I Is Detailed Balance Satisfied in a Two-State System in Quantum Mechanics?

[cozycoz]

In quantum mechanics, I've learned that when there are incident photons with energy spectral density $ρ(ω)$ , two states system with $E_i$ and $E_n$ ($E_n > E_i$) goes through transition (absorption and emission) with transition rate Γ_{i→n}=Γ_{n→i}=\frac {2π}{ħ}|V_{ni}|^2ρ(E_n-E_i).Is this detailed balance,? Wikipedia says that detailed balance definition is that "At equilibrium, each elementary process should be equilibrated by its reverse process."
Actually when I calculate transition rates with two states system, it seems that both emission and absorption process always have the same transition rate. Can I refer this as detailed balance? Thanks

 

[protonsarecool]

The probability in a Markov process for a particular transition from eg. i to n is given by (transition rate i->n)*(probability of i), or $\Gamma_{i \rightarrow n} P_i$. The detailed balance condition says that there is an equilibrium probability distribution $\Pi_i$ for the process (for instance a Boltzmann distribution) such that $\Gamma_{i \rightarrow n} \Pi_i = \Gamma_{n \rightarrow i} \Pi_n$, ie. the transition probabilities and their reverse cancel each other for each pair of states i and n. Taking the Boltzmann distribution in particular, this means that $\Gamma_{i \rightarrow n} e^{- \beta E_i} = \Gamma_{n \rightarrow i} e^{- \beta E_n}$.