite system/environment scenarios. Let $(| p\rangle)_p$ be a "pointer basis" for the system. Then any Hamiltonian of the form ...


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Commutation does become transitive, and thus an equivalence relation (reflexive and symmetric are trivial), when you impose an extra condition: nondegeneracy. If $A$, $B$, $C$ are Hermitian operators, and each of them has only unique eigenvalues, then $AB\! =\! BA\, \cap\, BC\! =\! CB$ implies $AC = CA$. Proof: for a nondegenerate operator, the eigenbasis ...


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No, it does not! Let me give you a counterexample: Consider the Hermitian operators $\mathsf{1}$ (identity operator), $p$ (momentum) and $x$ (position) in 1D. Now, the trivial commutation relations $[\mathsf{1},x]=0$ and $[\mathsf{1},p]=0$ do not imply $[x,p]=0$ as the correct relation is $[x,p]=\mathrm i\hbar\neq 0$.


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Commuting is not an equivalence relation. All components of angular momentum commute with $J^2$ but they don't commute with each other. How to find a complete set of mutually commuting observables is a difficult problem and I don't think you can give an algorithmic answer. It depends very much on the specific problem. An observable that commutes with the ...



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