 Solution

The Hamiltonian operator is invariant under the constant global phase symmetry $\psi \mapsto e^{i\theta }\psi$, $\bar{\psi }\mapsto \bar{\psi }{e}^{-i\theta }$. The Noether current of this unitary group $U(1)$ is the Dirac current obtained in b. For fields with vanishing flux at infinity, or periodic boundary conditions, the divergence theorem and that continuity equation implyThe canonical anticommutation relations also give $[Q,\psi ]=-\psi$ and $[Q,{\psi }^{†}]={\psi }^{†}$, hence $e^{i\theta Q}\psi e^{-i\theta Q}=e^{-i\theta }\psi$. Reversing the parameter sign recovers the stated phase convention. Thus $Q$ generates the global phase symmetry. With normal ordering, $Q$ counts fermions minus antiparticles, rather than the sum of both populations. Its conservation law is Dirac fermion number conservation. A position-dependent phase produces derivative terms in this free Hamiltonian operator; the symmetry under discussion is global.