Timeline for How do derivative couplings affect canonical quantization?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2016 at 18:59 | vote | accept | knzhou | ||
| S Apr 23, 2016 at 16:34 | history | bounty ended | AccidentalFourierTransform | ||
| S Apr 23, 2016 at 16:34 | history | notice removed | AccidentalFourierTransform | ||
| Apr 23, 2016 at 11:07 | history | edited | Qmechanic♦ |
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| Apr 23, 2016 at 10:50 | answer | added | Qmechanic♦ | timeline score: 4 | |
| Apr 20, 2016 at 7:15 | answer | added | Harold | timeline score: 0 | |
| Apr 17, 2016 at 17:57 | comment | added | Tomáš Brauner | @knzhou It looks like you actually have two questions: (i) how to derive the Feynman rule for this interaction, (ii) how to get the result using canonical quantization. While the latter can be tricky (cf. the above comments on higher-derivative theories), if you are just interested in the result, simply use the path integral. There are no issues associated with higher derivatives there. | |
| Apr 15, 2016 at 23:28 | history | tweeted | twitter.com/StackPhysics/status/721117986271203328 | ||
| Apr 15, 2016 at 21:47 | comment | added | Michael Seifert | @ACuriousMind: It is possible to obtain a Hamilton for higher-derivative theories; you can use the techniques that Ostrogradski used to prove that such theories are unstable. There's a nice explanation of how the derivation proceeds in Section 2 of this paper by R.P. Woodard. | |
| Apr 15, 2016 at 21:30 | comment | added | ACuriousMind♦ | I...don't think one can canonically quantize terms with higher derivatives because those do not have an associated Hamiltonian picture. | |
| S Apr 15, 2016 at 21:26 | history | bounty started | AccidentalFourierTransform | ||
| S Apr 15, 2016 at 21:26 | history | notice added | AccidentalFourierTransform | Draw attention | |
| Apr 4, 2016 at 1:25 | comment | added | Nahc | You just need to use the Feynman rule for free scalar field theory. Say the coupling constant in your interaction term is $\lambda$, then the first order correction is $\lambda \times (#)$, while the term (#) should come from the zero order result. | |
| Apr 3, 2016 at 19:48 | comment | added | knzhou | @HChan I know what the Feynman rule is, but it looks like actually deriving it is sort of tricky. Do you know how to? | |
| Apr 3, 2016 at 18:39 | comment | added | Nahc | This term has dimension 7 and is non-renormalizable. But in EFT, you should write down any terms that obey the symmetries of your theory, and this term should show up. Feynman rule for it is just $p^4$. | |
| Apr 2, 2016 at 0:45 | history | edited | knzhou | CC BY-SA 3.0 |
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| Apr 2, 2016 at 0:39 | history | asked | knzhou | CC BY-SA 3.0 |