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Let's say I have a time independent Hamiltonian so my system conserves energy. It's initially in an energy eigenstate with $E=1$ in whatever energy units you like. I measure a different observable that doesn't commute with $H$, then I measure $H$ again. I have some probability now of finding my system in an energy eigenstate with $E\neq1$.

What gives? If this was a harmonic oscillator for example, it could be the case that I end up in a state with hugely more energy. Where does this extra energy come from?

In this question, the accepted answer says that the probe particle imparts the energy: Energy conservation and quantum measurement

I get that idea, but after the probe interacts with the system and I measure the energy again, then my energy may go down, stay the same or go up. How does this relate to the energy of the probe particle? I feel I'm missing something fundamental here.

bRost03
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    Performing a measure means that you interact with the system, for example by sending photons and measure the emitted ones... Interacting with the system means that you perturb it – Kevin De Notariis Oct 11 '18 at 12:51
  • Yes I understand that, but looking specifically at the energy difference between the initial and final state, how can a simple measurement of (e.g.) position account for a potentially huge energy change? Where does that energy come from / go to? – bRost03 Oct 11 '18 at 12:54
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    My guess is that in reality when we measure, for example, position, we don't actually confine the system to a position eigenstate because our measurement is not infinitely precise. So the resulting linear combination of energy eigenstates may not necessarily contain the highest energy states. To truly confine the system to an eigenstate would require an infinitely precise measurement which would require infinite energy? But if I make a very precise measurement (lots of energy) then measure the energy again, I may end up in a high energy state but I may also end up in a lower energy state. – bRost03 Oct 11 '18 at 13:14
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    Possible duplicate of Energy conservation and quantum measurement – Hugo V Oct 11 '18 at 13:38
  • @HugoV I've edited the question to distinguish it from the linked question – bRost03 Oct 11 '18 at 15:49
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    @bRost03 In order to measure again, you must have another probe interact with the system. – probably_someone Oct 11 '18 at 15:53

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The fundamental is that you can’t know the effect of the measuring interaction on this system, so it might give it more energy or it might take some energy away from it, and you don’t know which is going to happen, or how it is going to happen. The interactions is not assumed to be with one particle, it is a complex interaction with a macroscopic object, so I don’t think it’s correct to talk about one probe particle in general. All you know are the possible states you could find your system in, and the probability of finding it in one of these states when measured. Quantum mechanics does not talk about probe particles, or else you would make the situation even more complex, since this probe particle would have to be treated quantum mechanically as well, and you wouldn’t be able to know it’s state without measuring, and you would end up having to perform an infinite number of measurements to get the information for any given system

Hugo V
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