Timeline for Intuition for spin 1/2 and 1 propagators
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 28, 2015 at 17:36 | history | tweeted | twitter.com/StackPhysics/status/659423559241093120 | ||
| Sep 10, 2015 at 19:41 | vote | accept | knzhou | ||
| Sep 9, 2015 at 22:37 | comment | added | Nikolaj-K | $\frac{1}{p^2-m^2}$ is $(\Box+m^2)^{-1}$ in Fourier space, in that it solves $(\Box+m^2)\,\phi=j$ for $\phi$. The spin 1/2 case $\frac{\gamma\,p+m}{N}$ solves $(\gamma\,\partial-m)\,\psi=j$, and so on. | |
| Sep 8, 2015 at 1:22 | comment | added | user73352 | I would suggest you read chapter 5 on this book to see where the propagators come from. nucleares.unam.mx/~alberto/apuntes/maggiore.pdf | |
| Sep 8, 2015 at 0:25 | comment | added | Winther | One can deduce some properties without explicitly calculating the propagator. The pole location is determined by the particle mass, the functional form follows from respecting relativity and not violating unitarity. The tensor structure is determined by the gauge choice + we only have two tensors to 'build with' and so on. | |
| Sep 8, 2015 at 0:24 | answer | added | Ali Moh | timeline score: 8 | |
| Sep 7, 2015 at 23:17 | comment | added | ACuriousMind♦ | I don't think this question is trivial - I don't understand what it is asking for. I don't get why your claimed "intuition" for the scalar propagator would lead you to believe $\frac{1}{p^2 - m^2}$ is the right expression for the Fourier transform of the propagator (i.e. "the particle likes to be on-shell" doesn't lead me to that expression in any way, intuitive or not), so I can't begin to tell you what "intuition" might be behind the non-scalar ones. | |
| Sep 7, 2015 at 23:15 | comment | added | knzhou | By intuition, I mean a couple of sentences that make me feel like I understand why these are the right expressions. You usually dismiss questions like this because you think they're trivial, but I don't find it trivial, so please tell me what you know. | |
| Sep 7, 2015 at 23:12 | comment | added | ACuriousMind♦ | I have no idea what you mean by "intuition" here. Why is the "intuition" for the scalar propagator "the particle likes to be on-shell"? Because it blows up to infinity there? (It does that for the other ones, too, and there are a bazillion other functions of $p$ and $m$ that also blow up near the shell values) | |
| Sep 7, 2015 at 22:55 | history | asked | knzhou | CC BY-SA 3.0 |