Exercise 6.1-1

Shuffling Fermi Functions

"Shuffling" Fermi distributions, while not really difficult, needs some practice and getting used to; it is good exercise to do it a few times (you may also consider to try it for some shuffling done before).

Show that the 1st laser condition

E_F^e – E_F ^h ³ hn ³ E_g

follows directly from

R_se R_fa	=	[f_{e in C}(E₁ + hn, E_F^e , T)] · [f_{h in V}(E₁, E_F^h, T)] [1 – f_{h in V}(E ₁, E_F ^h, T)] · [1 – f_{e in C}( E ₁ + hn, E_F^e, T)]

Link to the solution

With frame

6.1.1 Interaction of Light and Electrons; Inversion

Einstein Coefficients

Solution to Exercise 6.1-1