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What is the conclusion that we can draw from the Aharonov-Bohm effect? Does it simply suggest that the vector potential has measurable effects? Does it mean that it is a real observable in quantum mechanics? Does this effect has anything to do with the topology of the space?

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Beyond the polemics, I believe one can still answer this question.

The gauge-potential can not be an observable in quantum mechanics, since it is gauge covariant. Quantum mechanics is clear about that. In a first quantised version of QM, you measure only $\left\vert\Psi\right\vert^{2}$, and it is not affected by a gauge transform.

This paragraph does not really answer your question, but it may tell you something interesting. Note that in the double slit experiment the observable is the modulus squared of the electric field at a given position. So something which is gauge invariant. Next, if you look at one position only (on the screen say) you should notice nothing: you have some light reaching the screen. Fine! You need to look at different positions to record a change of the intensity. Since most of the time you see the full screen when you perform the expriment, the remark above is scarcely made. For the Aharonov-Bohm effect you record a current, not the gauge potential. From this current you infer that something gets strange when the magnetic field is changed. Obviously the interference pattern depends on the magnetic flux inside the loop. So for one flux value, you get one path, and no interference. Since the current is by construction gauge invariant, everyone is happy.

Of course the topology matters, since it is a magnetic effect. There exist more recent Aharonov-Bohm-like effects, where it is the topology of the Hilbert space itself which matters, not the topology in space. See e.g. Berry phase on Wikipedia. Sorry, the subject is now so vast that I prefer not to enter into details.

As for the interpretation of the AB effect, it is once again a matter of convenience: mathematics rules, words don't !

Yet, a strange fact: why do we record something like a magnetic flux when there is no magnetic field? Well, because a flux is non-local, since one needs to close the loop to observe it. That's the really strange effect which hurts most of the physicists born after Einstein, because it really look non-local. But I never really understood why, since a flux was a flux before him, and he never discussed charged matter (in opposition to charged particles, as in his 1905 article). I believe most of the polemics come from the subtle difference between a gauge field, a gauge potential and the integrals of them, which are equal thanks to the Stoke's theorem [see e.g. Can we measure an electromagnetic field? ] Moreover non-locallity (despite still debated) is well established for most of the physicists.

Next, this even more strange fact: why do we need to take the modulus square of the sum of the different amplitudes $\left\vert \sum_{i} \Psi_{i} \right\vert ^{2}$, and not the sum of the modulus square of the amplitude $\sum_{i} \left\vert \Psi_{i} \right\vert ^{2}$ ? This would be a question on the Huygens principle (or on Feynman path integrals if you wish to be pedantic), not on the AB effect. Still, the AB effect is a beautiful demonstration of the Huygens principle, as the double-slit experiment: the electromagnetic flux inside the loop makes the two paths having different length.

The AB effect is thus a

  • clear demonstration of the gauge principle
  • clear demonstration of the Huygens principle
  • clear demonstration that these two principles together give observable effects

To be sure: what I called a gauge principle here is the fact that a gauge transformation $A\rightarrow A + \nabla \chi$ obliges to transform the wave function as well $\Psi \rightarrow e^{i\chi} \Psi$.

The debate about the gauge potential vs. gauge field is clearly meaningless, since what you record can be expressed in term of both.

The debate about non-locality is much more subtle, and you should be warn of it when you learn the Aharonov-Bohm effect, or the Einstein-Podolsky-Rosen (pseudo-)paradox.

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  • $\begingroup$ This statement is odd : "The gauge-potential can not be an observable in quantum mechanics, since it is gauge covariant. Quantum mechanics is clear about that." It is not true that quantum mechanics implies gauge invariance. $\endgroup$
    – my2cts
    Commented Dec 25, 2019 at 22:58
  • $\begingroup$ @my2cts Sure you're right. That's why I add the next sentence : "The gauge-potential can not be an observable in quantum mechanics, since it is gauge covariant. Quantum mechanics is clear about that. In a first quantised version of QM, you measure only |Ψ|2, and it is not affected by a gauge transform." So to answer your comment: QM does not imply gauge invariance, but QM does not allow gauge-covariant quantity to be measurable. So the gauge-potential can not be an observable in quantum mechanics. $\endgroup$
    – FraSchelle
    Commented Dec 27, 2019 at 9:43
  • $\begingroup$ The Schrödinger equation gives generally gauge invariant results because of the so called minimal coupling, but this does not imply gauge invariance. $\endgroup$
    – my2cts
    Commented Dec 27, 2019 at 12:26
  • $\begingroup$ I removed some mildly inappropriate comments and responses to them. Please keep in mind our behavior expectations on this site. $\endgroup$
    – David Z
    Commented Dec 29, 2019 at 2:01
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The AB effect is gauge invariant. At the same time it shows that gauge invariance does not imply that the potential is unphysical.

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