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In the Schrödinger picture of QM, the time-evolution operator $\hat U_S(t_2,t_1)=e^{-i \hat H_S(t_2-t_1)}$ has the following action: \[\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\bra}[1]{\left<#1\right|} \newcommand{\braoket}[3]{\left<#1\right|#2\left|#3\right>} \ket{\psi(t_2)}=\hat U_S(t_2,t_1)\ket{\psi(t_1)}\] Now to transform to the Heisenberg picture (say) we do the following: $$\hat A_H=\exp(i\hat H_St/\hbar)\hat U_S(t_2,t_1)\exp(-i\hat H_St/\hbar)$$ Does this new operator have any meaning (i.e. is the above a valid expression)? If so what is it's action on states in the Heisenberg picture? What about if we transform to the Interaction picture - will we get $\hat U_I$?

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  • $\begingroup$ Assuming the Hamiltonian is time independent (which you have already done) you get the same operator again, because everything commutes. $\endgroup$ – Javier Aug 16 '17 at 15:51
  • $\begingroup$ @Javier Good point. I have actually just found these notes pa.msu.edu/~mmoore/TDPT.pdf (pg5) which explain the correct transformation is $\hat A_H=\exp(i \hat H_st/\hbar) \hat U_s(t_2,t_1)$ but I doesn't explain why this is different from other types of operators. $\endgroup$ – Quantum spaghettification Aug 16 '17 at 15:59

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