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A wavefunction has a particular energy and is spread over space. For example, an electron in the ground state (n=1) in hydrogen has an energy of -13.6 eV, but its possible that we measure the position of the electron far away from the Bohr radius from the nucleus. What if we found it much further way, like ten times the Bohr radius for example?

  1. Would it have the potential energy of an electron at the Bohr radius or the potential energy of an electron at ten times that radius?
  2. What energy would the electron actually have, -13.6 eV or something higher?
  3. What happens next? Will the new wavefunction just be the n=1 orbital from before?
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1) It would have a potential energy equal to that of an electron a distance of 10 Bohr radii. This is because the potential energy is just a function of position. Therefore it commutes with the position operator.

2) Now that we have made a position measurement, the state of the particle is now in a superposition of energy eigenstates. This superposition is found by writing the position state at a distance of 10 Bohr radii in the energy basis. It does not have a definite energy anymore

3) See above.

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    $\begingroup$ But if you measure the electron at $r=10 r_0$ it will stop being an energy eigenstate and it will be a position eigenstate with a non definite energy, right? $\endgroup$ – FrodCube Aug 18 '18 at 16:45
  • $\begingroup$ @FrodCube yep my mistake $\endgroup$ – Aaron Stevens Aug 18 '18 at 16:51
  • $\begingroup$ @FrodCube I have edited the question $\endgroup$ – Aaron Stevens Aug 18 '18 at 17:19
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    $\begingroup$ Hi @Joe, to answer the second part of your followup: "If it's a superposition of energy eigenstates now, it's possible that we measure it again and find it has -13.6 eV while it has a high potential energy (close to zero) due to its position. Won't it have a negative kinetic energy then?" You need to remember the uncertainty principle. If you measure it in a given energy eigenstate, it is now in a superposition of position eigenstates. You can't know the energy eigenstate and the position eigenstate at the same time. $\endgroup$ – SabrinaChoice Aug 19 '18 at 20:44
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    $\begingroup$ @SChoice I don't think the OP is asking about photon excitation specifically. He is asking about if you were to measure the position first, and then after this the expectation value of the energy would have to be larger since the electron was in the ground state. Do you mean to say the method of measurement would have to supply energy? This is what I would say is the case. During a measurement, the Hamiltonian is different than just the hydrogen atom, so it becomes more complicated. $\endgroup$ – Aaron Stevens Aug 19 '18 at 21:20

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