commutation relations on the field $\hat{A}_\mu$ and momentum $\hat{\Pi}^{t \mu }$ operators: $$\left[ \hat{A}_\mu(t,{\bf x}), \hat{A}_\nu(t,{\bf x}')\right]
= \left[ \hat{\Pi}^{t\mu}(t,{\bf x}), \hat{\Pi}^{t\nu}(t,{\bf x}')\right]= 0\,,
\label{CR1}$$ $$\left[ \hat{A}_\mu(t,{\bf x}), \hat{\Pi}^{t\nu}(t,{\bf x}')\right] = 
\frac{i\delta_\mu^\nu}{\sqrt{-g}}\delta^{p+1}({\bf x}-{\bf x}')\,,
\label{CR2}$$ where ${\bf x}$ and ${\bf x}'$ represent all spatial coordinates. The field $\hat{A}_\mu$ can be expanded using the modes we have obtained before: $$\hat{A}_{\mu}(t,{\bf x})
= \sum_{\rho} \int_{-\infty}^{+\infty}
\,\frac{d\omega}{\sqrt{4\pi|\omega|}} 
A_\mu^{(\omega\rho)}(t,{\bf x})a_{\omega\rho}\,,$$ where $A_\mu^{(\omega\rho)}$ is proportional to $e^{-i\omega t}$, $A_\mu^{(-\omega \rho )} \equiv \overline{A_\mu^{(\omega \rho)}}$, $a_{-\omega\rho} \equiv a^\dagger_{\omega\rho}$ and $\rho$ labels discrete quantum numbers. The commutation relations (\[CR1\])-(\[CR2\]) are equivalent to the following commutation relations in the “symplectically smeared" form as is the case for scalar fields [@Wald]: $$\left[ (A^{(\zeta)}, \hat{A}), (\hat{A}, A^{(\zeta')})\right]
= (A^{(\zeta)}, A^{(\zeta')})\,.  \label{smeared}$$ Since the inner product must be $t$ independent, the inner product of $A_\mu^{(\omega\rho)}$ and $A_\mu^{(\omega'\rho')}$ can be nonzero only if $\omega  = \omega'$. Thus, we can write $$(A^{(\omega\rho)}, A^{(\omega'\rho')})
= M^{\rho\rho'}\delta(\omega-\omega')\,. \label{Mdef}$$ By using Eq. (\[Mdef\]) in (\[smeared\]) one finds $$M^{\rho_1\rho_2}\left[ a_{\omega\rho_2},a^\dagger_{\omega'\rho_3}\right]
M^{\rho_3\rho_4} = 4\pi\omega M^{\rho_1\rho_4}\delta(\omega-\omega')\,.$$ Then, since $M^{\rho\rho'}$ is invertible in our case, we find $$\left[ a_{\omega\rho},a^\dagger_{\omega' \rho'}\right] = 4\pi\omega
(M^{-1})_{\rho\rho'}\delta(\omega-\omega')\,. \label{acommu}$$ Note here that we immediately have $[a_{\omega \rho},a_{\omega' \rho'}] = 0$ for $\omega, \omega' > 0$ 