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Consider Maxwell's equation (without source):

$$ \partial_\mu F^{\mu \nu} = 0 \implies \partial_\mu \partial^\mu A^\nu = \partial_\mu \partial^\nu A^\mu.$$

Can we find a pair of classical field configurations $A^\mu(x),A'^\mu(x)$ such that they both satisfy the equation above (assuming similar boundary conditions) but are not related to each other by a gauge transformation of the type:

$$A'^\mu(x) = A^\mu(x)+\partial^\mu \varphi(x) \quad ?$$

If it's impossible, how could we argue/show this?

Answer: Thanks to my2cts' answer, I've found 2 solutions $A$ and $A'$ not related by a gauge transformation : $A^\mu=(0,e^{-i(t-y)},0,0)$ and $A'^\mu=(0,0,e^{-i(t-x)},0)$. It makes sense since they both give rise to different EM fields, which are invariant under gauge transformations.

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    $\begingroup$ I don't think one can answer this question until the boundary conditions are defined. $\endgroup$
    – akhmeteli
    Feb 14 '19 at 5:52
  • $\begingroup$ I think that reading the comments and answer to my question posed in the following link answer this question. physics.stackexchange.com/q/452308 $\endgroup$ Feb 14 '19 at 7:20
  • $\begingroup$ @akhmeteli I completely agree and rarely if one ever sees them mentioned in the context of the vector potential but why? $\endgroup$
    – hyportnex
    Feb 14 '19 at 12:54
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    $\begingroup$ Just for the record: there an even more trivial example. $A=0$ is a solution, the set of all solutions that differ from it by a gauge transformation are $A=\mathrm{d}\phi$. All such solutions have $F=0$. Clearly there exist solutions with non zero $F$. $\endgroup$
    – MannyC
    Feb 14 '19 at 18:09
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Of course there are many solutions that are unrelated by a gauge transformation, namely plane waves of any frequency, propagation direction, polarisation and phase and superposition thereof.

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  • $\begingroup$ @my2cts If I understand your point, the idea is that we can find 2 linearly independent solutions $E,B$ and $E',B'$ that solve Maxwell's equation, and since the EM fields are unaffected by a gauge transformation we cannot relate them to each other? $\endgroup$
    – Jasmeru
    Feb 14 '19 at 15:24
  • $\begingroup$ It is not obvious that all plane waves satisfy the same boundary condition, as I wrote in my comment, this depends on the boundary condition. $\endgroup$
    – akhmeteli
    Feb 16 '19 at 3:22

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