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How I show that $$\Lambda(\textbf{x}')=\frac{q}{\hbar}\int \mathbf{A} \cdot d\mathbf{x'}$$ on $$ \tilde{\psi}(\textbf{x}',t)=e^{[\frac{iq\Lambda(\textbf{x}')}{\hbar c}]}\psi(\textbf{x}',t)$$ for Aharonov-Bohm effect? I don't know about path integral formulation, that was suggested on Sakurai,J.J. book... There is other way to show that? Using the gauge transform for $$\tilde{\mathbf{A}}=\mathbf{A}+\nabla\mathbf{\Lambda}~ ?$$

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You should be able to show by direct substitution that your proposed wave function $\tilde{\psi}(\mathbf{x},t)$ solves the Schroedinger equation, where the vector potential is included via the minimal coupling prescription $$ \mathbf{p} \to \mathbf{p} - \mathrm{i} q \mathbf{A}(\mathbf{x})$$ (up to a sign convention for $\mathbf{A}$ and some dimensionful constants which I may have got wrong), so long as $\psi(\mathbf{x},t)$ is the solution when $\mathbf{A} = 0$.

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