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20

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...


19

Maxwellian electrodynamics fails when quantum mechanical phenomena are involved, in the same way that Newtonian mechanics needs to be replaced in that regime by quantum mechanics. Maxwell's equations don't really "fail", as there is still an equivalent version in QM, it's just the mechanics itself that changes. Let me elaborate on that one for a bit. In ...


11

Absolutely. If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is $$k^\mu_{\text{in}} = ...


11

(I henceforth assume $c= \hbar=1$.) It is forbidden by the four-momentum conservation law. Put yourself in the centre of mass reference frame of the couple of massive particles (electron and positron). There $P_{e\overline{e}} = (2E,\vec{0})$ with $E\geq m_e>0$. Just because four momentum is conserved, this four-momentum must be the same as the one of the ...


9

I'm a bit rusty on my qed, but I'll give this a shot. The simplest case would be described by a diagram similar to: But the $e^--e^--\nu_e$ vertex doesn't exist (also note that I can't draw the required arrow on the neutrino) - the vertices of the standard model (with the exception of vertices involving the Higgs and neutrino oscillations) are: With ...


9

The fundamental particles we know today (of which the photon is one) are called fundamental exactly because they have no substructure, or indeed, spatial extent, we know of. They are point-like when localized. Note that these "particles" are quantum objects, not classical particles, so you should not imagine them as points whizzing about in space - they ...


9

Part b) is a big mathematical physics topic in its own right. The divergent tail of an asymptotic series is not garbage, rather it contains a lot of information that together with some additional information can be used to compute non-perturbative effects. A general introduction to this topic is given here. There are different approaches possible, some ...


8

The magnetic moment of the electron is a magnetic moment, so the right magnetic field around it is $$ \mathbf{B}({\mathbf{r}})=\nabla\times{\mathbf{A}}=\frac{\mu_{0}}{4\pi}\left(\frac{3\mathbf{r}(\mathbf{\mu}\cdot\mathbf{r})}{r^{5}}-\frac{{\mathbf{\mu}}}{r^{3}}\right). $$ The world is quantum mechanical – and so is any viable description of the spin – so we ...


7

The advantage of unitary gauge is that it completely removes unphysical fields, while adding additional degrees of freedom to the gauge bosons, which consequently become massive. This gauge works well for tree-level calculations, but complications arise when considering loops: The propagators of gauge fields and ghosts (which are needed to impose the ...


7

The extra term, in general $$\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{2 \xi}(\partial_{\rho}A^{\rho})^2 $$ is called gauge fixing term. This term is needed in order to be able to quantize the field $A_\mu$. Without this extra term the photon propagator is ill defined $$D^{\mu\nu}={-i\over k^2+i0}\left(g^{\mu\nu}\,+\,(\xi-1){k^\mu ...


7

The Lagrangian provided is Maxwell's Lagrangian, supplemented by a gauge fixing term: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$ The equations of motion are, $$\partial_\mu F^{\mu\nu} + \partial^\nu (\partial_\mu A^\mu) = \partial_\mu \partial^\mu A^\nu = 0$$ Instead of making a gauge fixing procedure a ...


6

The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles. Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is $$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p ...


6

The binding energy of the electrons in a silver atom is far less than the rest energy of an electron, so there is no ambiguity about the number of electrons in a silver atom. That makes adding up the spins a straightforward business. By contrast, the combined mass of the two up and one down quarks in a proton is about 10MeV (it isn't precisely known) but ...


6

This might not be quite the answer you are looking for, but one useful way to think about it is: the accelerating electron emits photons because nothing forbids it from doing so. By definition, because an electron has electric charge it is coupled to the electromagnetic field, and is able to produce excitations in this field which we can call photons. This ...


6

This comes from classical electrodynamics, there is no need to go to Standard model theory or quantum electrodynamics for this. The simple answer is that electric potentials, like electric fields, are just a way of characterizing the way charged particles interact with each other. So, charged objects create voltage analogous to the way that they create ...


5

There are very strict limits on the mass of the photon already, so it would only affect our understanding of physics on the largest scales. The cosmologists would have some hard thinking to do, for instance. However, contrary to a comment, it would not affect relativity beyond requiring us to reconsider the usual name for $c$: not "the speed of light" but ...


5

Regardless of renormalizability, the term that you wrote down $(gA_\mu A^\mu \phi)$ does not describe photons because it is not gauge invariant. This would be a theory of a massless vector boson with three dynamical propagating degrees of freedom (two transverse and one longitudinal), which is inconsistent with Lorentz invariance and irrelevant to ...


5

Disclaimer: This is answer is given from a mathematical physics point of view, and it is a little bit technical. Any comment or additional answer from other points of view is welcome. The classical limit of quantum theories and quantum field theories is not straightforward. It is now a very active research topic in mathematical physics and analysis. 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

There seems to be no empirical motivation to consider this possibility. But if we approach it as a pure "what if" anyway, it opens a big can of worms. (1) How do we even implement the idea in a principled way? We might take a given calculation in which the fine-structure constant is employed, and substitute a sampling of Gil's "very concentrated ...


5

In our modern understanding, every electron is thought to be a localized excitation of the electron (or Dirac) (spinor) field $\Psi(x^\mu)$, while every photon is considered to be an excitation of the photon (vector) field $A^\nu(x^\mu)$, which is the quantum field-theoretic counterpart of the classical four-potential. Thus, the answer to your questions ...


4

Quantum electrodynamics is needed to describe Nature instead of classical electrodynamics because quantum phenomena are observed – and have been observed at least since 1900 – which prove that classical physics in general and classical electrodynamics in particular is incorrect as a description of Nature and a better theory is needed. The quantum phenomena ...


4

There are, in general, no closed form solutions (aka formulas) for the spectra of multi-electron atoms. There are reasonably precise formulas for special cases, like approximate values of x-ray transitions from inner shell electrons, though. Unlike in case of hydrogen and Rydberg atoms, which can be treated as a non-relativistic one-body problems (i.e. for ...


4

Virtual particles are not real It's in the name. You may draw Feynman diagrams where there are internal lines, and we call these internal lines virtual particles. They are not real. You will never detect a virtual particle. They are not really exchanged between the real charged particles. Virtual particles are just-so stories designed to explain Feynman ...


4

Feynman diagrams are more than just the Lagrangian. They can be acquired by expanding the path integral of the theory into a perturbative series. There is a priori no reason to assume that all quantities needed in order to produce sensible results are consistent with gauge invariance. One possible issue is the problem of regularization: the way your ...


4

You should not imagine a virtual photon as an individual object wandering from one charged particle to another. This picture is simply inapplicable. Unfortunately, Feynman diagrams mislead people to imagine such things. Actually, Feynman diagrams are good for calculation and bad for imagination. Feynman diagrams have been introduced to help physicists to ...


4

There are two things to consider: What does the potential look like? Is the wave function of the qubit narrow in the flux or charge basis? Potential shape The Hamiltonian of the transmon (a junction in parallel with a capacitor) is $$H_{\text{charge qubit}} = - E_J \cos(\phi) + \frac{(-2en)^2}{2C}$$ where $E_J\equiv I_c \Phi_0 / 2\pi$, $I_c$ is the ...


4

I cannot answer to all the questions but would like to stress something regarding what the Casimir effect tells us and what it doesn't. If you look at how it is derived for the usual EM interaction, an experimental verification of the standard Casimir effect tells us that: the EM field can have standing waves between two plates and outside them There ...


3

The longitudinal mode decouples from all physical processes as a consequence of gauge invariance, which in turn forces the Ward identity $$ k^\mu \mathcal{M}_\mu = 0$$ where the S-matrix element decomposition $\mathcal{M}^\mu$ is obtained from the polarization vector $\epsilon^\mu(k)$ by $\mathcal{M} = \epsilon^\mu(k) \mathcal{M}_\mu$. This decoupling (and, ...


3

If there were a $t$-channel or $u$-channel diagram for this process, it would have to involve a vertex where an electron changes into a muon and some other particle. There is no such vertex in the standard model.



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