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In Quantum Mechanics (vol. 1) by Tannoudji et al., there are two pages discussing "perturbation created by a position measurement" (p.278-279). They consider a particle in an infinitely deep potential well of width $a$ and prepare the particle in the ground state $ ψ_1(x)$ = $ \sqrt{\frac{2}{a} }\sin \frac{\pi x}{a} $. It must be stressed that when the position is measured, the particle can never be an eigenstate of the position. Thus, once we say that the particle is found to be $x=\frac{a}{2}$, the position state can be chosen to be a long rectangle with small width $ \epsilon $, and the height is $\frac{1}{\sqrt{\epsilon}}$, which is determined by the resolution power of the apparatus of the position measurement. This step is exactly what measurement postulate indicates.

Now we hope to know what the energy of the position state, i.e., the long rectangle, was transferred during the position measurement above. Results show that the mean value of the energy is divergent. It is absurd but it is can be easily amended with the long rectangle being replaced by a sharp Gaussian function. Surely, the mean value of the energy is greater than ground state over which the position measurement is performed.

My question is: as far as I know, there is dynamical process within the current formalism of quantum mechanics which account for the energy transfer. WHY? does such a process would violate any principle of current quantum mechanics?

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    $\begingroup$ So your issue is basically how can we start in the state with the lowest possible energy, but then after measuring the position end up in a state whose mean energy is larger than the ground state? As if we added in extra energy through just the measurement process? $\endgroup$ – Aaron Stevens Aug 8 '18 at 3:03
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    $\begingroup$ @AaronStevens We can perform the position-measurement in any state, and obtain a sharply-localized-but-not-a-$ \delta $-function position state, and there is in general the energy transfer between the apparatus and the particle. MY QUESTION is: how does the energy transfer? what happens during the process? The measurement causes an discontinuous change, implying that the process happens instantly. That a process happens instantaneously is nonphysical. $\endgroup$ – Quanhui Liu Aug 8 '18 at 3:29
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    $\begingroup$ "Current quantum theory does not offer a dynamical mechanism". I hope you are aware that quantum theory is not a dynamical theory. The only absolute predictions possible from a quantum theory is probability distributions, distributions of ensembles of many interactions with the same boundary conditions. Also should be aware that Feynman diagrams used as an iconal representation of the integrals needed for an interaction have an instantaneous position for the interaction where the coupling constants reside. $\endgroup$ – anna v Aug 8 '18 at 5:25
  • $\begingroup$ @annav Thanks. I am well aware that the existence and evolution of a quantum state must satisfy the Schroedinger Equation, a dynamical equation. But, what I want: Clearly there is energy flow, why is there no dynamical equation describing such a flow? You may belong to the group of physicists who do not think that such an equation is necessary, but I belong to another group who expects such an equation. $\endgroup$ – Quanhui Liu Aug 8 '18 at 8:15
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    $\begingroup$ Wouldn't a definite energy transfer upon measurement of position violate the uncertainty principle? $\endgroup$ – Stéphane Rollandin Aug 8 '18 at 9:13
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I would say the answer is that any such measurement will necessarily involve some transfer of energy. For example, you might use a photon to measure the electron's position. But if that photon were smaller in energy than the gap between energy levels in the well, then it couldn't be absorbed, and it wouldn't measure anything. It needs instead to be a high energy photon, and thus carrying a lot of momentum. And in the process of scattering off the electron, it will transfer to it some of that momentum, leading to your localised high energy electron.

In other words, I think the problem is that you should describe the full quantum measurement process throughout suitable quantum equations, and you would find out that energy is overall conserved, and that the higher the precision in position you want, the higher the energy of the photon, the higher the energy transferred to the electron. The idea of a measurement and collapse are just approximations to these real physical processes that go all the way to the brain of the observer (and while there's still not a definite answer re: measurement problem, I am convinced that there is no true loss of unitarity involved).

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In quantum theory, psi function gives probabilities and from those expected mean values (expected for an ensemble of experiments) can be calculated. Quantum theory does not predict, track or analyze particle position, momentum or energy in a single experiment.

If you would like to do just that, you need to search for ideas outside the standard quantum theory.

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  • $\begingroup$ Perfect! But there is no such a theory for the ensemble, and it seems that have must have one. $\endgroup$ – Quanhui Liu Aug 8 '18 at 13:21

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