 Solution

The two oscillator families commute with each other. Their canonical commutation relations giveBoth terms in the mode expansion of a free field therefore acquire the same factor, givingThe Noether charge is the generator of the internal symmetry. In the active-operator convention $U(\alpha )=e^{i\alpha Q}$, repeated commutators giveThe sign agrees with part b. A negative sign for $[Q,\varphi ]$ would correspond to the opposite current convention, not to the current used here. Subtracting a multiple of the identity from $Q$ to fix the vacuum Noether charge has no effect on this generator relation.