# How to differentiate exponentials of operators?

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

They are the same, since any operator commutes with its exponential \begin{align} 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 \end{align} (and, in general, any operator $$A$$ commutes with every function $$f(A)$$ of it, for similar reasons).

• Thank you very much, your explanation is very helpful! Commented Jun 8, 2019 at 19:35

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..

• Thank you very much for your answer! Commented Jun 8, 2019 at 19:36