This question regards a paper by Zurek, Cucchietti and Paz from 2008 (DOI: https://doi.org/10.48550/arXiv.quant-ph/0611200), in specific how to arrive at (12) from (10) and (11). This seems to be straight forward math, but I don't get it. I will summarize what is written in the paper:
Equation (10) gives the so called decoherence factor for a special "simple" model, that the authors use:
$$r(t)=\sum_{l=0}^N {N\choose l}|\alpha|^{2(N-l)}|\beta|^{2l}e^{ig(N-2l)t}\tag{10}$$
Where $|\alpha|^2+|\beta|^2=1$ and both $\alpha$ and $\beta$ are complex numbers.
Equation (11) gives the Gaussian approximation of the factor
$${N\choose l}|\alpha|^{2(N-l)}|\beta|^{2l}\approx \frac{1}{\sqrt{2\pi N|\alpha\beta|^2}}e^{-\frac{(l-N|\beta|^2)^2}{2N|\alpha\beta|^2}}\tag{11}$$
This (according to the authors) immediately yields the result, given in equation (12):
$$|r(t)|=e^{-2N|\alpha\beta|^2(gt)^2}\tag{12}$$.
Substituting (11) in (10) and taking the absolute value doesn't seem to do it. Somehow (by arriving at (12)) there have been made several steps I seem to not fully understand. Can anyone explain to me what is happening here?
If $X \sim \text{Binomial}(N, |\beta|^2)$, then $r(t)$ is $\mathbb{E}[e^{ig(N-2X)t}] = e^{igNt} \mathbb{E}[e^{-2igtX}]$.
The normal approximation approximates this Binomial distribution with $\mathcal{N}(\mu, \sigma^2)$ where $\mu = N|\beta|^2$ and $\sigma^2 = N|\alpha \beta|^2$. For the rest of this answer, assume $X \sim N(\mu, \sigma^2)$.
The characteristic function of $X$ is $\mathbb{E}[e^{iuX}] = e^{i \mu u - \sigma^2 u^2 / 2}$. Plugging in the expressions for $\mu$ and $\sigma^2$ and plugging in $u = -2gt$, we have $\mathbb{E}[e^{-2igtX}] = e^{-2igt N|\beta|^2 - 2N |\alpha \beta|^2 (gt)^2}$.
Putting the pieces together, we have
$$|r(t)| = \left|e^{igNt} \mathbb{E}[e^{-2igt X}]\right| = \left|\mathbb{E}[e^{-2igt X}]\right| = \left|e^{-2igt N|\beta|^2 - 2N |\alpha \beta|^2 (gt)^2}\right| = e^{- 2N |\alpha \beta|^2 (gt)^2}.$$