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Consider a separable Hilbert space (over $\mathbb{C}$), and let $U(t)$ be a one-parameter group of unitary operators so that $$ U(t)=e^{iHt} \tag{1} $$ for some densely-defined operator $H$ as in Stone's theorem. Let $A$ be any bounded (everywhere-defined) operator on the Hilbert space, and define $$ A(t) = U(t)A U(-t). \tag{2} $$ For real numbers $\omega$ and $\epsilon$ with $\epsilon>0$, I want to define $$ B := \int_{-\infty}^\infty dt\ \exp(-i\omega t-\epsilon t^2) A(t). \tag{3} $$ Question: Is $B$ a well-defined operator on the Hilbert space? If not, is it at least densely defined? If the answer is "it depends," then is there a simple necessary-and-sufficient condition on $A$ and $H$ such that $B$ is at least densely defined for all $\omega$ and all $\epsilon>0$?

For whatever it's worth, here's the reason for the words "spectrum-shifting" in the title of the question: At least naively, equation (3) implies $HB=B(H+\omega)+O(\epsilon)$. In physics jargon, if $H$ is the energy operator, then applying $B$ to an "eigenstate" of $H$ shifts its energy by $\omega$, up to an arbitrarily small term of order $\epsilon$. That's the motive, but I don't know when (3) is actually well-defined.

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Because $A$ is bounded, then $\|A(t)\| \le \|A\|$ is bounded for all $t$. So the operator $B$ defines a bounded linear operator for $\epsilon > 0$, and $$ \|B\| \le \int_{-\infty}^{\infty}dt e^{-\epsilon t^2}\|A\|. $$

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  • $\begingroup$ Thank you. To make sure I understand: If $B=\int dt\ f(t) A(t)$ then $\|B\|\leq \int dt\ \|f(t) A(t)\|$? And that's sufficient to ensure that $B$ is well-defined on the whole Hilbert space, given that the integrand is for each $t$? $\endgroup$ Jun 16, 2019 at 0:49
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    $\begingroup$ @ChiralAnomaly Yes. Norm distributes over the integral in this way; and you have uniform bound for $\|A(t)\|$. So the resulting integral is a bounded linear operator. $\endgroup$ Jun 16, 2019 at 1:19
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    $\begingroup$ I believe that you also need to make sure that the integrand is measurable (in some sense) in order for the integral to be well defined. Here it is, as $t\mapsto U(t)$ is strongly continuous. $\endgroup$
    – Mogget
    Jun 17, 2019 at 16:16
  • $\begingroup$ @Mogget : That's correct, $A(t)$ is strongly continuous. $\endgroup$ Jun 17, 2019 at 17:03

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