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Let $$ \left.\begin{aligned} t'&=t\\x'&=x-vt \end{aligned}\right\} \quad \Longrightarrow\quad \dot{x}'=\dot{x}-v $$ and therefore $p'=p-mv$.

If $p'=-i\hbar\nabla' $, then $\nabla'=\nabla-iv/\hbar$. Shouldn't the partial derivatives $$\frac{\partial}{\partial x'}=\frac{\partial x}{\partial x'} \frac{\partial}{\partial x}+\frac{\partial t}{\partial x'} \frac{\partial}{\partial t}=\frac{\partial}{\partial x}-\frac{1}{v}\frac{\partial}{\partial t}$$ why $\nabla'=\nabla$?

If the Schrodinger equation exhibits Galilean invariance, how to prove the new wave function $\psi'(x',t')$ only times a factor ($\psi'(x',t')=a(x,t)\psi(x,t))$?

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