# Can we derive that the wavefunction changes as $\Psi'(\vec{r},t)=e^{i(q/\hslash c)\Lambda(\vec{r},t)}\Psi(\vec{r},t)$ under a gauge transformation?

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?

• 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 Commented Apr 20, 2023 at 0:54
• Commented Apr 22, 2023 at 5:19

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.
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.