Rate Equation

In pump prove time resolved spectroscopy, we assume reaction occurs just after pump pulse. So, for 1st order dynamics, what we should to solve is

\[\begin{equation*} \mathbf{y}'(t) = \begin{cases} 0& \text{if $t < 0$}, \\ A\mathbf{y}(t)& \text{if $t>0$}. \end{cases} \end{equation*}\]

with, \(\mathbf{y}(0)=\mathbf{y}_0\). The solution of above first order equation is given by

\[\begin{equation*} \mathbf{y}(t) = \begin{cases} \exp\left(At\right) \mathbf{y}_0 & \text{if $t\geq 0$} \\ \mathbf{0} \end{cases} \end{equation*}\]

, where \(\exp\left(At\right)\) is the matrix exponential defined as

\[\begin{equation*} \exp\left(At\right) = 1 + tA + \frac{1}{2!} t^2 A^2 + \frac{1}{3!} t^3 A^3 + \dotsc \end{equation*}\]

\(\mathbf{0}\) means, at \(t<0\) (i.e. before laser irrediation), there is no excited state species.

Suppose that the rate equation matrix \(A\) is diagonalizable. In general, it cannot be diagonalizable. Then we can write

\[\begin{equation*} A = V \Lambda V^{-1} \end{equation*}\]

, where \(V\) is eigen matrix of \(A\) and \(\Lambda = \mathrm{Diag}\left(\lambda_1,\dotsc,\lambda_n\right)\).

Then,

\[\begin{align*} \mathbf{y}(t) &= \exp\left(At\right) \mathbf{y}_0 \\ &= V \exp\left(\Lambda t \right) V^{-1} \mathbf{y}_0 \\ &= V \mathrm{diag}\left(\exp\left(\lambda_1 t\right),\dotsc,\exp\left(\lambda_n t \right)\right) V^{-1} \mathbf{y}_0 \end{align*}\]

Define \(\mathbf{c}\) as \(V\mathbf{c} = \mathbf{y}_0\) then

\[\begin{equation*} \mathbf{y}(t) = \sum_i c_i \exp\left(\lambda_i t\right) \mathbf{v}_i \end{equation*}\]

To model experimentally observed population, we need to convolve our model population \(\mathbf{y}(t)\) to instrumental response function \(\mathrm{IRF}\). Then we can model observed population \(\mathbf{y}_{obs}(t)\) as

\[\begin{equation*} \mathbf{y}_{obs}(t) = \sum_i c_i (\exp *_h {IRF})(\lambda_i t) \mathbf{v}_i \end{equation*}\]

, where \(*_h\) is the half convolution operator defined as

\[\begin{equation*} (f *_h g)(t) = \int_{0}^{\infty} f(x)g(t-x) \mathrm{d} x \end{equation*}\]

Relation between observed exponential component and species in rate equation

Suppose that the rate equation matrix \(A\) is lower triangular and assume that each diagonal element of \(A\) is different. Then \(A\) is diagonalizable and its eigenvalue \(\lambda_i = A_{(i,i)}.\) Define observed exponential component as

\[\begin{equation*} \mathbf{exp}_{(obs, i)} = (\exp *_h {IRF})(A_{(i,i)} t) \end{equation*}\]

Next define scaled eigen matrix \(V'\) of rate equation matrix \(A\) as

\[\begin{equation*} V' = [c_j V_{(i,j)}]_{(i,j)} \end{equation*}\]

Then observed exponential component \(\mathbf{exp}_{obs}\) and population of species in rate equation \(\mathbf{y}_{obs}\) satisfy following relation.

\[\begin{align*} \mathbf{y}_{obs} &= V' \mathbf{exp}_{obs} \\ \mathbf{exp}_{obs} &= V'^{-1} \mathbf{y}_{obs} \end{align*}\]

So, if one finds weigh vector \(\mathbf{w}\) from time delay scan fitting,

\[\begin{align*} {signal}(t) &= \sum_{i} w_i \mathbf{exp}_{(obs, i)} \\ &= \mathbf{w}^T \mathbf{exp}_{obs} \end{align*}\]

Then one can deduce chemical or physically meaningful difference absorption coefficient \(\Delta \mathbf{A}\) as

\[\begin{equation*} \Delta \mathbf{A} = V'^{-T} \mathbf{w} \end{equation*}\]

Above equation only holds when number of observed exponential decay component and excited species in rate equation are same.