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With a vacuum annihilated by all $a_{\mathbf{p}}^s,b_{\mathbf{p}}^s$, define the particle and antiparticle occupation operators of a Dirac fieldEach counts its corresponding occupied modes. The total positive particle count isCommuting it through either creation operator gives $[N_{\mathrm{total}},a^{†}]=a^{†}$ and $[N_{\mathrm{total}},b^{†}]=b^{†}$, so both kinds of one-particle states have eigenvalue one. In a normalized discrete mode, $(a^{†}a{)}^2=a^{†}a$ by the anticommutation relations, so its occupation eigenvalues are $0,1$.

For a Dirac field there is also a distinct conserved signed fermion number. Inserting the mode expansion into the normal-ordered local density givesSpatial integration makes the same-sector momenta equal. The cross terms vanish by $u^{†}(\mathbf{p})v(-\mathbf{p})=0$; normal ordering of $b{b}^{†}$ supplies the minus sign for the antiparticles. Thus the density integral is not the total positive particle count. It generates the global phase symmetry, and for Electron charge $-e$ the electric charge is $Q_{\mathrm{em}}=-e{Q}_D$.

The free normal-ordered Hamiltonian isso $N_e,N_{\bar{e}},N_{\mathrm{total}}$ and $Q_D$ are conserved in the free theory. In an interacting theory, pair creation can change the total count while preserving the signed charge. This distinguishes a positive occupation number operator from the charge behind Dirac fermion number conservation.