Thank you, guys. I was really confused about pictures. I have read more about this and now I think this is clearer, and your answers helped me. Sorry for the slightly confused question.
Yes, you are right, they don't comute in general. I confused it with another situation, I'm sorry.
I have already read it in a book, but I didn't understand what it means. The most books say that all operators are time-independent in this picture. So, how can they have an explicit dependence...
So, can I say that, in the Schrodinger picture, every operator is time independent, except (maybe) the hamiltonian? In this case, if I try to compute the expection value of the hamiltonian ##\hat{H}(t)## in the state ##| \psi (t) \rangle##, I get $$\langle \psi (t) | \hat{H}(t) | \psi(t) \rangle...
In the Schrodinger picture, the operators don't change with the time, but the states do. So, what happen if my hamiltonian depend on time? Should I use the others pictures in these cases?
I think I can write the density matrix as $$\rho = \frac{1}{2} ( |R \rangle \langle R | + |L \rangle \langle L | ).$$ The state of a linear polarized light in the direction ##\textbf{a}## can be write as $$|\theta \rangle = \frac{1}{\sqrt{2}} ( e^{-i \theta} |R\rangle + e^{i \theta} |L\rangle...