 Solution

For arbitrary linear operators, expansion of the anticommutators givesNow use the commutator product rule $[VX,W]=V[X,W]+[V,W]X$, with $W=YZ$, twice. The result isNo grading assumption is needed for these algebraic identities.

For the Dirac field, supplement the supplied equal-time canonical anticommutation relations with $\{{\psi }_a(x),{\psi }_b(y)\}=\{{\psi }_a^{†}(x),{\psi }_b^{†}(y)\}=0$. Define the fermion number density $n(\mathbf{x})={\sum }_a{\psi }_a^{†}(\mathbf{x}){\psi }_a(\mathbf{x})$. In the preceding identity take $V={\psi }_a^{†}(\mathbf{x})$, $X={\psi }_a(\mathbf{x})$, $Y={\psi }_b^{†}(\mathbf{y})$, $Z={\psi }_c(\mathbf{z})$ and sum over $a$. Only its first and last anticommutators survive, givingThis is an identity of operator-valued distributions. Point splitting, or smearing with test functions, gives a precise meaning to coincident bilinears. Normal ordering only subtracts a constant from the density and does not alter this commutator.