Timeline for What exactly is $\hat{\psi}^\dagger(x)$? An operator or a function?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 28, 2018 at 16:41 | comment | added | Ryan Thorngren | It was mostly a joke, since the Yoneda lemma says somehow that "everything is a function". :) I've never used it for anything except constructing classifying spaces. | |
| Mar 28, 2018 at 16:36 | comment | added | leftaroundabout | @RyanThorngren I've multiple times tried to understand what the Yoneda Lemma is supposed to be good for, but I've never considered it in the context of quantum mechanics. Sounds interesting, would you elaborate? | |
| Mar 28, 2018 at 15:31 | comment | added | Ryan Thorngren | en.wikipedia.org/wiki/Yoneda_lemma | |
| Nov 13, 2013 at 1:10 | history | edited | leftaroundabout | CC BY-SA 3.0 |
added 238 characters in body
|
| Nov 13, 2013 at 1:06 | comment | added | leftaroundabout | AFAICS, Wightman more or less defined "operator-valued distribution" as such a thing, but I'm not quite firm enough in axiomatic QFT to be able to tell. — That commutator does not prove anything: unbounded operators do not form a (semi)group, so even if you only put in well-defined unbounded operators the whole thing isn't necessarily a proper function. — Either way, your criticism definitely is right in at least one regard: I wrote "an operator is just a function", which is just not right in the usual sense for unbounded operators. | |
| Nov 12, 2013 at 20:07 | comment | added | jjcale | to "I'd need to think about it, but I believe the notion of an unbounded-operator valued distribution doesn't even make sense mathematically" : It makes sense, see e.g. Streater-Wightman axioms. $\hat{\psi}^\dagger(x)$ must be a distribution since $[\hat{\psi}(y),\hat{\psi}^\dagger(x)]=\delta(y-x)$ | |
| Nov 11, 2013 at 22:13 | comment | added | leftaroundabout | I'd need to think about it, but I believe the notion of an unbounded-operator valued distribution doesn't even make sense mathematically. No, $\hat{\psi}^\dagger$ is actually a function. What you probably mean: $y \mapsto \langle y | \hat{\psi}^\dagger(x) | \emptyset \rangle$ is not a function but a distribution. This has to do with $\hat{\psi}^\dagger(x)$ being an unbounded operator on the infinite-dimensional Hilbert space; but such an operator is still a perfectly well-defined function value (unlike "the infinite value" of $\delta(0)$). | |
| Nov 11, 2013 at 21:51 | comment | added | jjcale | $\hat\psi^\dagger$ is not an operator valued function but an operator valued distribution. | |
| Oct 14, 2013 at 11:29 | history | answered | leftaroundabout | CC BY-SA 3.0 |