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This is a question related to the Schrodinger and Heisenberg picture.

Consider a physical system. There are two states- initial and final. Now this is the explanation from Schrodinger and Heisenberg picture.

  1. Schrödinger Picture: States are time dependent. Operators are time independent. Hence initial state is evolved into final state because of the time dependence of states.

  2. Heisenberg Picture: State are time independent. Operators are time dependent. These time dependent operator act on the states and tell the states to evolve from initial to final.

Is this description right? Do the operators acting in the Heisenberg picture represent some interaction? If yes then can I say that interactions are not considered in Schrodinger picture while interactions are present in Heisenberg picture through the operators?

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The operators in the Heisenberg picture are just the usual operators, like $\hat x(t)$ or $\hat p(t)$. They do not represent any particular interaction. In many ways the Heisenberg picture is a little more natural since, in classical mechanics, the position and momentum are time-dependent during the evolution.

Suppose the Hamiltonian is time-dependent, and the time-evolution of the system given by $$ U(t)=e^{-it\hat H/\hbar} $$ Then, in the Schrodinger picture $\vert\psi(t)\rangle_S=U(t)\vert\psi(0)\rangle$ so \begin{align} \langle \psi(t)\vert \hat x_S \vert \phi(t)\rangle &= \langle\psi(0)\vert U^{-1}(t) \hat x_S U(t)\vert\phi(0)\rangle \, ,\\ &= \langle\psi(0)\left[\vert U^{-1}(t) \hat x_S U(t)\right]\vert\phi(0)\rangle\, ,\\ &= \langle\psi(0)\vert \hat x_H(t) \vert\phi(0)\rangle\, ,\\ \end{align} so $\hat x_H(t)$ does not represent any kind of "interaction": it's just another way of computing various matrix elements by displacing the time-dependence from the state to the observable.

The "advantage" is that the evolution of the observable has a form that resembles the classical evolution of the corresponding variable: $$ \frac{d}{dt}\hat {\cal O}_H=\frac{i}{\hbar}[\hat H,\hat {\cal O}_H] $$ for some arbitrary operator $\hat {\cal O}_H$.

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