let the population of degenerated lower states be n_i and the upper degenerated state be m_j . the excitation rate from state n_i to m_j is e_{ji} and the de-excitation rate from state m_j to n_i is d_{ij}. the notation of subscript is that, the first letter is target, and the later is the source.

the rate equations are:

\frac{d n_i}{dt} = - n_i \sum_j{e_{ji}}+\sum_j{d_{ij} m_j}

\frac{d m_j}{dt} = - m_j \sum_i{d_{ij}}+\sum_i{e_{ji} n_i}

combine these equations into one matrix equation by setting a population vector for h lower states and k higher states:

p=(n_1,n_2,...,n_h,m_1,m_2,...,m_k)

\frac{d p}{dt} = M.p

where M is the matrix.

M=\begin{pmatrix}Ln&D \\ E&Lm\end{pmatrix}

E = \begin{pmatrix} e_{11} & ... & e_{1h} \\ ... & ... & ... \\ e_{k1} & ... & e_{kh} \end{pmatrix}

D= \begin{pmatrix} d_{11} & ... & d_{1k} \\ ... & ... & ... \\ d_{h1} & ... & d_{hk} \end{pmatrix}

Ln= \begin{pmatrix} \sum_j{e_{j1}}& 0 & 0 \\ 0& ... & 0 \\ 0 & 0 & \sum_j{e_{jh}}\end{pmatrix}

Lm= \begin{pmatrix} \sum_i{d_{i1}}& 0 & 0 \\ 0& ... & 0 \\ 0 & 0 &\sum_i{d_{ik}}\end{pmatrix}

so the Matrix M is a (h+k) \times (h+k) matrix. the meaning of matrix E is the excitation matrix. D is de-excitation matrix. the matrix Ln is the matrix of the lost of lower states. the matrix Lm is the lost of the upper state. at equilibrium, the change will be zero. thus, the solution of population is:

M.p = 0

____________________example_____________________

3 upper states and 3 lower states with a circular polarization.

M = \begin{pmatrix}1/2&0&0&1/2&1/3&0 \\ 0&1/2&0&1/2&1/3&1/2 \\ 0&0&0&0&1/3&1/2 \\ 0&0&0&1&0&0 \\ 1/2&0&0&0&1&0 \\ 0&1/2&0&0&0&1 \end{pmatrix}

the solution is

p = ( 0, 0, 1 ,0, 0, 0)

which means, all population stay in n_3, the highest lower state.

 

Advertisements