# Tag Info

23

Let me begin with QED. I will subsequently connect with QCD. There are 4 kinds of divergence in QED: 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 ...

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. ...

13

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 ...

13

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 ...

12

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 ...

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.

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}} = ... 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 ... 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. 10 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 ...

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

Short answer: A virtual particle is not the opposite of a classical particle. While the other answer captures some aspects correctly, there are still a few flaws and inaccuracies which in the following, I will try to set straight. Wave-particle duality Strictly speaking, quantum objects are neither waves or particles. They are entities behaving like ...

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 ...

9

Virtual particles are not real. They come, as I've said in many answers on this site, from a naive interpretation of Feynman diagrams which should not be taken as an actual, exact description of how the physics works. The actual description of an interaction in the quantum field theory is more complicated than "photons are exchanged". In particular, ...

9

Would you dig a ditch with a surgeons scalpel? Yes, quantum mechanics ultimately underlies all physical observations but the mathematical expressions for large dimensions with respect to $\hbar$ become cumbersome and are replaced by the simplest ones for the appropriate study. Thermodynamics, for the study of bulk matter, blends smoothly with quantum ...

9

But what we never seem to see is why the electron and positron move the way that they do. Saying "they move like they do because of the force on them" doesn't explain anything at all. It's a non-answer. The equation of motion for charge particle (electron,positron) in magnetic field is  m\frac{d}{dt}\left(\frac{\mathbf ...

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