In the book about Schrödinger Operators by Cycon et al. there is a step in their calculations I don’t understand. When I pick two vector potentials $A_1$ and $A_2$ such that their curl (i.e. the magnetic field) is the same then their difference must be the gradient of some smooth function (since the curl of the difference vanishes):

$$ A_1 -A_2 = \nabla \lambda. $$

The conclusion is then the unitary equivalence of the magnetic Schrödinger operators $(-i\lambda - A_j)^2$, $j=1,2$. That is

$$ e^{i\lambda}(-i\nabla - A_1)e^{-i\lambda} = (-i\nabla - A_2) $$

I’ve tried showing this by expanding the exponentials up to first order but got stuck. I’m sure this is an age old argument but I don’t see it right now.


Hint (recall the $\nabla$ is an operator acting to everything on the irght of it): $$-\text{i}\nabla e^{-\text{i}\lambda}= -\text{i}\left(\nabla e^{-\text{i}\lambda}\right) - e^{-\text{i}\lambda}\,\text{i}\nabla\,.$$


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