Put $P\_j=\frac{\partial}{\partial \xi\_j}$ et $Q\_j=2 i \xi\_j$ with$\xi=\left(\xi\_1, \ldots, \xi\_n\right)$ et $x=\left(x\_1, \ldots, x\_n\right)$. How to prove :


1. $\exp \left(\sum\_{j=1}^n x\_j P\_j\right)(f)(\xi)=f(\xi+x)$.
2. $\exp \left(\sum\_{j=1}^n x\_j Q\_j\right)(f)(\xi)=\exp (2 i(x, \xi)) f(\xi)$.
where $(x, \xi)=\sum\_{j=1}^n x\_j \xi\_j$ and $i^2=-1$.
Thanks



Differential operators in $\Bbb R^n$
I want first to change your notations, sticking to the usual variables $x,\xi$ in the phase space. As a general statement about pseudo-differential operator with a symbol $a(x,\xi)$, I wish to write
$$
\bigl(\text{Op}(a) f\bigr)(x)=\int e^{2iπ x\cdot \xi} a(x,\xi) \hat f(\xi) \, d\xi.
$$
Let $y\in \mathbb R^n$ be given. Then we have
$$
f(x+y)=\int e^{2iπ (x+y)\cdot \xi}\hat f(\xi) \, d\xi
=
\int e^{2iπ x\cdot \xi}e^{2iπ y\cdot \xi}\hat f(\xi) \, d\xi=
\bigl(\operatorname{Op}(a\_y) f\bigr)(x), \quad a\_y(x,\xi)=e^{2iπ y\cdot \xi},
$$
which is your first formula.


Let $\eta\in \mathbb R^n$ be given. Then we have
$$
e^{2iπ x\cdot \eta}f(x)=\int e^{2iπ x\cdot \xi}e^{2iπ x\cdot \eta}\hat f(\xi) \, d\xi=\bigl(\text{Op}(b\_\eta) f\bigr)(x),\quad b\_\eta(x,\xi)=e^{2iπ x\cdot \eta},
$$
and this is your second formula.


