# Tag Info

## New answers tagged gauge-invariance

0

I can show this backwards and then explain the motivation. The uniqueness of the solution follows from the constraint on the transformation to be unphysical. Say, we take your equation and transform $\psi$: $$i \partial_t \psi = \left( \frac{\Pi^2}{2m} + e \phi \right)\psi$$ becomes  i \partial_t \left(e^{i\frac{e}{c} \Lambda}\psi\right) = \left( \...

2

This answer is motivated by the Aharonov-Bohm effect and proves what the OP asks for, but in the special case $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A} =\boldsymbol{0}=\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}' \quad \text{that is} \quad \mathbf{B} =\boldsymbol{0} \tag{01}$$ To simplify the expressions we : set ...

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