# Tag Info

## New answers tagged quantum-electrodynamics

2

Photoproduction is a process where something is produced by the interaction of a high energy photon. Something like $$\gamma + p \to p + \pi^0 \,.$$ Experimentally it is useful because the electromagnetic vertex is well understood, and photon taggers allow the creation of incident photons with well know energy and momentum.

1

The laws of conservations of momentum and energy combined forbid the reaction $$e^- + \gamma \rightarrow e^-$$ (Go ahead and do the math, is simple and enlightnening). But a completely different story is: $$e^- + \gamma \rightarrow e^- + \gamma$$ Where the incoming photon has a different energy that the outcoming one. And also, you can have an ...

0

"This illustration shows a Feynman diagram of a simple example of scattering. Two fast-moving electrons (e-) deflect each other via the electromagnetic force (represented in the diagram by the squiggly line of a photon)." @dmckee So that would be electrons interacting with each other via interaction with a virtual photon? ...

5

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 ...

2

I) The un-gauge-fixed QED Lagrangian density reads $$\tag{1} {\cal L}_0~:=~-\frac{1}{4}F_{\mu\nu}^2 + \bar{\psi}(iD\!\!\!\!/ \ \ -m)\psi.$$ The gauge-fixed QED Lagrangian density in the $R_{\xi}$-gauge reads $$\tag{2} {\cal L}~=~ {\cal L}_0 +{\cal L}_{FP}-\frac{1}{2\xi}\chi^2 ,$$ where the Faddeev-Popov term is $$\tag{3} {\cal L}_{FP}~=~ ... 2 Actually, in renormalization of QED, there is no demand to put the fermions of vertex diagram on mass shell. Renormalization procedure is usually performed on the level of Green functions with general four-momenta of outer legs. Note that the off-shell propagator you mention is connected to vertex function via Ward-Takahashi identity$$ (p'-p)_{\mu} ...

3

I understand that one can measure a single photon being absorbed using a photomultiplier tube or CCD. Can one measure a single photon being emitted by monitoring the current through an LED or the recoil of an emitting ion? The photon is a particle. It will have particle interactions, i.e. scattering off electrons and/or the spill over electric and ...

1

1) one way to find out is to measure motion of charged particles in their own fields on the microscale and infer which EM fields give best agreement with the observed motion. This was never done, because there are always EM fields of other sources and the motion is hard to measure accurately enough. 2) it depends on what you mean by "force is time-reversal ...

0

This can be solved by adding the non-electromagnetic energy $E_{p}$ of the Poincaré stresses to $E_{em}$, the electron's total energy $E_{tot}$ now becomes: $$\frac{E_{tot}}{c^{2}}=\frac{E_{em}+E_{p}}{c^{2}}=\frac{E_{em}+\frac{E_{em}}{3}}{c^{2}}=\frac{4}{3}\times\frac{E_{em}}{c^{2}}=\frac{4}{3}m_{es}=m_{em}$$ Thus the missing 4/3 factor is restored when ...

2

It can be seen to follow from a more general statement, namely: "if a parameter in the theory is such that the symmetry gets enhanced when it vanishes, then at every order in perturbation theory the corrections to this parameter will be proportional to its bare value". This is because perturbation theory respects the symmetry of the classical theory. If ...

1

You can only find the Hamiltonian if you do a so-called 'gauge fixing' procedure, since the Dirac field couples (minimally, but uniquely) to a gauge field. To get the Hamiltonian (density) you need to perform the full Dirac constraint analysis and at the end 'gauge fix'. See the books by Sundermayer or Henneaux+Teitelboim for details regarding the ...

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