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When we derive the Dyson series for usage as the time evolution operator in the case of a time dependent Hamiltonian, we start with the equation:

\begin{align}\hat{U}_I(t,t_i) = 1 - \frac{i}{\hbar}\int_{t_i}^t\hat{V}_I(t')\hat{U}_I(t',t_i)dt' \end{align}

And then we iterate the equation further by substituting in $U_I$ in the integral. Eventually, we do $N\longrightarrow \infty$ of these iterations, and we call that the solution. But my question is, even after we iterate this integral an infinity number of times, how can we consider that a solution? Won't the $N$th term in the Dyson Series still depend on $U$? Thus it would still continue to be an integral form of a differential equation, but not a solution itself?

I know the utility of this series is that we could, for example, use the first two or three terms in a perturbation expansion. But all sources I've read seem to imply that the infinite Dyson Expansion is THE unitary time evolution operator for time dependent Hamiltonian, when it just seems to be an integral form of the Schrodinger equation even for $N\longrightarrow \infty$?

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First, numerically one could never actually done the infinite summation.

Second, the "magic" of Dyson series was convergences, that if you did more and more closely step, discrete integration, the solution converges.

The first step may be found by CAF's answer here: Dyson series derivation However, if you sum it smaller and more, it would be more accurate. Read https://ocw.mit.edu/courses/nuclear-engineering/22-51-quantum-theory-of-radiation-interactions-fall-2012/lecture-notes/MIT22_51F12_Ch5.pdf 5.2.1 Dyson Time-ordering operator. If you are looking at convergence, read this https://en.wikipedia.org/wiki/Fixed-point_iteration.

For a more detailed answer, I found http://bolvan.ph.utexas.edu/~vadim/classes/11f/dyson.pdf to be very helpful, at the end of page four "time-ordered exponential".(It's the mathematical basis for the series.)

It looks to me that the main message of the Dyson series was to take care of the deviation of the operator's time dependence. By increasing the order, it resolves the time dependence, and when approach infinity, it's exact. It's basically integrating along with the time iteration in infinite decimal place in $t$. It doesn't matter which operator you are using, as long as it can be expressed as a valid Dyson series, which is important(if you can't write it as a valid Dyson series, this approach would nor apply, it may happen when you encounter some type of infinity). You don't need to worry about the solution unless you are thinking about approximation, Dyson series is an actual numerical solution in integration form.

Other useful information may be found here: http://farside.ph.utexas.edu/teaching/qm/lectures/node74.html or https://www.physicsforums.com/threads/derivation-of-dyson-series.465513/

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    $\begingroup$ But the point of my question is that even so, there still seems to be some dependence of U, on U itself. After doing some more reading my conclusion seems to be the following. For certain perturbation potentials V, we can assume that there exists some unique solution given by a unitary operator U. Under certain classes of operators such as bounded operators, we can then show that this series should indeed converge for any such U. But therefore, calling this a "general" solution seems flawed to me when in general it seems that it can only be used for bounded operators. $\endgroup$ – Electric to be Jul 16 at 16:42
  • $\begingroup$ @Electrictobe Check the reedited posts. bolvan.ph.utexas.edu/~vadim/classes/11f/dyson.pdf actually included proof in the first four pages if you read it carefully. $\endgroup$ – user9976437 Jul 28 at 18:55
  • $\begingroup$ Thanks. I actually found a few other sources as well. It seems this can work if the operator is bounded because this will guarantee that successive terms will get smaller and smaller, and that you may ignore terms of arbitrarily large order. But only for certain classes of operators, which does make sense. I haven’t looked over all of your sources yet but I will when I get the chance! $\endgroup$ – Electric to be Jul 29 at 23:20

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