Let $H$ be a self-adjoint operator on $\mathcal{H}$, $\psi\in D(H)$ and $\beta\geq 1/2$. How can I see that $$ L \colon \mathbb{R}\to \mathcal{L}(D(\mathcal{N^\beta}),D(\mathcal{N^{\beta-1/2}})), \quad t\mapsto a\left( e^{iHt}\psi\right)$$ is differentiable? Here $\mathcal{N}$ denotes the number operator and $a$ the annihilation or creation operator. (it should be true for both.)

  • $\begingroup$ Can you add for completeness the linear relations between $a, \mathcal{N}, H$? $\endgroup$ – DanielC May 31 at 14:09
  • $\begingroup$ I hope i understand you question right: we defined the annihilation and creation operators on the domain of the squareroot of the number operator: $(a^\dagger(\psi),D(\mathcal{N}^{1/2}))$ and $(a(\psi),D(\mathcal{N}^{1/2}))$. $\endgroup$ – uzizi_1 May 31 at 16:10

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