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Stimulated Emission = Stimulated Absorption

  1. Apr 8, 2015 #1
    I'm having a bit of trouble understanding this. I am going through the section in Griffith's regarding Einstein's Coefficients. For a system in equilibrium, the rate of particles undergoing emission needs to equal the rate of particles undergoing absorption in order to maintain equilibrium. When spontaneous emission is considered as a factor of emission as well, the rate of stimulated emission is still equivalent to the rate of stimulated absorption. I can see that this is proven mathematically, but I am trying to conceptualize this. Why wouldn't stimulated emission be less than stimulated absorption for a system in equilibrium?

    Thank you
  2. jcsd
  3. Apr 9, 2015 #2
    I'm curious as to why you would think stimulated emission would be less.
    If it makes sense mathematically what is making you think it should be any different?
  4. Apr 9, 2015 #3


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    Well, the proof in Griffiths book is circular, because he assumes the Bose-Einstein distribution as the equilibrium distribution to prove ##B_{ab}=B_{ba}##. What's really behind it is the principle of detailed balance, which can be derived from the unitarity of time evolution in quantum theory.
  5. Sep 6, 2015 #4
    Stimulated emission (B10) will equal stimulated absorption, but there is also normal absorption by unexcited molecules. The Einstein coefficient of absorption B01 is for both stimulated and normal absorption. In equilibrium total absorption equals total emission (B10 = A01 + B01, where A[10] is the coefficient of spontaneous emission.)

    Note, letters A and B were chosen by the German speaking Einstein who chose A for spontaneous Ausgang (emission), not to be confused with A for absorption.
    What I have called normal absorption N01 will equal spontaneous emission A10.
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