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In the case of stimulated emission we always see that one photon goes into the gain medium and two photons come out. How can this conserve energy?

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    $\begingroup$ Last I checked, lasers need a power source to operate... $\endgroup$ – Kyle Kanos Jul 23 '13 at 17:57
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I believe the key point here is that the electron is already excited. So yes, a photon enters and two leave, but the electron goes from an excited state to a lower energy level.

In a way, you could say that in a photon and an excited electron changes to 2 photons and an electron with lower energy than before.

Hence, conservation of energy!

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  • $\begingroup$ Let us suppose, one photon with energy (hν) is the input to a system. This photon will excite the electron from ground state to excited state i.e the process of absorption. In the second step after stimulated emission we will get two photons with the energy (hν) each plus the transition of atom from higher energy state to lower energy state. The point here is from where this second photon comes from ? $\endgroup$ – Yu Ze Jul 24 '13 at 2:45
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    $\begingroup$ If I understand your question correctly(do let me know if I haven't!), you assume that a photon excites the electron, and then the electron emits two photons: going into lower energy state as a result. That would not be an accurate assumption as the electron is already excited before this photon enters. We can ignore how the electron is excited, for the purpose of your question, as I believe it is the energy conservation afterward that concerns you? $\endgroup$ – mehfoos Jul 24 '13 at 13:48
  • $\begingroup$ thats right! I am still unable to understand how the second photon is generated ? $\endgroup$ – Yu Ze Jul 25 '13 at 2:34
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    $\begingroup$ Ok, in that case, let's assume a badly simplified version: Let's say the excited electron has energy hv+ energy at non-excited state. An incident photon of hv enters. Total energy in system? 2hv + electron energy at non-excited state. now stimulated emission occurs and two photon are emitted whilst the incident photon is annihilated. The electron is now in non-excited state. Total energy? 2hv(2 photons) + electron energy at non excited state. Obviously this is super simplified but what would happens instead is just some other combination where the energies add up. $\endgroup$ – mehfoos Jul 25 '13 at 9:28
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As far as I understand, an elementary process of stimulated emission includes both emission of an extra photon and a transition of an excited atom (or a molecule) into a state with lesser energy, so the total energy is preserved.

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  • $\begingroup$ Let us suppose, one photon with energy (hν) is the input to a system. This photon will excite the electron from ground state to excited state i.e the process of absorption. In the second step after stimulated emission we will get two photons with the energy (hν) each plus the transition of atom from higher energy state to lower energy state. The point here is from where this second photon comes from ? $\endgroup$ – Yu Ze Jul 24 '13 at 2:44
  • $\begingroup$ @Yu Ze: The photon that excites the system is annihilated (disappears) in the process. So if there is indeed stimulated emission, there should be another "external" photon to stimulate emission from the excited system. $\endgroup$ – akhmeteli Jul 24 '13 at 3:01
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When two photon come out of the atom, the atom goes back to its ground state from an excited state. Therefore energy is not created.

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