I was reading this paper http://venables.asu.edu/quant/Gas/gassupp16.pdf on the Aharanov-Bohm effect and became confused regarding the formula (16A-7) which reads(adsbygoogle = window.adsbygoogle || []).push({});

[tex] e \hbar \vec \nabla \Lambda (\vec r) + e \vec A = 0. [/tex]

Now i understand that (16A-7) can only be true if ## \vec B = 0 ##. This is solved to obtain the gauge transformation (16A-5)

[tex] \psi (\vec r) = e^{-i (e / \hbar) \int _{P} ^{r} \vec {dl} \cdot \vec A} \psi _{0} (\vec r). [/tex]

Now my source of confusion comes from equation (16A-12) which is by my calculation

[tex] \oint \vec {dl} \cdot \vec A = \iint \vec {dS} \cdot \vec \nabla \times \vec A = - \hbar \iint \vec {dS} \cdot \vec \nabla \times \vec \nabla \cdot \Lambda (\vec r) = 0. [/tex]

However in the paper this happens:

[tex] \oint \vec {dl} \cdot \vec A = \iint \vec {dS} \cdot \vec \nabla \times \vec A = \iint \vec {dS} \cdot \vec B = \Phi. [/tex]

I honestly cannot figure out the justification for using a result in which ## \vec B = 0 ## is required to calculate a result in which clearly ## \vec B \neq 0 ##. I mean, we have fixed ## \vec A ## to be a gradient of our gauge function ## \Lambda (\vec r) ##, so obviously ## \vec \nabla \times \vec A = 0 ## at all points. Having a non-zero ## \vec B ## should kill our (16A-7) and also (16A-5) destroying all results obtained with this for non-zero magnetic fields. What am i missing here?

Thanks.

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# The Aharanov–Bohm Effect

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