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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

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

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

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

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. ...


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

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

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

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

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

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

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

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

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 ...


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

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 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

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

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

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

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 ...


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 a just-so stories designed to explain Feynman ...


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

If I understand what you're asking, it's false: there are plenty of examples of theories that are asymptotically free and also weakly coupled in the IR. A QCD-like theory with more flavors of quarks would be an example. The phrase to search for is "Banks-Zaks fixed point." For the revised version of the question: there are certainly RG flows that are free ...


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


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 ...



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