Let $A$ be a self-adjoint operator on a Hilbert space , and let $d\Gamma(A)$ be the generator of the second quantization of $A$. Consider the following theorem from Segal's "Non-Linear Quantum Processes":
THEOREM 1. There exists a universal constant $k>0$ such that $e^{-t d \Gamma(A)}$ is a contraction from $[K_p]$ to $[K_{pe^{kt}}]$ for every self-adjoint operator $A$ in $H$ such that $A\geq I$, and all $p\in [2,\infty]$ and $t\in [0,\infty]$.
Segal next writes a lemma on the Ornstein-Uhlenbeck operator $(x-\partial_x)\partial_x$:
Lemma 1.1. Let $A$ now denote the self-adjoint operator in the Hilbert space $\mathbf{N}=L_2\left(R^1, \gamma\right)$, where $d \gamma=(2 \pi)^{-1 / 2} e^{-x^2 / 2} d x$, which multiplies the $n^{\text {th }}$ Hermite polynomial by $n$. Then $e^{-t A}$ is a contraction from $L_p$ to $L_p$, for all $p \in[1, \infty]$.
Note that this theorem is valid for Ornstein-Uhlenbeck operators, which satisfy all the properties of $A$ mentioned in the theorem.
I suppose the implication is that if $A$ is $(x-\partial_x)\partial_x$, then $d\Gamma(A)=A$. My questions are:
- How does one prove that $d\Gamma(A)=A$?
- How does one calculate $d\Gamma(A^2)$?
- The proof of Lemma 1.1 says that $e^{-tA}$ is given by an integral operator $K(x,y)$ whose kernel is non-negative. How does one see that? Is this also true for $A^2$?
