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1

From a classical point of view, if you look at the field lines created by charge at rest or at constant velocity you'll see straignt lines. Now if the charge changes its speed (i.e accelerating), a "ripple" will appear and propagate along those field lines. Check the animation on this page :http://www.tapir.caltech.edu/~teviet/Waves/empulse.html. You'll ...


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This might not be quite the answer you are looking for, but one useful way to think about it is: the accelerating electron emits photons because nothing forbids it from doing so. By definition, because an electron has electric charge it is coupled to the electromagnetic field, and is able to produce excitations in this field which we can call photons. This ...


0

To get accelerated, any body as well as any particle needs energy to be applied to this body or particle. This happens through photons or more common through electromagnetic waves. This is obvious in particle accelerators and it happens in every acceleration too. Throwing a ball with your hand you do not touch the ball in the meaning, that the molecules of ...


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If you want a simple answer: imagine electrons circling the atomic nucleus in fixed orbitals. The electrons have energy proportional to the distance of their orbital from the nucleus. When you energize an atom (an atom absorbs a photon), you knock the electron into a higher-energy orbital. When the electron falls back down from the higher-energy orbital to ...


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There are two type of emissions: Stimulated and Spontaneous. In the Stimulated emission the incoming photon of with a specific frequency interacts with an excited electron, causing it to drop to a lower energy level. The liberated energy transfers to the electromagnetic field, creating a new photon with identical phase, frequency, polarization, and ...


2

Let us make clear that the problem If proton spin emergence from quarks and gluons is mysterious, why is silver atom spin not? is a modelling problem. The spin both of the proton and the silver atom is measured and known to identify them. John's answer covers it, the energy carried by the virtual quarks and gluons within the proton are much larger ...


3

There are two things to consider: What does the potential look like? Is the wave function of the qubit narrow in the flux or charge basis? Potential shape The Hamiltonian of the transmon (a junction in parallel with a capacitor) is $$H_{\text{charge qubit}} = - E_J \cos(\phi) + \frac{(-2en)^2}{2C}$$ where $E_J\equiv I_c \Phi_0 / 2\pi$, $I_c$ is the ...


0

The transmon and Cooper pair box share the same design, but operate in opposite limits: Cooper pair box is operated under the condition $E_C\gg E_J$, so the charging energy dominates. While for transmon, it is $E_C\gg E_J$, and it is less sensitive to charge noise because of this parameter choice.


0

There are multiple ways to interpret Coulomb's law in quantum electrodynamics (QED). Interestingly, they don't lead to quite the same conclusion (but there is no inconsistency because they are not defined in the same way). The most commonly used way (that ACuriousMind refers to in his comment) consists in relating the notion of classical potential with that ...


1

If q is the momentum for positron, then the propagator for it is still $i\frac{\not{q}+m}{\not{q}^2 +m^2}$.


2

For reference, the fermion propagator is $$ \left\langle 0 \right| T\psi(x)\overline\psi(y) \left|0\right \rangle= S(x-y) = \int \frac{d^4k}{(2\pi)^4} \frac{i}{\not k-m}e^{-ik\cdot(x-y)}$$ Depending on the time ordering, this describes a particle moving from $y$ to $x$, or an antiparticle moving from $x$ to $y$. Now, consider a one-loop diagram in which ...


0

Putting the fermions on-shell or off-shell doesn't change the divergent part of counterterms. In the renormalization schemes, the counter terms are determined to cancel the divergencies. You can put $p^2=m^2$ or $p^2=-\mu^2$ or etc in the diagrams to determine the counter terms, but notice that the derived renormalization constants, $Z_1$, $Z_2$, etc, should ...


3

Schwartz is simply noting that the $\beta$-function has a generic expansion in QED of the form (29) where $\beta_{0,1,2}$ are some numbers that can be computed by explicitly calculating the various Feynman diagrams. For instance the leading $\epsilon/2$ is the tree-level result in $d=4-\epsilon$ dimensions. This can be easily seen as follows. In ...


1

He is writing the $\beta$ function as a perturbation series in the small parameter $\alpha$, $$\beta(\alpha) = \sum_{n=1}^\infty c_n \alpha^n.$$ By convention, it's common to write this expansion in the form he has given. The first coefficient is always $c_1 = - 4\pi\epsilon$ (where $\epsilon$ is the dimensional regularization parameter). The remaining terms ...


6

The binding energy of the electrons in a silver atom is far less than the rest energy of an electron, so there is no ambiguity about the number of electrons in a silver atom. That makes adding up the spins a straightforward business. By contrast, the combined mass of the two up and one down quarks in a proton is about 10MeV (it isn't precisely known) but ...


0

The Fermat principle does not say light ray follows the fastest path, it says when there is a light ray, the optical path (length divided by index of refraction) is stationary with respect to small variations in the shape of the ray that preserve the position of the boundary points. It is not as if light got everywhere the fastest way possible; it goes ...


1

Short answer Quantum optics is a field/branch of physics, with QED being a specific theory. Those working in quantum optics use a fully quantum theory (QED) when needed, but otherwise resort to using simpler tools such as the semiclassical approximation (classical field and quantum matter) which accurately describe a wide range of experiments (primarily ...


0

Technically no, because if something went faster than c it would be classified as a Tachyon. If they did however, they would only be a Tachyon by definition, physically they would still be Photons.



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