Energy conservation and quantum measurement

Consider a particle in a potential well. Let’s assume it’s a simple harmonic oscillator potential and the particle is in its ground state with energy E₀ = (1/2) ℏω₀. We measure its position (measurement-1) with a high degree of accuracy which localises the particle, corresponding to a superposition of momentum (and therefore energy) states.

Now we measure the particle’s energy (measurement-2) and happen to find that it’s E₁₀ = (21/2) ℏω₀. Where did the extra energy come from?

In the textbooks it’s claimed that the extra energy comes from the act of observation but I wonder how that could work. Measurement-1 which probed the position of the particle can’t have delivered to it a precise amount of energy, while measurement-2 might just have been passive. No doubt there is entanglement here between the particle state and the measuring device but where, and which measurement?

Answer

For a high degree of accuracy you would have to probe the particle with a high energy (short wavelength) photon so there is plenty of energy that can go into vibrational excitation. After such hard hit the particle will be smeared across a wide range of states $$\Psi=a_0*\Psi_0+a_1*\Psi_1+...$$This is not an entanglement but a simple superposition of eigenstates. The expectation value of the particle's energy $\bar{E}=\sum_{i}a_i^2E_i$ should not be equal to the energy of any particular state and the second measurement will yield $E_i$ with $a_i^2$ probability.

So, the extra energy comes from an interaction with a probe particle and it doesn't have to be precisely equal to the energy of a certain vibrational state.