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## New answers tagged quantum-electrodynamics

5

Spacetime-dependent symmetries are a problem for path integrals because they break the approximation scheme that's used to define them. In the finite-dimensional case, we can approximate an oscillatory integral by a sum over the stationary points of the action: $$\int_{\mathbb{R}^{n}} d^{n}x\, e^{\frac{i}{\hbar}S(x)} \simeq (2\pi\hbar)^{n/2}e^{i\pi/4}\sum_{... 0 I realise the question was posted a while ago, but there's an effect that hasn't been mentioned in the existing answers: the Ambjørn-Olesen instability. This is a classical result, that in Electroweak theory (or any theory with U(1) charged vector bosons), a constant, homogeneous magnetic field is unstable above the critical strength (in natural units)$$...

6

The residue of $$\frac{1}{k\cdot p}=\frac{1}{k^0p^0-\mathbf{k}\cdot\mathbf{p}}=\frac{1/p^0}{k_0-\frac{\mathbf{k}\cdot\mathbf{p}}{p^0}}$$ at $$k^0=\frac{\mathbf{k}\cdot\mathbf{p}}{p^0}$$ is $$\frac{1}{p^0}.$$

1

It's implicitly performing the modified minimal subtraction $\bar{MS}$ renormalization scheme, only in a sloppy way. One should retain the finite terms at the intermediate stage: $$e^2 = e^2_0\left(1-\frac{e_0^2}{12\pi^2}[\frac{2}{\epsilon} - \ln(m^2) - \gamma+\ln(4\pi)]+O(e_0^4)\right)$$ The usual minimal subtraction $MS$ renormalization scheme will ...

0

Let $Z=\exp (W)$, since $\langle\hat A\rangle=\frac{\delta W}{\delta j}$ we have that $$\frac{\delta Z}{\delta j}=\frac{\delta W}{\delta j}Z \tag 1$$ on the other hand $$\frac{\delta Z}{\delta j}=AZ \tag 2$$ Comparing (1) and (2) we arrive at $$\langle\hat A\rangle=\frac{\delta W}{\delta j}=A$$

0

I think the notation in OP is a bit mixed up. If you stay with the partition function $Z$ framework, you should add local and bi-local source terms such as $$\bar\psi j + \bar j\psi + \bar\psi\eta \psi + k_\mu A^\mu,$$ then in the functional integration you may substitute $$\bar{\psi_2}\psi_3 \rightarrow \frac{\delta}{\delta\eta_{23}}$$ or $$\bar{\... 2 Let's start by considering only flat surfaces i.e. specular reflection. The reflection coefficient is related to the refractive index by the Fresnel equations, so how reflective a surface is depends only on the refractive index. Metal reflects more light than glass because its refractive index is different. So your question: Does all macro reflections ... 0 It can't be bent however it can be rotated using the faraday effect. To say you can't affect light with magnets is false. What you can do is bend the medium light travels through and it will follow the path of density. You can bend almost any material with a strong enough magnet. 0 How to differentiate a pair-production from two interacting particles from a pair-annihilation ? (If it's different at all). That's the main question, The beauty of Feynman diagrams is that they can be read with different definitions of the time and space axes. It all depends on what one defines as input four vectors , and output four vectors. Your ... 1 "Pair creation" is usually something that happens in curved backgrounds (in gravity, cf Hawking radiation) or in electro-magnetic fields (Schwinger effect), because the background has enough energy so that the process of pair-creation, which is otherwise just a way to depict the vacuum of a quantum theory, can actually give rise to two observable particles. ... 0 Sometimes you can extend your model through the inclusion of the term of dipole interaction between atoms. But generally, the interaction between atoms is not taken in the original Tavis-Cummings model. 1 It's essentially just matrix algebra. I would refer you to appendix A.3 of Quantum Field Theory by Mandl and Shaw. It might be a good exercise to fully work through the algebra and compute the trace. 0 This is an interesting question. The clue is the in the acronym, Quantum Electro Dynamics. Usually in chemistry we average over the electrons to obtain an effective force on the nuclei. Situations where you might want to consider the dynamics of the electrons in a molecule generally involve strong radiation fields, such as higher harmonic generation or the ... 2 The question \textit{why it has spin 1} is inappropriate. Particles, by definition, are embedded into irreducible representations of the Poincaré group, i.e., a field. Fields with distinct Lorentz representations have distinct phenomenology and so we must \textbf{choose} the representation of the field in order to describe the correct phenomenology of ... 0 Photons are spin-1 particles, they can be polarized in two different ways, circular (left und right), like electromagnetic waves. They obey the Maxwell equation \partial_\mu F^{\mu \nu}=j^{\nu}. The question why is not so easy to answer. The photon is the U(1) gauge boson so it has to have integer spin. 4 First of all, let me say why photon obtain non-zero mass and how. It comes from one-loop correction to photon propagator, which is given by simple fermionic loop:$$\Pi_{\mu\nu}(k)\propto \int\frac{d^3p}{(2\pi)^3}\frac{\mathrm{Tr}\left[\gamma^{\mu}(\gamma\cdot p+m)\gamma^{\nu}(\gamma\cdot(p+k)+m)\right]}{(p^2-m^2+i\epsilon)((p+k)^2-m^2+i\epsilon)}.$$Trace ... 0 I just computed exactly this thing. My strategy was as follows: Introduce a coupling g to the interaction. With that in mind, we need to find the renormalization constant Z_\mathcal{O}. At one-loop we can extract it from the six diagrams + counterterm at the MS scheme by considering a f \bar{f} \to f \bar{f} process and keeping track of divergent terms ... 1 What you are looking for is a photon OAM switch. The OAM of light is the component of angular momentum of a light beam that is dependent on the field spatial distribution and not the polarization. It can be further split into an internal and an external OAM. The internal OAM is an origin-independent angular momentum of a light beam that can be associated ... 1 Upon reading the question carefully, I believe that the problems of the OP have nothing to do with the quantum nature of the interaction, but simply with the understanding of how modes work. To see this, let us simply write the interaction term in a different form which is in fact also mentioned in the question. Putting in the relevant functional dependences ... 1 We are talking about chiral invariance, right? First of all, the mass term$$ m\bar{\psi} \psi $$breaks the chiral symmetry. So if your professor demands chiral invariance, then we are dealing with massless QED. For massless QED, you can add a chiral symmetric mass dimension 6 term like (NJL 4-fermion interaction)$$ \Delta \mathcal{L} = g (\bar{\psi}\psi ...

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