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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 ...


20

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 ...


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 ...


18

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 ...


15

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 ...


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

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, ...


13

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 ...


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

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 ($m\to 0$) ...


12

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

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. ...


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 ...


11

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 ...


11

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 ...


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

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}} = ...


10

You have to be careful about use of the term bound state in QFT. There is a related notion of resonance, which is a state that looks approximately like a bound state for a while (possibly a long while!) but eventually decays. For example, if you consider the nonrelativistic hydrogen atom, it has bound states corresponding to the electron orbitals (the n,l,m ...


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

Virtual photon clouds are responsible for potentials, not electric and magnetic fields, and this is what makes the explanation of forces in terms of photon exchange somewhat difficult for a newcomer. The photon propagation is not gauge invariant, and the Feynman gauge is the usual one for getting the forces to come out from particle exchange. In another ...


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

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 ...


9

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 ...


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

Update: I went over this answer and clarified some parts. Most importantly I expanded the Forces section to connect better with the question. I like your reasoning and you actually come to the right conclusions, so congratulations on that! But understanding the relation between forces and particles isn't that simple and in my opinion the best one can do ...



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