I'm looking into a physical problem that involves the following sum:
$$r(t)=\sum_{\substack{z_l=0,1\\1\leq l \leq N}}|b_{z_1z_2\dots z_N}|^2e^{-2i\epsilon_{z_1z_2\cdots z_N}t}$$
where $\epsilon_{z_1z_2\cdots z_N}$ are random phases and $\sum |b_{z_1z_2\dots z_N}|^2=1$, where the $b$'s are normally distributed.
By trying to say something about it, one could look at it as a 2D random walk problem with fixed total lenth (sum of all step lengths). Apparently that yields
$$\langle|r(t)|^2|\rangle\propto \langle |b_{z_1z_2\cdots z_N}|^2\rangle=2^{-N}\tag{1}$$
Is it really necessary to look at it this way? Isn't it just a simple calculation involving some calculation rules of $\langle \cdot\rangle$?
For further information on the physics behind this problem, see the book from Schlosshauer on Decoherence: 978-3-540-35775-9 (p. 91).
I myself tried to do it with "simple math" although couldn't make it:
I calculated $$\mathbb{E}(|r(t)|^2)=\mathbb{E}\left(\left(\sum_{\substack{z_l=0,1\\1\leq l \leq N}}|b_{z_1z_2\dots z_N}|^2e^{-2i\epsilon_{z_1z_2\cdots z_N}t}\right)\left(\sum_{\substack{z_l=0,1\\1\leq l \leq N}}|b_{z_1z_2\dots z_N}|^2e^{2i\epsilon_{z_1z_2\cdots z_N}t}\right)\right)=\left(\left(\sum_{\substack{z_l=0,1\\1\leq l \leq N}}\mathbb{E}(|b_{z_1z_2\dots z_N}|^2)\mathbb{E}(e^{-2i\epsilon_{z_1z_2\cdots z_N}t})\right)\left(\sum_{\substack{z_l=0,1\\1\leq l \leq N}}\mathbb{E}(|b_{z_1z_2\dots z_N}|^2)\mathbb{E}(e^{2i\epsilon_{z_1z_2\cdots z_N}t})\right)\right)=2^{-N}\cdot 2^N \mathbb{E}(e^{-2i\epsilon_{z_1z_2\cdots z_N}t}) \cdot 2^{-N}\cdot 2^N \mathbb{E}(e^{2i\epsilon_{z_1z_2\cdots z_N}t})$$ with $$\mathbb{E}(e^{-2i\epsilon_{z_1z_2\cdots z_N}t})$$ the characteristic function of a random variable $\epsilon$ with (continuous) unform distribution on $[0,2\pi]$ (where no $N$ occurs). As you can see, the factor $N$ is not in the final formula, although $(1)$ says so. Am I making a mistake?
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