As it is well knowing, $U\left(1\right)$ is the group of the unitary matrix of the first order and that this group is connected with rotation operations. Under the complex scalar field perspective $$\phi\left(x\right)=\frac{1}{\sqrt{2}}\left[\phi_{1}\left(x\right)+i\phi_{2}\left(x\right)\right],\tag{A}$$ the $U\left(1\right)$ is associated with rotations in the 2-dimensional inner space (with $\phi_1$ and $\phi_2$ being the axis coordinates). In $U\left(1\right)$ representation, this rotation is given by $$\phi^\prime\left(x\right)=U\phi\left(x\right),\tag{B}$$ where $U=e^{-i\Lambda}$, with $\Lambda$ representing the rotation phase (angle) in clockwise. Clearly, $U$ is unitary because $$U^{\dagger}U=e^{+i\Lambda}e^{-i\Lambda}=1,\tag{C}$$ since that $\Lambda^{*}=\Lambda$.

Similarly, if Maxwell field is invariant under $U\left(1\right)$ symmetry, we should hope that there is a unit operator $U=e^{-iS}=1-iS$ such that $$ A^\prime_{\mu}\left(x\right)=UA_{\mu}\left(x\right)=\left(1-iS\right)A_{\mu}\left(x\right)\rightarrow \delta A_{\mu}\left(x\right)=-iSA_{\mu}\left(x\right),\tag{D}\label{D}$$ where $\delta A_{\mu}\left(x\right)=A^\prime_{\mu}\left(x\right)-A_{\mu}\left(x\right)$. However, we know that the gauge transformation to Maxwell field is given by $$A^\prime_{\mu}\left(x\right)=A_{\mu}\left(x\right)+\partial_{\mu}\chi\left(x\right)\rightarrow \delta A_{\mu}\left(x\right)=\partial_{\mu}\chi\left(x\right).\tag{E}\label{E}$$ Comparing Eq. \eqref{D} with \eqref{E}, we have $$-iSA_{\mu}\left(x\right)=\partial_{\mu}\chi\left(x\right),\tag{F}$$ and it is not clear as $S$ and $\chi$ relate such that the Maxwell field have $U\left(1\right)$ symmetry.

Follows from there the title question: How is possible to see that Maxwell's field have $U\left(1\right)$ symmetry?

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    $\begingroup$ The $U(1)$ symmetry doesn't act on the photon field through a phase (remember -- the photon field is real). Other fields, like the Dirac field, transform under $U(1)$ through a phase, and the photon field transforms in a way to make the $U(1)$ symmetry local. $\endgroup$ – Bob Knighton May 13 at 20:44
  • $\begingroup$ @BobKnighton , do you could to talk a little more about this last part? Perhaps with some practical example. $\endgroup$ – lucenalex May 13 at 21:29
  • $\begingroup$ @BobKnighton, based in your answer above I question: (a) The U(1) symmetry act on photon? and (b) In what way? $\endgroup$ – lucenalex May 13 at 23:27
  • $\begingroup$ @BobKnighton , so, actually, the gauge field doesn't have U(1) symmetry due not be possible to build a unity operator? If answer yes, so it is only said that it preserves the U(1) symmetry, not because he has such symmetry, but because it is necessary to preserve it. Right? $\endgroup$ – lucenalex May 14 at 12:46

The photon field does not transform in the same way that the electron fields, or any other fields, do.

This is easiest to see if we consider a non-abelian field $G$. The gauge field $A_{\mu}$ is now a field taking values in the lie algebra $\mathfrak{g}=\text{Lie}(G)$, and transforms under a general gauge transformation as

$$A_{\mu}\to g^{-1}A_{\mu}g+g^{-1}\partial_{\mu}g.$$

In particular, if $g\in G$ is constant, then $A\to g^{-1}Ag$. This shows that $A$ transforms in the adjoint representation.

The reason we don't see this in the case of the photon field is that $U(1)$ is abelian, and the adjoint representation of $U(1)$ is thus the trivial representation. The only way that we see that the total gauge group is $U(1)$ is because we impose the transformation laws

$$\psi(x)\to e^{i\alpha(x)}\psi(x),$$

where $g(x)=e^{i\alpha(x)}$ is a local $U(1)$ gauge transformation.

  • $\begingroup$ So, according to exposed above, if $S$ is an element belongs to $U(1)$, so any quantity which to be invariant by a representation self-adjoint $SAS=A$ is invariant by $U(1)$ symmetry. Right? In this case, ever same that $A_{\mu}$ to be displaced by a gradient $\partial_{\mu}\chi=i\left(\partial_{\mu}S\right)S^{-1}$, the $U(1)$ symmetry is not lost, i.e., $A_{\mu}$ continue being invariant by $U(1)$. $\endgroup$ – lucenalex May 14 at 14:45

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