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19

The main important idea of Feynman Wheeler theory is to use propagators which are non-causal, that can go forward and backward in time. This makes no sense in the Hamiltonian framework, since the backward in time business requires a formalism that is not rigidly stepping from timestep to timestep. Once you give up on a Hamiltonian, you can also ask that the ...


12

Here I'll try to basically connect some dots to guide you through the example of the second text you posted... Any quantum field theory of your choice associates certain integrals to observables, which you have to compute. The Feynman diagrams are representations of these integrals. The lines correspond to propergators, which encode the different field ...


10

The book Quantum dissipative systems by Weiss dedicates a subsection to the Feynman Vernon method, see also the original reference. See also this article and chapter 18.8 of the book by Kleinert. It's applied to the Caldeira-Leggett model, which is a toy model for a particle in contact with a heat bath. There are a number of mesoscopic systems out there in ...


10

:-) The best gentle introduction to basic twistor theory that I know of is the book by Huggett and Tod If you don't have access to that book and some other answers don't surface in the meantime I'm happy to write a few bits and pieces here, but will have to wait until the weekend. (I may be biased, but I think it's well worth learning, as the MHV ...


9

The MHV ideas are concerned, typically, with scattering amplitudes of gluons in Yang Mills theories. Most of the foundational work has been done with $\mathcal{N}=4$ supersymmetric Yang Mills theory, though I believe there have been extensions beyond this. The problem addressed is that you have n gluons meeting at a vertex, some incoming, some outgoing and ...


9

For nucleon-nucleon interaction please keep in mind that in this low-energy regime pertubative QCD breaks down and reactions are not really calculable. For the specific pion exchange you mention have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html as to why this QCD-process can be seen as an exchange of a pion. In general you ...


9

Your assumption that pair production is ruled out, rules out* that two photons interact through higher-order processes. Quantum electrodynamics tells us that two photons cannot couple directly. That leaves us with classical electromagnetism, which tells us that electromagnetic waves pass through each other without any interference. *Edit. The photons can ...


8

Well, it need not be a Lagrangian, you can work in Hamiltonian formalism too (which is often possible outside of particle physics, e.g. in condensed matter theory). But except for this caveat the answer is a resounding yes: the theory comes first and diagrams are just mnemonics. Diagram by itself doesn't make any sense if you can't reconstruct the ...


8

Drawing from Feynman's and Wheeler's memoirs: Feynman was originally motivated to produce a theory of EM without the infinities of self-interaction, but he then needed a mechanism to reproduce radiation reaction, the loss of energy of an accelerating electron. He thought that a nearby electron could back-react to achieve the effect, but his advisor ...


8

One of the avenues to search for an answer is the so-called Keldysh formalism which is used extensively in condensed matter, in particular in mescopic physics, to define and study steady-state and time-dependent quantum phenomena in systems with infinitely many degrees of freedom. A recent comprehensive review is given by Kamenev and Levchenko, ...


8

The first thing to notice, as pointed out in the comments, is that time increases going up. So if you are more familiar with viewing Feynman diagrams where time increases to the right, this problem is easily solved: just rotate the diagram by 90 degrees when you are interpreting it. If the problem is that you're not all that familiar with matter lines in ...


8

I would guess that the professor is explaining his/the(?) theory that dark matter is neutrinos, produced via a scattering process he calls "Witten's dog". It is funny because the neutrinos are coming out of the dog's butt. In the Standard Humor Classification, this is known as a "poop joke".


8

The fact that only connected Feynman diagrams contribute to the scattering amplitude can be interpreted in terms of the vacuum of the theory. Omitting disconnected diagrams amounts to shifting the vacuum: the vacuum of the interacting theory differs from that of the free theory. Regarding your second question: strongly connected (also called one-particle ...


8

This is a perceptive question. Consider the following from the Wikipedia article "Virtual Particle": As a consequence of quantum mechanical uncertainty, any object or process that exists for a limited time or in a limited volume cannot have a precisely defined energy or momentum. This is the reason that virtual particles — which exist only ...


7

There are, of course, a lot of codes floating around. Which of them you should choose, depends on what you want to calculate exactly. Here I mention four possibilities: 1) CALHEP - this package takes you from a given Lagrangian through its Feynmann rules to the calculation of cross sections. 2) xloops - this package calculates the 1-PI Feynman diagrams ...


7

The electron-positron pair can produce directly a Higgs boson, but this process is very suppressed, because the coupling between the leptons and the Higgs is proportional to the tiny mass $m_e$: $$g_{\rm Hee}=-i\frac{ m_e}{v},$$ where $v\approx 246 \,\rm{GeV}.$ On the other hand, the process $e^+ e^-\to H Z$ is more likely to happen, because the coupling ...


7

The reasons were given here. Essentially, at tree level you recover classical results. Loop corrections are proportional to powers of $\hbar$ and these are quantum terms.


6

In the normal usage, real and virtual are not properties of Feynman diagrams themselves, but of the particles depicted in them. The particles corresponding to external lines (attached to at most one vertex only) are real, the others (attached to two vertices) are virtual. A Feynman diagram may be considered as a repetitive part of a bigger diagram. This ...


6

If I understand your question correctly its just a matter of what you are calculating whether you put the external particles on shell or not. If you are, for example, calculating an amplitude to use for a cross section, you'll put the external particles on-shell and it will be what you call a 'real Feynman diagram'. If you are calculating an effective action ...


6

The main purpose of the space and time dimensions in Feynman diagrams is that the space dimension represents all possible spacial dimensions. 3D plots (which I assume you mean give two dimensions to space and one to time) would really only serve to give extra space on the diagram for interactions that would otherwise not fit on the page or become unreadable ...


6

A lowest order QED Feynman diagram for the process photon + photon $\rightarrow$ electron + positron looks like shown below (the time axis is the horizontal axis). From the point of view of energy conservation, this process is only possible if sum of the energy of the photons is above twice the electron mass. In the center of mass frame of the di-photon ...


6

For instance, how did he come up with interpreting the propagator as the propagation of particles? The path integral is usually introduced as a matrix element of the time evolution operator $$ \langle x_f\lvert\mathrm e^{-\frac{\mathrm i}{\hbar}\hat{H}(t_f-t_i)}\lvert x_i\rangle, $$ which is a measure of the probability of finding a system in final ...


6

In the first case, the vertex is a vertex in the common sense (used to construct diagrams). In the second case, the gauge field is not dynamic (in a path integral formulation, you do not integrate over), it is a background field that is fixed. In that case, we are interested on the effect of this non-dynamical field on the electron field. This is useful to ...


6

Assume that the generating functional is given by a sum of all possible diagrams, i.e. $$Z(J)=\Sigma_{n_i} D_{n_i}.$$ Furthermore, assume that each diagram D is given by a product of connected diagrams $C_i$, i.e. a diagram D can be disconnected. We will write this as $$D_{n_i}=\Pi_i\frac{1}{n_i!}C_i^{n_i},$$ where dividing by $n_i!$ amounts for a ...


6

The first process corresponds to $e^{-}e^{+}\to e^{-}e^{+}$ (Bhabha scattering), where the final and initial states are the same, consisting of an electron and positron. However, the second process is $e^{-}e^{+}\to \gamma \gamma$, where instead the final state is that of two photons. The scattering amplitudes will be different. Notice that the first diagram ...


6

I don't think that there would be any more diagrams. Having a total derivative term in the Lagrangian leads to derivative interaction vertex, which after symmetrising gives you something like \begin{equation} ig \sum_i p_i \ , \end{equation} where $g$ is some coupling and $p_i$ the momenta of the particles. This vertex, however, vanishes due to momentum ...


5

At the tree level (i.e. the simplest Feynman diagram) the both types of weak interaction result from the exchange of a weak boson. The weak bosons are the $Z^0$ (neutral) and the $W^\pm$ (charged). Guess how we assign the terms "neutral" and "charged" to weak interactions. Right, by the exchange boson. (We don't distinguish between interaction involving the ...


5

When drawing Feynman diagrams, it is important to fix the incoming and outgoing particles and their momenta. For the inexperienced, this is ideally done before drawing the rest of the diagram in order to avoid confusions like yours. So, let's assign each state some momentum: Let's give the electron in the upper left corner (4-)momentum $p_1$, lower left ...


5

The goal is to find the single particle propagator in the presence of interactions. This propogator will be the sum of all diagrams which have two external vertices. This sum of diagrams would be difficult to compute, but it turns out it easy to write this big sum of diagrams in terms of a sum of a smaller set of diagrams: the set of "one particle ...


5

The $^*$ notation does not mean excited in this case, it means "off shell" (i.e. virtual or having the "wrong" mass). At the second vertex the $Z^0$ is put "on-shell" by the emission of a Higgs (note, however, that it will decay very quickly in any case). The lepton pair can annihilate directly to the Higgs, but the event is experimentally identical to ...



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