Talk:Aharonov–Bohm effect

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The article states that "The underlying mechanism is the coupling of the electromagnetic potential with the complex phase of a charged particle's wave function". This requires a citation. Who says so and in what textbook or peer reviewed publication? Aoosten (talk) 00:33, 10 December 2013 (UTC)

You will be hard pressed to find a text that doesn't say it. Xxanthippe (talk) 04:47, 10 December 2013 (UTC).

This important statement should not be made anonymously, but should be backed up by one or more quotations. It is controversial because it does not explain the Aharonov-Bohm effect in terms of existing coupling of the electromagnetic potential to the charge-current. Thus it suggests that the effect involves some new electromagnetic effect.

Aoosten (talk) 14:26, 10 December 2013 (UTC) — Preceding unsigned comment added by (talk) 07:56, 10 December 2013 (UTC)

Help & Warning !

I have heard that the magnetic monopole business which is stated is far from being accepted by the majority of physicists. Is there any proof that this magnetic monople stuff has any existence ? Honestly I doubt. —Preceding unsigned comment added by (talk) 21:28, 4 November 2007 (UTC)

The monopoles described in this article are quasiparticles that crop up in condensed-matter physics quite often due to separation of equal but opposite magnetic charges, connected by a finite-width flux tube. They are thematically similar to 'true' monopoles but are in no way controversial. They cause a stir from time to time when someone reports experiments regarding Dirac strings as indicative of the discovery of real monopoles. A similar kerfuffle tends to occur when research regarding spin-charge separation is taken to indicate that electrons are composite particles. Note: I am not a licensed quantum mechanic. Eutactic (talk) 06:59, 4 June 2010 (UTC)

I don't think the magnetic monopole belongs in this article. Miguel (talk) 12:19, 4 January 2011 (UTC)

Comment: You are perfectly right, as there are no "magnetic monopole particles" (see Lorentz force)

LZob —Preceding unsigned comment added by (talk) 08:56, 23 May 2011 (UTC)

long solenoid

It is NOT TRUE that a long solenoid does not cause a magnetic field outside of itself. There is a tiny right-hand field, as well as a (comparatively) large up-down field (description is very bad, but try to understand).

An ideal solenoid would indeed have no magnetic field outside of the solenoid. This is of course impossible to acheive in real life. The original experiment didn't even use a solenoid so far as I know, but rather a thin whisker of magnetic material. In theory the ideal AB experiment would have zero magnetic field outside, but here we just get into the experimental practicalities of negligible versus zero. --Laura Scudder | Talk 20:09, 18 August 2005 (UTC)

Of course, as Laura S. and the earlier unsigner pointed out, it is not strictly zero. But it is very small. And just to dispel lingering doubts, people have redone the experiment with toroidal solenoids, which give an even better approximation to 0, and tested for dispersion in the AB effect, which dispersion should show up if the effect were due to any lingering, very mall E-M field. No dispersion has been found.

See for an example of such testing. (talk) 18:49, 29 July 2008 (UTC)


Could someone add something which explains this effect in layman's terms. I have a strong sense that this could have immense siginifance but as soon as the technical terms and algebraic equations turn up I'm left in the dust. Please could someone elucidate for a non-specialist, arts-background person like me? :-) ThePeg 28.7.2006

Another, in some ways better, summary is available at . I am working on gauge fixing right now; perhaps when I am at a good stopping point there I will try to improve this article. Cheers, Michael K. Edwards 03:23, 20 October 2006 (UTC)

I think ThePeg may be asking for too much. How do you explain the AB effect without using the terms "E-M field", and "vector potential" and 'gauge'? And how do you describe these without some E-M coursework in your background, and at least as much vector calculus as in the marvellous little book, "Div, Grad Curl and all That"?

After all: the whole point of the AB effect is that the vector potential plays such a surprising role, instead of being the a convenience for computing E & B, as it usually appears in junior-level E&M courses. But without that background, how do you know to be surprised?

In fact, what I almost get from the article, but still don't quite get, is how the AB effect does not contradict gauge invariance. So clarification of that point would be welcome.

One way to do this is to express the interference effect between the two electron paths in the solenoid/double-slit setup as the flux of the magnetic field across any surface bounded by the two paths. The flux of the magnetic field is a gauge-invariant quantity - in fact, it can be measured independently by Faraday's Law of Induction. The Aharonov-Bohm effect is not so different after all from the electric field induced on a closed loop of wire going around a solenoid when the current on the solenoid is varied. The time-rate of change of the AB phase is Faraday's electromotive force. Miguel (talk) 12:33, 4 January 2011 (UTC)

Now for magnetic monopoles: the consensus now is that magnetic monopoles existed in the early universe, but have all decayed. So no, there aren't any of them anymore. But even the fact that they once did exist is enough to explain (per Dirac) why charge is quantized. (talk) 18:51, 29 July 2008 (UTC)

Another Interpretation

The Feynman Lectures on Physics (Vol II, Sec 15-5), has a different interpretation from the one stressed here. The article says that the AB effect shows that knowledge of the classical electromagnetic field acting locally on a particle is not sufficient to predict its quantum-mechanical behavior. Feynman says that the AB effect shows that A is a "real" vector field. The effect on the phase of the interference pattern arises because the electron passes through regions of non-zero A. In my teaching, I have repeated Feynman's argument that A deserves to be thought of as a real field. It is A (and phi) that appear in Schroedinger's equation, not E and B. Another argument, by the way, is that A and phi form the four-vector, not E and B. Can anyone comment on this? I think it is worth noting. Timb66 (talk) 05:00, 11 April 2008 (UTC)

If you want to argue Feynman, remember that Feynman teaches quantum mechanics from the double-slit experiment, in which you cannot pretend to know the trajectory of the particle unless you measure it, which destroys the interference pattern. However, in the path integral approach to quantum mechanics you still have to consider the value of the action over any possible trajectory, and the action of a trajectory cannot be written in terms of the electromagnetic field but only in terms of the electromagnetic potential. A "deserves to be thought of as a real field" (and appears in the action) because it is the connection form of the electromagnetic gauge bundle. However, all its physical effects can be described in terms of the electromagnetic tensor F which is the curvature form of the bundle. And this includes the Aharonov-Bohm effect since, by Stokes' theorem, the Aharonov-Bohm phase shift between two paths with equal endpoints is proportional to the flux of the magnetic field through a surface bound by the two paths. Miguel (talk) 14:29, 4 January 2011 (UTC)
There is no contradiction between the article and Feynman. Classically, the vector potential only exerts a force on a particle through its curl (= B), and in regions of zero curl there is no classical possibility of any effect on the particle. So, knowledge of the classical field acting locally on the particle is not sufficient to understand its quantum effect, as the article says. (Much the same point can be found in the book by Peshkin and Tonomura, if I recall correctly.)
What this means is that not all physical effects of the electromagnetic field manifest themselves as forces. A shift in an interference pattern is a physical effect, but it is not interpretable as the effect of a force acting locally. Miguel (talk) 15:11, 4 January 2011 (UTC)
Feynman interprets this somewhat metaphysically as implying that the vector potential is "real" in a sense he defines (that it acts locally on the particle). In this sense, there are more "real" fields in quantum mechanics than in classical electrodynamics. My own inclination (and, I suspect, that of many physicists), is that questions of what fields are "real" lead to the same philosophical confusion as 19th-century arguments about whether negative numbers and imaginary numbers are "real" things. A is useful in describing physical laws that make accurate predictions, and that is enough.
Yes, A and φ form part of a four-vector. And E and B form parts of a rank-2 4-tensor. These are both useful descriptions, but whether it makes one of the descriptions "more real" seems like another unresolvable metaphysical diversion.
This is not to say that Feynman's interpretation should not be noted in the article; it's another way of pointing out the significance of this phenomenon. I just don't think it should be put forth as a contrasting view, rather than as another spin on the same underlying idea.
—Steven G. Johnson (talk) 17:50, 11 April 2008 (UTC)

Hi Steven. Thanks for the response, which sounds sensible to me. I wrote an essay on the AB effect in my final year of undergrad (1987), and I recall that the Feynman position was the prevailing one at the time. I will try to dig out my essay. These days I am a humble astronomer and, though I teach senior-level EM, I have not looked into the AB effect since then. I agree this should go into the article. Feel free to do that, if you have the time! (I don't at the moment, I'm afraid). Timb66 (talk) 22:50, 18 April 2008 (UTC)

I think Timb66's point that usual interpretation of AB is that it demonstrates the reality of the A potential is correct, and is not sufficiently stressed by the article. Feynman deplored metaphysics, and if he thought that AB implied the reality of the underlying potentials (which he does -- he is quite clear about this in his lectures (II, 15-5)) then I think we should take his views seriously and represent them more clearly. --Michael C. Price talk 10:32, 16 August 2009 (UTC)

As the article points out, either the A field is real, or the physics is nonlocal. Which leads us down the EPR path... Miguel (talk) 14:29, 4 January 2011 (UTC)

More examples needed

I am currently trying to understand this phenomena, starting out from the example given in chapter 18 (By McEuen) in Kittel's Introduction to Solid State Physics, 8th Edition. This Wikipedia article does not seem to help very much, unfortunately. More examples would help - also external links would help, not just references to articles. —Preceding unsigned comment added by (talk) 12:01, 25 April 2008 (UTC)

I second the motion: we need references to something more accessible than Physics Review articles. And the references should be numbered, at least some of them. I suggest ArchiveX as a possible source: not all the papers there are trash;) (talk) 18:52, 29 July 2008 (UTC)

I agree that theb article is confusing. Luckily I already undertsand the AB effect, because I would never have understood it from this article. The language is a mess; when it talks about the "classical fields". What does that mean? The E and B fields, but not the A field? The A field was invented before QM, by Maxwell and Heaviside and others, and so is a classical field as well.

Classically, the E and B were "fields" and the φ and A were "potentials", so no confusion there unless you want to really split hairs with the more general meaning of "field" as "any scalar, vector or tensor-valued function varying across space and/or time". Miguel (talk) 14:33, 4 January 2011 (UTC)

Also, the very first sentence is confused.

The Aharonov–Bohm effect, sometimes called the Ehrenberg–Siday–Aharonov–Bohm effect, is a quantum mechanical phenomenon by which a charged particle is affected by electromagnetic fields in regions from which the particle is excluded.

Now what do they mean by unlinked term, electromagnetic fields? I guess they mean the E and B fields. But I have to guess, which is not a good sign. --Michael C. Price talk 11:02, 16 August 2009 (UTC)

Copyvio removed

Copyright problem removed

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See [1]

If someone wants to rewrite the nano-ring section, please do, it's pretty bare right now. --Steve (talk) 18:52, 19 December 2009 (UTC)

Feynman's views

This text has been reverted with the claim that it is incorrect:

in contrast, the effect depends on A only in the region where the test particle is allowed. Therefore we can either abandon the principle of locality (which most physicists are reluctant to do) or we are forced to accept the realisation that the electromagnetic potential offers a more complete description of electromagnetism than the electric and magnetic fields can. In classical electromagnetism the two descriptions were equivalent. With the addition of quantum theory, though, the electromagnetic potential A is seen as being more fundamental or "real"[1]; the E and B fields can be derived from the potential A, but the potential can not be derived from the E and B fields. Furthermore some physical effects are dependent only on A, which the E and B fields are simply unable to account for.

Which bit is incorrect, aside from the last sentence, which is a bit simplistic? It comes from the Feynman lectures, Vol II, Sec 15 (10-14). --Michael C. Price talk 01:00, 14 February 2010 (UTC)

"The potential can not be derived from the E and B fields", for starters. The quoted sentence from the Feynman Lectures ("knowledge of the classical electromagnetic field acting locally on a particle is not sufficient to predict its quantum-mechanical behavior") is factually correct. It's phrased in such a way that one might conclude that some important information about A is missing from E and B, but it doesn't actually say that, and it happens to not be true.
What we're talking about here is just the nature of gauge fields. Here's a perhaps more understandable example from general relativity, which is a gauge theory of a slightly different kind. Around an infinite cylindrical cosmic string, the spatial geometry is a cone. Spacetime is Riemann-flat. A test particle initially at rest near the cosmic string will not fall toward it. In effect, there's no gravitational field. There is, however, an angular deficit. If two initially comoving objects pass the string on opposite sides, they will be moving towards each other on the other side, even though neither one encountered any curvature.
You can measure the angular deficit by integrating the metric on a loop around the string. Alternately, you can integrate the scalar curvature over a surface bounded by the loop (which necessarily intersects the string, which is the only place the curvature is nonzero). If you use the metric, you're integrating only over the region where the test objects travel, but you're using a field with unphysical degrees of freedom. If you use the curvature, you're using only the physical degrees of freedom, but integrating over a region where the test objects don't travel.
This is interesting stuff. But I would object if someone wrote, "this forces us to conclude that either general relativity is intrinsically nonlocal, or curvature alone is insufficient to understand gravity and the metric must be treated as real." I hope you see why. -- BenRG (talk) 02:20, 14 February 2010 (UTC)
I have a problem with your first sentence; you're saying that the potential A can be derived from E and B? This is impossible by virtue of the gauge symmetry of A. It's interesting, you seem to be using the gauge freedom of A to argue against the reality of A, whereas Feynman / myself are explicit that this is irrelevant. Feynman is explicit that gauge fields are "real".
The potential A can be derived from the E and B fields, known globally, up to an integration constant which is a manifestation of gauge freedom and which is not observable. The formulas for obtaining A from E and B should be in any decent classical E-M textbook. In classical newtonian gravity you can also deduce the gravitational potential from the gravitational accelaration field up to an additive constant of integration, which has no physical effects. But in fact, that constant adds to the quantum mechanical action of a non-relativistic particle in a newtonian gravitational potential, and so affects the frequency of oscillation of the wavefunction. It's just that there are no interference effects that can reveal the value of the additive constant. Though the newtonian gravitational phase can be observed in an interference experiment (the famous Colella-Overhauser-Werner neutron interferometry experiment). Miguel (talk) 14:45, 4 January 2011 (UTC)
Do you have the Feynman lectures handy? - I can supply more detailed quotes about the supremacy of A over E and B; Feynman is quite clear about this, makng the point repeatedly.--Michael C. Price talk 02:48, 14 February 2010 (UTC)
Some quotes: the vector potential a "real" field? ... a real field is a mathematical device for avoiding the idea of action at a distance. .... for a long time it was believed that A was not a "real" field. .... there are phenomena involving quantum mechanics which show that in fact A is a "real" field in the sense that we have defined it.... .... E and B are slowly disappearing from the modern expression of physical laws; they are being replaced by A and [the scalar potential]
--Michael C. Price talk 03:09, 14 February 2010 (UTC)
I'm not especially interested in whether A or E/B are real or not, and neither is Feynman. That's why he consistently puts scare quotes around "real". (I do have the Lectures handy.) He's taking a pragmatic view of things: instead of talking about what's "real", talk about what works well in the theory. A works better in QED and the Standard Model. I have no objection to that. I think you believe that I have something against A, and by mentioning that E/B contain the same information I'm trying to make A "look bad". It has nothing to do with that. I just want to mention that E/B contain the same information as A because it seems to be a common misconception that they don't. I don't care whether people use E/B; I don't want them to use E/B; I just want to clear up that misconception.
Furthermore, you still seem to want to include something in the article that's of the form "A is more real than E and B because E and B are derivable from A while A is not derivable from E and B". That's ridiculous and has nothing to do with what Feynman was talking about. Gauge symmetry is like Lorentz symmetry. Suppose I had a field P, and from that I could derive some object Q, and from Q I could rederive P up to a Lorentz transformation—in other words, I couldn't quite reproduce P because Q only tells me the relative motion of things, while P contains absolute velocities. Is P more real in that situation? Of course not. When I say that A can be rederived from E and B, I mean of course "every aspect of A that is relevant in any physical context whatsoever, including Aharonov–Bohm-type experiments". Are you sure you understand that there's no extra information about the world in A? It looks like you didn't understand it last August when you wrote the disputed text.
I would appreciate some comment on the bulk of my last reply, the GR part. -- BenRG (talk) 01:57, 15 February 2010 (UTC)
The GR argument is the same as the EM one, and my views are the same about both cases. I don't want to get distracted by GR, since all the issues are covered by the EM case, which is what this article is about, but I will add that yes, I would regard P as more real, especially since you need "spooky" action-at-a-distance if you just use Q.
As for how you are using "derived" / "rederived", it packs a lot of implications that will confuse readers. The fact remains that Au CANNOT be derived from E & B; this is just a mathematical truth that no amount of redefinitions can get around - you know this as well as I. If you wish to say something else by this statement then say it explicitly.
Feynman couldn't be clearer on the subject, it's a straight choice between using the E & B fields (or Fuv) and action-at-a-distance or use local potentials, Au. The current wording in the article makes that clear.
As for your statement:
I just want to mention that E/B contain the same information as A because it seems to be a common misconception that they don't.
this is not true, for the reason I stated; you can't derive Au from E & B. But I know what you mean by this, you mean that Au contains no extra physical information. But even this is debatable, given the "spooky" action-at-a-distance twist. Mathematically, though, the statement is plain wrong, and we have to be clear about this.
--Michael C. Price talk 08:24, 15 February 2010 (UTC)

I got here from an announcement at WT:PHYS. What is the disagreement about, precisely? ― A._di_M. (formerly Army1987) 23:21, 15 February 2010 (UTC)

In the Coulomb gauge 3-vector A can be determined uniquely and explicitly from 3-vector B. Of course, for a general gauge one has to add the gradient of an arbitrary scalar gauge function. That is the reason for the indeterminacy. Xxanthippe (talk) 23:33, 15 February 2010 (UTC).
That's a good point; restrict the gauge enough and A becomes determinable from E & B. But Feynman's point (which is independent of the gauge principle) remains: using A avoids invoking action-at-a-distance and so, by Feynman's terminology, the A field is still more real than the E & B fields.--Michael C. Price talk 00:34, 16 February 2010 (UTC)
Actually, A is determined from B alone and not E. Xxanthippe (talk) 00:40, 16 February 2010 (UTC).
I'm using A as a short-hand for the four vector, Au, which includes the scalar potential. --Michael C. Price talk 00:59, 16 February 2010 (UTC)
In the Coulomb gauge the vector potential is determined by B alone and the scalar potential by the (instantaneous) charge density so that E does not get a look-in, formally at least. Xxanthippe (talk) 01:46, 16 February 2010 (UTC).
Yes, I should have said, restrict the gauge enough and Au becomes determinable from E, B and the charge density.--Michael C. Price talk 06:53, 16 February 2010 (UTC)
I think Xxanthippe's point is that in the Coulomb gauge, in order to find A0, E only appears in its divergence. ( or something like that, right?) ― A._di_M. (formerly Army1987) 15:57, 16 February 2010 (UTC)
Yes, with the qualification that in the equation for A the del operates on r (so it is best taken outside the integral) and in the equation for phi it acts on r prime. To be specific, like this (with all fields taken at the same and in gaussian units):

Xxanthippe (talk) 01:32, 17 February 2010 (UTC).

If I may act as devil's advocate: since, as seen above, A in the Coulomb gauge (and the Aharonov-Bohm effect is invariably discussed in the Coulomb gauge) is obtained explicitly from B and the potential in any other gauge is obtained by adding an arbitrary pure gauge term it is hard to see how A contains any more information than B. Xxanthippe (talk) 11:27, 19 February 2010 (UTC).
The fact is, to know A here you need to know B everywhere, so that argument only works if you admit action-at-a-distance. ― A._di_M. (formerly Army1987) 11:49, 19 February 2010 (UTC)
More clearly, if you consider a doughnut-shaped region of space, knowing Aμ there gives more information than knowing Fμν there, doesn't it? To calculate the phase difference a charged particle picks when going round the ring, the former suffice, but the latter doesn't. That is the point. ― A._di_M. (formerly Army1987) 12:21, 19 February 2010 (UTC)
There is certainly an asymmetry in that to calculate A you have to know B everywhere but to calculate B you have to know A only in the immediate vicinity, rembering, however, that A itself encodes B everywhere. However, I don't think that this leads to 'spooky action at a distance' for A any more than does the calulation of, say, the energy of a field by integrating its energy density over all space at the same time, although special relativity states that an observer can only know events that occur in his own light cone. Xxanthippe (talk) 01:40, 20 February 2010 (UTC).

An explicit example desirable; confusing description at present

It would be great if somebody could work out specific examples, with an actual model for B(r)/E(r), resulting A(r) and the phase change, for both the magnetic and particularly the electric effects.

The present description is not immediately obvious for the magnetic case, and look plain impossible for the electric effect!

Specifically: 1) Electric case: How can one "construct a situation" where the electric field (grad phi) is zero everywhere along a closed loop (the too paths with common start and end), yet the electric potential (phi) is different at different points along the paths??? The description of the speicific experiments looks quite cryptic, but whatever they were I doubt they fit the present description of the electrical AB effect, which is essentially trying to say one can make a=b and a!=b at the same time.

2) Magnetic case: To begin with, the statement that "ideal solenoid" has zero field outside of it is puzzling: shouldn't an ideal cylindrical solenoid have dipole-like field infinitely far away? Now, I'm OK with assuming that the field is zero either because it can be ideally shielded, or that it is zero because the solenoid in between the two paths is actually toroid. But then there appear fields in opposing directions, so I don't see as totally obvious that their contributions do NOT cancel! Say, if you say calculate the net flux across any cross-section you will get zero.

The only more or less obvious scenario I can think of is an ideal toroid solenoid with one path going through the center of the torus and the other fully outside the torus -- but even then the field is non-zero due to the net current flowing along the _outer_ radius of the torus! Besides, this is not what the article presently describes.

So if anyone could specify a particular model geometry, with clear behavior of magnetic field both inside and outside the plane of drawing (NOT with magnetic field lines disappearing into we don't care where), and show why within a specific gauge A(r) becomes nonzero along the paths -- that would greatly help!!! —Preceding unsigned comment added by (talk) 09:34, 23 October 2010 (UTC)

Philosophical issues not about relativity but about gauge invariance.

These paragraphs from the introduction are rather confused. I intend to rewrite them and I copy them here for discussion in case my version is later challenged:

The Aharonov–Bohm effect is important philosophically: the "classical" approach to physics is completely based on Newton's equation F = ma; so for centuries physics was all about forces, and the electric field E and the magnetic field B can be used to calculate the force on a charged particle. However, force is not a valid 4-vector in special relativity. Because of this, in special relativity it is difficult to calculate with forces. Since the E and B fields are force fields, they are also not valid special relativity 4-vectors: one can easily see this because the E and B fields one calculates or measures depend on velocity. Similarly in quantum mechanics particles are represented as waves, and it's difficult to describe how waves interact with forces. So in quantum mechanics one prefers to work with potential energy. In relativistic quantum mechanics, including quantum electro-dynamics, it's difficult to even speak of a force.
When formulating electrodynamics, one may choose to use either E and B, or the electric potential Φ and the magnetic (vector) potential A. Together, these quantities form a valid special relativistic 4-vector. However, this 4-vector is not uniquely determined, one may always add certain scalar or vector functions, derived from so-called gauge functions, to this 4-vector, and the resulting E and B fields will not change. For a long time people wondered if the A field was fundamental, even though it cannot be uniquely determined, or if it was just a mathematical trick.

The conceptual issues around the AB effect are not fundamentally about relativity, as the Aharonov-Bohm effect is observable experimentally at ordinary speeds and can be easily formulated in terms of the nonrelativistic Schrödinger equation.

The conceptual issue is properly about forces, fields, potentials and gauge invariance. I am editing accordingly. My first version is

The Aharonov–Bohm effect is important conceptually because it illustrates the physicality of electromagnetic potentials whereas previously it was possible to argue that only the electromagnetic field was physical and the electromagnetic potential was a mathematical construct (being non-unique, in addition to not appearing in the Lorentz Force formula). The non-uniqueness of the electromagnetic potential is a manifestation of electromagnetic gauge invariance, with the electric and magnetic fields and forces being gauge invariant and therefore directly observable in principle. The complex phase of the electron wavefunction is coupled with the electromagnetic potential and the Aharonov-Bohm effect can be illustrated by interference experiments.

If that is not clear enough it can be developed in the body of the article and explained a little better in the intro. It also needs to be wikified. Miguel (talk) 21:51, 1 January 2011 (UTC)

I shall delete the clause:

or more precisely: a certain gauge invariant quantitiy, ΦB

from the introduction. It has bothered me for awhile and no one had clarified it. Do you agree? -- cheers, Michael C. Price talk 08:07, 2 January 2011 (UTC)

Yes, I don't know what that was about. Miguel (talk) 11:08, 4 January 2011 (UTC)

Un-needed mystification

To return to my ramble of a year ago: It seems to me that there is some un-needed mystification about the magnetic vector potential in this article. In its irreducible form (i.e. one that cannot be reduced to a gradient) the vector potential is given by its Coulomb gauge version written above explicitly in my comment of 17 February 2010. This form shows that the vector potential A at a given point encodes (to the gradient of an arbitrary scalar function) the magnetic field B at every point in space, not just the point at which the vector potential is evaluated. So the phase shift of the electron wavefunction that give rise to the A-B effect depends on the magnetic field at every point in space. This result is hardly surprising because the electron wavefunction also exists at every point in space (with finite or zero amplitude). The objection that the expression for the Coulomb vector potential contains "action at a distance" because it contains an integral over B everywhere at the same time holds little weight because B everywhere is calculated from the Maxwell equations which themselves are causal. If B (and E) are known everywhere then everything is known about the system (at least so far as the A-B effect is concerned). No extra information is contained in A (or the scalar potential phi). It is only necessary to discuss the Coulomb gauge, because making a gauge transformation from the Coulomb gauge to any other gauge has no effect on the A-B phase shift, as we all know. Xxanthippe (talk) 05:47, 17 January 2011 (UTC).

What changes are you proposing? -- cheers, Michael C. Price talk 06:23, 17 January 2011 (UTC)
Not much, for the time being, as my views are not entirely in the mainstream. But the statement in the article that "the 4-potential cannot be derived from the E and B fields" needs to be clarified as it can be derived up to the gradient of an arbitrary scalar field. The arbitrary term is known as a pure gauge field and contributes nothing to any physical observable like the A-B phase shift. Xxanthippe (talk) 22:37, 17 January 2011 (UTC).
Well, "up to the gradient of an arbitrary scalar field" is a rather major caveat. Your final statement is just a restatement of the gauge principle, which (hopefully) is explained in its own article. -- cheers, Michael C. Price talk 07:29, 17 January 2011 (UTC)
The caveat is no more major that that for the constant of integration of an indefinite integral. Xxanthippe (talk) 08:41, 17 January 2011 (UTC).
A little bit more major, I think. But, regardless, forgetting about the constant can have serious consequences. Anyway, I stuck a link to gauge freedom in the statement in the article. -- cheers, Michael C. Price talk 08:58, 17 January 2011 (UTC)
Forgetting about the gradient of the arbitrary scalar field has no consequences at all for the A-B phase difference or any other physical observable. The magnetic vector potential (known everywhere) contains the same amount of physical information as the magnetic field (known everywhere). The A and B fields are just different ways of packaging the same physical information. Seen in this way, much of the mumbo-jumbo talked about gauge evaporates. Xxanthippe (talk) 09:10, 17 January 2011 (UTC).
People seem to have a problem with the requirement to "know the field everywhere". The point of the Aharonov-Bohn effect is that knowing the field locally is not sufficient. Under the Lorentz Force law all effects are purely pointwise. The Aharonov-Bohn effect requires you to know the field in an extended region and people are bothered by that. It's the famous EPR philosophical prejudice of naïve realism. Miguel (talk) 11:12, 20 January 2011 (UTC)
Yes, that is the point entirely. -- cheers, Michael C. Price talk 12:06, 20 January 2011 (UTC)
But you'll agree this has also nothing to do with potentials vs. fields, then. In the case of the Aharonov-Bohm double slit experiment you need to either know the A field over the entire loop formed by the two paths, or know the B field over an entire surface bound by the loop, because the observable quantity is the phase change around the loop. Miguel (talk) 20:02, 24 January 2011 (UTC)
Incorrect. There are not two paths. There are an infinite number of them as the electron wave function spreads everywhere. See Path integral formulation. Xxanthippe (talk) 02:26, 25 January 2011 (UTC).
Hair-splitting. With narrow slits and perfectly absorbing barriers, and under a saddle point approximation to the path integral, the interference pattern is dominated by two paths. Miguel (talk) 15:02, 25 January 2011 (UTC)
Irrelevant. -- cheers, Michael C. Price talk 08:12, 25 January 2011 (UTC)
Eh? AFAICS Miguel, your first sentence is contradicted by the second.-- cheers, Michael C. Price talk 00:47, 25 January 2011 (UTC)
It is not. Whether you describe the Aharonov-Bohn phase shift as the line integral of the potential or the flux of the field, it does not suffice to know the potential/field locally. The effect is not pointwise like the Loretz force, it is a global quantity associated to an extended loop or surface. Miguel (talk) 15:02, 25 January 2011 (UTC)
You said
But you'll agree this has also nothing to do with potentials vs. fields, then.
which I disagree with. But this seems a issue with the terminology which is not worth pursuing. We agree on the physics. -- cheers, Michael C. Price talk 17:35, 25 January 2011 (UTC)
Only in the sense that the Stokes' theorem of calculus on manifolds is "a little bit more major" than the Fundamental theorem of calculus. Given a function in 1D, is there another function whose derivative is the original function? The answer is locally yes, up to a function whose derivative is zero (that is, a constant). Globally this may fail to hold. Given a vector field in 3D, is there another function whose grad/curl is the original function? The answer is locally yes, up to a function whose grad/curl is zero (that is, a constant/gradient). Globally this may fail to hold.
This has nothing to do with the gauge principle, yet. The gauge principle comes in in the following way. An Aharonov-Bohm potential is not globally a gradient, only locally. Is there any possible physical effect of this? On particles obeying the Lorentz force law, obviously not. The Aharonov-Bohm effect manifests itself as a shift in an interference pattern, not as a local force on a point particle. So it's only with quantum mechanics that there can be a physical effect in the absence of a nonzero electromagnetic tensor (thinking just right now I couldn't guarantee classical waves interacting with the electromagnetic field couldn't exhibit an Aharonov Bohm effect but it seems unlikely).
And here's where the gauge principle comes in: the line integral of the vector potential around a loop is a mysterious quantity physically, but mathematically it is the holonomy of a connection on a line bundle. And gauge invariance is a physical interpretation of the mathematical theory of bundles. It is actually not necessary to talk about gauge invariance or the gauge principle as a way to deduce electromagnetism from first principles. It can just as well be considered a property of the constructs used to describe experimental data. But it seems philosophically to have a lot explanatory power to reduce lots of disparate facts about electromagnetism to "quantities are physically observable if and only if they are gauge invariants of the electromagnetic line bundle". And it so happens that in the theory of line bundles not all gauge invariants are local quantities. Some of them are holonomies around noncontractible loops, hence the Aharonov Bohm effect in addition to the (fully local) Lorentz Force law. Miguel (talk) 18:23, 19 January 2011 (UTC)

The case that the solenoid is far from the two slit

I imagine the case that the solennoid is very far from the two slit.
In that case, Aharonov-Bohm effect continues or disappears?.
Custum quantum theory does'nt deal with the that case. .
Qauntum theory does'nt consider the distance between the solenoid betweeen thw slit.
If one knows the answer, contact me, — Preceding unsigned comment added by Younghun park (talkcontribs) 03:20, 5 October 2011 (UTC)

The case of the solenoid far from the slit isn't interesting because then the particle can pass through the solenoid. The cool thing about the effect is that the particle responds to changes in a region it can't pass through. — Laura Scudder | talk 21:02, 9 November 2012 (UTC)
Tthe whole point of the experiment is that the walls of the solenoid are made thick and dense enough that they are effectively impenetrable by electrons, so regardless of the distance the particles cannot pass through the solenoid. What matters is whether they can pass around the solenoid by two different paths, and in principle this could occur regardless of the distance from the slits to the solenoid. — Steven G. Johnson (talk) 16:35, 10 November 2012 (UTC)

Article is misleading

See here why. Count Iblis (talk) 18:07, 28 October 2011 (UTC)

Very interesting. Tell me if I have understood this: the magnetic AB effect is caused by a temporary rotational differential in the solenoid, caused by the passage of the electron, which induces an entangled B field outside the solenoid? This entanglement vanishes after the electron has passed.-- cheers, Michael C. Price talk 06:09, 29 October 2011 (UTC)

What is the correct gauge?

Consider the classical electromagnetic maxwell equations: before this gauge-dependent effect was observed, people were free to choose their gauge (the specific choice of gauge (Lorentz gauge, Coulomb gauge, ...) was considered physically irrelevant); then this effect showed that the gauge does influence the physics (i.e. part or all of the degrees of freedom in choosing the gauge are eliminated). A pressing question is: which gauge or subclass of gauges is correct? Its like saying we can measure in some sense the gauge, but the page does not say what gauge/subclass of gauges was determined through the effect... — Preceding unsigned comment added by (talk) 23:58, 21 April 2012 (UTC)

The IP spa misunderstands. The effect is gauge-invariant. Xxanthippe (talk) 00:18, 22 April 2012 (UTC).
"The effect is gauge-invariant." if I change my gauge such that E,B along the paths of electron dont change (by changing the B where the electron doesnt pass) but A does change our measurement changes => not gauge-invariant. The pre AB effect maxwell equations are gauge-invariant, The effect is not. — Preceding unsigned comment added by (talk) 05:59, 22 April 2012 (UTC)
bump — Preceding unsigned comment added by (talk) 10:12, 25 April 2012 (UTC)
The effect is gauge invariant. We say the effect as being caused by the four-potential, which has gauge freedom, because otherwise somehow the particle knows about electromagnetic fields in places it never goes, but when we do the calculations, all the observables end up depending only on gauge invariant quantities like the magnetic flux. This is analogous to working a classic electrostatics problem, where you have some scalar electric potential that you are free to add any number to, because all the physical effects end up only depending on differences in potential.
I think the thing you're missing is that if you make a gauge transformation, E and B don't just stay the same along the particle's path, but everywhere, including inside the solenoid. — Laura Scudder | talk 21:00, 9 November 2012 (UTC)


This article could use a few more diagrams. Please see here for some. I'll draw another soon for the double-slit interference of an electron beam around a solenoid, a separate one complete with a "magnetic wisker" (assuming that's the correct name for the thing to deflect the beam and shift the interference pattern...). Thanks, M∧Ŝc2ħεИτlk 06:34, 9 May 2013 (UTC)

  1. Richard Feynman, The Feynman Lectures on Physics Vol II, Sec 15-5, "knowledge of the classical electromagnetic field acting locally on a particle is not sufficient to predict its quantum-mechanical behavior."