Timeline for Why is $\hat{p} \circ \hat{p}$ the operator corresponding to $p^2$?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 29, 2014 at 16:59 | comment | added | Valter Moretti | I agree, the definition of $f(T)$ via power series is almost always impossibly difficult to handle and should be avoided....The case of the exponential is very particular as there are several results in the literature (the "theory of analytic vectors"), by Nelson in particular... | |
| Sep 29, 2014 at 15:30 | comment | added | Mateus Sampaio | You are right, I made a mistake in my claim. But the definition via power series has complicated issues, like the radius of convergence of the series, even if $\psi \in P^T_{[a,b]}$ (what is not an issue in your example, since the radius is $\infty$) and cannot handle the definition of Borel functions of operators in general. | |
| Sep 29, 2014 at 14:58 | comment | added | Valter Moretti | You (@Mateus Sampaio) are right about the general problem, these things must be handled by means of the general spectral theory. However for $T$ unbounded self-adjoint (also for unbounded closed normal operators), $\cap^\infty_{n=1} dom(T^n)\neq \emptyset$ and that intersection always includes a dense subspace of vectors! For instance all vectors in the subspaces $P^{(T)}_{[a,b]}({\cal H})$ ($P^{(T)}$ being the PVM of $T$) for every $a,b \in \mathbb R$, $b>a$. For these vectors $\psi$, for instance $e^{itT}\psi = \sum_{n=0}^{+\infty} \frac{(it^n)}{n!}T^n \psi$... | |
| Sep 29, 2014 at 12:39 | comment | added | Mateus Sampaio | The idea here is right, but not the way how it's done with rigor. In general for, Self-adjoint operators, we have $\text{dom} \, T^2 \subset \text{dom} \, T$, and may be the case that $\bigcap_{n=1}^\infty \text{dom} \, T^n =0$, so that the operator defined by the power series is only trivialy defined. This only works in the case that the operator $T$ is bounded. The right way to define functions of operators is via the spectral theorem. Being $\mu_\psi^T$ the spectral measure of $T$ with respect to $\psi$, we define $\langle f(T)\rangle=\int_\Bbb{R} f(t)\mu_\psi^T(t)$. | |
| Sep 29, 2014 at 12:36 | vote | accept | David Zhang | ||
| Sep 29, 2014 at 6:51 | history | answered | Everett You | CC BY-SA 3.0 |