While studying the Ehrenfest theorem or Heisenberg equation of motion in quatum mechanics, a question has come to my mind, which is written as follows: Suppose, I have a classical equation of motion of an electorn such that:
$m_e\frac{d^2x}{dt^2}=-eE_0\sin (\omega t)$
Where, $eE_0\sin (\omega t)$ is the time dependent external force acting on the electron, while placed in an oscillating electric field.The solution of this differentil equation $x(t)$ is already known. Can we represent this equation inserting the expectation value of $\langle x(t) \rangle$ in the place of $x(t)$ i.e. to say
$$m_e\frac{d^2\langle x(t)\rangle }{dt^2}=-eE_0\sin (\omega t)$$ Where, using the Heisenberg picture, it can be written that $$\frac{d\langle x(t)\rangle }{dt}=\frac{1}{i\hbar}\langle[x(t),H]\rangle +\langle \frac{\partial x(t)}{\partial t}\rangle$$ If the hamiltonian of the system can be considered with an interaction term such as: $$H(t)=\frac{p^2}{2m_e}+eE_0 \sin (\omega t) x(t)$$ does the above differential equation make any sense? If so, then what is its solution? Would you please tell me where is my conceptual problem lies or what is the exact way to find this kind of phenomena. Thanking you...
I corrected your hamiltonian to agree with the classical equation of motion you posited. You are badly misreading the Ehrenfest theorem, which, as you know, has the same canonical equations of motion for the Heisenberg and the Schrödinger pictures, $$ \frac{d\langle x\rangle }{dt}= \langle p\rangle /m ,\\ \frac{d\langle p\rangle }{dt}= -eE_0 \sin (\omega t) , $$ by the linearity of the potential in x.
Combining the two, you end up with the classical equation of motion, a very exceptional situation, $$ m\frac{d^2\langle x\rangle }{dt^2}= -eE_0 \sin (\omega t). $$ If you solved the classical equation, you solved this one as well, $$ \langle x\rangle= c_0 + c_1t+\frac{eE_0}{m\omega^2}\sin (\omega t), $$ but perhaps you need to tweak the ICs you opt for.