Stimulated Emission = Stimulated Absorption

[zuukr]

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

 

[Silver_rose]

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?

 

[vanhees71]

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.

 

[Alastair McD]

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.