My problem is about the optical pumping of Alkali atoms by circularly polarized pump light. Consider a circular polarized light ($\Delta m=+1$) $$\vec{E}(z,t)= \vec{E}^{(+)}_0 e^{-i\nu t}+c.c. $$ incident on a $\Lambda$-type atom (shown below). The optical bloch equation (OBE) $\dot{\rho}=[H_0+V,\rho]/(i\hbar)-1/2\{\Gamma,\rho\}$ (where $V=-e \vec{r}\cdot \vec{E}$) from standard textbooks is

$$\dot{\rho}_{11}=\frac{1}{i\hbar}(V_{13}\rho_{31}-\rho_{13} V_{31})+\gamma_{31}\rho_{33},$$
$$\dot{\rho}_{22}=\gamma_{32}\rho_{33}, $$
$$\dot{\rho}_{33}=-\frac{1}{i\hbar}(V_{13}\rho_{31}-\rho_{13} V_{31})-\gamma \rho_{33},$$
$$\dot{\rho}_{31}=-(i\omega_{31}+\gamma/2)\rho_{31}+i(\rho_{33}-\rho_{11})\frac{V_{31}}{\hbar}.$$
From the last equation we get the solution for $\rho_{31}$:
$$\rho_{31}=\frac{(\rho_{33}-\rho_{11})V_{31}}{\omega_{31}-\nu-\frac{i\gamma}{2}}$$ where $\gamma=\gamma_{31}+\gamma_{32}$.
Inserting this into the other equations and finding static solutions $\dot{\rho}_{11}=\dot{\rho}_{22}=\dot{\rho}_{33}=0$ results in $\rho_{22}=1, \rho_{11}=\rho_{33}=0$. We immediately see the problem: the atomic ensemble is completely pumped into the $m=+1/2$ ground state even when the pump light is very weak (as long as it's nonzero!!). 

That's quite ridiculous. We know from intuition that when the pump light is sufficiently weak, the atoms are distributed according to Boltzmann law $$\rho=e^{-\beta H_0}/Z$$. Seems that we need to include other terms in OBE so that its solution could reproduce Boltzmann law at weak field limit. How to do this?
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  [1]: https://i.sstatic.net/4AtSD.png