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Read somewhere that potential is a more fundamental quantity than EM field because if the latter is more fundamental then gauge transformation will reduce to nothing more than a mathematical trick. I am thinking it another way. Please point out whether my assumption is right or wrong. Since the action principle is the most fundamental of all, anything that is involved in it has to be more fundamental. 4-potential is involved in the action integral in the term describing the interaction of the particle and the field. So doesn't that make it in that sense more fundamental? After all EM field tensor itself is constructed from the potential term in the EOM.

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Independently of Quantum Mechanics, if you believe in the variational method, then the $A$ field is more fundamental than the $\vec{E}, \vec{B}$ fields since the lagrangian cannot give the Maxwell equations without the 4-potential. You need the potentials to make a variation of the fields that will give you their equations of motion (i.e. Maxwell's equations).

Also, even at the classical level, Statistical Mechanics need an hamiltonian, which asks for the potentials. How would we do Classical Statistical Mechanics without the electromagnetic potentials ?

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That the vector potential A of the electromagnetic field is not just a mathematical trick and has some physical significance has been demonstrated experimentally by the Aharonov-Bohm effect. Formerly the opinion prevailed that the the vector potential was a mathematical convenience to derive the electric amd magnetic fields.

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  • $\begingroup$ Isn't it the line integral of A that is important in the A-B effect? Not quite the same thing. $\endgroup$ – Rob Jeffries Sep 21 '16 at 6:49
  • $\begingroup$ @ Rob Jeffries - This is correct. But it is the line integral over a space region where there is no B or E-field. Thus some sources claim that this is evidence that the A-field has direct physical significance. $\endgroup$ – freecharly Sep 21 '16 at 12:55
  • $\begingroup$ The "proper" way to think about what the A-B effect says about the vector potential as an observable is in terms of the bundle formulation of gauge theory, see here physics.stackexchange.com/questions/77368/… $\endgroup$ – Robin Ekman Oct 22 '16 at 16:22
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The vector potential is necessary for describing QED. You might wonder if you can get away with writing your theory only in terms of the trivially gauge invariant tensor $F_{\mu\nu}$. The problem with that is that photons are spin one particles, and I don't know of a way to quantize $F_{\mu\nu}$ that makes just spin 1 degrees of freedom propagate. On the other hand it is not (too) hard to make a theory of spin one particles from a vector field $A_\mu$.

As another complication, if $F_{\mu\nu}$ is to annihilate asymptotically free particles then it must have dimension 2, and so it has no renormalizeable couplings with fermions which would make writing down any sensible QFT problematic.

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I think it becomes clear that the $4$-potential is the more fundamental quantity when one views the issue from a geometric perspective. It can be said that $A$ is a $\mathfrak{u}(1)$-valued connection one-form, and in this case, $F=dA$ has the interpretation of the curvature; it is after all a commutator of derivatives.

This view of $F$ as curvature is even more apparent when examining gauge anomalies which turn out to be proportional to Chern classes involving $F$, and the Chern class of the tangent bundle of a manifold is indeed in terms of the curvature two-form.

Analogously in GR, one thinks of the metric $g_{ab}$ as more fundamental than any of the curvature tensors one may derive from it; after all one would not say $R_{ab}$ is more fundamental.

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The following paper (I am one of the authors) explain the foundamental role of electromagnetic four potential:

Maxwell's equations and Occam's razor

from the abstract: "In this paper a straightforward application of Occam's razor principle to Maxwell's equation shows that only one entity, the electromagnetic four-potential, is at the origin of a plurality of concepts and entities in physics"

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  • $\begingroup$ Hi Stefano. Welcome to Phys.SE. Are you in any way related to the author of the link? For your information, Physics.SE has a policy that it is OK to cite oneself, but it should be stated clearly and explicitly in the answer itself, not in attached links. $\endgroup$ – Qmechanic Jun 21 '18 at 7:43
  • $\begingroup$ Hi, welcome to Physics SE! I think this may be better as a comment on the question instead... $\endgroup$ – user191954 Jun 21 '18 at 8:23

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