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19

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 ...


12

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 ...


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 ...


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 ...


7

Yes. Consider quantizing electromagnetic fields in a box. This corresponds to photons being trapped inside of said box since photons are just the mode quanta of the EM fields. The Hilbert space (called Fock space in this case) of the quantized radiation is found to be spanned by states $$ |\mathbf k_1, \mu_1; \dots, ; \mathbf k_N, \mu_N\rangle, \qquad ...


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 ...


7

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 ...


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

The group $\mathrm U(1)$ can be described as the set of complex numbers of unit modulus with the group multiplication given by multiplication of complex numbers. Given this characterization, notice that the transformation $$ \Psi \to e^{iq\chi/\hbar} \Psi $$ constitutes an action of $\mathrm U(1)$ on the Dirac field $\psi$. If $\chi$ is then promoted to ...


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

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 ...


5

The two equations would lead to equivalent physics but only one of them may be right: the correct equations of motion for Dirac spinor fields are first-order in derivatives. The convention is that only the first equation is right for the Dirac field $\Psi$. The second equation is simply incorrect and doesn't follow from the Lagrangian etc. If we wanted ...


5

Because you can prepare a state with an arbitrarily long wavelength, hence arbitrarily low energy, photon. That's essentially the definition of a massless particle. If you put in an IR regulator, by putting the system in a box for example, a gap appears since there is now a largest possible wavelength. This can be mimicked by giving the photon a small mass. ...


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

$U(1)$ 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

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

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 ...


4

The bottom line of my entire question is whether the quantum field theory of an electron is a direct consequence of the fact that the particle producing the field is a quantum particle (and not a classical one) or does it involve much more than that? It involves "much more than that": If I understand correctly, you're taking the classical ...


4

I guess this was prompted by one of my comments on this question. The point I was making is that quantization, even of a system which is defined by a Lagrangian encapsulating a nonlinear set of equations of motion (such as in interacting QED) proceeds by defining a Hilbert space of states and operators on this space which evolve unitarily $$ |\Psi(t)\rangle ...


4

On Unification I presume you're asking whether just classical gravity & classical EM can be unified. They sure can! Classical General Relativity and Classical Electromagnetism are unified in Kaluza-Klein-Theory, which proves that 5-dimensional general relativity is equivalent to 4-dimensional general relativity plus 4-dimensional maxwell ...


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

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

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

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 ...


4

The same way you determine if the interval is space-like or time-like. In fact you do it by computing the square of the four-momentum and examining the sign. Which sign is space-like and which time-like is a matter of convention, and varies from source to source. I like to compute the squared-interval as $$ (\Delta s)^2 = (\Delta t)^2 - (\Delta \vec{x})^2 ...


4

Suppose a scattering process with a 3- particle vertex : $A \to B + \gamma$. Here we suppose that particles $A$ and $B$ are massive, with the same mass $m$, and $\gamma$ is the "virtual" photon. Let $a,b, q$ be the momenta of the particles $A,B,\gamma$. You have : $q^2 =(a-b)^2 \\= (a_0 - b_0)^2 - ( \vec a - \vec b)^2 \\=(a_0^2 - \vec a^2) + (b_0^2 - ...


4

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 ...


4

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 ...



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