0$. The variance of the wave packet mode could be smaller than one if the third term of Eq. (\[VarianceGaussian\]) is larger than the second term. In order to show that single-mode squeezing is possible, we consider a Gaussian wave packet centered at $V_0 = 0$. Eq. (\[VarianceGaussian\]) considerably simplifies, yielding

$$\begin{aligned}
\big(\Delta X^{min}\big)^2 = 1 - 4\sqrt{\frac{2}{\pi}}\frac{\sigma}{k_0} \int_0^{\infty} d \Omega 
\exp \bigg(-\frac{2\sigma^2\Omega^2}{k_0^2}\bigg) \times (1-\cos \theta_{\Omega})\frac{e^{\pi \Omega}}{(e^{\pi \Omega}+1)^2} <1\end{aligned}$$

for the minimum of $\big(\Delta X({\phi})\big)^2$, which is at $\phi = 0$.

The variance of the quadrature beats the quantum shot noise, showing that the Gaussian wave packet mode is squeezed. When the center of the Gaussian wave packet is away from the past horizon $V_0 = 0$, the mode is squeezed at a different quadrature phase angle. According to Eq. (\[VarianceGaussian\]), the minimum of the variance is reached when $\phi_s + k_0 V_0 =0$ is satisfied, that is $$\label{phase}
\phi_s = - k_0 V_0.$$ The squeezing phase angle $\phi_s$ depends on both the central frequency and central position of the Gaussian wave packet. Other than the rotation of the squeezing phase angle, the squeezing amplitude decreases when the center of the wave packet is away from the past horizon. Fig. 