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I am studying the Hartree equation for N-particles for the first time and things are not clear to me. Given the density matrix

$$\gamma_{N, t}^{(n)} (x_1,..,n_n; y_1,...,y_n) = \begin{cases} \int \gamma_{N,t}(x_1,...,x_n,x_{n+1},...,x_N; y_1,...,y_n,x_{n+1},...,x_N) dx_{n+1}...d_{x_N},\,\,\,\,\, 1 \leq n \le N\\ 0,\,\,\,\,\, o.w. \end{cases}$$

by which we define the Wigner transform

$$w_N(x;v):= \frac{1}{(3 \pi)^{3 N}} \int e^{- i v. y} \gamma_N \left( x + \frac{y}{2}, x- \frac{y}{2}\right) dy.$$

and the rescaled Wigner transform

$$w_{N,\varepsilon}(x;v)=\varepsilon^{-3N}w_N(x;v):= \frac{1}{(3 \pi)^{3 N}} \int e^{- i v. y} \gamma_N \left( x + \frac{\varepsilon y}{2}, x- \frac{\varepsilon y}{2}\right) dy.$$

The author says that using the Heisenberg equation: $$i\varepsilon \partial_t \gamma_{N,t} =[H_N, \gamma_{N,t}],\,\,\,[A,B]=AB-BA,$$

And $H_N$ is the Hamiltonian of the N particle system, we can (easily) derive the time evolution of Wigner transform, namely

$$\partial_t W_N (t; x,v) +\sum_{j=1}^N v_j . \triangle_{x_j} W_N (t;x,v)=\frac{- i \varepsilon^2}{(2 \pi)^{3N}} \int \left[ U\left(x+ \frac{\varepsilon y}{2}\right)- U\left(x- \frac{\varepsilon y}{2} \right) \right] e^{i y (u-v)} W_N(t; x,v) du dy.$$

I can not find how the author gets this result. Could you please enlighten me?

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    $\begingroup$ I think the Hamiltonian in this case has no 2-body interactions. It just has the kinetic energy represented by the Laplacian plus a potential energy represented by the multiplication operator by the function U. In Fourier analysis, when you multiply two functions like h=f g then for the Fourier transform you take the convolution of the Fourier transforms. It looks like that is what is going on here. The Wigner transform is like a Fourier transform in 1 variable, not $x_j$ nor $x_k$ but $(x_k-x_j)$. Also derivatives go over into multiplication by the Fourier transform variable. $\endgroup$ Commented Feb 1, 2022 at 11:52
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    $\begingroup$ you may want to first learn from a simpler case, such as the harmonic oscillator; this is worked out in some detail at physics.stackexchange.com/a/574704/38462 -- the Hartree equation proceeds very similarly, with a self-consistent potential instead of the quadratic potential. $\endgroup$ Commented Feb 1, 2022 at 12:03

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