Is the Aharanov-Bohm Effect Independent of External Potential?

[say_cheese]

Gauge theory says that Potential not field strengths are fundamental in a force field. Aharanov- Bohm experiment showed that charged particles changed characteristics (phase) when placed inside a shielded box with zero electric and magnetic fields, but a non-zero potential outside.

But gauge theory (and routine electromagnetic calculations) says that potential in an isolated system can be changed by a constant arbitrary potential.

Does it mean that the Aharanov effect is independent of the potential outside the box? That does not make sense either. Because then there should be an effect even for zero potential.

 

[The_Duck]

I think the gauge freedom in the potentials does not affect the results of any calculations you might perform. For instance, consider the interference experiment in the diagram http://en.wikipedia.org/wiki/Aharonov-Bohm_Effect#Magnetic_solenoid_effect". The phase difference and thus the interference effects seen depend on the line integral of A around the loop down along "path one" and back along "path two." In a gauge transformation you can add to A the gradient of any scalar field, but the curl of a gradient is zero, so the line integral of the gradient around the loop is zero, so the change in A does not effect any physical predictions.

 

[Matterwave]

One can explicitly see what "The Duck" is saying by converting the line integral of A into a surface integral of B. You see that because B=curlA, then the line integral of A around contour C is actually just the surface integral of B across the surface bounded by C. This is the flux of the B field through that surface. Therefore, if the B field doesn't change, neither does that integral.

 

[say_cheese]

Thanks. I believe the experiment implies that the phase shift is proportional to the change in the vector potential when the solenoid is turned on. One may add an arbitrary value to the vector potential everywhere, including inside and outside the solenoid before the solenoid is turned on.

Though, I don't quite understand the context of Duck's explanation. I understand that the vector potential is in the direction of current which is in the direction of circle around the letter B. The magnetic field is perpendicular to the paper. The magnetic flux enclosed by a path is non-zero, because it would include the solenoid. For example, the line integral around a circle of a radius R with vector potential A around the solenoid is 2piAR. If the vector potentail is zero before the solenoid is turned on, when the solenoid is turned on, the vertical component of the vector potential must be +/- on the left of solenoid and -/+ on the right side. The wave passing through different slits experience different halves- on each side of the solenoid, where the vector potentials have opposite directions and therefore see different flux and hence the shift in the interference fringe.

I believe the gauge freedom implies that we need to add the arbitrary value of vector potential before conducting the thought experiment.

 

[tom.stoer]

Gauge symmetry does not mean that the gauge potential is completely arbitrary, but only that the space of all gauge field configurations on spacetime (fibre bundle) is devided into equivalence classes. The line integral in the Aharonov-Bohm effect is a functional w.r.t. these equivalence classes.

The topology of the fibre bundle reflects the geometry of the gauge group together with the geometry of spacetime which is non-trivial due to the singular flux tube affecting the gauge equivalence classes (compared to trivial spacetime geometry).

 

[strangerep]

Thanks. I believe the experiment implies that the phase shift is proportional to the change in the vector potential when the solenoid is turned on.

One crucial point to understand is that we only observe the (equivalent of) a line
integral of the potential around a closed loop. For more detail, look at Stokes'
theorem on Wikipedia:

http://en.wikipedia.org/wiki/Stokes_theorem

In particular,

<br /> \oint_{\partial\Sigma} A \cdot dr ~=~ \int_{\Sigma} \left(\nabla \times A\right) \cdot d\Sigma<br />

where \Sigma is the (oriented) area enclosed by the loop, and \partial\Sigma
is its boundary (i.e., the loop itself). The rhs is gauge invariant because the curl in
the integrand is. Therefore so is the lhs.

This is basically what The_Duck was explaining.

I believe the gauge freedom implies that we need to add the arbitrary value of vector potential before conducting the thought experiment.

A second crucial point is that gauge transformations are unphysical. No observations
depend on the choice of gauge. Gauge freedom is just a mathematical artefact of
splitting kinetic and potential terms in the full Hamiltonian.

 

[tom.stoer]

A second crucial point is that gauge transformations are unphysical. No observations depend on the choice of gauge.

Yes.

Gauge freedom is just a mathematical artefact of splitting kinetic and potential terms in the full Hamiltonian.

Why? I don't see any connection to a "splitting". If you look at the Lagrangian there is no "splitting"; can you please explain?

 

[strangerep]

Gauge freedom is just a mathematical artefact of splitting kinetic and
potential terms in the full Hamiltonian.

Why? I don't see any connection to a "splitting". If you look at the Lagrangian
there is no "splitting"; can you please explain?

Yes, if you keep it all together it's gauge invariant.

But if one considers separately a kinetic term like

<br /> \bar\psi (x) \, \dots \partial_{\mu} \, \psi(x) ~~~~~~ (\gamma&#039;s~\mbox{not shown})<br />

and an interaction term like:

<br /> j_{\lambda}(x) \, A^{\lambda}(x)<br />

then each term is not separately gauge invariant,
although their sum is.

Alternatively, one could say that the gauge freedom is an
artefact of splitting the combined electron-photon system
into electron (without Coulomb field) + photon.

 

[tom.stoer]

Hm, if you would like to explain it that way, ...

... but the gauge symmetry is always related to unphysical degrees of freedom which have to be eliminated - even if you do not split the covariant derivative into kinetic and interaction term

 

[dextercioby]

Alternatively, one could say that the gauge freedom is an
artefact of splitting the combined electron-photon system
into electron (without Coulomb field) + photon.

There's no splitting. It's the other way around. It's a reunion of 2 free fields, one massive (aka <matter field>) and the other massless (aka <gauge field>). The coupling of the 2 is usually unique, if a restriction on the no. of derivatives in the coupling is imposed.

Gauge freedom is not a consequence, but a root cause. Putting classical fields in a flat spacetime exhibits this option: gauge invariance. Once the gauge group is given, you usually have little to no room to build the rest of the classical theory (field tensor, gauge algebra, lagrangian, coupling to matter or with other gauge fields). Quantum theory follows then easily in a path-integral approach.

 

[strangerep]

There's no splitting. It's the other way around. It's a reunion of 2 free fields, one massive (aka <matter field>) and the other massless (aka <gauge field>).

Except that the "free" matter field (here I'm talking about electrons) is unphysical
because it's unaccompanied by its Coulomb field. Dressing an electron so as to
become gauge invariant also gives it the physically correct Coulomb field.

P.A.M. Dirac,
"Gauge-Invariant Formulation of Quantum Electrodynamics",
Can. J. Phys., vol 33, (1955), p. 650.