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I believe I frequently see it stated that a photon is absorbed and an identical photon is emitted.

How can the energy in equal the energy out, with no loss?

user40910
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    Energy is conserved. – Puk May 25 '22 at 05:12
  • my answer here may be relevant https://physics.stackexchange.com/questions/503445/does-a-photons-wavelength-and-energy-change-when-reflecting-off-a-mirror/516959#516959 or here https://physics.stackexchange.com/questions/248726/how-does-qed-explain-reflection/248741#248741 . has to do with center of mas versus lab considerations – anna v May 25 '22 at 05:53
  • It's not that reflected photon is ideally the same as the input photon. Reflected photon changes phase by $180^{\circ}$ and is partially (or fully polarized if falls in a Brewster angle) – Agnius Vasiliauskas May 25 '22 at 07:22
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    It's also worth noting that the outgoing photon can have less or more energy than the incident photon, e.g. due to interactions with molecular vibrations. See Raman scattering. – Puk May 25 '22 at 09:43
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    Please provide a reference source for your statement so we can explain under what conditions it's true -- or not true. Example: Electron absorbs photon and rises to a different orbital. electron emits same photon and returns to original orbital. But electron might emit a different photon and transit to a third orbital level. – Carl Witthoft May 25 '22 at 12:51

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