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Suppose we have $$e^{At}e^{Bt}=F(t),$$ where $$A, B$$ - operators that do not commute. Now I need to take the derivative $$dF(t)/dt.$$ In which order do I write the operators?

$$dF(t)/dt = Ae^{At}e^{Bt} + e^{At}Be^{Bt}$$ or $$dF(t)/dt = e^{At}Ae^{Bt} + e^{At}e^{Bt}B~?$$

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They are the same, since any operator commutes with its exponential $$ Ae^{tA} = A\left(1+tA+\frac{1}{2!}t^2A^2+\dots\right) = A+tA^2+\frac{1}{2!}t^2A^3+\dots = \left(1+tA+\frac{1}{2!}t^2A^2+\dots\right)A = e^{tA}A $$ (and in general with any function of it, for similar reasons).

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  • $\begingroup$ Thank you very much, your explanation is very helpful! $\endgroup$ – prividenie Jun 8 at 19:35
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Expanding on Cosmas Zachos' comment Suppose $G(x)$ is some function which has an expression as a Taylor series.

$G(x) = \sum_i g_i x^i$

It can be proven* by induction that

$$ [X,G(X)]=0 $$

This means $[X,e^{Xt}$]=0$

In particular

$$ Ae^{At} = e^{At}A $$ $$ Be^{Bt} = e^{Bt}B $$

so both of your expressions are equal. What is important is to take care that you don't swap any terms involving $A$'s with any terms involving $B$'s. The answer may have been more complicated.

*At least in a physics usage of the word..

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  • $\begingroup$ Thank you very much for your answer! $\endgroup$ – prividenie Jun 8 at 19:36

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