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I know that under $SU(2) \times SU(2)$, the left-handed electron transforms under $ ( \frac{1}{2},0 ) $ representation and the vector gauge field $A_\mu$ under $ ( \frac{1}{2},\frac{1}{2}) $.

Since the electron transforms under $U(1)$, there must be a represenation under which it transforms. What is this representation? Does it have a name?

Apparenly $A_\mu$ does not transform under the same representation, which would mean $e^{\alpha(x) Q} A_\mu$, but instead as $A_\mu + i \partial_\mu \alpha(x)$ ? What representation is this?

Of course I realize that the transformation of $A_\mu$ can't be different for the Lagrangian to be invariant, but that shouldn't be used to define the it.

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    $\begingroup$ Mathematically, the gauge field are not the same kind of object than the other fields, they represent connections en.wikipedia.org/wiki/Connection_(mathematics) $\endgroup$ – Adam Nov 11 '13 at 21:39
  • $\begingroup$ Can you state in an abstract way how their transformation properties are obtained? $\endgroup$ – Konstantin Schubert Nov 11 '13 at 21:40
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The electron field transforms under the $\mathbf 1$ of $U(1)$, i.e., the generator is $i$ or $1$ depending on your convention/notation.

The gauge fields transform in the adjoint representation , but they transform as a connection, as @Adam mentioned. In other words, if $\psi \to g \psi$, then $D_\mu \psi \to g D_\mu \psi$ implies that $A_\mu \to g D_\mu g^{-1}$. It's a bit misleading for $U(1)$ because you don't see the non-Abelian structure, but you get the idea.

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  • $\begingroup$ I think this is the answer to my question. It is more complicated than I thought and I will have to read up on adjoind representations until I will hopefully understand it. $\endgroup$ – Konstantin Schubert Nov 11 '13 at 22:53
  • $\begingroup$ You should read up on gauge theory. There's a lot of beautiful mathematics and connections (no pun intended) to differential geometry and gravity. You start out with a field and a Lagrangian that has some global symmetry $\psi \to g \psi$. You then promote the global symmetry to a local one, by turning derivatives into covariant derivatives (just a mathematical trick). As long as $A$ is a "pure gauge" you've done nothing. But let $A$ be dynamical and out pops all of E&M :) I think it's very beautiful. $\endgroup$ – lionelbrits Nov 11 '13 at 23:46
  • $\begingroup$ I am trying to read up, but most physics literature starts nice and slow on the Lagrange group but then skips ahead when it comes to gauge symmetries. Do you have a recommendation? $\endgroup$ – Konstantin Schubert Nov 12 '13 at 6:35
  • $\begingroup$ Quantum Field Theory by Ryder, perhaps? You should be able to request it from your library. Also Topology and Geometry for Physicists by Nash and Sen if you want to get down and dirty, although my printing had a few typos. Geometry, Topology and Physics by Nakahara is good too. $\endgroup$ – lionelbrits Nov 12 '13 at 11:36
  • $\begingroup$ Do you really mean $A_\mu \to g D_\mu g^{-1}$? Or rather $D_\mu \to g D_\mu g^{-1}$? Otherwise I don’t see how we would get $A_\mu \to A_\mu + i \partial_\mu \alpha(x)$ in the end. $\endgroup$ – Socob May 31 '17 at 9:35
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$U(1)$ is an Abelian group. Abelian groups only have 1-dimensional irreducible representation. Namely, transformation by a phase (in the case of the electron). The charge of fermion field is proportional to the coefficient of the phase. In particular, a field of charge $q$ transforms as $\Psi \to e^{i q \theta(x)} \Psi$

EDIT: As pointed out in the comments, the earlier answer was incorrect.

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    $\begingroup$ Usually, the (most) trivial representation is left untouched by the transformation, right ? For U(1), there is only two possibilities, either the field is untouched (not charged) or take a phase (what you called the trivial representation). For O(2) (which is more or less U(1)), you would have scalar $\phi$, such that $g\phi=\phi$ and the vector ${\bf V}$, such that $g{\bf V}={\bf V'}=R{\bf V}$, where $R$ is a rotation. $\endgroup$ – Adam Nov 11 '13 at 23:15
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    $\begingroup$ @Konstantin, keep in mind that what we call the singlet state in $SU(2)$ would transform by a phase in $U(2)$. $\endgroup$ – lionelbrits Nov 11 '13 at 23:39
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    $\begingroup$ @Prahar, I don't think that's true, even for finite groups. For finite groups, the number of inequivalent irreps is the number of conjugacy classes, which for an abelian group is the order of the group. For U(1), there are infinitely many representations, labeled by $\mathbb{Z}$ (homomorphisms of U(1) into U(1) of the form $\rho_n(e^{i \theta})=e^{in\theta}$). See en.wikipedia.org/wiki/Circle_group#Representations $\endgroup$ – Dan Nov 12 '13 at 2:45
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    $\begingroup$ Sorry. I think I meant to say that all irreducible representations are 1 dimensional. My statement is indeed incorrect. I have edited it. $\endgroup$ – Prahar Nov 12 '13 at 2:48
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    $\begingroup$ @Adam and others, does that mean that different charges (0,1/3,2/3,1,...) denote different representations of U(1)? And summing up those charges in a term is the process of calculating the representation under which the term transforms? $\endgroup$ – Konstantin Schubert Nov 12 '13 at 6:30

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