Timeline for Subtraction scheme invariance in QFT
Current License: CC BY-SA 4.0
8 events
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| Jan 9, 2019 at 10:04 | comment | added | AfterShave | Okay, I see. But don't different schemes alter the momentum depnendecy of the dressed propagator? It doesn't seem obvious to me that these should all drop out when we calculate observable effects. | |
| Jan 8, 2019 at 0:59 | comment | added | InertialObserver | It could seem so at first. But the correction to the coulomb potential is $\propto \frac{e_P^2}{m_P^2} \delta^3(x)$, and so it only depends on physical quantities, which are just numbers. | |
| Jan 8, 2019 at 0:43 | comment | added | AfterShave | Okay but doesn't the Green's function contribute to physical effects such as corrections to Coulomb's law? I was under the impression that the renormalized green's functions were strictly physical since they appear in finite results of the theory. | |
| Jan 8, 2019 at 0:41 | history | edited | InertialObserver | CC BY-SA 4.0 |
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| Jan 8, 2019 at 0:34 | comment | added | InertialObserver | In your original question you asked about the observables. And the only observable here is the physical mass, which is defined by making sure there is a pole in the propagator at the physical mass. I should note that there is one more condition I didn't mention, which I will put in my answer. | |
| Jan 8, 2019 at 0:32 | history | edited | InertialObserver | CC BY-SA 4.0 |
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| Jan 8, 2019 at 0:32 | comment | added | AfterShave | Thanks for the reply. How does the one condition nail down the renormalized greens' function seeing as we have multiple free parameters in the renormalization scheme? Am I correct in understanding that the renormalized green's function is supposed to be independent of subtraction schemes across its entire domain? As in literally being the same function for every four-momentum argument? | |
| Jan 8, 2019 at 0:19 | history | answered | InertialObserver | CC BY-SA 4.0 |