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22

In QED there are 4 kinds of divergences: Ultraviolet divergences. Naive calculations depend on the cut-off in such a way that they go to infinity as the cut-off do. However, QED is a perturbatively renormalizable theory so that non-naive, well-done computations (see regularization and renormalization) give sensible results. Landau pole. The coupling ...


18

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 ...


15

Comments to the question: First it should be stressed, as OP does, that the Euler-Lagrange equations (= classical equations of motion = Maxwell's equations) are unaffected by scaling the action $S[A]$ with an overall (non-zero) constant. So classically, one may choose any overall normalization that one would like. As Frederic Brünner mentions a ...


14

Let us take standard fluid dynamics as a model of continuum media physics. It is obvious what is its "short-distance" physics isn't it? These are atoms and molecules. Let's apply your questions to this case -- suppose that we know nothing about atoms and molecules. in that case we can still use hydrodynamics, don't we? Do we replace the unknown/wrong ...


14

Richard Feynman's PhD thesis was about just this topic, if I am understanding your question rightly. Here is an earlier question about Feynman's thesis that addresses some of the fascinating issues involved with this. At the suggestion of his thesis adviser John Wheeler, Feynman explained photon emission as a two-way interaction in which the regular photon ...


12

The factor is there so that once you add a source term, i.e. $J^\mu A_\mu, $ you get the correct equations of motion, namely Maxwell's equations: $\partial_\nu F^{\mu\nu}=J^\mu.$ Furthermore, this convention produces the usual $1/2$ in front of the kinetic term of the gauge fields.


12

Frequency is not quantized, and has a continuous spectrum. As such, a photon can have any energy, as $E=\hbar\omega$. However, quantum mechanically, if a particle is restricted by a potential, i.e. $$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2 + \hat{V}$$ for $V\neq 0$, the energy spectrum is discrete. For example, in the case of the harmonic oscillator, ...


12

There is only one kind of photon. Indeed, when we describe elementary interactions between two electrons for example, we call the photon "virtual" as opposed to a physical photon that might exist outside of this process. Still, these are the same particles, i.e. excitations of the same fundamental field, as the photons that make up light for example. ...


12

Photons do not exhibit the property of virtual particles, but it is not your reasoning that is faulty, you have simply fallen prey to an imprecise use of terminology. Let me start with my view of the wave/particle duality. Most of the images of "particles" and "waves" comes from a time when we really didn't understand the quantum world, and some ...


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 ...


11

Oddly, polarizing sunglasses provide a quite solid proof that photons are spin 1. That's because if you rotate polarizers by only 90$^\circ$, you will find that you can break photons down into two mutually exclusive populations of photons. That is geometrically possible only if the particle in question is a vector boson, that is, a spin 1 particle. In ...


11

A "plane wave" generally refers to an infinite and perfectly flat wavefront, which cannot exist in reality, of course. However, there is nothing at all impossible about a plane wave of finite extent. Such a wave will experience diffraction at its edges, of course, but can still propagate over long distances before losing its planar nature. The problem with ...


10

Hwlau is correct about the book but the answer actually isn't that long so I think I can try to mention some basic points. Path integral One approach to quantum theory called path integral tells you that you have to sum probability amplitudes (I'll assume that you have at least some idea of what probability amplitude is; QED can't really be explained ...


10

P&S almost seem to argue that they need to be included for the simple reason that, if they didn't include them, they would get a nonsensical (i.e. infinite) result. Well, I am confident that Peskin and Schroeder not only "seem" to argue in this way but they explicitly and comprehensibly enough write this fact because it is both true and important. ...


10

The other answers explain that there's no paradox but they don't explain why the particular particle called photon is massless. It's massless because it is the messenger particle responsible for electromagnetism which is a long-range force. Its range is infinite so the mass has to be zero. One may view the Coulomb potential as the zero-mass limit ($m\to 0$) ...


10

Yes, the effect you're looking for is called Schwinger pair production. It requires immensely strong electric fields (of the order of $10^{18}$ V/m) for a constant field. One of the methods for computing the rate is the worldline method, described briefly here. To follow it, some knowledge of effective action methods are required.


9

Dear Claude, you are extrapolating electromagnetism way too high. You're going from low energies to the Planck scale, assuming that nothing qualitatively changes, but this assumption is wrong. The fine-structure constant is essentially constant below the mass of the electron - the lightest charged particle - which is 511,000 eV or so. You are extrapolating ...


9

One method is based on the conservation of angular momentum. The electronic transition must follow the selection rule $\Delta l=\pm 1$. So the first thing to do is to choose an atom with zero total angular momentum, then let the atom absorb a photon and make a transition to $l=1$ state. Secondly, we use the Stern-Gerlach experiment to detect the magnetic ...


9

(1) The completeness relationship for a basis of vectors orthonormal with respect to $\eta_{\mu\nu}$ is \begin{equation} \eta_{ij}\epsilon^{(i)}_\mu \epsilon^{(j)}_\nu = \eta_{\mu\nu} \end{equation} This normalization convention is picked for Lorentz invariance... I know you said you didn't want that answer but the point is that the normalization of these ...


9

This is a very good question, but it is really two completely different questions in one. Feynman's propagator The probability amplitude for a photon to go from x to y can be written in many ways, depending on the choice of gauge for the electromagnetic field. They all give the same answer for scattering questions, or for invariant questions involving ...


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 ...


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 ...


8

Both the wave theory of light and the particle theory of light are approximations to a deeper theory called Quantum Electrodynamics (QED for short). Light is not a wave nor a particle but instead it is an excitation in a quantum field. QED is a complicated theory, so while it is possible to do calculations directly in QED we often find it simpler to use an ...


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 ...


8

Worrying about the walls can be misleading. See A blackbody is not a blackbox for an illuminating account of the derivation of the Planck spectrum without enclosing the field in a box. If you cant get the published version, see the arxiv version. EDIT (25 March 2012) Planck's Radiation Law: A Many Body Theory Perspective discusses blackbody radiation ...


8

Michael Dine's response, quoted with permission: I now have to think back, but the argument in QED is based on the spectral representation ("Kallen-Lehman representation"). The argument purports to show that the wave function renormalization for the photon is less than one (this you can find, for example, in the old textbook of Bjorken and Drell, ...


8

$\chi$ is a real-valued function. This is part of the definition of the gauge transformation, since $U(1)$ is a one (real) dimensional group. In general, when talking about gauge transformations in particle physics, group parameters are restricted to be real by convention. In principle, I suppose you could perform a transformation on the wavefunction that ...


8

Just because $F^{\mu\nu}$ has two indices does not mean that it represents a spin-2 particle. Note that the metric $g^{\mu\nu}$ is a symmetric two indexed object while the EM field strength $F^{\mu\nu}$ is antisymmetric. In fact, the metric $g^{\mu\nu}$ is analogous to potential $A^\mu$ in EM and the field strength of gravity is the four indexed Riemann ...


8

The possibility of spontaneous Lorentz symmetry violation due to the infrared problem of the Dirac-Maxwell equation was conjectured a long time ago by Frohlich, Morchio and Strocchi, in references [1,2] mentioned in the given Balachandran and Vaidya article. In perturbative QED, we usually assume that the scattering states are free eigenstates of the number ...



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