Timeline for Mathematical proof for virtual particle can be an off-shell particle
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2021 at 15:40 | comment | added | QFT addict. | Where you from @ Buraian | |
| May 12, 2021 at 15:31 | vote | accept | QFT addict. | ||
| May 12, 2021 at 14:23 | comment | added | Brian | Nice to see another user from Kerela :) | |
| May 12, 2021 at 14:16 | answer | added | Vladimir Kalitvianski | timeline score: 0 | |
| May 12, 2021 at 14:04 | answer | added | QFT addict. | timeline score: 1 | |
| May 10, 2021 at 18:41 | comment | added | QFT addict. | I made a small edit and make the question as Feynman propagator's particular case such that $x^0>y^0$, could you please tell now how the virtual particles associated with the above propagator be off-shell? @ Zack | |
| May 10, 2021 at 18:35 | history | edited | QFT addict. | CC BY-SA 4.0 |
added 9 characters in body
|
| May 9, 2021 at 21:42 | comment | added | QFT addict. | Please refer Eq 2.50 and 2.58 of Peskin and Shroeder@ Zack | |
| May 9, 2021 at 21:29 | comment | added | Zack | It's because you incorrectly evaluated the contour integral. Recall from Jordan's lemma that you must close integrals of the form $f(z) e^{i k z}$ in the upper-half plane, while you must close integrals of the form $f(z) e^{-i k z}$ in the lower-half plane. For you, this mean that you pick up a pole with residue $+E_{\vec{p}}$ or $-E_{\vec{p}}$ depending on whether $x^0 - y^0$ is positive or negative. The details are readily available in any QFT textbook -- I recommend chapter 2 of Peskin and Schroeder. | |
| May 9, 2021 at 21:12 | comment | added | QFT addict. | Could you please write, between which step the integration has an error? if need I can add more steps in between the integrals. @ Peter Kravchuk | |
| May 9, 2021 at 20:49 | comment | added | Peter Kravchuk | The two functions are clearly different: while the first integral solves the wave equation $(\partial^2+m^2)\Delta_F(x-y)\propto\delta(x-y)$, the second solves $(\partial^2+m^2)\Delta_F(x-y)=0$. (Because $a\times 1/a=1$ and $a\times \delta(a)=0$.) | |
| May 9, 2021 at 20:32 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
edited body; edited tags
|
| May 9, 2021 at 20:01 | comment | added | QFT addict. | Where it is?@Zack | |
| May 9, 2021 at 20:00 | comment | added | Zack | Oops, I made a typo -- I meant to say that your integration is INcorrect! | |
| May 9, 2021 at 19:58 | comment | added | QFT addict. | Yeah, It's the matter of flipping the contour in the upper and lower half-planes, the real question here is regarding the virtual particles and it's Mass shell connections, could you give some comments regarding this? @Zack | |
| May 9, 2021 at 19:45 | comment | added | Zack | Your integration of the Feynman propagator is correct -- you've implicitly assumed that $x^0 > y^0$. | |
| May 9, 2021 at 18:51 | history | asked | QFT addict. | CC BY-SA 4.0 |