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When deriving the electric dipole form of the semiclassical hamiltonian from the minimal coupling form, we define a new state vector as:

$$\psi_{old}=e^{i\textbf{A}\cdot\textbf{x}/\hbar}\psi_{new}=\textit{u}(\textbf{x},t)\psi_{new}$$

we now put this into the SE:

$$i\hbar\frac{\partial}{\partial t}\psi_{old}=[\frac{\textbf{p}^2}{2m}-\frac{e}{m}\textbf{A}\cdot\textbf{p}+\frac{e^2}{2m}\textbf{A}^2]\psi_{old} $$

and obtain the following:

$$i\hbar \textit{u}(\textbf{x},t)[\frac{ie}{\hbar}\dot{\textbf{A}}\cdot\textbf{x}+\frac{\partial}{\partial t}]\psi_{new}=[\frac{\textbf{p}^2}{2m}-\frac{e}{m}\textbf{A}\cdot\textbf{p}+\frac{e^2}{2m}\textbf{A}^2]\textit{u}(\textbf{x},t)\psi_{new}$$

My question is relatied to the left side of the previous equation. Namely, after inserting the expression for the $\psi_{old}$ into the second equation, why wasn't $\frac{\partial}{\partial t}$ applied to the coordinate operator $\textbf{x}$, to give an additional term - $\frac{ie}{\hbar}\textbf{A}\cdot\dot{\textbf{x}}$?

So, I guess that the gist of the question is what does the time derivative operator $\frac{\partial}{\partial t}$ do when applied to the coordinate operator $\textbf{x}$?

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Here you work in the Schrödinger representation, where the time dependence is encoded in the wave function. (This is opposed to the Heisenberg representation, where the time evolution is shifted to the operators.) Thus the derivative of any operator in respect to time is zero, unless it contains explicit time dependence, as the vector-potential in this case.

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  • $\begingroup$ that fully clarifies it!! Thanks a lot! $\endgroup$ – nezelim Mar 13 at 13:26

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