- Can anyone explain this in layman's words?
I'm new to QED, so I want to have a general grasp of what's going on. I just want to understand it conceptually. Can anyone explain it in a way so a layman can understand?
Yes, (if I remember correctly) the probability of a route is the square of the sum of the amplitudes of all possible routesAs an extension question,
I read parts of the QED lecture. It said that light can take an infinitely different route, but some routes are more likely for light to take.
It is also a sum of amplitudes but the lines on Feyman diagrams are not routes. The diagram shows all the ways in which given initial states can become (proposed ) output states. To calculate the probability of this one must sum the amplitudes for all the ways over all space. It is very technical as this paper ( 'Feynman diagrams for beginners') shows. The diagram that you think is an electron anihilating with a being deflected is not that. It is probably an electron anihilating with a positron.My question is that how does that idea relate to the Feynman diagrams (where 2 electrons deflect off each other because of a photon)
When we talk about light "taking different routes", we are basically talking about one Feynman diagram: the one that has just one photon line coming in and one photon line going out, with nothing happening in between. In other words, this one Feynman diagram represents an infinite set of possibilities: all of the possible paths that a photon could take between two fixed points in spacetime. Integrating over all those paths gives the probability amplitude that QED, or more precisely the pure photon part of QED (i.e., no electrons or interactions, just photon propagation), predicts for a photon that is emitted from the first point in spacetime to be detected at the second point.It said that light can take an infinitely different route, but some routes are more likely for light to take. My question is that how does that idea relate to the Feynman diagrams (where 2 electrons deflect off each other because of a photon)
Nobody knows. We don't even know if the question is meaningful. Electrons and photons aren't little machines. They're just electrons and photons. There might be no answer to the question of how they do what they do, other than "that's what they do". At some point, the search for a "mechanism", which basically means explaining something in terms of something else, has to stop; there has to be something fundamental, that just does what it does without being explainable in terms of something else.What's the mechanism behind how an electron emits (or absorbs) a photon? How does that work?
We don't measure electrons emitting single photons. The "electron emits a photon" process is a virtual process; it's part of the theoretical model shown in Feynman diagrams. It's not something we directly measure.Do electrons just go off an emit photons randomly? Or only when a circumstance happens? Is there a pattern to it?
So how did Feynman arrive at "electron emits a virtual photon?" And it carries momentum which causes electrons to recoil and deflect off each other.We don't measure electrons emitting single photons. The "electron emits a photon" process is a virtual process; it's part of the theoretical model shown in Feynman diagrams. It's not something we directly measure.
From the math of QED, which contains terms that, when translated into Feynman diagrams, can be interpreted that way. But the interpretation is not important; the predictions are. QED would make the same predictions if you didn't even try to interpret Feynman diagrams at all, but just did the integrals they told you to do and got numbers out. All the talk about electrons emitting and absorbing photons is just a way many physicists like to describe the diagrams in ordinary language to help them think about the calculations.how did Feynman arrive at "electron emits a virtual photon?"
We don't observe this either. We observe electrons appearing to repel each other via an "electromagnetic force", but we don't observe individual photons traveling between them and pushing on them. The photons are virtual photons. And the virtual photons appearing in the Feynman diagrams for things like the static Coulomb repulsion between electrons aren't even on shell, meaning they don't obey the relativistic energy-momentum relation for photons (another way to put it is that they don't travel at the speed of light).it carries momentum which causes electrons to recoil and deflect off each other.
There are an infinite number of Feynman diagrams for this case. The simplest one has two electron lines coming in, two electron lines going out, and a single internal photon line between them. But there are more complicated diagrams that have more internal lines. (Actually, even that's not the simplest possible diagram--see below.)Coming back to the basic scenario of two electrons shot at each other and bounce off. That is what the most basic Feynman diagram represents I think.
Made of virtual photons, which we never directly measure.In quantum physics, those electric field supposedly are made of individual discrete photons (i think).
This is actually another way to view the pure photon part of QED; the "path of least time" described in that chapter is the same as the "path of greatest amplitude" when you calculate the integral over all possible paths for the simplest one-photon Feynman diagram.In the 2nd chapter, it talks about light and how it is most likely to take the path of least time required to travel
Not really, because, as I said above, when we are talking about electrons repelling each other, the photons they are exchanging are virtual photons; whereas when we are talking about light propagation and the various phenomena described in QED chapter 2, we are talking about real photons that we actually observe (more precisely, we observe the light propagation phenomena being described).If any of that information relevant to the electron situation I mentioned above?