Aharonov-Bohm Effect

Eric Cordian emc at wire.insync.net
Sun Dec 6 09:35:41 PST 1998



Tim Griffiths <griffith at wis.weizmann.ac.il> writes:
 
> What do you actually want explaining? Although the magnetic field is
> zero along the paths traveled (so that electrons never feel the
> magnetic field and the Lorentz force that affects a moving charge in
> one), the vector potential field is not, and is different for the two
> paths. This causes the AB-effect. See, for example G.Baym, 'Lectures
> on Quantum Mechanics' (Benjamin, NY, 1969) page 77-79.
 
Why do I find this odd?  Because the electromagnetic field acts
locally, the vector potential is not directly observable and depends
on ones choice of gauge, and wavefunction phase is also not directly
observable.  The integral of the vector potential around the closed
path is of course well-defined, but this is global, not local.
 
So it seems strange, and hints at some sort of "action at a distance"
thing occuring.  The wavefunction is being perturbed according to the
magnetic flux enclosed by a path, even while propagating in a region
distant to the flux which contains no electromagnetic field at all.
 
> As for relevance, it is left as an exercise for the reader to work out
> why it might be useful to in principle measure fluctuations in a
> magnetic field which is heavily screened, and it's application to
> cryptography and privacy.
 
It still seems counterintuitive to suggest that I can make a tour
around a magnetic field an arbitrary distance away, and get a precise
reading of its strength.  Would it work a mile away?  A light year
away?  Does this result follow from QED?  Some equations might be
helpful.

--  
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http://www.cyberspace.org/~enoch/crakfaq.html
 






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