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The relevant time-dependent Schrodinger equation, for a spinless charged particle in an EM field, reads $$ i\hslash\frac{\partial \Psi}{\partial t}=\left[\frac{1}{2 m}\left(\vec{p}-\frac{q}{c}\vec{A}\right)^{2}+q\phi\right]\Psi $$ where $\vec{p}=-i\hslash\nabla$. The wavefunctions, $\Psi$ and $\Psi^\prime$, due to a pair of potentials $(\phi,\vec{A})$ and $(\phi',\vec{A}')$ respectively, that are related to each other by a gauge transformation is related as $$ \Psi'(\vec{r},t)=e^{i(q/\hslash c)\Lambda(\vec{r},t)}\Psi(\vec{r},t).\tag{1} $$ where $\Lambda$ is the gauge function. This can be verified by direct substitution. But can we derive this result? QM textbooks always verify this result knowing it beforehand. But is there a way to derive it?

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  • $\begingroup$ Maybe begin with Ψ'=Ψ+δψ (correction)... substitute this in Schrodinger equation , manipulate the kinetic term in RHS by adding and subtracting p²/2m and see how δψ is related to Ψ $\endgroup$
    – paul230_x
    Apr 20, 2023 at 0:54
  • $\begingroup$ Related: physics.stackexchange.com/q/562654/2451 $\endgroup$
    – Qmechanic
    Apr 22, 2023 at 5:19

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What you can do is derive it backwards. Start with a pure global phase, convert it to local phase, see how to modify that so that it would fit the requirements of gauge symmetry. i.e. start with $$e^{i\varepsilon\Lambda(\vec r,t)}$$ and derive what const $\varepsilon$ will fit what you want.

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Well a gauge transformation can't have any physical consequences so the wavefunction must transform with an overall phase, which means that $$\psi\to e^{im(x)}\psi.$$ Plug this into the SE equation together with the transformation for $A$ and requiring all the new terms to vanish gives you the desired result.

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