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In addition to link, may I ask you for details of the Hamiltonian transformation. Knowing that: $[\textbf{x},\textbf{p}]=i\hbar$, $[\textbf{x},\textbf{p}^2]=2i\hbar\textbf{p}$ and $e^{-i\alpha A}Be^{i\alpha A}=B-\alpha c$, where $c=[A,B]$ and it is a c-number.

Step by step, it goes:

$$e^{-ie\textbf{A}\cdot\textbf{x}/\hbar}\frac{\textbf{p}^2}{2m}e^{ie\textbf{A}\cdot\textbf{x}/\hbar}= \frac{1}{2m}(\textbf{p}^2-\frac{ie}{\hbar}[\textbf{A}\cdot\textbf{x},\textbf{p}^2])=\frac{1}{2m}(\textbf{p}^2-\frac{ie}{\hbar}\textbf{A}\cdot2i\hbar\textbf{p})=\frac{1}{2m}\textbf{p}^2+\frac{e}{m}\textbf{A}\cdot\textbf{p}$$

$$e^{-ie\textbf{A}\cdot\textbf{x}/\hbar}(-\frac{e}{m}\textbf{A}\cdot\textbf{p})e^{ie\textbf{A}\cdot\textbf{x}/\hbar}=-\frac{e}{m}\textbf{A}\cdot(\textbf{p}-\frac{ie}{\hbar}[\textbf{x},\textbf{p}])=-\frac{e}{m}\textbf{A}\cdot(\textbf{p}+e\textbf{A})=-\frac{e}{m}\textbf{A}\cdot\textbf{p}-\frac{e^2}{m}\textbf{A}^2$$

while the term $\frac{e^2}{2m}\textbf{A}^2$ remains the same, as it commutes with both $\textbf{x}$ and $\textbf{A}$.

Summing all of these leaves an extra term of $-\frac{e^2}{2m}\textbf{A}^2$. This is clearly a mistake.

May I ask you where did I make the mistake?

Ref: Minimal coupling to electric dipole form

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