Conservation of energy in quantum measurement

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Okay, so my question was about what he calls "Type 1" violations. Unfortunately his treatment of this type (at the end) is very brief and I don't quite get what he means.
Is or is not the total <Jz> for "source +particles + detector" conserved in each macroscopic branch?
Regardless, his point about measurement errors eliminating the need for cross-terms is fascinating! Is this something that has been discussed before in the literature?
 
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Demystifier said:
A recent related paper:
http://lanl.arxiv.org/abs/1609.05041
It points out that the standard conservation law is only a statistical law, which, by itself, is not sufficient to understand conservation of energy at the individual level.
Thanks for sharing this! In the case where the particle is measured with a high final energy, I think this shows the same effect we were discussing: the total energy after the measurement is higher than the initial energy, with the difference made up by a loss of energy in "branches" of the WF where the particle was not measured- if you believe in those.
The article itself focuses on the change in the probability distribution for the particle's energy, under unitary evolution. I would point out one detail that they didn't mention: If we look for eigenstates of the full Hamiltonian (particle + opener +interaction) , rather than of the particle and opener separately, it is clear that some of these will have superpositions of the free and trapped particle. In those eigenstates that make up the initial conditions in the article, and that are such superpositions, the free element in fact has high energy, which, I think, implies that the trapped element does as well. Meanwhile other, purely trapped eigenstates have lower energies. So this gives us an alternative decomposition of the initial WF of the trapped particle, one which does have high-energy terms, in contrast to the simple Fourier decomposition which does not.
In other words, you cannot know what the possible energies are for the particle in the initial state without knowing the details of interactions it may have in the future, because even if it's currently not entangled with anything, there are many ways to write its WF as a sum of frequencies, and the "correct" one depends on all possible interaction Hamiltonians. Is this correct?
 
Correction: the standard Fourier series if of course unique as the way to decompose a function on an interval as a "sum of frequencies"- sine functions that vanish at the ends. What I should have said is that the eigenstates of the full Hamiltonian are entangled states between the particle and the "opener", and therefore their projections on the space of particle positions, for a given opener position, are not simple sines.
The bottom line is the same though: the existence of a possible future interaction makes relevant a new decomposition of the particle WF, which may include energies that are very different from any in the Fourier decomposition that gives the particle eigenfunctions.
According the the article, there should nevertheless be no effect on any of the moments of the probability distribution for the particle energy. This seems intuitive, but I'd love to see it proven for the general case of eigenstate decompositions of states corresponding to separate subsystems with a possible future interaction.
 
@Demystifier so conservation of energy could be violated at small microscopic scale due to uncertainty principle. Did I get it right?
 
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Demystifier said:
A recent related paper:
http://lanl.arxiv.org/abs/1609.05041
It points out that the standard conservation law is only a statistical law, which, by itself, is not sufficient to understand conservation of energy at the individual level.

The experiment described in the paper depended on the opening timing (the box was open only for a time T), but they didn't talk about uncertainty relation ΔTΔE:
https://arxiv.org/pdf/quant-ph/0105049v3.pdf

Am I missing something here?
 
Demystifier said:
One can talk about ΔE without talking about ΔT.
One must talk about ΔE when talking about T.