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Lets define a $6\times6$ matrix $M(t)$ with entries $m_{ij}$ as

m11 = (g-r)/2;
m12 = 0;
m13 = (r+2)/2* Exp[2 I J t];
m14 = 0;
m15 = I k E1 Exp[I(J-w)t];
m16 = I k E1 Exp[I(J+w)t];

m21 = 0;
m22 = m11;
m23 = 0;
m24 = (r+2)/2* Exp[-2 I J t];
m25 = -I k E1s Exp[-I(J+w)t];
m26 = -I k E1s Exp[-I(J-w)t];

m31 = (r+2)/2* Exp[2 I J t];
m32 = 0;
m33 = m11;
m34 = 0;
m35 = I k E1 Exp[-I(J+w)t];
m36 = I k E1 Exp[-I(J-w)t]; 

m41 = 0;
m42 = m31;
m43 = 0;
m44 = m11;
m45 = -I k E1s Exp[I(J-w)t]; 
m46 = -I k E1s Exp[-I(J+w)t];

m51 = I k E1s Exp[-I(J-w)t];
m52 = I k E1 Exp[I(J+w)t];
m53 = I k E1s Exp[I(J+w)t]; 
m54 = I k E1 Exp[-I(J-w)t];
m55 = -gamma
m56 = 0;

m61 = -I k E1s Exp[-I(J+w)t];
m62 = -I k E1 Exp[I(J-w)t];
m63 = -I k E1s Exp[I(J-w)t]; 
m64 = I k E1 Exp[-I(J+w)t];
m65 = 0;
m66 = -gamma;

where

 w = 22.8; k = 5*10^-5; gamma = 0.037; J = w/2; E0 = 2.5*10^5; d = -3 J; g = 0.5; r = 1;
 E1 = (I E0/3.2)(1/(d+J) Exp[-I J t]+ 1/(d-J) Exp[I J t] - 2 d/(d^2-J^2)Exp[I d t]);
 E1s = (I E0/3.2)(1/(d+J) Exp[I J t]+ 1/(d-J) Exp[-I J t] - 2 d/(d^2-J^2)Exp[-I d t]);

One can formally write the following expression as a $6\times6$ matrix with entries $d_{ij}(s,T)$: \begin{equation} \exp{\left[\int_{s}^{T} dt\; M(t)\right]}= \begin{pmatrix} d_{11}(s,T) & d_{12}(s,T) & \cdots & d_{16}(s,T) \\ d_{21}(s,T) & d_{22}(s,T) & \cdots & d_{26}(s,T) \\ \vdots & \vdots & \ddots & \vdots \\ d_{61}(s,T) & d_{62}(s,T) & \cdots & d_{66}(s,T) \end{pmatrix} \end{equation}

And I want to find \begin{equation} \int_{0}^{T} ds\; d_{12}(s,T)\,d_{23}(s,T) \,E(s) \,e^{-iJs} \end{equation}

MarixExp for this problem is not fast and efficient, so I thought about the following trick:

\begin{equation} \exp{\left[\int_{s}^{T} dt\; M(t)\right]} \approx \exp{\left[\sum_{i = 1}^{N} \Delta t_i\; M(t_i)\right]}; \quad \Delta t_i = (T-s)/N = h \end{equation} \begin{equation} = \exp{\left[M(t_1) h\right]} \exp{\left[M(t_{2}) h\right]} ... \exp{\left[M(t_N) h\right]}= \displaystyle\prod_{i = N}^{1} M(t_i) h \end{equation}

The reverse order for the limits is due to time ordering. If $h$ is small enough (hence, N large, say $N = 1000$), one can write $\prod \exp{\left[M(t_i) h\right]}\approx \prod (1+ M(t_i) h)$. Although $T$ is a number (say, $T = 50$), $s$ is an integration variable, and I have no idea how to calculate the product and then integrate: $$\int_{0}^{T} ds\; d_{12}(s,T)\,d_{23}(s,T) \,E(s) \,e^{-iJs}$$.

In other words, I need two specific entries of the matrix ($d_{12}$ and $d_{23}$), and then numerically integrate. Do you have any idea about the product when one limit is itself an integration variable?

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  • 2
    $\begingroup$ D = -3 do not use D - that's the derivative function, use d instead. Avoid capital letters in your variables. $\endgroup$ – flinty Jun 14 at 0:26
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    $\begingroup$ @flinty I fixed it. $\endgroup$ – Saeid Jun 14 at 1:30