Tag Info

Hot answers tagged

13

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 ...


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

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

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 ...


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

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 ...


7

We assume a square box, because it simplifies the argument. Yes, in the limit of $L_1, L_2, L_3 \to \infty$ this is equivalent to a square box in the limit $L \to \infty$ (we can't measure the difference between infinities). Also, in the limit $L \to \infty$ the quantized momenta will eventually cover all of momentum space, making the distinction ...


7

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 ...


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 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 ...


6

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 ...


6

Yes, you would have to introduce another gauge field. For example in the Standard Model there is gauge invariance under $SU(3)\times SU(2) \times U(1)$, and so there are three gauge fields: the gluons, the $W^\pm, Z$ weak gauge bosons and the photon. In general terms, it is simpler to argue like this: if you have gauge invariance under a Lie group $G$, the ...


6

A major difference between real and virtual photons is that virtual particles are not required to have energy and momentum on the "mass shell". That is, virtual photons may have $E^2-p^2 \neq m^2$, while real photons must obey $E^2-p^2=m^2=0$. My memory disagrees with Neuneck (v1): I think that a coherent superposition of real photons is a laser, while ...


5

For instance, how did he come up with interpreting the propagator as the propagation of particles? The path integral is usually introduced as a matrix element of the time evolution operator $$ \langle x_f\lvert\mathrm e^{-\frac{\mathrm i}{\hbar}\hat{H}(t_f-t_i)}\lvert x_i\rangle, $$ which is a measure of the probability of finding a system in final ...


5

The QFT for the scalar is considered to be massive for a very good reason: it is infinitely unlikely for the mass to vanish. There is no symmetry principle that would protect the scalar field from acquiring a generic mass. (The gauge symmetry is the principle that protects the masslessness of the photon but the scalar fields can't sacrifice to lose ...


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

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

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 ...


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

Classical electrodynamics and quantum electrodynamics don't agree with each other in general. They are distinct, inequivalent theories. Your observation that classical electrodynamics is deterministic, unlike QED, is one sufficient proof that they're inequivalent. At most, the expectation values of some observables in quantum electrodynamics obey the same ...


3

There's no difference between plasmon and plasmon polariton. Both of them indicate the resonant excitations involving electromagnetic wave and collective electronic motions simultaneously. "surface" stresses that the excitation in many cases occurs at the interface of a metal and a dielectric. However, there exist bulk plasmons as well. So "surface ...



Only top voted, non community-wiki answers of a minimum length are eligible