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Just open any string text which has a discussion of the relativistic point particle. http://arxiv.org/abs/0908.0333 - Section 1 for example or Green, Schwartz, Witten Volume 1 Punchlines: 1) Time can be introduced as an operator but you need to introduce a 'proper time' parameter for which the system evolves with. In doing this you introduce a gauge ...

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This is one of the open questions in Physics. J.S. Bell felt there was a fundamental clash in orientation between ordinary QM and relativity. I will try to explain his feeling. The whole fundamental orientation of Quantum Mechanics is non-relativistic. Even though, obviously, QM can be made relativistic, it goes against the grain to do so, because the ...

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$$\langle{q}|\hat{p}|\alpha\rangle = \frac{\hbar}{i}\frac{\partial}{\partial q}\langle q|\alpha \rangle$$ So, $$\langle{q}|\hat{p}|p\rangle = \frac{\hbar}{i}\frac{\partial}{\partial q}\langle q|p\rangle$$ This is equivalent to $$p\langle{q}|p\rangle = \frac{\hbar}{i}\frac{\partial}{\partial q}\langle q|p\rangle$$ then its just a first order linear ...

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The situation here is a bit complicated. The problem is that with a measurement process, you are not allowed to consider only the evolution of the measured system, but you have to take into account the evolution of the global system formed by the measured system and measurement apparatus. For that system, the (global) wavefunction always evolves by means of ...

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You can have a repeatable measurement process (i.e. a measurement process that, roughly speaking, gives the same result if done twice in a row) only for discrete observables. A discrete observable is an observable whose spectrum is purely discrete. So with the Hamiltonian it is possible to have repeated measurements, provided it is a system with purely ...

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Energy is a bit of a special case because the eigenfunctions of the Hamiltonian are time independent (assuming a time independent Hamiltonian). So when you make an energy measurement and collapse the system to an eigenfunction of the Hamiltonian it stays there. However the position operator does not commute with the Hamiltonian so when you measure the ...

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Answer to this question should start from why we want the physical observables to be represented by linear operators. Theoretical physics is about constructing a mathematical model which we hope describes the phenomena it's being modeled for and hence helps predicting stuff. In classical physics this mathematical model is based simply on the real numbers ...

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According to this paper, an experiment was performed that measured the single electron's physical wave function by causing it to interfere with itself. The interference pattern matched the predictions of the Schrodinger equation. So, apparently this was a direct measurement of an electron's wave. Hydrogen Atoms under Magnification: Direct Observation of ...

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Actually, if the energy of particles is high enough to take relativity into consideration, the concept of particles in quantum mechanics is no longer as valid. For example, the uncertainty relationship $\Delta E \cdot \Delta t \approx \hbar$ and energy-mass relation $E=mc^2$ suggest that there will be new particles created and annihilated in those cases. So ...

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