Timeline for What's the "propagator for real particles"?
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16 events
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| Oct 30, 2019 at 10:59 | comment | added | don't train ai on me | The vacuum in the interaction picture with interactions is not the same as in the free theory so the propagator is still different in the interacting theory. I mean one way to see it, is that if you draw the diagrams which give you the 2-point function (propagator), the first order diagram is the free propagator. The next orders are non zero. | |
| Oct 30, 2019 at 10:15 | review | Close votes | |||
| Nov 14, 2019 at 3:05 | |||||
| Oct 30, 2019 at 10:07 | comment | added | jak | @doublefelix In the interaction picture the field evolution is generated by the free Hamiltonian even in the presence of interactions. Hence the argument that we construct the Feynman propagator by using the free Lagrangian and thus has nothing do with interactions is not correct. As mentioned below, this also follows immediatly from the fact that the Feynman propagator is a Green function. A Green function describes how a field reacts to a delta source and is our most basic tool when we want to describe interactions. | |
| Oct 30, 2019 at 10:04 | history | edited | jak | CC BY-SA 4.0 |
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| Oct 30, 2019 at 9:57 | comment | added | ohneVal | Perhaps understand virtual particles more, elucidates the issue. Check possible duplicate of What actually are virtual particles? | |
| Oct 30, 2019 at 9:56 | comment | added | don't train ai on me | So by using the free lagrangian, we assumed it is a free particle. And the derivation of the propagator you wrote above doesn't work if you change the lagrangian. So it is only the propagator for free particles. | |
| Oct 30, 2019 at 9:54 | comment | added | don't train ai on me | Not because we use the Equation of Motion at all, but rather because the EOM that we use comes for the Lagrangian for free particles, $\partial_\mu \phi \partial ^\mu \phi - m^2 \phi ^2$. Therefore the results we get only apply to the case that we have a free particle. And if we add an interaction term like $\phi^4$ to the lagrangian, the old solution for $\phi$ does not solve the EOM that we get from the interacting lagrangian | |
| Oct 30, 2019 at 9:49 | comment | added | jak | @doublefelix I'm not sure I understand you correctly. You say the propagator is for free particles because we use the EOM and thus it describes on-shell particles? I don't think this is necessarily the case. Moreover, how then do you explain that it's not a solution of the EOM? | |
| Oct 30, 2019 at 9:39 | comment | added | don't train ai on me | The propagator you wrote above is only for free particles. It comes from plugging in the solution to the EOM that you get from the free lagrangian (which is no longer the solution with interaction terms) | |
| Oct 30, 2019 at 9:17 | history | edited | Qmechanic♦ |
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| Oct 30, 2019 at 8:57 | comment | added | jak | @doublefelix I don't think the Feynman propagator is a "free propagat". See my answer below. | |
| Oct 30, 2019 at 8:56 | answer | added | jak | timeline score: -1 | |
| Oct 30, 2019 at 8:55 | history | edited | jak | CC BY-SA 4.0 |
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| Oct 30, 2019 at 0:24 | comment | added | don't train ai on me | They probably just meant that this is only the free propagator. Particles in real life are interacting, so the propagator is not the same. | |
| Oct 29, 2019 at 16:48 | history | edited | jak | CC BY-SA 4.0 |
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| Oct 29, 2019 at 13:28 | history | asked | jak | CC BY-SA 4.0 |