# Does $U_S(t_2,t_1)$ have any meaning in the Heisenberg or interaction pictures?


• Assuming the Hamiltonian is time independent (which you have already done) you get the same operator again, because everything commutes. – Javier Aug 16 '17 at 15:51
• @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. – Quantum spaghettification Aug 16 '17 at 15:59