# What determines the charge of a field?

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## Main Question or Discussion Point

I understand what we classically know as the charge of a particle is actually the parameter of the local phase symmetry of the field the particle belongs to, the Noether current of which permits its coupling to the electromagnetic field. But when a field has phase symmetry it is symmetric under any phase transformation ψ → ψe, not just the one with a specific α as parameter.

So why is it that fields just couple with the current corresponding to one charge? Would it be possible for the Dirac field to couple to the electromagnetic field or to the scalar field with a coupling constant other than e?

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vanhees71
Gold Member
2019 Award
The charge is given by the coupling constant in the Lagrangian. It appears in the local gauge transformation at some place too, but it's not the parameter in the gauge transformation that determines the coupling constant.

The charge is given by the coupling constant in the Lagrangian. It appears in the local gauge transformation at some place too, but it's not the parameter in the gauge transformation that determines the coupling constant.
Then there's some detail I misunderstood. To me the local phase symmetry ψ → ψeiq gives rise to a conserved current eψ*γ0γμ ψ, and when we use that current to couple the Dirac field to the EM field the parameter e plays the role of coupling constant. Did I get that wrong?

vanhees71
Gold Member
2019 Award
No, it's all right. You introduce the coupling constant in the principle of minimal substitution $\partial_{\mu} \rightarrow \partial_{\mu} + i g A_{\mu}^a T^a$ to make the so far only global symmetry local.

No, it's all right. You introduce the coupling constant in the principle of minimal substitution $\partial_{\mu} \rightarrow \partial_{\mu} + i g A_{\mu}^a T^a$ to make the so far only global symmetry local.
Ok, so the parameter of the transformation doesn't end up bein the coupling constant then?

vanhees71