 Solution

Use the combinations $a_{\pm }=a\pm b$ at each momentum. Part a gives $[Q,a_{+}]=0$, $[Q,a_{-}]=-2a_{-}$, while their creation operators obey $[Q,a_{+}^{†}]=0$, $[Q,a_{-}^{†}]=2a_{-}^{†}$. Applying b givesThe creation operators transform by the Hermitian adjoint formulas with $e^{+2i\lambda }$. At $\lambda =\pi /2$, both exponentials are $-1$, soSubstituting into the fixed-time mode expansion of a free field of the complex scalar field givesThus $C$ exchanges particles and antiparticles, as required for charge conjugation. To see the charge reversal explicitly, the conserved charge $Q_{\mathrm{em}}={\int }_{\mathbf{p}}(a_{\mathbf{p}}^{†}{a}_{\mathbf{p}}-b_{\mathbf{p}}^{†}{b}_{\mathbf{p}})$ obeys $C{Q}_{\mathrm{em}}{C}^{†}=-Q_{\mathrm{em}}$. The generator $Q$ in the question is distinct from this charge.

The two occupation terms in the Hamiltonian are exchanged, with the same relativistic energy $E_{\mathbf{p}}$, and thereforeIn fact $[Q,H]=0$ because its mixing terms exchange equal-energy modes. With normal ordering, $:Q:=2{\int }_{\mathbf{p}}{c}_{-,\mathbf{p}}^{†}{c}_{-,\mathbf{p}}$ for $c_{-}=(a-b)/\sqrt{2}$. Then $C=e^{i\pi N_{-}}$ fixes the vacuum state and satisfies $C^2=1$. The construction is a bosonic species-swap generator.