at worst a factor of two ($(1+\delta-1) / (1+\delta+1) \sim \delta/2$)   $$\begin{aligned}
\alpha_{12}'^* - \alpha_{21}'
&=&
      \frac{1}{ \hbar}
          \exp \left[ +\imath W_{21} t \right]
  \sum_i
      d_{1i}^* 
      d_{i2}^*  
    \frac{ W_{i2} }
         {W_{i2}^2-\omega_0^2}
~~~~
-
      \frac{1}{ \hbar}
          \exp \left[ -\imath W_{12} t \right]
  \sum_i
      d_{2i} 
      d_{i1} 
    \frac{ W_{i1} }
         {W_{i1}^2-\omega_0^2}
\\
&=&
      \frac{1}{ \hbar}
  \sum_i
      d_{1i}^*  
      d_{i2}^*  
    \left[
      \frac{ W_{i2} }
           {W_{i2}^2-\omega_0^2} 
     -
      \frac{ W_{i1} }
           {W_{i1}^2-\omega_0^2}
    \right]
          \exp \left[ +\imath W_{21} t \right]
\\
&\approx&
0\end{aligned}$$

This approximation allows me to equate $\alpha_{21}'$ to $\alpha_{12}'^*$, and hence change $- \alpha_{21}' c_1^* c_1
+ \alpha_{12}'^* c_2 c_2^* \longrightarrow 
\alpha_{21}' \left( c_2^* c_2
- c_1^* c_1 \right)$ in the $\rho_{12}$ equation below. This is a vital step in getting to a simple form equivalent to the Bloch equations.

### Sum is double

This follows from the above assumption (as per HPB) that the forward and backward transitions (between levels 2 and 1) have nearly the same amplitude. This approximation allows me to replace $\alpha_{21}'$ with $\alpha_{12}'$ in the $w$ equation below, which simplifies my notation and makes the coupling term match that in the $\rho_{12}$ equation, important in getting to true Bloch equations.

  $$\begin{aligned}
\alpha_{12}' + \alpha_{21}'^*
&=&
      \frac{1}{ \hbar}
          \exp \left[ -\imath 