The <bosonic annihilation operator> $a^\lambda(p)$ removes a vector particle of <four-momentum> $p$ and <polarization vector> $\varepsilon^{\mu,\lambda}(p)$. The <bosonic creation operator> $c^{\lambda\dagger}(p)$ 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 gives
$$
\hat C V^\mu(x)\hat C^{-1}
=\eta_C\sum_{p,\lambda}\left[\varepsilon^{\mu,\lambda}(p)c^\lambda(p)e^{-ip\cdot x}+\varepsilon^{\mu,\lambda *}(p)a^{\lambda\dagger}(p)e^{ip\cdot x}\right].
$$
Comparison with the <Hermitian conjugation> of the original expansion therefore gives
$$
\boxed{\hat C V^\mu(x)\hat C^{-1}=\eta_C V^{\mu\dagger}(x).}
$$
The factor $\eta_C$ is the <intrinsic charge-conjugation phase>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 305 / 1 / b / Solution

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 Incoming links: Bosonic annihilation operator