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A fermionic operator is odd under fermion parity: if $P=(-1{)}^N$ is the parity operator, then $PAP=-A$. A fermionic creation operator or fermionic annihilation operator is an example. Interchanging odd insertions in a time-ordered product contributes the corresponding fermionic sign. Odd parity alone does not imply that every pair of such operators has zero anticommutator.

For two odd fermionic operators, time ordering contributes a minus sign whenever the chronological rearrangement exchanges them. For several fields the sign is the sign of a permutation restricted to the odd fields. This convention makes the free Dirac propagator a Green function of the Dirac equation with its contact term fixed by the canonical anticommutation relations.

First conjugate the Dirac equation: $(-i{\gamma }^{\mu \ast }{\partial }_{\mu }-m){\psi }^{\ast }=0$. A matrix $\Gamma$ taking ${\psi }^{\ast }$ into another solution of the same Dirac equation must obeyBecause ${\gamma }^1$ anticommutes with the real ${\gamma }^0,{\gamma }^2$ and commutes with the purely imaginary ${\gamma }^1$, a unitary choice isAny constant overall phase is equally valid. To verify invariance of the action, rather than only of its equation, keep the fermion exchange signs. Define the charge-conjugation matrix $C=\Gamma {\gamma }^0={\sigma }_2$. It obeysFor a classical Grassmann field, interchanging two components contributes a minus sign; equivalently use normal ordering for quantum fermionic operators. ThusandThe last term integrates to a boundary term, vanishing with the usual decay or compatible boundary conditions. Both the kinetic term and the Dirac mass term therefore leave $S$ invariant under charge conjugation. Treating the components as commuting numbers would give the wrong mass-bilinear sign; the fermion statistics is essential.

For a single species both panels of the figure in c contribute, now with $j=1$. They are the $s$-channel annihilation and $t$-channel scalar exchange, where the Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$ and $s+t+u=4m^2$. Fix the external-state order consistently with c. The exchange of identical external fermionic operators gives a relative minus sign:Changing the convention for ordering all external states can change the overall sign, but not this relative fermionic sign.

The exchange $p_1\leftrightarrow -p_4$ must include crossing symmetry the associated external spinors, as well as continuing the four-momenta. It interchanges $s$ and $t$, leaves $u$ fixed, and interchanges the two spinor pairings. Hence the analytically continued invariant scattering amplitude is antisymmetric:The first relation is understood with consistently chosen spinor crossing phases. It is not a substitution of negative-energy arguments into the original positive-energy $u$ spinors without changing external states.

Without repeating the spin sum calculation, c fixes the square of the first diagram. Crossing symmetry fixes the second by $s\leftrightarrow t$. Consequently the strongest immediate inference isThe interference function is not determined by the separate diagram squares. In particular adding the two squares and dropping interference is unjustified. The Cauchy-Schwarz inequality applied to the arrays of spin amplitudes also gives $|C(s,t)|\le 2|(s-4m^2)(t-4m^2)|$ on the physical scattering region.

For completeness, its value can be checked with one contracted trace, without another diagram calculation. The cross product isThe relative minus in the invariant scattering amplitude gives $C=-2K$. Thus the fully evaluated result, beyond the symmetry inference alone, isEvery term respects the inferred $s,t$ symmetry. In the massless limit the interference numerator reduces to $st$, which is nonzero at a generic scattering angle.

For three odd fermionic operators, the identityconverts the commutator with a quadratic Hamiltonian into the given canonical anticommutation relations. Writing a mode label as $i=(\mathbf{p},s)$, this yieldsThe opposite-species number terms commute with each fermionic creation operator or fermionic annihilation operator, because two anticommutations give a positive sign. Restoring the continuous normalization, the Dirac delta distribution in the mode brackets cancels $d^3\mathbf{q}/(2\pi {)}^3$. ConsequentlySubstitute these into the mode expansion of a Dirac field. The result isSince $i{\partial }_t{e}^{-ip\cdot x}=E_{\mathbf{p}}{e}^{-ip\cdot x}$ and $i{\partial }_t{e}^{ip\cdot x}=-E_{\mathbf{p}}{e}^{ip\cdot x}$, this is exactlyThe Heisenberg equation of motion uses an ordinary commutator with the even Hamiltonian, despite the odd fermionic operators obeying anticommutators among themselves.

A time-ordered product sorts operator insertions from latest time on the left to earliest time on the right. Reordering odd fermionic operators contributes the sign of their permutation. For bosonic $A,B$, $T[A(t)B(t^{\prime })]=\theta (t-t^{\prime })A(t)B(t^{\prime })+\theta (t^{\prime }-t)B(t^{\prime })A(t)$. Vacuum expectations of free time-ordered products are evaluated by the Wick theorem.