Aharonov-Bohm topological explanation

In summary, the Aharonov-Bohm effect shows something that cannot be explained by classical physics in the sense that makes observable a classical EM global gauge transformation that shouldn't be observable within classical EM, where only the effects of the fields are observable but not the effects of the potentials. However, in QM where potentials are physical, and so it was theoretically predicted and later experimentally confirmed that charged particles going thru a a region where a magnetic field is negligible show a shift in their diffraction pattern caused just by the vector potential and varying with the flux thru the solenoid.
  • #36
Jano L. said:
OK, I think I understand your question. You assume hypothetical world which has an infinite cylindrical hole inside which the field is B is nonzero, and worry whether the explanation of the B-A shift really works for such setup (because perhaps there is no vector potential).
Right.
Jano L. said:
I believe the answer is no, but for a different reason.

First, I think it is not necessary for the standard quantum-theoretical explanation of the shift. One should get it for ordinary simply-connected space, otherwise there should not have been so much fuss about it.
If this was the case of course my objections dissolve.
Jano L. said:
But let's adopt the above assumption to see where the argument for the shift fails.
...

So, there is vector potential mathematically, but we have to copy the boundary conditions from the real situation where the coil is part of the system. For this reason, I think that the idea that the effect has something to do with breakdown of simple-connectedness is wrong.
Well, it is widely used.
 
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  • #37
TrickyDicky said:
It doesn't forbid its existence per se, but in a physical experiment you have to justify where it comes from, and usually it is related to a magnetic field, or at least in this case , as usually explained, is. Do you mean that in the AB effect experiment the A-field has nothing to do with the magnetic field?

as I said, you do not need the B-field; it's a derived quantity, and all physics can be formulated entirely based on the A-field; of course physically the source of the A-field is a current i.e. the solenoid and one has to check whether this works (this is where Maxwell's equations enter the stage; everything else can be formulated w/o any reference to these equations) of course the solenolid solution is an idealization, but besides that there should be no physical problem
 
  • #38
Jano L. said:
For this reason, I think that the idea that the effect has something to do with breakdown of simple-connectedness is wrong.

The not simply connected region is needed so that dA=0 doesn't lead to A being a constant which gives trivial physics and no effect(no change). So I guess it is a necessary condition.
 
  • #39
not really; dA=0 does not automatically imply A=const; dA=0 means that A is pure-gauge, but here it cannot be gauged away due to the non-trivial topology of the vector bundle defined by R³ * A

but I agree, there is some non-trivial topology involved; whether it's due to R³ \ R or due to the vector bundle or physically due to the solenoid is a matter of taste; but you need something like that
 
  • #40
tom.stoer said:
as I said, you do not need the B-field; it's a derived quantity, and all physics can be formulated entirely based on the A-field;

And as I said this statement has nothing to do with what I'm explaining in this thread.
tom.stoer said:
of course physically the source of the A-field is a current i.e. the solenoid and one has to check whether this works (this is where Maxwell's equations enter the stage; everything else can be formulated w/o any reference to these equations)
Sure you can calculate the vector potential from the source current geometry, but a magnetic field is always implied since the existence of A comes from ∇.B=0→∇XA=B , but this derivation needs the additional requirement that the space is simply connected. The opposite way:∇XA=B→∇.B=0 needs no topological assumptions, solenoidal vector fields are divergnce free. Do you agree with this?
 
  • #41
tom.stoer said:
not really; dA=0 does not automatically imply A=const; dA=0 means that A is pure-gauge, but here it cannot be gauged away due to the non-trivial topology of the vector bundle defined by R³ * A

but I agree, there is some non-trivial topology involved; whether it's due to R³ \ R or due to the vector bundle or physically due to the solenoid is a matter of taste; but you need something like that

Right, so what's with the "not really"? ;)
 
  • #42
Here's one of the sources I'm using for the AB effect:"Topology and Geometry for Physicists (Dover Books (Paperback) by Charles Nash, Siddhartha Sen", pages 301-302. They are freely available in the amazon.com reader page.

One thing I note in their explanation is that they use differential forms rather than vector fields and say(calling F the magnetic field): F=0→dA=F which assumes the form is exact but that requires trivial topology and however the AB effect is based on the nontrivial topology.

Besides they are assuming the magnetic field to be conservative, when I was always told the magnetic field is solenoidal but not conservative.
 
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  • #43
the 'not really' means that it's the gauge field itself which induces the non-trivial topology, not necessarily the underlying manifold; but as I said this is a matter of taste

one can discuss this based on differential forms, but of course the results do depend on the formulation; I would recommend Nakahara which seems to better accessable than Nash & Sen
 
  • #44
I have a hard time accepting that simple-connectedness has something to do with this. One gets non-zero loop integral of A easily in normal space. For example, imagine a finite cylinder with surface currents, whose field approaches that of a coil. Let the magnetic field inside is B and the cross section S. The symmetry of the current distribution is such that the vector potential in Coulomb gauge

[tex]
\oint \mathbf A \cdot d\mathbf l = BS \neq 0
[/tex]

over any reasonable curve drawn around the cylinder. One gets the same result even with continuous 3D distribution of currents in the thick wires. It does not matter whether the coil is finite or infinite or whether it creates some mathematical division of space into two parts.
 
  • #45
For a finite cylinder we run into difficulties:

The flux is

[tex]\Phi[C] = \oint_C A [/tex]

Now let's use a pure-gauge A-field

[tex]U = e^{-i\chi(x)}[/tex]
[tex]A = U^\dagger \;id\;U = d\chi[/tex]

where U has winding number n

[tex]U = e^{-in\phi};\;\phi\in [0,2\pi];\;n=0,\pm 1,\pm 2,\ldots[/tex]

The integral is related to this winding number which is a topological invariant; b/c of periodicity of U the integer n cannot be changed continuously (*)

[tex]\Phi[C] \sim n[/tex]

Now assume that the loop C is deformed smoothly and moved continuously through space. Therefore we know that
a) n changes continuously -- but is ruled out b/c of (*)
b) n changes discontinuously -- but is ruled out b/c we allow only for smooth deformations of C
c) n does not change at all
b/c (a) and (b) are rules out we know that no matter how we move C, n must not change.

But with a finite cylinder the loop C could be moved freely which may contradict integer n ≠ 0. Therefore moving C freely is ruled out by topology of the pure gauge A-field.
 
  • #46
Can you please explain what you did above? I am sorry, but I do not see relevance of your calculation. We can calculate the loop integral through any closed contour by using standard parametrization. Deforming the contour into point will result in continuous change of flux from initial value into 0. The contour relevant for calculation is such that the flux is non-zero; apart from that, there is no reason to insist it is integer multiple of something else...
 
  • #47
TrickyDicky said:
It doesn't forbid its existence per se, but in a physical experiment you have to justify where it comes from, and usually it is related to a magnetic field, or at least in this case , as usually explained, is. Do you mean that in the AB effect experiment the A-field has nothing to do with the magnetic field?

I think in principle nontrivial A fields need not be associated with B fields (of course, in the case of a solenoid we always get both). Suppose we *actually* lived in R^3 \ (an infinite cylinder). Then we might well find that when we passed charged particles around the defect, we got an Aharonov-Bohm interference effect. This would demonstrate the existence of an A field around the defect. Perhaps someone will correct me on this, but I think this gives a perfectly consistent theory of electrodynamics. Physicists in this universe should include a parameter in their models which they might poetically call "the magnetic flux through the defect," even though no B field can be detected. They would have to determine this parameter by experiment.
 
  • #48
The_Duck said:
I think in principle nontrivial A fields need not be associated with B fields (of course, in the case of a solenoid we always get both). Suppose we *actually* lived in R^3 \ (an infinite cylinder). Then we might well find that when we passed charged particles around the defect, we got an Aharonov-Bohm interference effect. This would demonstrate the existence of an A field around the defect. Perhaps someone will correct me on this, but I think this gives a perfectly consistent theory of electrodynamics. Physicists in this universe should include a parameter in their models which they might poetically call "the magnetic flux through the defect," even though no B field can be detected. They would have to determine this parameter by experiment.

Yes, to me the existence of an A field is not actually in doubt. But I'm finding it paradoxical under the usual assumptions with respect to Maxwell equations, closed line integrals and Stokes theorem dependence on trivial topology and the effect itself dependence on nontrivial one, so far I haven't found a way out.
 
  • #49
A question: what do you mean by "non-existence of the B-field"; do you mean that B=0? or do you mean that the B-field does not exist mathematically?
 
  • #50
tom.stoer said:
A question: what do you mean by "non-existence of the B-field"; do you mean that B=0? or do you mean that the B-field does not exist mathematically?

No, I'm of course referring to the region where B=0, outside the solenoid, otherwise it wouldn't make sense to talk about the A-field either.
But I don't recall saying those words.
 
  • #51
TrickyDicky said:
My entire point was that the Stokes theorem requires simply-connectednes to hold.
And mine is (partly) that regarding the AB setup as a plane with a point (or small disc) removed (hence not simply connected) is an unphysical idealization. The space of the (physical) experiment does not have a hole. It has a region where curl A is nonzero. The topological arguments are a distracting waste of time.
[...] I'm finding it paradoxical under the usual assumptions with respect to Maxwell equations, closed line integrals and Stokes theorem dependence on trivial topology and the effect itself dependence on nontrivial one, so far I haven't found a way out.
You could try solving the Maxwell equations expressed in terms of just A and a source current j. I.e., for an AB-like current configuration, find A. Then integrate your A solution around a closed loop...
 
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  • #52
Jano L. said:
Can you please explain what you did above? I am sorry, but I do not see relevance of your calculation.
I'll come back to it later
 
  • #53
Let's summarize:
- the B-field is neither required nor sufficient to explain the Aharonov-Bohm effect
- the B-field can be regarded as a derived entity, whereas the A-field is fundamental
- this holds not only in QM but already in classical electrodynamics (Maxwell's equiations can be formulated w/o using the B-field)
- it is wrong (although it is often claimed) that the A-field is unphysical or unobservable; the loop integral represents a (gauge invariant) classical observable

strangerep said:
... regarding the AB setup as a plane with a point removed is an unphysical idealization. The space of the (physical) experiment does not have a hole. ... The topological arguments are a distracting waste of time.
Not really; I agree that physically the situation must be described using the solenoid rather than the punctured plane; but for the calculation of the phase shift there is no difference; the electron can't distinguish between the "physical" and the "topological" scenario b/c it can't penetrate the solenoid. Therefore the topological consideration is relevant.
 
  • #54
Don't the mechanics of "spin" enter into consideration here also? A rotating particle containing an extended charge or sub-charge will generate a magnetic dipole protruding from both ends of the rotation axis. But if the rotation is complex, possibly consisting of several rotations about several axes, then magnetic monopoles or quasi-monopoles would exist in the microscopic realm. The fields would no longer be simply-connected.

There seems to be a history, starting from Heaviside, to allow for the possibility of the existence of magnetic monopoles in an initial, primitive invocation of the Maxwell equations for particular problems where those monopoles are removed prior to the finalization of a particular solution to the EM problem. Doing so could be interpreted as invoking SU(n) topology in the beginning and resolving it into U(1), couldn't it?
 
  • #55
strangerep said:
And mine is (partly) that regarding the AB setup as a plane with a point (or small disc) removed (hence not simply connected) is an unphysical idealization. The space of the (physical) experiment does not have a hole. It has a region where curl A is nonzero. The topological arguments are a distracting waste of time.
Hmm, your opininion contradicts all the texts I've consulted.


You could try solving the Maxwell equations expressed in terms of just A and a source current j. I.e., for an AB-like current configuration, find A. Then integrate your A solution around a closed loop...
Integrating around the closed loop independently of path requires the trivial topology, so only in case you were right and topology was irrelevant would you have a point.
 
  • #56
tom.stoer said:
Let's summarize:
- the B-field is neither required nor sufficient to explain the Aharonov-Bohm effect
...

What do you mean is not required, there is no effect without a magnetic flux inside the solenoid, is there?
 
  • #57
I think it doesn't make sense to repeat everything umpteen times; the B-field is a derived quantity; the A-field is fundamental and sufficient; the A-field is pure gauge locally, so no B-field is assoiciated with it; the B-field is zero outside; the electron can't "look" into the solenoid and therefore does not see any B-field; the magnetic flux is not the same as the magnetic field, but it's a loop-integral over the A-field, the flux around the loop does not require a B-field inside b/c the electron cannot distinguish between the B-field in the solenoid and a pure gauge w/o any B-field ...

As long as you insist on using the B-field you will never be able to understand the AB effect, local gauge theory with global gauge effects and other consequences.
 
  • #58
TrickyDicky said:
Integrating around the closed loop independently of path requires the trivial topology, so only in case you were right and topology was irrelevant would you have a point.
No!

Integrating around closed loops tells you something regarding topology, winding number etc. The integral is NOT path-independent in general but only path-indep. for homotopic paths. That's one key lesson: the magnetic flux IS a topological quantity related to the first homotopy group of the base manifold.
 
  • #59
So do we agree at least that the solenoid cross section acts as a hole of the space of the paths of the electrons?
If so, how does the independent of path closed line integral of A work?
It is usually stated there must be no holes for the Green and Stokes theorems to hold. Is this wrong?
 
  • #60
tom.stoer said:
I think it doesn't make sense to repeat everything umpteen times; the B-field is a derived quantity; the A-field is fundamental and sufficient; the A-field is pure gauge locally, so no B-field is assoiciated with it; the B-field is zero outside; the electron can't "look" into the solenoid and therefore does not see any B-field; the magnetic flux is not the same as the magnetic field, but it's a loop-integral over the A-field, the flux around the loop does not require a B-field inside b/c the electron cannot distinguish between the B-field in the solenoid and a pure gauge w/o any B-field ...

As long as you insist on using the B-field you will never be able to understand the AB effect, local gauge theory with global gauge effects and other consequences.
I agree with all the points you list, that shows to me you are not really getting my point.
 
  • #61
tom.stoer said:
No!

Integrating around closed loops tells you something regarding topology, winding number etc. The integral is NOT path-independent in general but only path-indep. for homotopic paths. That's one key lesson: the magnetic flux IS a topological quantity related to the first homotopy group of the base manifold.
Once again you start with a negative and then go on to say what I'm saying in other words rather than answering specific questions.
 
  • #62
TrickyDicky, I don't get your point b/c there seems to be no point at all!

I do not start with the negative! I simply say that the integral is not path-independent in general (this is contary your statement). The path-dependence related to the first homotopy group is is one key factor in understanding the topology of the AB effect. So I do not simply repeat what you are saying "in other words", there is a fundamental difference

And I do not answer specific questions b/c I don't see them. So what are your questions? (and what is missing in our explanations?)
 
  • #63
tom.stoer said:
TrickyDicky, I don't get your point b/c there seems to be no point at all!
mfb, strangerep and Jano L. got what I was saying perfectly well. For the last two the solution came from discarding the topology as relevant to the effect, an opinion I don't agree with.

tom.stoer said:
I do not start with the negative! I simply say that the integral is not path-independent in general (this is contary your statement). The path-dependence related to the first homotopy group is is one key factor in understanding the topology of the AB effect. So I do not simply repeat what you are saying "in other words", there is a fundamental difference
Well, then you dindn't understand my wording, I was precisely underlining that the path independence is NOT general but TOPOLOGY DEPENDENT, are you actually reading what I write?
tom.stoer said:
And I do not answer specific questions b/c I don't see them. So what are your questions? (and what is missing in our explanations?)
Have you not spotted posts #56 and #59 for instance?
 
  • #64
TrickyDicky said:
I was precisely underlining that the path independence is NOT general but TOPOLOGY DEPENDENT

For a particle that behaves symmetrically with respect to external fields, path independence would be broken if the curl of the particle's internal field (projected outward) were interrupted by an obstacle of some kind, right? Are there other obvious reasons for a breakdown of path independence?
 
  • #65
This is how the magnetic AB effect is described in wikipedia just for reference:"The most commonly described case, sometimes called the Aharonov–Bohm solenoid effect, takes place when the wave function of a charged particle passing around a long solenoid experiences a phase shift as a result of the enclosed magnetic field, despite the magnetic field being negligible in the region through which the particle passes and the particle's wavefunction being negligible inside the solenoid. This phase shift has been observed experimentally.
The magnetic Aharonov–Bohm effect can be seen as a result of the requirement that quantum physics be invariant with respect to the gauge choice for the electromagnetic potential, of which the magnetic vector potential A forms part.

Electromagnetic theory implies that a particle with electric charge q traveling along some path P in a region with zero magnetic field B, but non-zero A (by [itex] \mathbf{B} = 0 = \nabla \times \mathbf{A}[/itex]), acquires a phase shift [itex] \varphi[/itex], given in SI units by

[itex] \varphi = \frac{q}{\hbar} \int_P \mathbf{A} \cdot d\mathbf{x},[/itex]

Therefore particles, with the same start and end points, but traveling along two different routes will acquire a phase difference [itex] \Delta \varphi[/itex] determined by the magnetic flux [itex]\Phi_B[/itex] through the area between the paths (via Stokes' theorem and [itex] \nabla \times \mathbf{A} = \mathbf{B})[/itex], and given by:

[itex] \Delta\varphi = \frac{q\Phi_B}{\hbar}. [/itex]
In quantum mechanics the same particle can travel between two points by a variety of paths. Therefore this phase difference can be observed by placing a solenoid between the slits of a double-slit experiment (or equivalent). An ideal solenoid (i.e. infinitely long and with a perfectly uniform current distribution) encloses a magnetic field B, but does not produce any magnetic field outside of its cylinder, and thus the charged particle (e.g. an electron) passing outside experiences no magnetic field B. However, there is a (curl-free) vector potential A outside the solenoid with an enclosed flux, and so the relative phase of particles passing through one slit or the other is altered by whether the solenoid current is turned on or off. This corresponds to an observable shift of the interference fringes on the observation plane." End quote

I'm only centering on this part of the quote: "but non-zero A (by [itex] \mathbf{B} = 0 = \nabla \times \mathbf{A}[/itex])", to make the addition that is not mentioned in the wikipedia that what is between parenthesis only hold when the space is simply connected, and that requirement is precisely the one demanded by the explanation of the effect that not be fulfilled.
Is this apparent contradiction so hard to see?
 
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  • #66
PhilDSP said:
For a particle that behaves symmetrically with respect to external fields, path independence would be broken if the curl of the particle's internal field (projected outward) were interrupted by an obstacle of some kind, right? Are there other obvious reasons for a breakdown of path independence?

I'm not sure what you mean by "behaves symmetrically with respect to external fields" and " internal field (projected outward)".
Anyway here the obvious reason for a breakdown of path independence is the nontrivial topology.
 
  • #67
TrickyDicky said:
... then you didn't understand my wording, I was precisely underlining that the path independence is NOT general but TOPOLOGY DEPENDENT, are you actually reading what I write?
Where have you written this statement?

TrickyDicky said:
Have you not spotted posts #56 and #59 for instance?
I saw these posts and I already commented them:
In #56 you are asking "What do you mean is not required, there is no effect without a magnetic flux inside the solenoid" I said (quite often in the meantime) that there is a difference between the magnetic FIELD (which is zero outside and which is NOT required) and the magnetic FLUX as calculated via the line integral.

Please try to get the following statement: the AB effect does not require any B-field; it can be expressed purely in terms of the A-field; and the A-field is evaluated purely outside the solenoid where the B-field is zero. The electrons do not interact with an B-field but with an A-field!

In #59 you are asking "how does the independent of path closed line integral of A work?" I think we agree that there is no path independence in general.

OK; so what are you specific questions that have not been addressed? (sorry for insisting on these questions, but I want to avoid answering the wrong questions or addressing issues that are already clear to you)
 
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  • #68
tom.stoer said:
Where did you wriute this?

Tom, it is the key to my argument so it is all over my posts but for instance "Integrating around the closed loop independently of path requires the trivial topology" meaning that it requires simply connectednes, in post #55.
 
  • #69
I am not sure whether I get this point (which was not addressed to me, I guess) "Integrating around the closed loop independently of path requires the trivial topology".

The phase shift measured is

[tex]\Delta\phi \sim \oint_C A[/tex]

This does neither require trivial topology (or Stokes theorem to calculate the flux) nor any B-field.

Let's cite Wikipedia:

The Aharonov–Bohm effect ... is a quantum mechanical phenomenon in which an electrically charged particle is affected ... despite being confined to a region in which both the magnetic field B and electric field E are zero. The underlying mechanism is the coupling of the electromagnetic potential with the complex phase of a charged particle's wavefunction ...
This is the AB-effect!

The next sentence refers to the very specific case of solenoids, but this is only one possible experimental setup, not the general case:
The most commonly described case, sometimes called the Aharonov–Bohm solenoid effect, ...

So all you need is

[tex]\oint_C A \neq 0[/tex]
 
  • #70
See #65
 

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