Let $A$ be a self - adjoint operator on a Hilbert space $\mathcal{H}$ and let $D(A)$ be its domain. If $\psi \in D(A)$ then $exp(-itA) \psi \in D(A)$ iff $A$ is bounded ?

Thank you guys, sorry if the question is too trivial ! ;)

Physics beginner

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    $\begingroup$ This question would be better for math.stackexchange.com, as it is not really "research level". $\endgroup$ – Nate Eldredge Oct 13 '11 at 5:00
  • $\begingroup$ In fact, the implication $\psi \in D(A) \Rightarrow e^{-itA}\psi \in D(A)\ \forall t\in \mathbb R$ is true regardless of boundedness and self-adjointness of $A$. You only need that $iA$ generates a group at all, cf. the book of Engel-Nagel. $\endgroup$ – Delio Mugnolo Dec 5 '16 at 22:38

No. If A is self-adjoint, then exp(-itA) maps D(A) to D(A) regardless whether A is bounded. You should read a text on semigroup theory for linear operators, for instance Pazy's book.

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    $\begingroup$ No semigroup theory is needed here. Just note that the spectral theorem reduces the question to the case where $\mathcal{H}$ is an $L^2$ space and $A$ is a multiplication operator. $\endgroup$ – Nate Eldredge Oct 13 '11 at 4:59
  • $\begingroup$ The perhaps only interesting thing here is that $\exp(-iA)\psi$ can not be defined by the exponential series for all vectors in $D(A)$. One really needs the definition via the spectral calculus, i.e. the integral over the projection valued measure of $A$. $\endgroup$ – Stefan Waldmann Oct 13 '11 at 7:37

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