Klein-Gordon inner product [@Birrell82], $$\begin{aligned}
\label{MinkowskiUnruh}
\hat{a}_k &=& \int d \omega \bigg(  \langle u_k, G_{\omega}  \rangle\hat{c}^{\prime}_{\omega} + \langle u_k, \bar{G}_{\omega}  \rangle\hat{d}^{\prime}_{\omega} \bigg) \nonumber \\
&\equiv& \int d \omega ( A_{k \omega} \hat{c}^{\prime}_{\omega} + B_{k \omega} \hat{d}^{\prime}_{\omega} ),\end{aligned}$$ where $A_{k\omega} =  \langle u_k, G_{\omega}  \rangle$ and $B_{k\omega} = \langle u_k, \bar{G}_{\omega}  \rangle$ are the Bogoliubov transformation coefficients. Since we only consider left-moving modes here, without introducing any confusion, we have omitted the subscript $``1"$. Using the relation between Unruh modes and Rindler modes Eq. (\[UnruhRindler\]) and the relation between Rindler modes and Minkowski modes [@Crispino08], We can find the transformation between Unruh modes and Minkowski modes. A more straightforward way is to directly calculate the Klein-Gordon inner product using the explicit expressions of Unruh modes Eq. (\[Unruhmodes\]). The result is $$\begin{aligned}
A_{k\omega} = B^*_{k\omega} =  \frac{i \sqrt{2 \sinh(\pi \omega/a)}}{2\pi \sqrt{\omega k}} \Gamma(1-i\omega/a) \bigg(\frac{k}{a}\bigg)^{i\omega/a}, \nonumber \\
$$ where $\Gamma(z)$ is the Gamma function. In realistic quantum optics experiments a detector normally detects localized wave packet modes. In order to take this into account we consider Gaussian wave packet modes defined as $$\label{GaussianMode}
\hat{a}(f) = \int_0^{\infty} d k f(k; k_0, \sigma, V_0) \hat{a}_k, 
$$ where $$f(k; k_0, \sigma, V_0) = \bigg(\frac{1}{2\pi \sigma^2}  \bigg)^{1/4} 