-1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.