# Minimal coupling to electric dipole form

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 $$\textbf{x}$$?