(\[MinkowskiUnruh\]), (\[GaussianMode\]), (\[A\]), (\[B\]) and (\[B1\]), we have

$$\begin{aligned}
\label{CroCorrelation}
&&\langle 0_M |\hat{a}(f) \hat{a}(f) |0_M \rangle  = \int dk \int dk^{\prime} f(k) f(k^{\prime})  \int d\omega \int d\omega^{\prime} 
 \big[A_{k\omega} B_{k^{\prime} \omega^{\prime}}\langle 0_M |\hat{c}^{\prime}_{\omega} \hat{d}^{\prime}_{\omega^{\prime}} |0_M \rangle
+  B_{k\omega} A_{k^{\prime} \omega^{\prime}}\langle 0_M |\hat{d}^{\prime}_{\omega} \hat{c}^{\prime}_{\omega^{\prime}} |0_M \rangle\big] \nonumber \\
\nonumber \\
&=& - \sqrt{\frac{8}{\pi}}\frac{\sigma}{k_0} e^{-2ik_0V_0} \int_0^{\infty} d \Omega 
\exp\bigg[-\frac{\sigma^2(\Omega-k_0 V_0)^2}{k_0^2}\bigg] \exp\bigg[-\frac{\sigma^2(\Omega+k_0 V_0)^2}{k_0^2}\bigg]
(1-\cos \theta_{\Omega})e^{\pi \Omega}\frac{e^{2\pi \Omega}+1}{(e^{2\pi \Omega}-1)^2}. \nonumber \\\end{aligned}$$

The quadrature observable of the localized wave packet mode $\hat{a}(f)$ is defined as $$\hat{X}(\phi) \equiv \hat{a}(f)e^{-i\phi} + \hat{a}^{\dagger}(f)e^{i \phi},$$ where $\phi$ is the quadrature phase. From Eqs. (\[ParticleNumber\]) and (\[CroCorrelation\]), we find that for a narrow bandwidth Gaussian wave packet the variance is

$$\begin{aligned}
\label{VarianceGaussian}
&&\big(\Delta X({\phi})\big)^2 = \langle 0_M | \hat{X}^2({\phi}) |0_M \rangle - \langle 0_M | \hat{X}({\phi}) |0_M \rangle^2 
 = 1+ 2 \langle 0_M | \hat{a}^{\dagger}(f)\hat{a}(f) |0_M \rangle + 2~ \text{Re} \bigg[\langle 0_M | \hat{a}(f)\hat{a}(f) |0_M \rangle e^{-2i\phi} \bigg] \nonumber \\
&=& 1+ 4\sqrt{\frac{2}{\pi}}\frac{\sigma}{k_0} \int_0^{\infty} d \Omega 
\bigg\{\exp\bigg[-\frac{2\sigma^2(\Omega-k_0 V_0)^2}{k_0^2}\bigg]+\exp\bigg[-\frac{2\sigma^2(\Omega+k_0 V_0)^2}{k_0^2}\bigg] \bigg\}
(1-\cos \theta_{\Omega})\frac{e^{2\pi \Omega}}{(e^{2\pi \Omega}-1)^2}\nonumber \\
&&- 4\sqrt{\frac{2}{\pi}}\frac{\sigma}{k_0} \cos(2\phi+2k_0V_0)\int_0^{\infty} d \Omega 
\exp\bigg[-\frac{\sigma^2(\Omega-k_0 V_0)^2}{k_0^2}\bigg] \exp\bigg[-\frac{\sigma^2(\Omega+k_0 V_0)^2}{k_0^2}\bigg]
(1-\cos \theta_{\Omega})e^{\pi \Omega}\frac{e^{2\pi \Omega}+1}{(e^{2\pi \Omega}-1)^2}, \nonumber \\ \end{aligned}$$

where we have used the fact that in the Minkowski vacuum state, $\langle 0_M | \hat{X}({\phi}) |0_M \rangle = 