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I have a question. I have seen Heisenberg's equation of motion for observables: $$\frac{dA}{dt}=\frac{1}{i\hbar}[A,H]+\frac{\partial A}{\partial t}.$$ However if I want to calculate for example the mean time dependence of an operator in the Schrödinger picture then I arrive at the following: $\langle A(t)\rangle=\langle \psi|e^{iHt/\hbar}Ae^{-iHt/\hbar}|\psi\rangle$ One can then use the Baker Hausdorff identity to calculate the time dependence of the operator. My question is this the solution to the Heisenberg's equation of motion?

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  • $\begingroup$ I don’t get what you mean exactly. Left and right hand side of Heisenberg’s equation of motion are operator quantities, while the expectation value as a function of time you wrote down is a scalar quantity. $\endgroup$
    – Milarepa
    Oct 18, 2020 at 13:08
  • $\begingroup$ yes, but in the second equation. One can relate the Heisenberg picture to the Schrödinger picture by transforming the operator with the exponentials of the Baker Hausdorff identity. However the time dependence of an observable is given by solving the first equation. So shouldn't the transformed operator be a solution of the first equation? I hope I am able to make my question understandable. $\endgroup$
    – eeqesri
    Oct 18, 2020 at 13:39
  • $\begingroup$ You are right about the first thing, that’s how you get from the Heisenberg picture to the Schrödinger picture. However, <A(t)> is just a number, not an operator, regardless whether you are reasoning in the Heisenberg or Schrödinger picture, and can therefore not be plugged into Heisenberg’s equation of motion $\endgroup$
    – Milarepa
    Oct 18, 2020 at 13:44
  • $\begingroup$ Yes I understand. But what if I just Took the middle of the bracket. Would that be a solution to the Heisenberg equation of motion? Perhaps I didn't really explain my question ideally. Is there a connection between the Heisenberg equation of motion for the operator and that transform of the operator? $\endgroup$
    – eeqesri
    Oct 18, 2020 at 17:52
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    $\begingroup$ If you neglect the bracket in your second equation, than you have to decide whether considering the Heisenberg or Schrödinger picture, which means to decide whether the two exponentials are left when taking away the brackets or if they go with the bra on the left and the ket on the right. Once you decide in which picture you are working in, the equation of motion modifies accordingly. And you need to solve the latter separately. $\endgroup$
    – Milarepa
    Oct 18, 2020 at 17:57

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