Charge conjugation of a Dirac field (${\psi }^c=C{\bar{\psi }}^T$)

Choose the negative-frequency Dirac spinors as the charge conjugates of the positive-frequency spinors. Transposing $\bar{u}(/p-m)=0$ and using the charge-conjugation matrix identity gives $(/p+m)C{\bar{u}}^T=0$, so the displayed choice solves the correct antiparticle equation. In the mode expansion of a Dirac field, the unitary charge-conjugation operator then exchanges particle and antiparticle annihilation operators, with conjugate phases on their creation operators. A different spinor basis can insert conventional phases or a spin-index relabeling; the displayed equality specifies a compatible choice.

For ${\psi }^c=C{\bar{\psi }}^T$, use a unitary charge-conjugation matrix with $C^{-1}{\gamma }^{0}C=-({\gamma }^0{)}^T$ and Hermitian ${\gamma }^0$ of square one. The Dirac adjoint then satisfiesIf the unitary charge conjugation operator sends $\psi$ to ${\eta }_C{\psi }^c$, it sends $\bar{\psi }$ to $-{\eta }_C^{\ast }{\psi }^T{C}^{-1}$. Antisymmetry $C^T=-C$ also gives $({\psi }^c{)}^c=\psi$. The unitary operator acting on fields is not an antiunitary operator acting on numerical coefficients.

In the complex-conjugate convention for charge conjugation of a Dirac field, a unitary matrix $B$ obeys the displayed gamma matrix identity. For $M^{\rho \sigma }=\frac{1}{4}[{\gamma }^{\rho },{\gamma }^{\sigma }]$, two minus signs cancel to give $M^{\rho \sigma }B=B{M}^{\rho \sigma \ast }$. Real Lorentz transformation parameters therefore give $SB=B{S}^{\ast }$, so $B{\psi }^{\ast }$ has the same Spinor representation of the Lorentz group transformation law as $\psi$. Complex conjugating the free Dirac equation and multiplying by $B$ proves that the conjugate also solves it for real mass. This $B$ is the matrix multiplying ${\psi }^{\ast }$; the convention multiplying ${\bar{\psi }}^T$ instead uses $B({\gamma }^0{)}^T$.

The Dirac adjoint of a chiral field has opposite projector placement because ${\gamma }_5$ anticommutes with ${\gamma }^0$: $\bar{\psi }{\gamma }_5=-\eta \bar{\psi }$. Since $C{\gamma }_5^T{C}^{-1}={\gamma }_5$, the algebraic charge conjugate satisfies ${\gamma }_{5}C{\bar{\psi }}^T=C(\bar{\psi }{\gamma }_5{)}^T=-\eta {\psi }^c$. This is conjugation of a projected spinor, distinguished from operator conjugation of a chiral field component, where the numerical projector is kept outside the conjugation operator.

The unitary charge-conjugation operator acts on field operators and commutes with numerical chiral projectors, so $\hat{C}(P_L\psi ){\hat{C}}^{-1}=P_L{\psi }^c$. This component is left-chiral. The algebraic conjugate of an already projected spinor instead satisfies $(P_L\psi {)}^c=P_R{\psi }^c$, because its adjoint carries the opposite projector. Keeping the two operations separate avoids a false chirality conclusion.

If $C{\gamma }^{\mu T}{C}^{-1}=-{\gamma }^{\mu }$, transposition shows that $C^T{C}^{-1}$ commutes with every gamma matrix. Irreducibility of the complex Dirac representation and the Schur lemma make it scalar. Transposing $C^T=zC$ again yields $z^2=1$, so $C^T=\pm C$.

In the Weyl representation of the gamma matrices, $C=i{\gamma }^2{\gamma }^0=\mathrm{diag}(\epsilon ,-\epsilon )$ with $\epsilon =i{\tau }_2$. The antisymmetry of $\epsilon$ gives $C^T=-C$, and direct multiplication gives $C{\gamma }^{\mu T}{C}^{-1}=-{\gamma }^{\mu }$ and $C^{†}C=I$. Any two such intertwining matrices differ by a scalar: their ratio commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. Changing spinor basis sends $C$ to $SC{S}^T$, so transpose antisymmetry is basis-independent. Its sign is therefore fixed, not merely one of the two alternatives supplied by the general transpose symmetry of a charge-conjugation matrix argument.

For ${\psi }^C={\eta }_{C}C{\bar{\psi }}^T$ and ${\gamma }^{0T}={\gamma }^0=-C{\gamma }^0{C}^{-1}$, the Dirac adjoint transforms as ${\bar{\psi }}^C=-{\eta }_C^{\ast }{\psi }^T{C}^{†}$. Anticommuting the components of the fermion bilinear gives $(\bar{\psi }\psi {)}^C=\bar{\psi }(C^{†}C{)}^T\psi$. Invariance of the scalar bilinear therefore fixes $C^{†}C=I$.