Timeline for What is $\phi(x)|0\rangle$?
Current License: CC BY-SA 4.0
13 events
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| Sep 27, 2022 at 22:20 | history | edited | CommunityBot |
replaced http://www.mat.univie.ac.at/~neum with http://arnold-neumaier.at
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| Sep 5, 2021 at 17:22 | comment | added | reverendjamesm | Great, thank you very much! | |
| Aug 29, 2021 at 16:19 | comment | added | Arnold Neumaier | @jmg: I address this now in an addition to my answer. | |
| Aug 29, 2021 at 16:18 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
added details prompted by comment
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| Aug 28, 2021 at 14:45 | comment | added | reverendjamesm | Hi, do you happen to have a reference that explains this measure-theoretic POV? I couldn't find it in your FAQ. I am working on the Dirac equation and I am interested in its probabilistic position interpretation. If I have understood it correctly, one has to choose the position operator in the Foldy-Wouthuysen (FW) representation to have an adequate interpretation. What I haven't understood is what would that mean exactly for the density. Is it still the usual $\|\psi\|^2=\psi\cdot\bar{\psi}$ or does one need to take the FW transform somehow? $\|U_{FW}\psi\|^2$ maybe? Or perhaps these agree? | |
| Jul 12, 2015 at 13:13 | comment | added | Arnold Neumaier | The main point to take home from this discussion is that the particle concept is a semiclassical notion, valid only in an approximation that fails at short distances. For example, the photon concept works intuitively correct for the quantized electromagnetic field in exactly those situations where geometric optics is applicable. | |
| Jul 12, 2015 at 13:10 | comment | added | Arnold Neumaier | Feynman diagrams and their conventional interpretation as processes are only for building intuition for the perturbative series. They can in no way be identified with actual processes happening in space-time. That's why one talks about ''virtual'' particles and processes. Virtual means nonreal, unphysical, imagined. | |
| Jul 11, 2015 at 22:00 | comment | added | JLA | Do you know why then the Feynman propagator (say for the Klein-Gordon field) is described as the amplitude for a particle to travel from $x$ to $y$? It seems this isn't the case then; at best it's the amplitude for a particle which is most likely to be found near $x$ to end up in a state for which it is most likely to be found at $y$. This kind of kills the whole "mystery" of why the Feynman propagator is nonzero outside the light cone, since initially the particle had an amplitude to be outside of the light cone and so never needed to travel faster than $c\,.$ | |
| Jul 11, 2015 at 20:09 | comment | added | Arnold Neumaier | I meant to say (but wasn't allowed to edit it): Covariant position is never fixed, always uncertain with an uncertainty of the Compton length, due to Zitterbewegung. | |
| Jul 11, 2015 at 20:06 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
added remark of electromagnetic field
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| Jul 11, 2015 at 20:03 | comment | added | Arnold Neumaier | Position is never fixed, always uncertain with an uncertainty of the Compton length. This reconciles the different points of view. Note that probabilities are associated with observations, which always happen in the eigenframe of the observer. What you describe is a simplified version of the Unruh effect, which even says that the notion of particle is frame dependent. | |
| Jul 11, 2015 at 16:46 | comment | added | JLA | OK so to be clear, you are saying that despite the common occurrence of QFT books/notes claiming that $\phi(x)$ operating on the vacuum creates a particle at $x\,,$ it does not? I've read about the Newton-Wigner position operator, though it does seem a bit weird to me. So in one frame the particle can be located at a point, but at another its wave function is spread out... | |
| Jul 11, 2015 at 7:50 | history | answered | Arnold Neumaier | CC BY-SA 3.0 |