Timeline for LSZ reduction formula valid for any type of observables?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 20, 2023 at 21:40 | comment | added | Leonid | @AccidentalFourierTransform: Could you please give a more detailed answer? I just don't see how you reached the conclusion that the pole is invariant based on what you wrote. If you want you can write an answer and I can accept it. | |
| Dec 18, 2023 at 20:51 | comment | added | AccidentalFourierTransform | poles do not depend on arbitrary choices (such as normalization, etc). E.g., the state $|\vec p\rangle$ enters LSZ twice: one in $1=|\vec p\rangle\langle\vec p|$ and one in $\langle 0|\phi(x)|\vec p\rangle=e^{ipx}$. The pole is invariant under $|\vec p\rangle\mapsto f(\vec p)|\vec p\rangle$. | |
| Dec 18, 2023 at 19:13 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 23 characters in body; edited tags
|
| Dec 17, 2023 at 18:55 | comment | added | Leonid | @AccidentalFourierTransform: Thanks. It seems you're saying that this is indeed true for any kind of correlation functions. The thing that still confuses me though is that the pole structure can be traced back to the resolution of the identity which can be traced back to our convention of normalizing states. If we chose a different normalization convention, say $\langle p|p\rangle=1$ (non Lorentz invariant), then wouldn't the two point function have no pole structure at all? And hence it won't be the green's function. | |
| Dec 17, 2023 at 18:29 | comment | added | AccidentalFourierTransform | related: physics.stackexchange.com/q/784436/84967 | |
| Dec 17, 2023 at 14:56 | history | asked | Leonid | CC BY-SA 4.0 |