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What about positivity? The product of bounded positive operators is positive if they commute (see proof below), otherwise there is no guarantee. If your initial POVMs are not compatible, in general, the operators of the final candidate POVM is not made of positive operators and thus they do not define a POVM. Proposition. If $A,B \geq 0$ where $A,B :\cal ... 2 To "see another World" would require doing a measurement that involves (partially) reversing the effect that led to the split. In practice this is impossible to realize because the observer is a macroscopic object itself and it will decohere very fast. Decoherence means that the system becomes correlated with the environment and that poses a big problem ... 1 The question is "Why are the possible outcomes the same for all directions?" It happens also for observables of classical physics! QM does not matter here, the truly relevant idea is the fact that in a inertial system physics appears to be isotropic. In practice, it is not possible to physically distinguish different directions with physical experiments. ... 1 Although I see already two answers, I’ll add mine. The situation, as original poster described, is such for a massive spin-½ particle and is not symmetric in this sense for a massless spin-½ particle (some people say this case is properly named “helicity”, but it is a question of terminology). Why does it matter? Because a massive particle has its ... 0 I'm surprised that none of the other answers (from last year) mention interaction-free measurement. The way we know that the uncertainty principle is not a side effect of interactions during measurement is that it applies even when there is no interaction (and even when the measuring device is nowhere near any part of the system). 1 We have a pretty good understanding of every operationally meaningful aspect of measurement. The key difference between quantum mechanics and other wave theories is that the wave interference happens in phase space, rather than physical space. In other words, quantum interference happens between different quasi-classical histories of the entire world that ... 1 I would generally support Ayesha's answer, in that it explains that decoherence is instigated by the microscopic interacting with the macroscopic. As with many things in Quantum Physics, this is evidently true at the extremes (e.g. putting a detector in the path of a photon in the dual slit experiment), but it is not clear when something is considered ... 0 To also give a more direct reply to your photon in a slit example, in simple words: Observing the photon passing through the slit is not just a conceptual act, it means placing a photon detector at the slit and as the photon passes through, its state is measured and since a measurement in quantum mechanics comes at the cost of perturbing the original state ... 1 Suppose we represent the wavefunction of the photon as$\psi_p$and the wavefunction of the observer as$\psi_o$. As long as the photon and observer do not interact in any way we can write the total wavefunction as a product: $$\Psi = \psi_p\psi_o \tag{1}$$ The trouble is that you can't make any measurement of the photon without interacting with it, and ... 1 I'll illustrate by referring to the Schrodinger's cat thought experiment. The experiment consists of a cat in a box, to which is attached a small amount of a radioactive substance. In the course of an hour, the substance may or may not emit a particle. If it does, then a Geiger counter triggers a can of cyanide, killing the cat. After an hour, the cat's wave ... 2 Luboš Motl's answer gives you all you need, so I'll just add some basics that might also help you (in future readings of similar texts): In order to interpret the statement you've copied: ...finding anywhere in space... First remember that the electron is represented by a wavefunction$\Psi(x,t)$(simplest 1D case for now) that describes the ... 11 Well, the wave function of the electron in the ground state of a hydrogen atom (and very similarly in other atoms) behaves like $$R(r) \sim \exp(-r / a)$$ where$a$is the Bohr radius, effectively the radius of the atom. The exponential is in principle nonzero for an arbitrarily large$r\$, so the electron may be found arbitrarily far from the nucleus at a ...
Consider a system in the state $$|\psi\rangle=\sum_kc_k|\psi_k\rangle$$ Suppose that this system couples to another system. The coupling will leave unchanged some set S of commuting observables of the original system and will change the others. In general, unless the system happens to be in an eigenstate of that observable this will change the expectation ...