 Solution

Take a periodic comoving box of volume $L^3$, with $\mathbf{k}=2\pi \mathbf{n}/L$. A real field has the expansionwhere $[{\hat{a}}_{\mathbf{k}},{\hat{a}}_{\mathbf{q}}^{†}]={\delta }_{\mathbf{kq}}$ and the other ladder-operator commutators vanish. The coefficient of $e^{i\mathbf{k}\cdot \mathbf{x}}$ is therefore $w_k{\hat{a}}_{\mathbf{k}}+w_k^{\ast }{\hat{a}}_{-\mathbf{k}}^{†}$. This opposite momentum is required by reality; the single-oscillator shorthand with the same label on both operators must not be read literally as every traveling-wave Fourier coefficient.

The quadratic action has canonical momentum $\hat{\pi }=a^3\delta \dot{\hat{\varphi }}$. Its equal-time field commutator requiresChoose the vacuum annihilated by all $a_{\mathbf{k}}$ and the subhorizon positive-frequency condition. For constant $H$ and negligible effective mass, put $x=k/(aH)$, $C_k=L^{-3/2}H/\sqrt{2k^3}$. The candidate is $w_k=C_k(i+x)e^{ix}$. Since $\dot{x}=-Hx$,Direct substitution makes ${\ddot{w}}_k+3H{\dot{w}}_k+(k^2/a^2)w_k$ identically zero. Its Wronskian is $2iH{C}_k^2{x}^3=i/(L^3{a}^3)$, which has the required normalization. ThusThis is the existing Bunch-Davies mode normalization in a finite comoving volume. It is exact for a massless minimally coupled test field in de Sitter and the leading approximation for a light slowly evolving field. The homogeneous zero mode needs separate treatment.