I am not a mathematician, but would like really like to get some confirmation on the things I am doing here.


Let $-\Delta: H^2(\mathbb{R}) \subset L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$ then every physicist knows that the evolution operator $U(t)=e^{i\Delta t} \in L(L^2,L^2)$ preserves the $L^2$ norm, i.e.
$$||U(t)\psi||\_{L^2}= ||\psi||\_{L^2}.$$


Taking the Fourier transform reveals that $$\widehat{(U(t)\psi)}(k) = e^{-i|k|^2t}\widehat{\psi}(k) $$
also preserves the $H^n-$norms (take for example the characterization of the Sobolev spaces via the Fourier transform, there the evolution is cancelled by an absolute value).


Thus, $U(t) \in L(H^n,H^n).$ It is now clear that the generator is still the Laplacian, but this time defined as a map $-\Delta: H^{n+2} \subset H^{n} \rightarrow H^{n}.$ 


**Question 1**: Is the Laplacian also in this case self-adjoint (on $H^n$)?


Certainly, it would be interesting to have $T$ defined on $H^{-n}$. Now, afais there are two ways to do it. Namely, if you take $\psi \in H^{n}$ and $l \in (H^{n})^\*$, then there is either $\phi \in H^{-n}$ :
 $$l(\psi) := (\psi, \phi) = \int\_{\mathbb{R}} \hat{\psi} \hat{\phi} $$ or 
$\phi \in H^{-n}$ such that $$l(\psi) = (\psi, \phi) = \int\_{\mathbb{R}} \psi \overline{\hat{\phi}}.$$


So far, I only found the first definition in the literature, but I like the second better, as this is more natural for this case:


We could then define $(\psi, U(t)^\* \phi) := (U(t)\psi,\phi)$ and thus define $U$ on $H^{-n}$ as $U(t):=U(-t)^\*.$ (sounds natural to me, as this is also how you would do it for $\phi \in H^n.$


In the first one, we could define $U(t)$ as a Banach-adjoint, right?. That is, 
$$ (U(t)' \psi)(\phi) = \psi (U(t)\phi).$$ But then we would have to set $U(t):=U'(t)$ on $H^{-n}$, right?


**Question 2**: Are both definitions correct and give eventually the same operator?



Free quantum evolution operator on Sobolev space
Question 1: yes, I think so. The generator of any one-parameter strongly continuous unitary group is self-adjoint. This is [Stone's theorem](https://en.wikipedia.org/wiki/Stone%27s_theorem_on_one-parameter_unitary_groups).


Question 2: frankly, it's hard for me to follow your notation here since you use the same symbol $U$ for everything and ${}^\*$ sometimes seems to be Banach space adjoint and sometimes Hilbert space adjoint (since you write both $U(t)^\*\phi$ and $U(t)^\*\psi$). But I think when you straighten out the notation --- call the two pairings $(\cdot,\cdot)\_1$ and $(\cdot,\cdot)\_2$ and the two results $U\_1$ and $U\_2$ --- then you'll be right and the calculation which shows this is easy. You just have to check that $(U\_1(t)\psi,\phi)\_1 = (U\_2(t)\psi,\phi)\_2$ for all $\phi$ and $\psi$, which you can do just by plugging in the definitions.


