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In the build up to Aharonov-Bohm effect, one has to represent the gauge covariant form of STIE. We need to consider two things; the vector potential A changes`under gauge transformation and as a consequence, the wavefunction picks up an exponential coefficient to transform to a new wavefunction. $$\vec{A}\longrightarrow\vec{A}+\nabla\chi=\vec{A}' $$ $$\psi \longrightarrow \psi '=e^{\dfrac {ie\chi } {\hbar }}\psi$$ Starting with the TISE for a charged particle in EM field $$-\dfrac {\hbar ^{2}} {2m}\left( \nabla -\dfrac {ie} {\hbar }\vec{A'}\right) ^{2}\psi' =\left( E-q\phi \right) \psi' ,$$ and assuming that the $\phi$ is time independent. How one can show that by substituting $\vec{A'}$ and $\psi'$ into the given TISE, we will get $$-\dfrac {\hbar ^{2}} {2m}\left( \nabla -\dfrac {ie} {\hbar }\vec{A}\right) ^{2}\psi =\left( E-q\phi \right) \psi .$$

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closed as off-topic by ACuriousMind, user36790, Kyle Kanos, Sebastian Riese, Qmechanic Feb 28 '16 at 1:08

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