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An observable basically is something you can measure. However compared to just any general measurement, an observable has one specific property: If you repeat the measurement without the system changing in between, you get the same result (this is obviously not true for all possible measurement procedures). Such measurement have indeed been experimentally ...


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Double slit experiment with electrons is a demonstration of a quantum behavior. When we say observe we mean expose to some kind of interaction. So when electron travels from its source towards the double slit and then passes through, and hits the detector, we see that it is a particle. But, when many electrons pass through the slit we observe a interference ...


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A free moving electron does not emit photons. If one want to see an electron you have to illuminate it. Sending photons, the electron reflect (absorb and re-emit) photons which one can detect then. By doing this the electron get disturbed and changes his direction. The fringes on the observation screen will be destroyed. Since one cannot synchronize the ...


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When you send the photons through the double-slited wall, they form a diffraction pattern on the other side, which is a wave-like phenomenon. It makes no sense to think of the photons as "particles" anymore, since they would need to cross both slits simoultaneously in order to create the diffraction pattern. So a way to find out what's happening is to place ...


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Your question is not silly. In the two slit expariment you measure the position if the photon, that is a particle-like quality. What happens is that a photodetector records when it absorbs a photon. That gets recorded in the memory. Some schools of physics interpret the interaction of the photon with the detector as a measurement. Others interpret that the ...


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A Hermitian operator is diagonalizable and has real eigenvalues. If you have an operator with real eigenvalues and such that all its eigenstates are orthonormal and, moreover, form a complete basis, then this operator is Hermitian. This is because you can pick its eigenbasis as an orthonormal basis and in that basis it will be represented by a manifestly ...


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The spectral theorem only holds for normal operators. Self-adjoint operators are normal, symmetric ones not necessarily so. In physicist-speak, we want the generalized eigenvectors to from a 'complete basis' of the Hilbert space. For example, the generalized eigenvectors of the momentum operator in position representation are plane waves, and even though ...


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I'm pretty sure there exists an answer for this here already, but I can't find it (it's always about unboundedness). For the Hamiltonian, the answer is basically given by Stone's theorem on one-parameter unitary groups. There is a one-to-one correspondence between self-adjoint operators and strongly continuous one-parameter families of unitaries. Why is ...


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Hermitian observables yield real eigenvalues. Also, for an observable to be a good observable, it must be diagonal in some basis, in which case it has real diagonal elements. So your observable is related to a real, diagonal matrix by a unitary transformation. That pretty much restricts what you can have. This isn't axiomatic, of course. It's more from a ...


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If we have two arbitrary quantum-mechanical operators $\hat A$ and $\hat B$, then the corresponding observables have to satisfy the Robertson-Schrödinger uncertainty relation: $$\Delta A \Delta B \geq \frac12 \lvert\langle [\hat A,\hat B] \rangle\rvert$$ This equation implies that it is impossible to measure both $A$ and $B$ to arbitrary precision at the ...


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While I agree with Noah Steinberg's answer, there are some other points. Usually when quantizing a system, we label our states using quantum numbers, which are as close to classical parameters (also called c-numbers or commuting numbers) as you can get. The calculation is usually easiest when you can find the largest set of commuting observables, and ...


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If two observables are compatible it means the eigenstate of one observable is also an eigenstate of the other and that the commutator of the two operators is 0. This means that if two observables are compatible one can make a measurement of the first observable and then measure the second observable without changing the state of the particle.


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The electron does not move - it has no well-defined position in the orbital state, and hence no well-defined momentum. Neither does it "teleport" around - as long as it is not interacting with something that forces it to be at a definite position, its state is "smeared" all over the electron as an electron cloud. Yes, this is essentially the Bohr model, ...


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There are many confuse things in this question: 1) "The multiple paths the particle simultaneously travels, interfere with each other" It's not the paths that interfere, when these paths cross one another, i.e. the wave-packet on one path and the wave-packet on the other path pass through the same region at the SAME TIME. 2) "but as it is absorbed, it ...


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you should look for the theory of "Covariant Quantum Mechanics" originally introduced by M. Modugno and J. Jadcyk, in particular the "special algebra of quantizable functions". There is not so much literature since the theory is mathematically quite hard to enter, but afterwards it is worth. I have worked in that field for several years. The basic idea is to ...



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