# Could you explain this flow of calculation?

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.

• There is not much to unterstand. The 2nd equation is iteratively substituted into itself (i.e., $w(t_1)$ on the rhs of the 2nd equation is substituted with the integral formula for $w(t)$, etc.) – Norbert Schuch Oct 6 '15 at 5:50
• Yup, the 2nd equation is substituted into itself an infinite number of times. LOL! Kind of mind-boggling to look at the equation that resulted from that. Also, it brings up the question of whether this resulting infinite series actually converges. Of course, we physicists tend to be a bit carefree about questions of that nature and leave it to mathematicians to worry about those details. – Samuel Weir Oct 6 '15 at 5:53
• Errr! I have not a single clue to what you are saying. Please do it in detail. – TBBT Oct 6 '15 at 5:58
• Look at the second equation above. It gives an expression for the function w(t). But that expression itself has the function w(t) in it. So take the expression for the function w(t) and substitute it for the function w(t) in the expression. Repeat an infinite number of times. – Samuel Weir Oct 6 '15 at 6:05
• Wait a minute, please. – TBBT Oct 6 '15 at 6:09

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 $$w(t)=w(0)+\sum_{n=1}^{\infty}\left(-\frac{i}{\hbar}\right)^n\int_{0}^{t}\mathrm{d}t_1\int_{0}^{t_1}\mathrm{d}t_2\dots\times\int_{0}^{t_{n-1}}\mathrm{d}t_n\left[V(t_1),\left[V(t_2),\dots\left[V(t_n),w(0)\right]\right]\right].$$
• I think the convergence of the series is a separate question. The solution is just a formal solution with some flavor of perturbation theory. It may or may not converge determined by the commutator of $\left[V(t_n),\omega(0) \right]$. – Xiaodong Qi Oct 6 '15 at 6:12
• I don't think that $V(t_n)$ and $w(0)$ commute. – TBBT Oct 6 '15 at 6:17