The Eigensolution and its Transformation

 

• In a diagonalizing basis e_i' our equation looks as follows:
  

  $$\frac{\textrm{d} a^{l'}(t)}{\textrm{d} t} = -\lambda_{(l')} a^{l'}(t)$$ (3.56)

  
  
  where a bracketed index implies that there is no summation. The solution is, for every l':
  

  $$a^{l'}(t) = a^{l'}(0)e^{-\lambda_{(l')} t}$$ (3.57)

  
  

• In order to obtain a solution in terms of a^l(t) we simply need to rotate the basis back to the original one:
  

  $$\Lambda^l{}_{l'}a^{l'}(t) = \Lambda^l{}_{l'}a^{l'}(0)e^{-\lambda_{(l')} t}$$ a^l(t) =

  $$\Lambda^l{}_{l'} e^{-\lambda_{(l')} t}\Lambda^{l'}{}_k a^k(0)$$ =

  $$\sum_{l'=1}^n \Lambda^l{}_{l'} e^{-\lambda_{(l')} t}\sum_{k=1}^n \Lambda^{l'}{}_k a^k(0)$$ = (3.58)

  
  
  The presence of $e^{-\lambda_{(l')} t}$ between the two big lambdas implies that we cannot simply contract $\Lambda^l{}_{l'}\Lambda^{l'}{}_k$ and get Kronecker delta. The evolution of the system derives from those terms $e^{-\lambda_{(l')} t}$.