- #1
PRB147
- 122
- 0
- TL;DR
- I encountered difficulty in deriving the the second order nonlinear optical signal from the book of Mukamel "Principles of Nonlinear Optical Spectroscopy"; Two different approaches results in different results, please help.
P119, Mukamel's book "Principles of Nonlinear Optical Spectroscopy"
Eq.(5.21) in Liouville space
$$ S^{(2)}(t_2,t_1)=\left(\frac{i}{\hbar}\right)^2 \left\langle \left\langle V\left|\mathscr{G}(t_2)\mathscr{V}\mathscr{G}(t_1)\mathscr{V}\right|\rho(-\infty) \right\rangle\right\rangle $$
in Hilbert space
Eq.(5.22) $$S^{(2)}(t_2,t_1)=\left(\frac{i}{\hbar}\right)^2 \left\langle \left[\left[V(t_2+t_1),V(t_1)\right],V(0)\right]\rho(-\infty)\right\rangle $$
Eq.(5.24) $$Q_1(t_2,t_1)=\left\langle V(t_1+t_2)V(t_1)V(0)\rho(-\infty)\right\rangle; Q_1(t_2,t_1)=-\left\langle V(t_1)V(t_1+t_2)V(0)\rho(-\infty)\right\rangle$$
I derived the Eq (6.18) in Page 150 in Liouville space approach and in Hilbert space:
In Liouville space, The drivation as follows:
$$
\begin{aligned}
~&S^{(2)}(t_2,t_1)=\left(\frac{i}{\hbar}\right)^2 \left\langle \left\langle V\left|\mathscr{G}(t_2)\mathscr{V}\mathscr{G}(t_1)\mathscr{V}\right|\rho(-\infty) \right\rangle\right\rangle \\
&=\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,2\cdots, 5} \left\langle \left\langle V| a_1b_1\right\rangle\right\rangle\rangle\left\langle \left\langle a_1b_1\left|\mathscr{G}(t_2)\right|a_2b_2\right\rangle\right\rangle
\left\langle\left\langle a_2b_2 \left|\mathscr{V}\right|a_3b_3 \right\rangle\right\rangle \left\langle \left\langle a_3b_3 \left|\mathscr{G}(t_1)\right|a_3b_3\right\rangle\right\rangle \\ &
\left\langle \left\langle a_3b_3\left|\mathscr{V}\right|a_5b_5\right\rangle\right\rangle\left\langle \left\langle a_5b_5\left|\rho(-\infty) \right.\right\rangle\right\rangle \\&
=\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\times\left[\mu_{a_1a_3}\delta_{b_1b_3}-\mu_{b_3b_1}\delta_{a_1a_3}\right]\times\left[\mu_{a_3a_5}\delta_{b_3b_5}-\mu_{b_5b_3}\delta_{a_3a_5}\right]\rho(a_5)\delta_{a_5b_5} \\ &
\end{aligned}
$$
11 (The product of the first term in the first parentheses and the first term in the second parentheses in the above equation)
$$
\begin{aligned}
~& \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\times\left[\mu_{a_1a_3}\delta_{b_1b_3}\right]\times\left[\mu_{a_3a_5}\delta_{b_3b_5}\right]\rho(a_5)\delta_{a_5b_5} \\
& =\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5}\rho(a_5) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\times\left[\mu_{a_1a_3}\mu_{a_3a_5}\right]\left(\delta_{b_1b_3}\delta_{b_3b_5}\delta_{a_5b_5}\right) \\
&= \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1}^{a_3a_5}\rho(a_5) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_1} (t_1)
\times\left[\mu_{a_1a_3}\mu_{a_3a_5}\right]\left(\delta_{b_1a_5}\right) \\
&= \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1}^{a_3}\rho(b_1) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_1} (t_1)
\times\left[\mu_{a_1a_3}\mu_{a_3a_5}\right] \\
&\xlongequal[b_1\rightarrow a,a_1\rightarrow b]{a_3\rightarrow c}\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{abc}\rho(a) \left[\mu_{ab}\mu_{bc}\mu_{ca}\right] I_{ba}(t_2) I_{ca} (t_1)
\end{aligned}
$$
12(The product of the first term in the first parentheses and the second term in the second parentheses
$$
\begin{aligned}
~& \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1) \times\left[\mu_{a_1a_3}\delta_{b_1b_3}\right]\left[-\mu_{b_5b_3}\delta_{a_3a_5}\right]\rho(a_5)\delta_{a_5b_5} \\ &
=-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \rho(a_5)\mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1) \times\left[\mu_{a_1a_3}\mu_{b_5b_3}\right]\left[\delta_{b_1b_3}\delta_{a_3a_5}\delta_{a_5b_5}\right] \\ &
-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{(b_5)a_ib_i}^{i=1,3} \rho(a_3)\mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1) \times\left[\mu_{a_1a_3}\mu_{b_5b_3}\right]\left[\delta_{b_1b_3}\delta_{a_3b_5}\right] \\ &
-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1}^{a_3} \rho(a_3) I_{a_1b_1}(t_2) I_{a_3b_1} (t_1) \times\left[\mu_{b_1a_1}\mu_{a_1a_3}\mu_{a_3b_1}\right] \\ &
\xlongequal[b_1\rightarrow a,a_1\rightarrow b]{a_3\rightarrow c}-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{abc}\rho(c) I_{ba}(t_2) I_{ca} (t_1) \times\left[\mu_{b_1a_1}\mu_{a_1a_3}\mu_{a_3b_1}\right]
\end{aligned}
$$
21 (The product of the second term in the first parentheses and the first term in the second parentheses
$$
\begin{aligned}
~& \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[-\mu_{b_3b_1}\delta_{a_1a_3}\right]\times\left[\mu_{a_3a_5}\delta_{b_3b_5}\rho(a_5)\delta_{a_5b_5}\right] \\ &
=-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \rho(a_5) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[\mu_{b_3b_1} \mu_{a_3a_5}\right]\times\left[\delta_{a_1a_3}\delta_{b_3b_5}\delta_{a_5b_5} \right]\\ &
=-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{(a_5)a_ib_i}^{i=1,3} \rho(a_5) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[\mu_{b_3b_1} \mu_{a_3a_5}\right]\times\left[\delta_{a_1a_3}\delta_{b_3a_5} \right]\\&
=-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1b_3}\rho(b_3) \left[\mu_{b_1a_1}\mu_{b_3b_1} \mu_{a_1b_3}\right] I_{a_1b_1}(t_2) I_{a_1b_3} (t_1)
\\&
\xlongequal[b_1\rightarrow a,a_1\rightarrow b]{b_3\rightarrow c}-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{abc}\rho(c) \left[\mu_{ab}\mu_{bc}\mu_{ca} \right]
I_{ba}(t_2) I_{bc} (t_1)
\end{aligned}
$$
22(The product of the second term in the first parentheses and the second term in the second parentheses
$$
\begin{aligned}
~& \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[-\mu_{b_3b_1}\delta_{a_1a_3}\right]\times \left[-\mu_{b_5b_3}\delta_{a_3a_5}\right]\rho(a_5)\delta_{a_5b_5} \\ &
=\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \rho(a_5) I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[\mu_{b_1a_1}\mu_{b_3b_1}\mu_{b_5b_3}\right]\times\left(\delta_{a_5b_5}\delta_{a_1a_3}\delta_{a_3a_5}\right) \\&
=\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1b_3b_5} \rho(a_1) I_{a_1b_1}(t_2) I_{a_1b_3} (t_1)
\left[\mu_{b_1a_1}\mu_{b_3b_1}\mu_{b_5b_3}\right]\times\left(\delta_{a_1b_5}\right) \\&
=\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1b_3} \rho(a_1) I_{a_1b_1}(t_2) I_{a_1b_3} (t_1)
\left[\mu_{b_1a_1}\mu_{b_3b_1}\mu_{a_1b_3}\right] \\&
\xlongequal[a_1\rightarrow a,b_3\rightarrow b]{b_1\rightarrow c}\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{abc} \rho(a) I_{ac}(t_2) I_{ab} (t_1)
\left[\mu_{ab}\mu_{bc}\mu_{ca}\right]
\end{aligned}
$$
But I found that the equations obtained by Liouville space approach is different from the Hilbert space by directly finding the trace of Eq.(5.24). and my Liouville results are different from Mukamel's results in Eq.(6.18)
$$
\begin{aligned}
& Q_1(t_2,t_1)=\left\langle V(t_1+t_2)V(t_1)V(0)\rho(-\infty)\right\rangle=\sum\limits_{abcd} \left \langle a| V(t_1+t_2)|b \right\rangle
\left\langle b|V(t_1) |c\right\rangle \left\langle c|V(0) |d\right \rangle \left\langle d| \rho(-\infty)| a\right\rangle \\ &
=\sum\limits_{abc} \rho(a) I_{ba}(t_1+t_2) I_{cb} (t_1) \mu_{ab}\mu_{bc}\mu_{ca}
\end{aligned}
$$
Who can give some suggestions and hints, many manty thx!
Eq.(5.21) in Liouville space
$$ S^{(2)}(t_2,t_1)=\left(\frac{i}{\hbar}\right)^2 \left\langle \left\langle V\left|\mathscr{G}(t_2)\mathscr{V}\mathscr{G}(t_1)\mathscr{V}\right|\rho(-\infty) \right\rangle\right\rangle $$
in Hilbert space
Eq.(5.22) $$S^{(2)}(t_2,t_1)=\left(\frac{i}{\hbar}\right)^2 \left\langle \left[\left[V(t_2+t_1),V(t_1)\right],V(0)\right]\rho(-\infty)\right\rangle $$
Eq.(5.24) $$Q_1(t_2,t_1)=\left\langle V(t_1+t_2)V(t_1)V(0)\rho(-\infty)\right\rangle; Q_1(t_2,t_1)=-\left\langle V(t_1)V(t_1+t_2)V(0)\rho(-\infty)\right\rangle$$
I derived the Eq (6.18) in Page 150 in Liouville space approach and in Hilbert space:
In Liouville space, The drivation as follows:
$$
\begin{aligned}
~&S^{(2)}(t_2,t_1)=\left(\frac{i}{\hbar}\right)^2 \left\langle \left\langle V\left|\mathscr{G}(t_2)\mathscr{V}\mathscr{G}(t_1)\mathscr{V}\right|\rho(-\infty) \right\rangle\right\rangle \\
&=\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,2\cdots, 5} \left\langle \left\langle V| a_1b_1\right\rangle\right\rangle\rangle\left\langle \left\langle a_1b_1\left|\mathscr{G}(t_2)\right|a_2b_2\right\rangle\right\rangle
\left\langle\left\langle a_2b_2 \left|\mathscr{V}\right|a_3b_3 \right\rangle\right\rangle \left\langle \left\langle a_3b_3 \left|\mathscr{G}(t_1)\right|a_3b_3\right\rangle\right\rangle \\ &
\left\langle \left\langle a_3b_3\left|\mathscr{V}\right|a_5b_5\right\rangle\right\rangle\left\langle \left\langle a_5b_5\left|\rho(-\infty) \right.\right\rangle\right\rangle \\&
=\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\times\left[\mu_{a_1a_3}\delta_{b_1b_3}-\mu_{b_3b_1}\delta_{a_1a_3}\right]\times\left[\mu_{a_3a_5}\delta_{b_3b_5}-\mu_{b_5b_3}\delta_{a_3a_5}\right]\rho(a_5)\delta_{a_5b_5} \\ &
\end{aligned}
$$
11 (The product of the first term in the first parentheses and the first term in the second parentheses in the above equation)
$$
\begin{aligned}
~& \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\times\left[\mu_{a_1a_3}\delta_{b_1b_3}\right]\times\left[\mu_{a_3a_5}\delta_{b_3b_5}\right]\rho(a_5)\delta_{a_5b_5} \\
& =\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5}\rho(a_5) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\times\left[\mu_{a_1a_3}\mu_{a_3a_5}\right]\left(\delta_{b_1b_3}\delta_{b_3b_5}\delta_{a_5b_5}\right) \\
&= \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1}^{a_3a_5}\rho(a_5) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_1} (t_1)
\times\left[\mu_{a_1a_3}\mu_{a_3a_5}\right]\left(\delta_{b_1a_5}\right) \\
&= \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1}^{a_3}\rho(b_1) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_1} (t_1)
\times\left[\mu_{a_1a_3}\mu_{a_3a_5}\right] \\
&\xlongequal[b_1\rightarrow a,a_1\rightarrow b]{a_3\rightarrow c}\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{abc}\rho(a) \left[\mu_{ab}\mu_{bc}\mu_{ca}\right] I_{ba}(t_2) I_{ca} (t_1)
\end{aligned}
$$
12(The product of the first term in the first parentheses and the second term in the second parentheses
$$
\begin{aligned}
~& \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1) \times\left[\mu_{a_1a_3}\delta_{b_1b_3}\right]\left[-\mu_{b_5b_3}\delta_{a_3a_5}\right]\rho(a_5)\delta_{a_5b_5} \\ &
=-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \rho(a_5)\mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1) \times\left[\mu_{a_1a_3}\mu_{b_5b_3}\right]\left[\delta_{b_1b_3}\delta_{a_3a_5}\delta_{a_5b_5}\right] \\ &
-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{(b_5)a_ib_i}^{i=1,3} \rho(a_3)\mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1) \times\left[\mu_{a_1a_3}\mu_{b_5b_3}\right]\left[\delta_{b_1b_3}\delta_{a_3b_5}\right] \\ &
-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1}^{a_3} \rho(a_3) I_{a_1b_1}(t_2) I_{a_3b_1} (t_1) \times\left[\mu_{b_1a_1}\mu_{a_1a_3}\mu_{a_3b_1}\right] \\ &
\xlongequal[b_1\rightarrow a,a_1\rightarrow b]{a_3\rightarrow c}-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{abc}\rho(c) I_{ba}(t_2) I_{ca} (t_1) \times\left[\mu_{b_1a_1}\mu_{a_1a_3}\mu_{a_3b_1}\right]
\end{aligned}
$$
21 (The product of the second term in the first parentheses and the first term in the second parentheses
$$
\begin{aligned}
~& \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[-\mu_{b_3b_1}\delta_{a_1a_3}\right]\times\left[\mu_{a_3a_5}\delta_{b_3b_5}\rho(a_5)\delta_{a_5b_5}\right] \\ &
=-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \rho(a_5) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[\mu_{b_3b_1} \mu_{a_3a_5}\right]\times\left[\delta_{a_1a_3}\delta_{b_3b_5}\delta_{a_5b_5} \right]\\ &
=-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{(a_5)a_ib_i}^{i=1,3} \rho(a_5) \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[\mu_{b_3b_1} \mu_{a_3a_5}\right]\times\left[\delta_{a_1a_3}\delta_{b_3a_5} \right]\\&
=-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1b_3}\rho(b_3) \left[\mu_{b_1a_1}\mu_{b_3b_1} \mu_{a_1b_3}\right] I_{a_1b_1}(t_2) I_{a_1b_3} (t_1)
\\&
\xlongequal[b_1\rightarrow a,a_1\rightarrow b]{b_3\rightarrow c}-\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{abc}\rho(c) \left[\mu_{ab}\mu_{bc}\mu_{ca} \right]
I_{ba}(t_2) I_{bc} (t_1)
\end{aligned}
$$
22(The product of the second term in the first parentheses and the second term in the second parentheses
$$
\begin{aligned}
~& \theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \mu_{b_1a_1} I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[-\mu_{b_3b_1}\delta_{a_1a_3}\right]\times \left[-\mu_{b_5b_3}\delta_{a_3a_5}\right]\rho(a_5)\delta_{a_5b_5} \\ &
=\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_ib_i}^{i=1,3,5} \rho(a_5) I_{a_1b_1}(t_2) I_{a_3b_3} (t_1)
\left[\mu_{b_1a_1}\mu_{b_3b_1}\mu_{b_5b_3}\right]\times\left(\delta_{a_5b_5}\delta_{a_1a_3}\delta_{a_3a_5}\right) \\&
=\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1b_3b_5} \rho(a_1) I_{a_1b_1}(t_2) I_{a_1b_3} (t_1)
\left[\mu_{b_1a_1}\mu_{b_3b_1}\mu_{b_5b_3}\right]\times\left(\delta_{a_1b_5}\right) \\&
=\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{a_1b_1b_3} \rho(a_1) I_{a_1b_1}(t_2) I_{a_1b_3} (t_1)
\left[\mu_{b_1a_1}\mu_{b_3b_1}\mu_{a_1b_3}\right] \\&
\xlongequal[a_1\rightarrow a,b_3\rightarrow b]{b_1\rightarrow c}\theta(t_2)\theta(t_1)\left(\frac{i}{\hbar}\right)^2\sum\limits_{abc} \rho(a) I_{ac}(t_2) I_{ab} (t_1)
\left[\mu_{ab}\mu_{bc}\mu_{ca}\right]
\end{aligned}
$$
But I found that the equations obtained by Liouville space approach is different from the Hilbert space by directly finding the trace of Eq.(5.24). and my Liouville results are different from Mukamel's results in Eq.(6.18)
$$
\begin{aligned}
& Q_1(t_2,t_1)=\left\langle V(t_1+t_2)V(t_1)V(0)\rho(-\infty)\right\rangle=\sum\limits_{abcd} \left \langle a| V(t_1+t_2)|b \right\rangle
\left\langle b|V(t_1) |c\right\rangle \left\langle c|V(0) |d\right \rangle \left\langle d| \rho(-\infty)| a\right\rangle \\ &
=\sum\limits_{abc} \rho(a) I_{ba}(t_1+t_2) I_{cb} (t_1) \mu_{ab}\mu_{bc}\mu_{ca}
\end{aligned}
$$
Who can give some suggestions and hints, many manty thx!
Last edited: