Timeline for What are the spaces in which quantum fields belong and how does that affect the hermitian conjugate of $\partial_{\mu}$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 13, 2022 at 22:09 | comment | added | Shiki Ryougi | @Janik Thank you for the response! This is pretty much what I was looking for, i.e. mathematical rigor. But I need some time to grasp what you've said - I will come back when I'm more informed. | |
| May 13, 2022 at 14:54 | comment | added | Cosmas Zachos | @Vangi Sorry, I don't wish to get involved in cross-link "you know what I mean" conversations. I stand by all my formal statements. You obviously appreciate how misframed and misbegotten the entire conversation is. Any decent QM text clears up the hermiticity and application rules for the momentum operator, which only differs from the gradient by a -i factor... | |
| May 13, 2022 at 14:42 | comment | added | Vangi | @CosmasZachos So, you do agree with the answer which OP gave in his first link. The context he was talking in was $(\partial \phi)$, but gave (wrong) justification by writing down the transformation of $\partial$, which you have corrected in your comment on the question. | |
| May 13, 2022 at 13:30 | comment | added | Cosmas Zachos | @Vangi Just follow the rules: $(\partial \phi)^\dagger= \phi^\dagger \partial^\dagger =\phi^\dagger \overset{_\gets}\partial = \partial \phi^\dagger$. | |
| May 13, 2022 at 8:51 | comment | added | Janik | If the field is hermitean, then, for every test function $f$, we have $\phi(f)^\dagger = \phi(\overline{f})$. Thus, $\partial \phi(f)^\dagger = - \phi(\overline{\partial f})$. | |
| May 13, 2022 at 1:29 | comment | added | Vangi | Could you explicitly type what $(\partial \phi)^\dagger$ can be rewritten as? | |
| May 12, 2022 at 22:13 | history | edited | Janik | CC BY-SA 4.0 |
added 138 characters in body
|
| May 12, 2022 at 22:02 | history | answered | Janik | CC BY-SA 4.0 |