# Tag Info

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 ...

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, ...

11

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. ...

11

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 ...

9

(1) The completeness relationship for a basis of vectors orthonormal with respect to $\eta_{\mu\nu}$ is $$\eta_{ij}\epsilon^{(i)}_\mu \epsilon^{(j)}_\nu = \eta_{\mu\nu}$$ 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 ...

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 ...

7

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 is to choose an atom with zero total angular momentum, then lets the atom absorbs a photon and make a transition to $l=1$ state. Secondly, we use the Stern-Gerlach experiment to detect the magnetic ...

6

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 ...

6

There is a ward identity that links the charge renormalization to the photon's wave function renormalization. Ward identities are relationships between correlation functions that follow from the quantum theory having a symmetry. In this case the gauge invariance of QED relates (among other things) the electron's two point function (propagator) to the ...

6

Forward scattering need not be equivalent to "no scattering" - and, indeed, will only rarely be indistinguishable from it. In the usual scattering-theory setup, you have an electron coming in in a plane wave $$\psi(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}=e^{ikz}$$ and impinging on some short-range potential. This will add to the wavefunction a scattered ...

6

Pure QED, unlike Pure Yang Mills ('pure' in the sense that there is only an $F^2$ term in the lagrangian, and it doesn't couple to matter) is a free theory. That means that it's boring, there's no need for renormalization or perturbation theory or anything. So the coupling constant (in this case the wave function renormalization of the photon) doesn't run ...

6

The Lagrangian has many parts that are each guessed at according to symmetry principles, requirements that the theory be well behaved, and reproduce experimental results. It's not something you can do from first principles, because the first principles aren't known. But the aforementioned process took about a 75 years and many Nobel prizes and PhDs were ...

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

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 ...

6

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 ...

5

First of all, virtual particles are indeed a consequence of the uncertainty principle – without any quotation marks. Virtual particles are those that don't satisfy the correct dispersion relation $$E = \sqrt{m^2 c^4 +p^2 c^2}$$ because they have a different value of energy by $\Delta E$. For such a "wrong" value of energy, they have to borrow (or lend) ...

5

In this link there exists a mathematical explanation of how an ensemble of photons of frequency $\nu$ and energy $E=h\nu$ end up building coherently the classical electromagnetic wave of frequency $\nu$. It is not simple to follow if one does not have the mathematical background. Conceptually watching the build up of interference fringes from single photons ...

5

The electron field transforms under the $\mathbf 1$ of $U(1)$, i.e., the generator is $i$ or $1$ depending on your convention/notation. The gauge fields transform in the adjoint representation , but they transform as a connection, as @Adam mentioned. In other words, if $\psi \to g \psi$, then $D_\mu \psi \to g D_\mu \psi$ implies that $A_\mu \to g D_\mu ... 5$U(1)$is an Abelian group. Abelian groups only have 1-dimensional irreducible representation. Namely, transformation by a phase (in the case of the electron). The charge of fermion field is proportional to the coefficient of the phase. In particular, a field of charge$q$transforms as$\Psi \to e^{i q \theta(x)} \Psi$EDIT: As pointed out in the comments, ... 5 In the first case, the vertex is a vertex in the common sense (used to construct diagrams). In the second case, the gauge field is not dynamic (in a path integral formulation, you do not integrate over), it is a background field that is fixed. In that case, we are interested on the effect of this non-dynamical field on the electron field. This is useful to ... 5 A major difference between real and virtual photons is that virtual particles are not required to have energy and momentum on the "mass shell". That is, virtual photons may have$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

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 ...

4

The reason can be found in the masslessness of photons. What this means is that the rest mass of the photon vanishes. This can be seen by analyzing the framework of special relativity, which is based on the observation that light is moving at the same velocity in all frames of reference. The relativistic energy-momentum relation of a general particle is ...

4

LSZ reduction for photons is discussed e.g. in (little) Chapter 56 of Srednicki's book http://adsabs.harvard.edu/abs/2007qft..book.....S

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In particle physics there exists elastic scattering for all interactions: change of direction but not of energies. When a photon penetrates into a medium composed of particles whose sizes are much smaller than the wavelength of the incident photon, the scattering process, also known as Rayleigh scattering, is also elastic. In this scattering process, ...

4

These are just my thoughts as someone who studied the subject for a while: The concept of virtual photons that mediate interaction should not be seen as "what really happens". A virtual photon is not a real object (hence the name "virtual"), but an artifact of perturbation theory. If we knew an effective way (or even "a" way) to do the calculations without ...

4

The photon polarization directions are only transversal when it is free in space . The polarization of an interacting photon or a photon with nonfree boundary conditions is not transversal in general. One example is that electromagnetic waves possess longitudinal polarizations in waveguides. Another relatively simple example where (a certain combination ...

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