Could you explain this flow of calculation?

Question

I am reading this book, Quantum Optics by Walls and Milburn. I am working on Chapter 6 which is about the Stochastic Methods. I don't understand a calculation in this chapter.

Let $w(t)$ be the total density operator and $V(t)$ is the Interaction Hamiltonian of the system. The equation of motion is, $$ \frac{dw(t)}{dt}=-\frac{i}{\hbar}\left[V(t),w(t)\right]. $$ Integrating this equation, we obtain $$ w(t)=w(0)-\frac{i}{\hbar}\int_{0}^{t}\mathrm{d}t_1\left[V(t_1),w(t_1)\right]. $$ Iterating this solution, we find $$ w(t)=w(0)+\sum_{n=1}^{\infty}\left(-\frac{i}{\hbar}\right)^n\int_{0}^{t}\mathrm{d}t_1\int_{0}^{t_1}\mathrm{d}t_2\dots\times\int_{0}^{t_{n-1}}\mathrm{d}t_n\left[V(t_1),\left[V(t_2),\dots\left[V(t_n),w(0)\right]\right]\right]. $$

Could you explain the last calculation please? I don't understand about that integral expansion. Also, if you could prove it to me whether this infinite series is actually converges or not.

Answer 1

The last step is just substitute the solution of $\omega(t)$ into the equation of itself. For details, you do the following:

Insert $$ w(t)=w(0)-\frac{i}{\hbar}\int_{0}^{t}\mathrm{d}t_1\left[V(t_1),w(t_1)\right]$$ when $t=t_1$ on its right side, one can obtain \begin{align} w(t) &=w(0)-\frac{i}{\hbar}\int_{0}^{t}\mathrm{d}t_1\left[V(t_1), \omega(0)\right] + \left(-\frac{i}{\hbar}\right)^2 \int_{0}^{t}\mathrm{d}t_1\int_{0}^{t_1}\mathrm{d}t_2\left[V(t_1), \left[V(t_2),\omega(t_2) \right]\right] \end{align} Now, you plug in $$ w(t_2)=w(0)-\frac{i}{\hbar}\int_{0}^{t_2}\mathrm{d}t_3\left[V(t_3),w(t_3)\right]$$ to the equation above, you will get \begin{align} w(t) &=w(0)-\frac{i}{\hbar}\int_{0}^{t}\mathrm{d}t_1\left[V(t_1), \omega(0)\right] + \left(-\frac{i}{\hbar}\right)^2 \int_{0}^{t}\mathrm{d}t_1\int_{0}^{t_1}\mathrm{d}t_2\left[V(t_1), \left[V(t_2),\omega(0) \right]\right] \\ &\quad + \left(-\frac{i}{\hbar}\right)^3 \int_{0}^{t}\mathrm{d}t_1\int_{0}^{t_1}\mathrm{d}t_2 \int_0^{t_2}\mathrm{d}t_3\left[V(t_1), \left[V(t_2),\left[V(t_3),\omega(t_3) \right] \right]\right] \end{align}

By going through this process for infinite times, Iterating this solution, and ignoring the last term regarding $\omega(t_\infty)$, you can find the last step $$ w(t)=w(0)+\sum_{n=1}^{\infty}\left(-\frac{i}{\hbar}\right)^n\int_{0}^{t}\mathrm{d}t_1\int_{0}^{t_1}\mathrm{d}t_2\dots\times\int_{0}^{t_{n-1}}\mathrm{d}t_n\left[V(t_1),\left[V(t_2),\dots\left[V(t_n),w(0)\right]\right]\right]. $$