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We know that an operator A in quantum mechanics has time evolution given by Heisenberg equation:

$$ \frac{i}{\hbar}[H,A]+\frac{\partial A}{\partial t}=\frac{d A}{d t} $$

Can we derive from this that $$ A(t)=e^{\frac{i}{\hbar}Ht}A(0)e^{-\frac{i}{\hbar}Ht} \qquad ? $$

L.M.: I added $i/\hbar$ in front of $[H,A]$.

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Yes, we can. However, even though this solution may look very explicit, it may be hard to evaluate - exponentials of operators on an infinite-dimensional space - in the interesting examples, so we often have to use different methods than the "most straightforward" evaluation, anyway. I have added the $i/\hbar$ factor before the commutator. – Luboš Motl Feb 16 '11 at 6:37
Is the question about the differentiability of exponential functions of operators? Or about the definition of exponential functions of operators? Or what can be done with the "formal" solution to get more information about the time dependent behaviour of the system? – Tim van Beek Feb 16 '11 at 10:17
@Luboš Motl thank you for the correction and the comment. +1 – Boy Simone Feb 16 '11 at 18:29
This is in general only possible if $H$ and $A$ do not depend explicitly on time $t$. – Qmechanic Mar 22 '11 at 12:55
up vote 5 down vote accepted

We have to consider the operator, that doesn't explicitly depends on time.

$\frac{\partial A}{\partial t} = 0$

Let's apply commutator formula recursively:

$\frac{d^2 A}{d t^2} = \left(\frac{i}{\hbar}\right)^2[H,[H,A]]$

$\frac{d^3 A}{d t^3} = \left(\frac{i}{\hbar}\right)^3[H,[H,[H,A]]]$


Then we combine those derivatives in a series for $A(t)$

$A(t)=A(0)+\frac{d A}{d t}t+\frac{1}{2!}\frac{d^2 A}{d t^2}t^2+\frac{1}{3!}\frac{d^3 A}{d t^3}t^3+...$


And then you use this formula to arrive at the result:

$e^XYe^{-X} = Y+\frac{1}{1!}[X,Y]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+...$

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So you used Hadamard formula. Thank you. +1 – Boy Simone Feb 16 '11 at 18:34

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