# Is this spectrum-shifting operator well-defined?

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.

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 $$\|B\| \le \int_{-\infty}^{\infty}dt e^{-\epsilon t^2}\|A\|.$$
• 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 $t$? Jun 16, 2019 at 0:49
• @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. Jun 16, 2019 at 1:19
• 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. Jun 17, 2019 at 16:16
• @Mogget : That's correct, $A(t)$ is strongly continuous. Jun 17, 2019 at 17:03