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A photon is emitted by an electron (which is in a bound state). Is the energy of the electron lost immediately, or is the energy emitted during the complete transition time? I think my second assumption is correct but confirmation would be greatly appreciated.

In other words, should I view photon emission as part of the transition process or as merely the cause of it? I am aware that during the transition the election can be viewed as transitioning through a multitude of highly unstable orbitals until it finally settles into the lower level.

This post What is the Quantum Transition Time for Photon Emission? is very useful and really brings home the subtleties involved in quantum mechanics, but I don't think it addresses my question directly.

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    $\begingroup$ I edited your question. Please note that proper spelling, puctuation, etc. make the question easier to read and therefore more likely to get a good answer. Also note that apologizing for e.g. your current state of knowledge is never relevant for any question on this site. $\endgroup$ – DanielSank Mar 14 '15 at 21:19
  • $\begingroup$ excellent points, the main lesson here to me is never try to write questions using a mobile phone on a moving bus. So am I checking the composition of this comment very carefully now? you bet.... thanks for your advice and time much appreciated $\endgroup$ – user74893 Mar 14 '15 at 21:59
  • $\begingroup$ As per your request, check out the accepted answer of physics.stackexchange.com/questions/174450/… $\endgroup$ – Chris L. Apr 7 '15 at 19:21
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A photon is emitted by an electron (which is in a bound state). Is the energy of the electron lost immediately, or is the energy emitted during the complete transition time? I think my second assumption is correct but confirmation would be greatly appreciated

One has to let go of the classical framework. Immediately in time has a meaning for a ball falling in the Newtonian gravitational field and mathematics can give you the rate of energy loss per delta(t) because in principle every (x,y,z) point is reached at a specific t. This does not hold in the quantum mechanical framework of an electron bound to an atom.

An individual atom with its electron in an excited state may emit the photon at an arbitrary time t. One has to take a large sample of atoms with the electrons at that energy level and measure the time the photon hits the detector . One then will have a curve characterizing the lifetime of that bound state's collective time behavior

Now in the post you have quoted the answers are indicating the mathematical formulation within the theory of Quantum Mechanics that reproduce this experimental observation. Quantum mechanical calculations give probability distributions for the variables under consideration, time in this instance, to fit the experimental observations.

What is really happening at the individual atom's decay from a higher energy level to a lower one is random to first order ( it is the assumption of calculating the half life curves) . The time will be within the Heisenberg Uncertainty of delta(E)*delta(t) is all that can be safely claimed , and can be presumed "instantaneous" , in the sense that there are no experimental tools to explore further, other than the mathematical ones discussed in the answers of the other question.

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  • $\begingroup$ If I thought about it in terms of the replacement of a classical definite trajectory by an average (expectation value) of the possible paths the electron takes, which I do understand and am used to that concept, then used the formula t = d/v , this would imply that t is also based on an expectation values , which would explain (to me) your second paragraph , would I be correct? In other words, follow through the implications of no definite trajectory into every classical notion and the equations underlying them, which I failed to do in my question ? Thanks and regards $\endgroup$ – user74893 Mar 15 '15 at 7:52
  • $\begingroup$ In the case of the photon, yes the averages would work, though for time the half life is used to describe a decay time, as in the link. $\endgroup$ – anna v Mar 15 '15 at 8:02
  • $\begingroup$ This can be a separate question if you think it's more appropriate than in this comment section, but does your statement "in the case of the photon" imply that massive particles need a different treatment, or just a reflection of the fact that massive particles are not involved in energy loss or gain in bound states? This will be my last imposition on your time on this question, lots to think about as it is...Thank you $\endgroup$ – user74893 Mar 15 '15 at 8:19
  • $\begingroup$ Well, the electrons are described by the orbitals. To ask "where the electron is during the emission of a photon" in terms of orbitals needs the full mathematical treatment, as sketched in the answers of the other question. $\endgroup$ – anna v Mar 15 '15 at 9:30
  • $\begingroup$ In nuclear physics one can have emission of massive particles and the lifetimes will again be described by the half life. The QM theoretical models used are various, as the shell model. The corresponding bound states of the nuclear levels again will require particular study ( before and after the decay) $\endgroup$ – anna v Mar 15 '15 at 9:49

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