Timeline for Is an elementary particle traveling through a vacuum the *same* particle at points A and B?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 31, 2021 at 18:00 | comment | added | anna v | @Edouard no, those are detector dependent effects | |
| May 31, 2021 at 17:53 | comment | added | Edouard | @anna v --This may be a little off-topic, but I'd read somewhere, I think in a paper by Veneziano that NASA had posted on the web but has since removed, that real particles may be produced by the gravitational field thru the separation of virtual particles from their partners in virtual particle/antiparticle pairs by more than the Compton wavelength for more than the Compton time. Might that process account for some of the dashed or broken lines in the photograph that you've posted in your answer? | |
| Oct 19, 2016 at 21:32 | comment | added | IamZack | @annav Well, as i said the field operator is composed of those functions a and a+. And I'm pretty sure of that. Because the fields are expansions (fourier expansions) and the operators are the coeficients of that expansion. So i stated that, it should be said: "Operators into theese fields." instead of "operators ON theese fields". But whatever | |
| Oct 16, 2016 at 5:53 | comment | added | anna v | @IamZack well, a creation/annihilation operatpr a+ acts on the ground state wave function which represents the field, free electron wave function for example, for electrons acted by the a+ is one electron at that (x,y,z,t) for a wavepacket this covers a moving volume in four space, with annihilation operators bringing back the vacuum | |
| Oct 16, 2016 at 2:01 | comment | added | IamZack | Hell of an answer! Anyway, this statement is a little bit confuse, or it is just my english , or my bad understanding of QFT ( lol) but... "...are the result of creation and annihilation operators on these fields," The fields themselves are expansions of creation/annihilation operators... so it should be : "are result of creation and annihilation operators into these fields." Shouldn't? | |
| Oct 12, 2016 at 12:40 | comment | added | MissMonicaE | Thank you! This is the question I spent a semester trying to get a straight answer to. :) | |
| Oct 12, 2016 at 12:35 | comment | added | anna v | @MissMonicaE yes. in physics we define the terms we use dependent as closely as possible to observations and measurements. The "duckness" is a reminder of this, that one has to start with definitions based on measurements/observations. | |
| Oct 12, 2016 at 12:27 | comment | added | MissMonicaE | "If it looks like a duck and it quacks like a duck it is a duck by definition"--so to a physicist, a "duck" is a thing that looks and quacks in the manner we call "ducklike"? I.e., saying "X is a duck" is a shorthand for "X looks and quacks like a duck"--is that right? (This comment sounds sarcastic to me but I'm honestly just trying to clarify.) | |
| Oct 12, 2016 at 12:22 | history | answered | anna v | CC BY-SA 3.0 |