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I know that a system's energy can be measured with an energy that can be below or above the expectation value, if the system was not in an energy eigenstate, so that energy is only conserved on average over multiple measurements.

Does that mean that energy is not conserved in individual cases and if so, does that mean that the extra or missing energy is being created/destroyed for those individual cases?

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    $\begingroup$ Energy is absolutely conserved for every single quantum, even in quantum mechanics, at least locally. It can not be measured with arbitrary precision because of time-energy uncertainty. The difference between the quantum system's energy and the measured energy does, in that case, come from the measurement system. If the system was not in an energy eigenstate (i.e. stationary), then the uncertainty comes from the preparation, i.e. the source. Excited states are, of course, never truly stationary. They have to decay with a time constant that depends on the coupling to the external field. $\endgroup$
    – FlatterMann
    Commented Jan 29 at 5:25
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    $\begingroup$ energy conservation is part of the axioms en.wikipedia.org/wiki/Axiom in the theories of physics which has been never falsified. $\endgroup$
    – anna v
    Commented Jan 29 at 5:39
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    $\begingroup$ The only energies that are in play are the measured energies. The only measured energies are those of the source and those at the detector and they are, for all known kinds of spectroscopy, identical. For optical measurements on free atoms you do have to consider the recoil of the atom, of course, but it's very small. In high energy physics that "recoil" is also part of the energy budget and the total energy is known to be conserved on a single quantum process level. We actually calculate it all the time to make sure that we aren't missing things like the neutrino. $\endgroup$
    – FlatterMann
    Commented Jan 29 at 5:42
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    $\begingroup$ @FACald I read the paragraph, it is not a point of contest, it is a hypothesis of the author that maybe in the future .... That is true for all axioms in our axiomatic main stream theories $\endgroup$
    – anna v
    Commented Jan 29 at 5:53
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    $\begingroup$ @FACald Wikipedia? OK.... In Moessbauer spectroscopy we can check the agreement between the emission and absorption energy using Doppler effect on a 14.4keV gamma with a precision of better than 1 part in 10^8. We can probably do better than that in optical spectroscopy on atoms used in atomic clocks. So IF nature were to violate energy conservation on a case by case basis, then it would have to do so by telling both the emitter and the absorber atom/nucleus what that violation is going to be. Now we are in the fairy-realm of superdeterminism. That's not even a testable hypothesis. $\endgroup$
    – FlatterMann
    Commented Jan 29 at 6:34

2 Answers 2

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Energy is not created or destroyed during interactions between quantum systems, including measurement interactions. When two systems interact and one of them changes its energy the energy is transferred to the other system. When the measured system has a very small energy in comparison to that of the measurement device, which is true in many measurements, the change in the energy of the measurement device will be small in comparison to its original energy and will be difficult to detect:

https://arxiv.org/abs/2108.08342

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The measured energy of the quantum system is not conserved across measurements unless the quantum system only has a single energy eigenstate.

This should be a boring statement. To measure a quantum system is to entangle its state with the state of something our macroscopic selves can actually measure, with which it could exchange an unknown amount of energy.

As to whether total system energy including the entangled macroscopic system is conserved, I'm fairly certain it's logically impossible to even ask the question. If you know the total system energy before you do the experiment, and you know the total energy of the macroscopic system before it is coupled, then you already know the energy of the quantum system, which means that it must already be in a particular energy eigenstate, not in a superposition of energy eigenstates. But energy conservation across measurements for a system in a superposition of energy eigenstates was the entire thing we wanted to ask about in the first place.

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  • $\begingroup$ I understand that before the measurement we cant know the total system energy, my question more focus on the gap between the expectation value and the measured value and does the difference between these values (when there is one) comes about $\endgroup$
    – FACald
    Commented Jan 29 at 6:28
  • $\begingroup$ The expectation value is the amount of energy in each energy eigenstate weighted by the the probability of finding the system in each of those eigenstates. It's not out of the question that the expectation value is never the energy we measure. For a toy example, if a system is in a superposition such that A and B are equally likely and exhaustive, then the expectation value is (A+B)/2, but of course we never find (A+B)/2, we only find A or B. $\endgroup$
    – g s
    Commented Jan 29 at 6:31
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    $\begingroup$ so I guess my question was if finding A or B and not (A+B)/2 supposes a violation of conservation of energy for that individual case $\endgroup$
    – FACald
    Commented Jan 29 at 6:40
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    $\begingroup$ That the total system energy is conserved is known with exquisite precision. We never had more energy come out of a nuclear/muon/e-e+ decay than is present in the total mass-energy of the system. The only uncertainty is that of the measurement system. So unless we assume that nature tells the emitting system to be "nice" and always "hide its violation" underneath the actual noise level of the measurement system, the proposition of a violation is very questionable. The latter is, of course, known as "superdeterminism", but that's not even science. $\endgroup$
    – FlatterMann
    Commented Jan 29 at 6:45

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