I don't quite understand what is actually meant by a field charged under a $U(1)$ symmetry. Does it mean that when a transformation is applied the field transforms with an additional phase? More specifically, why is it called "charged"?

  • $\begingroup$ If you remove the word "additional" (additional compared to what?) your definition is correct. The "phase" can mean mixing of two real fields into each other, the fields don't have to be complex (although, of course, the mixing defines how they become complex if you wish to represent them as complex fields with phases). $\endgroup$ – Ron Maimon May 16 '12 at 6:56

"Charged", in the context of a symmetry, means the field is/carries a representation of the corresponding symmetry group.

For instance, $U(1)$ representations are labeled by a real parameter $\alpha$ (the charge). Consider the group element $e^{ig}\in U(1)$. A field with charge $\alpha$ transforms according to

$$ \psi \rightarrow e^{i\alpha g} \psi$$

In general $\psi$ can be a vector (say, for $SU(2)$) and it transforms as $\psi_a\rightarrow \sum_b\rho(g)_{ab}\psi_b$. Here the mapping $\rho$ is called the representation and $\psi$ is an element of the representation space.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.