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The bosonic annihilation operator $a^{\lambda }(p)$ removes a vector particle of four-momentum $p$ and polarization vector ${\epsilon }^{\mu ,\lambda }(p)$. The bosonic creation operator $c^{\lambda †}(p)$ creates its antiparticle with the same four-momentum and polarization label. The polarization vector describes the corresponding spin-one mode expansion of a free field.

Because charge conjugation acts through a unitary operator, it leaves the numerical polarization vector coefficients unchanged. Applying the specified charge conjugation to the operators givesComparison with the Hermitian conjugation of the original expansion therefore givesThe factor ${\eta }_C$ is the intrinsic charge-conjugation phase.

For a real vector relativistic quantum field, $V^{\mu }=V^{\mu †}$. Its transformed field must also be a Hermitian operator, so the intrinsic charge-conjugation phase satisfies ${\eta }_C={\eta }_C^{\ast }$; combined with $|{\eta }_C|=1$, this givesThe quantum electrodynamics interaction is $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. Since the Dirac electromagnetic current is odd under charge conjugation, charge conjugation invariance of this interaction requires the photon field to be odd too: