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64

In short, the answer is: because gluons behave in a way that makes them useless for this purpose. To understand why, let's back up a little and look at how photons are useful, and then see how gluons behave differently. We (animals pretty broadly) evolved to see photons because they allow us to move around in and respond to our environment more ...


16

Very reasonable question. I will try to answer it in an intuitive way. If you have a scattering medium, photons are reflected in random directions; but when you have a refractive medium, something else happens. The photon is not absorbed and re-emitted: instead, the photon interacts with the electrons in the medium, and since these electrons are somewhat ...


9

The fact that the theory is not gauge invariant implies that all degrees of freedom of $A_\mu$ must have physical meaning: This is not the theory of photons where only transverse degrees of freedom make sense. This way you must tackle some non-trivial issue like the negative norm associated with temporal modes. This could be avoided by adding a mass to ...


3

In quantum mechanical domain these type of question does not have meaning. Every single photon is associated with a wave and vice versa. But to talk whether an electromagnetic wave contains a single photon or not is an ambiguous statement. When people say an electromagnetic wave necessarily contains many photons it only means that a incident beam of ...


3

It does involve bound states. For an electron in the Coulomb field $A(r)=-\frac{Ze}{r}$ that arises from a nucleus (of mass $m$), the lowest energy is $E_0=m\sqrt{1-(2Z\alpha)^2}$ where $\alpha$ is the fine structure constant. This is positive for $Z\le\frac{1}{2\alpha}$, which is what Weinberg means by "moderate" fields. The assumption so far was that the ...


3

We define positive chirality to be right-handed. Ultimately, this was an arbitrary sign choice (like the choice of which charges are negative versus positive), and (like the choice of charge sign) it was probably not the best choice. However, the choice of chirality, which is really just our choice to use right-handed coordinates, and which goes back ...


2

Short answer: because the Sun emits photons, not gluons. Having a long range sense is vital for finding food and recognizing predators. Seeing light and forming an image of our surroundings is one of the three long range senses we have (the others are hearing and smelling). Gluons are extremely short range; they don't even exist as naked particles. How ...


2

A wave plate is a passive component, and can be modelled as a unitary operator on the quantum state. The state is a superposition of left and right circular polarized photons, and the operator gradually alters the relative phases. The total distance then determines the final polarization state.


2

'Crossover diagrams' exist when the final products are indistinguishable. When using Feynman diagrams for computations, you need to add all distinct diagrams that depict the same process, i.e., have the same external lines. Both the crossed and uncrossed diagrams for the electron-electron process have the same final products, since you cannot tell the ...


2

QED uses Feynman diagrams and they are written in the center of mass. In the case of a photon hitting a mirror, it is hitting the field of electrons in the outer band of the mirroring substance. For reflection there should be elastic scattering in the center of mass so that the phases of the photons are kept and the image emerges intact on reflection. So it ...


1

First of all, for non-interacting particles, there's no reason one couldn't pass right through the other. However electrons and protons do interact via the electromagnetic force, and an attractive force can also produce scattering (for a classical example, think of how a comet's trajectory would change if it passes close to a planet). Also, note that not ...


1

What force prevents particles from penetrating other particles This is the basic question that gave rise to the need for a different mechanics than newtonian mechanics. In different words: why does not the electron fall into the proton (and for higher Z atoms the nucleus) and charge disappear? It led to the development of quantum mechanics. Quantum ...


1

It's possible that I do not understand this theory correctly, but it seems to me that it was disproved by experiment. Indeed, it is possible to strip an atom of its electron cloud. If the nucleus was purely an effect of the electron field, nothing would be left once the electrons are removed. This is not the case. See e.g. this post. Note that you can never ...


1

Huygens worked with scalar, longitudinal waves. This was proved to be incorrect in 1821 by Fresnel, who showed that polarization requires transverse waves. A (relevant but incomplete) history of light and how well different models explain certain experiments is given in my recent lecture ''Classical models for quantum light''; see ...


1

Of course in the spectrum of QED there is hydrogen atom. The problem is what to compute in the case of bound states and how to compute. The usual approach to a general QFT is perturbative, meaning that you start from free fields (in this case electron and photon) and then you think the compulg coupling between these fields (in this case $\bar\psi ...


1

It comes about by assuming that the wavelength ($\sim k^{-1}$) is much larger than the typical atomic length scales ($r$).


1

According to the rules of qft there are virtual photons in the vacuĆ¼m. No, according to QFT the vacuum is static, in the sense that $P^\mu|\Omega\rangle=0$. Or put it another way, The vacuum at a time $t$ is exactly the same vacuum at a time $t+\Delta t$ for any $\Delta t$. This means that the picture of particles constantly appearing and disappearing ...


1

The Lagrangian for Dirac's equation is $$ \mathcal L=-mc^2\psi^2+\cdots \tag{1} $$ As we know that $H\sim\mathrm d^3\boldsymbol x\ \mathcal L$ has units of energy, we conclude that $$ \psi^2\sim x^{-3}\tag{2} $$ and therefore $\psi$ has units of $[\mathrm{length}]^{-3/2}$. If you use a different convention for $\mathcal L$ instead of $(1)$ you'll get a ...



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