 Solution

Divide the infinitesimal variation by its constant parameter $\alpha$, so the transformation generator is $(\delta {\varphi }_1,\delta {\varphi }_2)=({\varphi }_2,-{\varphi }_1)$. The variation of the kinetic term isand that of ${\varphi }_1^2+{\varphi }_2^2$ is $2{\varphi }_1{\varphi }_2-2{\varphi }_2{\varphi }_1=0$. Thus the Lagrangian density is invariant pointwise, without a nonzero boundary current. The finite transformation is a rotation of the two real scalar fields, giving an SO(2) global symmetry in field theory.

Noether's theorem supplies the parameter-independent Noether currentIts divergence is ${\partial }_{\mu }{j}^{\mu }={\varphi }_2\square {\varphi }_1-{\varphi }_1\square {\varphi }_2$, since the first-derivative terms cancel. Both fields obey the same Klein-Gordon equation, so this vanishes on solutions. With spatial boundary conditions giving zero flux at infinity, the Noether charge isA finite box with periodic boundary conditions works as well. The spatial integral is conserved because the local continuity equation converts its time derivative into a boundary flux; neither vanishing local current nor constant individual fields is required. Equivalently, $\Phi =({\varphi }_1+i{\varphi }_2)/\sqrt{2}$ transforms as $\Phi \mapsto e^{-i\alpha }\Phi$, exhibiting the equivalent global U(1) symmetry and fixing the sign of the generator.