 Solution

For a constant real parameter, ${\partial }_{\mu }(e^{i\alpha }\varphi )=e^{i\alpha }{\partial }_{\mu }\varphi$. The phases therefore cancel separately in the kinetic term and in ${\varphi }^{\ast }\varphi$, leaving any $f({\varphi }^{\ast }\varphi )$ unchanged. This is the global phase symmetry of a complex scalar field, with circle group $U(1)$.

Use infinitesimal variations per unit parameter, $\delta \varphi =i\varphi$, $\delta {\varphi }^{\ast }=-i{\varphi }^{\ast }$. The Lagrangian density is exactly invariant, so no total-derivative correction is needed in Noether's theorem. The Noether current isIt depends only on derivatives of the Lagrangian density with respect to field derivatives, so the derivative-free scalar potential cannot alter its form. Direct differentiation cancels the mixed first-derivative terms:Thus it is conserved on solutions of the Euler-Lagrange equations. Its spatial integral is a conserved Noether charge when the flux through the boundary vanishes. Its overall sign is fixed here by $\delta \varphi =+i\varphi$; reversing the orientation of the phase transformation reverses the current.