Derivative of Time Evolution Operator: Exp(-iHt)

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SUMMARY

The discussion focuses on the differentiation of the time evolution operator, represented as exp(-iHt), with respect to time (t). The derivative is established as ##\frac{d}{dt}e^{-iHt}=-iHe^{-iHt}##, leveraging the commutation property of the Hamiltonian operator (H) with the exponential operator. For non-commuting operators, such as in the case of ##e^{At}e^{Bt}##, the product rule is applied, resulting in the expression ##\frac{d}{dt}\left(e^{At}e^{Bt}\right)=Ae^{At}e^{Bt}+e^{At}Be^{Bt}##.

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For the time evolution operator:

exp(-iHt)

How do I take the derivative of an operator like this keeping the order correct? I mean I of course know how to differentiate an exponential function, but this is the exponential of an operator.
 
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You mean differentiating that with respect to ##t##? It's just ##\frac{d}{dt}e^{-iHt}=-iHe^{-iHt}=-ie^{-iHt}H##, because ##H## commutes with ##e^{-iHt}##. (you can see this by expanding the exponential to power series and noting that ##H## commutes with any power of itself)

If the problem were to differentiate ##e^{At}e^{Bt}##, where ##A## and ##B## are non-commuting operators, you would have to be more careful:

##\frac{d}{dt}\left(e^{At}e^{Bt}\right)=\left(\frac{d}{dt}e^{At}\right)e^{Bt}+e^{At}\left(\frac{d}{dt}e^{Bt}\right)=Ae^{At}e^{Bt}+e^{At}Be^{Bt}##.
 

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