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## Main Question or Discussion Point

Hi all,

This is likely a naive question, following up on something @vanhees71 posted some time ago in another thread:

If so, this to me would imply that any change of the wavefunction from exactly zero to non-zero would mean that, were we able to detect it, one could measure the electron in the higher energy state and then measure the electron in the lower energy state (and with a photon produced of course)

Clearly the above is incorrect, since QFT respects special relativity - I'm just wondering how the change in electron energies (and the exact place/time at which a photon is emitted) in QFT squares with special relativity. @vanhees71 has provided part of the answer above, I was just looking for further clarification (since I think I'm stuck in the 'particle' picture).

Thanks.

This is likely a naive question, following up on something @vanhees71 posted some time ago in another thread:

My question is the following - if we take an electron that has, for example, absorbed a photon, is the portion of the wavefunction representing the electron in a lower energy state ever exactly zero (and hence the probability of finding it in the lower energy state also zero)?... the emission of a photon is a continuous process, described by unitary time evolution. However, in relativistic QFT it is not possible to give the transient state a proper meaning in terms of a particle interpretation. The only observable quantities are S-matrix elements, describing the transition probabilities from one asymptotically free state to another.

If so, this to me would imply that any change of the wavefunction from exactly zero to non-zero would mean that, were we able to detect it, one could measure the electron in the higher energy state and then measure the electron in the lower energy state (and with a photon produced of course)

*essentially instantaneously.*That is, if the wavefunction changes from zero to non-zero, would this not imply that the measured position of the electron, say, could change in a superluminal manner?Clearly the above is incorrect, since QFT respects special relativity - I'm just wondering how the change in electron energies (and the exact place/time at which a photon is emitted) in QFT squares with special relativity. @vanhees71 has provided part of the answer above, I was just looking for further clarification (since I think I'm stuck in the 'particle' picture).

Thanks.