Electric charge and spontaneous symmetry breaking

[Neitrino]

Hi,

If I have a Lagrangian of complex scalar field (just U(1) local invariance).
And I know that phi^star describes field with -e electric charge and phi describes field with e electric charge. How do I apply "charge issue" when I write Lagrnangian after spontaneous symmetry breaking in terms of Goldstone (which are afterwards adsorbed) and Higgs modes ?
They are charged not charged ? why how ? they become real fields and to have suppose electric charge fields shoud be complex...


Thanks a lot

 

[Neitrino]

Silly question... ? :(

 

[DrDu]

Silly question... ? :(

No, not at all, but you could be more specific.
I would proceed as follows:
The current operator is obtained as the functional derivative of the Lagrangian with respect to the magnetic vector potential A. And the new Lagrangian is quadratic in the vector potential which has eaten the Goldstone boson. I.e. if I rewrite the original field as
$$\phi=\phi_0 +\rho \exp(i\sigma)$$ then the new lagrangian depends only on rho and $$\tilde{A}=A-\frac{1}{e} \rho \nabla \sigma$$, or $$L=e^2|\tilde{A}|^2+ \ldots$$ hence the current should be something like $$e\tilde{A}$$. Thus the current is carried entirely by the Higgs field.