I am studying Rodnianski and Schlein - Quantum Fluctuations and Rate of Convergence Towards Mean Field Dynamics. Everything was clear for me and I reproved everything before inequality (3.22) (except 3.20 is still not clear). I feel there is something I am overseeing. I can not reach to same inequality in the paper. Can some one give a hint, solution, or recommend some reference to read?
This is a description for my question incase one does not want to go into paper:
Lemma 2.1. Let $ f \in L^2(\mathbb{R}^3)$. Then \begin{align*} \|a(f) \psi\| &\leq \|f\| \|\sqrt{\hat{N}} \psi\|, \\ \|a^*(f) \psi\| &\leq \|f\| \|\sqrt{\hat{N} + 1} \psi\|, \quad (2.4) \\ \|\phi(f) \psi\| &\leq 2\|f\| \|\sqrt{\hat{N} + 1} \psi\|. \end{align*}
Lemma 2.2. Let $f, g \in L^2(\mathbb{R}^3)$.
i) The Weyl operator satisfies the relations: \begin{equation} W(f)W(g) = W(g)W(f)e^{-2i \mathrm{Im}\langle f,g \rangle} = W(f + g)e^{-i \mathrm{Im}\langle f,g \rangle}. \end{equation}
ii) $W(f)$ is a unitary operator, and \begin{equation} W(f)^* = W(f)^{-1} = W(-f). \end{equation}
iii) We have: \begin{equation} W(f)a_xW(f)^* = a_x + f(x), \quad W(f)a_x^*W(f)^* = a_x^* + f(x). \end{equation}
iv) From iii), we see that coherent states are eigenvectors of annihilation operators: \begin{equation} a_x \psi(f) = f(x) \psi(f) \Rightarrow a(g) \psi(f) = \langle g, f \rangle_{L^2} \psi(f). \end{equation}
v) The expectation of the number of particles in the coherent state $\psi(f)$ is given by $\|f\|^2$, that is \begin{equation} \langle \psi(f), \hat{N} \psi(f) \rangle = \|f\|^2. \end{equation} Also, the variance of the number of particles in $\psi(f)$ is given by $\|f\|^2$ (the distribution of $N$ is Poisson), that is \begin{equation} \langle \psi(f), \hat{N}^2 \psi(f) \rangle - \langle \psi(f), \hat{N} \psi(f) \rangle^2 = \|f\|^2. \end{equation} Coherent states are normalized but not orthogonal to each other. In fact, \begin{equation} \langle \psi(f), \psi(g) \rangle = e^{-\frac{1}{2} (\|f\|^2 + \|g\|^2 - 2\langle f, g \rangle)} \Rightarrow |\langle \psi(f), \psi(g) \rangle| = e^{-\frac{1}{2} \|f-g\|^2}. \end{equation}
\begin{align*} U^*_N(t;s) \hat{N} U_N(t;s) &= \int{dx} U^*_N(t;s) a^*_x a_x U_N(t;s) \\ &= \int{dx} W^*(\sqrt{N} \varphi_s) e^{i H_N(t-s)} (a_x^* - N \varphi_t(x)) \\ &\quad \times (a_x - N \varphi_t(x)) e^{-i H_N(t-s)} W(\sqrt{N} \varphi_s)\\ &= W^*(\sqrt{N} \varphi_s) \left( \hat{N} - \sqrt{N} e^{i H_N(t-s)} \phi(\varphi_t) e^{-i H_N(t-s)} + N\right) W(\sqrt{N} \varphi_s). \quad (3.21) \end{align*}
Recall that:
$$\phi(\varphi) = a^*(\varphi) + a(\varphi) = \int dx(\varphi(x) a_x^* + \varphi(x) a_x).$$
From Lemma 2.1 and Lemma 2.2, we get
\begin{align*} \langle \psi, U^*_N(t;s) \hat{N} U_N(t;s) \psi \rangle &\leq 2 \langle \psi, W^*(\sqrt{N} \varphi_s) (\hat{N} + N + 1) W(N \varphi_s) \rangle \\ &= 2 \langle \psi, (\hat{N} + \sqrt{N} \phi(\varphi_s) + 2N + 1) \psi \rangle \\ &\leq 6 \langle \psi, (\hat{N} + N + 1) \psi \rangle, \end{align*}
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