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There are (at least) two ways to perturbatively solve a matrix initial value problem: the Dyson expansion and the Magnus expansion

To be explicit, suppose you're solving for a (interaction-picture) density matrix $\rho(t)$ at time $t$ characterized by a Hamiltonian $H = H_0 + H_{\mathrm{interaction}}$, where your initial state is $\rho(0)\equiv \rho_0$. The Dyson expansion gives a perturbative solution: $$ \rho(t) = \rho_0 - i \int_0^t ds\ \left[ H_{\mathrm{I}}(s), \rho_0 \right] + (-i)^2 \int_0^t ds \int_0^s dr\ \left[ H_{\mathrm{I}}(s), \left[ H_{\mathrm{I}}(r), \rho_0 \right] \right] + \ldots $$

Where $H_{\mathrm{I}}(t)=e^{+iH_0 t} H_{\mathrm{interaction}} e^{-iH_0 t}$ in the interaction picture.

I have heard that the Magnus expansion is useful in that it preserves the symplectic form of the time evolution at every order. I heard that this means that at every order, the volume of phase space is preserved.

What does this mean precisely?

(In contrast, the utility of the Dyson series is that probabilities are preserved at every order, meaning the trace of each contribution order-by-order remains $1$, this I can understand)

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  • $\begingroup$ $\uparrow$ Heard where? $\endgroup$
    – Qmechanic
    Commented Sep 18, 2018 at 17:56
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    $\begingroup$ Isn't the expansion you wrote down is Dyson expansion in contrast to magnus expansion which should be something else. $\endgroup$
    – Sunyam
    Commented Sep 18, 2018 at 17:57
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    $\begingroup$ I think the OP has it backward in that the Dyson expansion is not unitary when truncated to finite order, while the Magnus expansion always is. See for example: personales.upv.es/serblaza/2010EJP.pdf $\endgroup$
    – Rococo
    Commented Jun 25, 2020 at 14:15
  • $\begingroup$ Related: physics.stackexchange.com/q/736194/226902. Example of use of Magnus expansion: physics.stackexchange.com/a/118877/226902 $\endgroup$
    – Quillo
    Commented Nov 12, 2022 at 15:47

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